ALT_1 Teaching mathematics About this free course This version of the content may include video, images and interactive content that may not be optimised for your device. You can experience this free course as it was originally designed on OpenLearn, the home of free learning from The Open University – https://www.open.edu/openlearn/education-development/teaching-mathematics/content-section-overview?active-tab=description-tab There you’ll also be able to track your progress via your activity record, which you can use to demonstrate your learning.
1.1 Learning to count and counting effectively Young children learn about quantity as they interact with their environment. They recognise that it is possible to have one of something, or to have more of something. They can judge whether there is more of one item, say, sweets, than there is of another item, say, biscuits, by comparison (for example, by using one-to-one correspondence). Small children also learn to count, usually helped by the adults and older children in their life. For example, it is common to count the number of stairs you are climbing, or the number of biscuits you have. Counting seems to be an easy skill to learn but it is actually more complex than many people think. Activity 1 How do children learn to count? Allow 5 minutes Take a minute to think about what counting entails and the skills needed to be able to count. Did you think of the following? Children need to learn the names for the numbers and the order in which they come. In the English language this is one, two, three, four, etc. When counting objects, children learn to refer to each object (perhaps by pointing) and to say the next number in the sequence. This is called one–to–one correspondence. The last number in the counting sequence gives the number of objects being counted (one, two, three, four, so there are four objects). This is known as cardinality. The cardinal number is the last number in the counting sequence and this gives you the quantity of the objects. Clearly, you need to start with the number 1 if you are to do this accurately. Finally, children learn that the order in which the objects are counted does not matter. They still get the same final number. This is the conservation principle.
1.2 Making the connection between quantity and number Young children learn about quantity in a natural, intuitive way and are able to understand that you can have more of one thing than another. Quantity is a measure of how much you have of something. In order to be able to indicate the quantity, you need to use numbers (and a measure, so that you can say you have a number of kilograms or metres of something; but, for our current purposes, you are thinking of discrete objects which can be counted). Children are taught to count and to name the numbers used in counting. The next stage of their development in understanding number is to put quantity and number together. Specifically, children need to be able to link the final number in the counting sequence with the size of the quantity of objects which they have counted. This is not an easy concept to develop and can take several years in primary school.. Children learn that adding 1 to a number gives the next number in the sequence. However, it is more difficult to grasp the effect of adding 2 to a number. Early calculation skills which are taught in primary school Counting forwards from 1 and backwards to 1. Starting from different numbers and then counting forwards and backwards. Addition by counting. For example, you want to add 7 biscuits to 4 biscuits. In order of sophistication, you could do this in three ways: Count all – you could count out 7 biscuits, then you could count out 4 biscuits. Then, after you have put the two groups together, you could count the total number of biscuits. Count on – you could keep the number 4 in our head and then count on 7 from there. Count on from larger number – you could save yourself some effort by beginning with the larger number, i.e. 7, and count on 4 from there. This all seems very easy, but you need to remember that young children do not have the ready familiarity with numbers that you have. To try to imagine how tricky it can be for a young child to use counting skills in addition, try the following activity (without converting the letters into numbers!). Activity 2 Adding numbers when you are unfamiliar with their names Allow 5 minutes Suppose the counting sequence is Y, J, A, P, K, T, B, M. Count from A to B. What is J + K? Count back from M to P. Count back from B to J. What is T – A? What is K – Y? From A to B is P. J + K = B From M to P is P From B to J is K T – A = A K – Y = P How did you do?
1.3 Part-whole relationships and number bonds You can make the number 8 in different ways: $1+7$, $2+6$, etc. Children can learn these by manipulating physical objects such as bricks and tiles. Starting with 8 objects, a child can split these objects into two or more groups(Figure 3). This can also be done pictorially using a part-whole diagram or a bar diagram (Figure 4). Part-whole relationships are important for learning about number as they develop the understanding of how numbers are made up of other numbers. That is, numbers can be partitioned into smaller constituents in the same way that 8 can be made up of 2 and 6. The number bonds refer to pairs of numbers which sum to a given number. The important sums are 10, 20, 100, etc. Learning the number bonds to 10 is an important first step in being able to calculate. For example, adding 6 + 7 can be done as 6 + 4 + 3 if it is known that 6 and 4 are numbers which sum to 10. This particular method is known as bridging because you are making the number up to 10 and then adding the remaining part of the number to go past it. In this way, addition is facilitated by using the number bonds to 10. So, knowing that 4 is added to 6 to make 10, and then knowing the part-whole relationships for 7, i.e. that 7 is also 4 + 3.
1.4 Understanding that addition and subtraction are inverses of each other Children first learn to apply inverse relationships to quantities before they can apply them to numbers. For example, five-year old children understand that if you have 6 biscuits and you add 3 more biscuits and then take those same 3 biscuits away, then you have the same number of biscuits that you started with. However, if you were to take away 3 different biscuits, the child may not realise that the result is the same number of biscuits that you started with. This suggests that the child is thinking in a very concrete way about the biscuits themselves, rather than abstractly thinking about numbers. When children eventually understand that addition and subtraction are inverses of each other, this is a big step towards them developing number sense. Multiplying using additive reasoning Children learn how to multiply in the first instance by using repeated addition. For example, if there are 3 flowers in each of 4 vases, children can work out the total by counting 3 for each vase using a one-to-one correspondence. This underpins the concept of multiplication and paves the way for developing secure methods of multiplication calculations later on. If children do not understand the concept of multiplication as repeated addition then learning their multiplication tables will not really help later in their school career. The basic foundational concept is important. Laying the foundations for future mathematics learning If children are able to understand that addition and subtraction are inverses of each other, and if they can use one-to-one correspondence to solve early multiplication problems, this gives them a secure foundation for later learning of mathematics, not just in arithmetic. It is important that children who struggle with this in the early school years are given extra support to make these developmental strides.
1.5 Moving on from counting and working out relationships between quantities Suppose I have 7 biscuits and you have 4 biscuits. I have 3 more than you. Here, the number 3 gives the relationship between the two different quantities of biscuits, between the number of my biscuits and the number of your biscuits. It is harder for children to work out problems like this, which involve relationships, than problems that just involve quantities. It is not so much the number crunching that is the difficulty but the conceptual understanding of relationships which is difficult. Children can count biscuits but they cannot count the relationship between the numbers of biscuits. Skills which help children to grasp the concept of relationships between quantities are: adding and subtracting, understanding that these are inverses of each other and part-whole relationships. Moving from using manipulatives to pictures and on to abstract numbers When children first begin learning about numbers in the early years, their teachers will give them concrete materials to manipulate and count. These are referred to as manipulatives. Using manipulatives such as counters, bricks and specialist educational equipment is necessary for young children to develop their number sense. As children become more proficient at using manipulatives, the teachers will begin to link the manipulatives to pictures, and to use pictorial images to represent number. As the children become more proficient at using images, they will be carefully introduced to more abstract ways of representing a number problem. Whenever children need to make sense of new knowledge, teachers will often return to the chosen manipulatives or pictures and start the process of abstraction again.
2 Developing mental methods Being able to calculate mentally helps learners to develop a secure understanding of numbers and the number system, allows learners to develop their own intuitive methods and provides a firm foundation for written methods of calculation. The practice of doing mental calculations is an important skill in everyday life. It also helps to develop fluency and confidence with numbers. How can you help your learners to develop effective mental methods? What follows are examples that you can use or modify for use with your learners.
2.1 Using number bonds to 10, 20, 100 Learning number bonds is an important skill which supports learners to be able to answer more difficult questions. For example, learning how to quickly calculate number bonds to 100 can help with bridging over 100. A quick way of calculating 100 – 38 is to make the tens number up to 90 and the units number up to 10. $30\to 90\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}60$ $8\to 10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2$ So $38\to 100\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}62$ You can also say that 62 is the complement to 38. Note that we have not used an ‘equals’ sign because these are not strictly mathematical sentences but they mirror what the learner may be saying in their head or aloud. Similar quick ways can be used to find numbers that sum to 1000. For example, to work out 1000 – 159: $100\to 900\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}800$ $50\to 90\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}40$ $9\to 10\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1$ So $159\to 1000\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{i}\mathrm{s}\text{\hspace{0.17em}}\text{\hspace{0.17em}}841$ or $1000-159=841$ This method also works for reckoning the change from £10.00.
2.2 Mental addition and subtraction strategies Using number bonds can help with mental addition calculations because numbers can be split up to make the tens or hundreds number and then the extra can be added on. Activity 3 Adding large numbers mentally Allow 5 minutes Try these questions and reflect on your methods. $43+32$ $47+38$ $142+286$ Did you partition the numbers into smaller numbers? For example, for question (a) did you work out 40 + 30 and then work out 3 + 2? Did you bridge across a multiple of ten? For example, in question (b) did you mentally change 47 + 38 into 47 + 3 (to make 50) and then add 35? Once a learner can calculate what to add to 38 to make 100 they can answer the question 100 – 38 and move on to calculations such as 402 – 238 by bridging through 300. This mental arithmetic can be illustrated using an empty number line. The actions on the number line can act as a way to record the stages in a mental calculation. This is shown in the following video. Video 1 Using the empty number line to subtract 238 from 402 INSTRUCTOR: I'm going to calculate 402 takeaway 238 using the number line. And I've drawn a number line ready for this below. The way I'm going to do this is to actually work out what needs to be added to 238 to make 402. So I'll start with 238, and I'll end with 402. This is a process known as complementary addition. First of all, I'm going to go to the first tens number after 238-- so that's 240. And to do that, I shall have added 2. Next I'll work up to the next hundreds number. That's 300. And to make that jump, I should have added 60. And finally, I can get from 300 to 402 in one jump by adding 102. So altogether, I will have added 102, 60, and 2, and that is 164. So 402 take 238 is 164. We can use this method to answer questions such as £4.02 – £2.38 or 4.03 m – 2.38 m, taking us into the realm of using measures. This method also helps with time calculations where learners need to remember that time is not a decimal system. For example, a train journey which starts at 2.38 pm and goes on until 4.02 pm takes 22 minutes + 1 hour + 2 minutes which comes to 1 hour 24 minutes (bridging through 3 pm and 4 pm). Activity 4 Subtraction problems for you to solve Allow 5 minutes Try answering the following in your head using whichever method you prefer. $73-56$ $820-552$ $£10.00-£4.75$ You may check your answers using a calculator.
2.3 Mental multiplication and division strategies Doubling and halving are easy for most people, so make use of these tactics: doubling and doubling again is the same as multiplying by 4 doubling and doubling and doubling again is the same as multiplying by 8. Use the inverse of doubling (i.e. the opposite of doubling) which is halving: halving is the same as dividing by 2 halving and halving again is the same as dividing by 4 halving and halving and halving again is the same as dividing by 8. Once we begin talking about halving being the same as dividing by 2 this connects to fractions, decimals and percentages. There will be more of this in Week 2 but for now: half of 100% is 50% → to find 50% of an amount just half it half of 50% is 25% → to find 25% of an amount half and half it again. Activity 5 Finding halves and quarters Allow 5 minutes Try the following questions. $\frac{1}{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}62$ $\frac{1}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}62$ $25%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}124$ $25%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}18$ $\begin{array}{cc}\frac{1}{2}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}62& =31\\ \frac{1}{4}\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}62& =15.5\end{array}$ $\begin{array}{cc}25%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}124& =31\\ \text{25%\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}18& =4.5\end{array}$ Listed below are some useful mental multiplication strategies: Multiply by 5 by multiplying by 10 then halving $14×5=140÷2=70$ Multiply by 20 by multiplying by 10 then doubling (or vice versa) $24×20=240×2=480,\text{\hspace{0.17em}}\mathrm{alternatively}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}24×20=48×10=480$ Multiply by 50 by multiplying by 100 and halving (or vice versa) $26×50=2600÷2=1300,\text{\hspace{0.17em}}\mathrm{alternatively}\text{\hspace{0.17em}}\text{\hspace{0.17em}}26×50=13×100=1300$ Multiply by 11 or 9 by multiplying by 10 and adding or subtracting $13×11=1300+13=143$ Multiply by 21 or 19 by multiplying by 20 and adding or subtracting $13×21=13×20+13=273$ Understanding division problems Many learners (and adults) have problems with division because they do not exactly understand what it means. A straightforward division can be ambiguous without the context of the problem. Learners need to understand the difference between sharing and grouping which are the two main types of division problem. For example: $12÷3$ can mean there are 12 sweets shared between 3 people, so they each get 4 sweets. 12 has been divided (or shared) into 3 equal parts of 4. 12 ÷ 3 can mean that I have 12p, so how many 3p sweets can I buy? In this case, you can count up in threes to find how many sweets you can buy. You are counting up in groups of 3, (i.e. 3, 6, 9, 12). So there are four 3s in 12. Division by partitioning can be a useful strategy Partitioning the number into a multiple of the divisor plus the remainder is a useful way if a learner is familiar with their multiplication tables. For example, 84 ÷ 7: Activity 6 Division by partitioning Allow 10 minutes Calculate 136 ÷ 4 using the partition method. Then watch Video 2 for a demonstration of one way to do this. Video 2 Dividing 136 by 4 using partitioning INSTRUCTOR: I'm going to work out 136 divide by 4 using the partition method. That means I'm going to split 136 into parts. And then I will work out how many 4's go into those parts. I know that 4 goes into 100, so I'm going to split 136 into 100, and then the amount left over is 36. 4's into 100, there are 25 of these. And then into 36, 4 goes 9 times. So altogether, I've got 25 4's and 9 4's. That makes 34. Finally, 136 divided by 4 is 34. Multiplication tables While there is no doubt that knowing the multiplication tables is a useful skill, learning by rote without an understanding of the concept of multiplication and of division is less than useful. Mental maths, which includes multiplication and division, helps learners to form the appropriate concepts. Many successful mathematicians do not know their tables by rote. However, they do know how to work them out as needed. Developing mental arithmetic with your learners Practise frequently with your learners. The beginning and end of lessons are often good times to practise mental arithmetic – maybe spend two or three minutes doing this. Starter activities could include working out mental arithmetic questions as part of a quiz, an activity, etc. Make it into a game. Do ask learners to say how they worked something out. They could tell their method to the person sitting next to them. Or they could tell their method to the class. Validate mental calculations. Children will not see the point in having to write down a method of arithmetic for a calculation they can do in their head. Nor should they have to: mental calculation is a perfectly good way of working out the answers to number questions, so encourage its use.
3 Developing formal written calculation methods This course mainly focuses on teaching children in the middle school years, so they have probably met all of the four rules (addition, subtraction, multiplication and division) and been taught formal methods of calculation. This does not guarantee that learners have learned these methods effectively! Nor is there any guarantee that if a learner can operate a method and get lots of right answers that they understand how the method works. As a teacher, therefore, you are striving to teach effective methods, procedural understanding of how those methods work, and conceptual understanding of the mathematics which underpins the methods.
3.2 Formal written methods for subtraction Subtraction is experienced as being more difficult than addition. Subtraction is the inverse of addition and knowing this can help learners to use complementary addition in informal methods of subtraction. Video 4 demonstrates the decomposition method (or column method) of subtraction and the empty number line method which uses complementary addition. This latter method is a visible example instantiation of the method which many people use mentally. Video 4 Choosing the best method for subtraction INSTRUCTOR: So a lot of learners in school are taught to subtract in columns. So for example, if we've got 86 and I want to subtract 39, those would be put in columns, so the units are underneath each other, the tens underneath each other. And I'm going to subtract. Now, I can't do 6 take away 9. So I'm going to go over to the 8, that's eight 10s. I'm going to take one away, which will leave me with 7. And then that 10 over here becomes 10 units. So together that makes 16 units, take away 9. 16 take 9 is 7. And then 7 take 3 is 4. So that seems quite straightforward. That's known as the decomposition method because we decompose the 8 into 7 tens and then 10 units over here. This can sometimes be quite problematic. Sometimes, learners can just simply do the 9 take 6 and provide me with 3. And then 8 take 3 would be 5. So that's the very common mistake. They know they've got to subtract something so they just take the smaller from the larger and don't take any notice of whether the number's actually on the top or not. Consider this. If I've got 16 take away 9 and I tried to do that using composition composition, I can't do 6 take 9. So I come here, take a 10 away, that leaves me with none. I put the 10 there is 10 units. That's now 16 take 9. And actually, I've got the same problem I had before, haven't I? So 16 take 9 doesn't really make terribly much sense using the decomposition method. Or again, if I had a number like this 1,004 and supposing I wanted to subtract 99 and I wanted to use the decomposition method, I can't do 4 take 9. So I go to that, but there aren't any n or to there. So I've got to go all the way to the thousands. I'll take that 1,000 off and leaves with 0. Then I can put that there as 10 hundreds. I can now take 1 away from there and leave it with 9. I've now got 10 tens. And I'll take one of those off and leave me with 9. And then I've got that 10 units. finally there. 14 take nine is 5. 9 take 9 is nothing. 9 take nothing is 9. And there isn't anything left in the thousands. Now I've done that accurately, but it is very messy. And it's fraught with difficulties, because you've got to go all the way to the thousands to decompose and put the numbers across. So in the both of these cases, it would actually be better to think about those in a different way. Certainly, if we think of the number line, which was demonstrated in the previous section-- so if I was thinking about nine and what I need to add to make 16 I could say to myself, well to get to 10, I've added 1. And then 10 to 16, I've added 6. So that's 7. 16 take 9 is 7. And over here, I could also use a number line. So if I were to decide what I need to add to 99 to make 1,004, so going from 99 to the next 100, I've added 1. 100 to 1,000, I've added 900. I should set out 1 there. And finally, that last little bit I added 4. So altogether, I've added 900, 4, and 1. So I've added 905. So 1,004 take 99 is 905. Both of these subtractions made much better sense to do using a number line. Or you might think for this one, 99 is quite close to 100, so take 100 off and then add 1 back on. So 1,004 take off 100, 904, add 1 back on 905. So what it shows is that it rather depends on the problem you're trying to solve as to which is the best method to use. Activity 8 Choosing the most appropriate method of subtraction Allow 8 minutes Consider the following subtractions and decide whether it is more appropriate to use the decomposition method or the empty number line method. $465-253$ $102-98$ The length of a journey which starts at 9.55 a.m. and ends at 11.05 a.m. If the numbers in the subtraction are close together then it makes more sense to ‘count up’ from the smaller to the larger, the method known as known as complementary addition. This is especially useful if the numbers bridge over a 10s or 100s sort of number or over the hour in a time calculation.
3.3 Multiplication and division The next video demonstrates two methods for long multiplication and discusses the pros and cons of each method. Video 5 Two methods for long multiplication INSTRUCTOR: I am going to find the result of multiplying 74 by 36 first of all by using the grid method. This links to finding the area of a rectangle 74 units long and 36 units wide. I'm going to partition 74 and 36 into that tens and units, which will mean splitting the length of the rectangle into 70 and 4 and the width of the rectangle into 30 and 6. I'm now then to find the areas of the four parts of the rectangle. So the top left, 30 times 70, three 7's are 21. So 30 times 70 is 2,100. Top right, 30 times 4 is 120. Bottom left six times 70 is 420. 6 times 4 is 24. So the area of the rectangle is the area of those four parts added up. I'll work out some of the numbers on the top row, that's 2,220. And the bottom row will be 444. And if I add those two up, I'll get the answer to 74 times 36 which is 2,664. The beauty of the grid method is that the learner works out four products for a two digit by two digit calculation. And if any mistakes are made, it is easy to check where those mistakes are. The learner is also dealing with those numbers as the sum of their parts, 70 and 4, 30 and 6 in this case. Another method of working out the same multiplication would be to use the standard column method. This is the traditional method, which has been taught for decades. So you would think that every adult would be proficient at it but surveys show that very few people in the general population are proficient at using the standard column method. So we'll look at that method now. The numbers are written one underneath the other, 74, 36, and I put a multiplication at the side to remind us that that's what we're doing. We're going to have three lines of working. First of all, I'll multiply 74 by 6. Six 4's are 24. So I'll put down the 4 and carry the 2 very small in the next column. six 7's are 42 plus the two that I carried makes 44. Then I come to multiply by the 3, remembering that it is actually 30. So to multiply by 30, I put down the 0 then carry on as if I were multiplying by 3. Three 4's are 12. So I put down the 2 and carry the 1 very small into the next column. Three 7's are 21 plus the one that I carried makes 22. And if I add those two lines up I got 2,664, the same as I did for the grid method. You might notice that they've got the same lines of working. Now the standard column method is neat and quick if the learner knows how to do it correctly. However, for many learners, this method is fraught with difficulties. It treats each number as a string of digits, unlike the grid method which treats each number as the sum of its parts. This can mean the learner operates the standard column method according to the set of rules, and they can be easily forgotten or confused. Some of the procedural problems I've seen are learners not knowing what to do with the carrying figures or forgetting to put down the 0 or not putting it in the units column but putting it at the side and getting lost during the working as they don't know their multiplication tables and need to work them out, whereas in the grid method, they still have to work out the multiplication table facts, but at least the other three boxes can wait for them. And there are no carrying figures to worry about. Division Division is the arithmetic operation where learners experience the most difficulty. Division is the inverse of multiplication which means that it is possible to approach division by complementary multiplying. For example, to answer the question 78 ÷ 6 you can ask ‘How many 6s make 78?’. One way to answer the question in this way is to use the ‘chunking’ method where you add ‘chunks’ of 6 until you reach 78. It saves having to add 6 at a time, which could be laborious. For example: $10×6=60$ (you know that 78 is more than 6 lots of 10, so start there) $3×6=18$ (you have 18 left to make the 60 up to 78, that’s 3 lots of 6) $13×16=78$ So $78÷6=13$ Or you could start with 78 and keep subtracting 6 until you have nothing (or a number smaller than 6) left: For example: 78 78 ÷ 6 calculating by subtracting sixes $78-60=18$ $18-6=12$ $12-6=6$ $6-6=0$ $\mathrm{So the answer is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}13$ Whether working up to 78 or down from 78, the method is basically finding how many 6s there are in 78. This method clearly becomes harder with larger numbers and requires knowledge of multiplication tables and good mental arithmetic. It is, however, an intuitive method of dividing. The traditional method of division is known among teachers as either the bus stop method (because the working looks like a bus stop),or the ‘gazinta’ method (because you ask how many times 6 ‘goes into’ each number). Figure 5 shows that 6 goes into 7 once with 1 left over. So, put the 1 (for the once) above the 7 and the 1 left over next to the 8. Now read it as 6 goes into 18, which is 3. Put the 3 over the 8. This is an easy method to operate if you can remember the procedure. It is not so easy to understand why it works. Yet many learners prefer it to having to use the ‘chunking’ method. The bus stop method is efficient to use with larger numbers as well, as long as your multiplication tables are good, and you can do some mental calculation. When dividing by two-digit numbers, I prefer to use and to teach a version of this short division method. Activity 9 Solving a division problem Allow 10 minutes Suppose you have 1960 pupils and teachers in a school to ferry to an event in buses which each carry 43 passengers. How many buses will be needed? So, you are trying to find out how many times 43 goes into 1960 and you need an extra bus for the remainders. Spend ten minutes ‘having a go’, using whichever method you prefer. Then watch the video. Video 6 Two ways to solve a division problem INSTRUCTOR: How many buses do I need to carry 1,960 passengers if each bus carries 43? First, I will begin by using an intuitive method whereby I will add up multiples of 43 until I can carry all 1,960 people. So I'm going to start by working out how many 10 buses would carry. 10 times 43 is 430. I obviously need a lot more than 10 buses. I will try 4 lots of 10 buses, which is 40 buses. That will be 4 times 430-- I could double and double again, that will give me 1,720 passengers. This is closer to 1,960, but adding another 10 buses will take me well over the 1,960, so I'm now going to add smaller numbers of buses. I could add 5 buses. That'll be half of 430. So 5 times 43 will be 215. I'm going to add those together, just the last two lines, to give me 45 lots of 43. So that will be 1,935. So that's not quite enough passengers yet can be taken, so I need to add another bus. Let's try 46 buses by adding 43 onto my last figure. So that will give me 1,978. So I can definitely carry all 1,960 people, and they will also be 18 spaces left. So I need 46 buses. The traditional way to work out how many buses are needed is to do a division. So I'm doing 1,960 divided by 43. I'm going to use a short division method even though I have a two-figure number as a divisor. Years of teaching experience shows me that learners struggle with the long division method with its complicated procedure. So 43 into 1,960, I'm going to have to work out how many 43's are in 196. So I'm going to write the 43 times table down the side. So two of them will be 86, three will be 129, four will be 172. So that's below 196, but another 43 would take it over. So 43 is into 196. It must be 4, then of course, it's actually 443-- this is 172, so there's a remainder of 24, which I put next to the 0, and read it as 240. So 43 is into 240, I just need to go a bit further with my 43 times table, add another one of those on. So that's 215. That's just less than 240. Another 43 would take it over, so 43 is into 240, go 5. And of course, five 43's a 215. So there's a remainder of 25. So we've got an answer-- 45 remainder 25. What does remainder 25 mean? Do 25 people have to walk? We can't have that, so I will order them another bus. So altogether, I will need 46 buses.
4 The concept of place value and decimals Place value underpins our number system, so understanding it is important in being able to calculate with numbers. Even learners in secondary education can be unsure of the place value system, so it is important to provide activities which support their conceptual understanding of it.
4.1 The decimal point and calculations with decimals The decimal point is a symbol which is placed between the whole part of the number and the fraction part of the number. It signifies that the digit to its left is a units digit and the digit to its right is a tenths digit. All the rest of the digits fall into place to the left and right. Nothing more, nothing less. The decimal point is not some sort of bridge to jump over or obstacle to be overcome but many learners treat is as if it is. Here is an example of a common mistake. $1.5+1.5=2.10$ Since 5 + 5 makes 10, the learner has written 10 in the tenths column, presumably because they do not think that carrying over the decimal point is allowed. They see the decimal point as splitting the numbers into two parts which need to be dealt with separately. However, a knowledge of place value means that the 1.5 is seen as 1 unit + 5 tenths, which when doubled is 2 units + 10 tenths (convert to 1 unit) which is 3 units. $1.5+1.5=3.0$ One way of helping learners to appreciate this is to practise decimal times tables. Here are the first ten multiples of 0.3: 0.3, 0.6, 0.9, 1.2, 1.5, 1.8, 2.1, 2.4, 2.7, 3.0 It is tempting to say 0.12 after 0.9 but think about what it means in terms of the value of each digit. Bridging over the whole numbers is always the crunch point for the decimal times tables. They provide a useful exercise for helping learners see that numbers are carried into the next column on the left in just the same way as in whole numbers. They also look the same (as in having the same digits) as their related multiplication tables but including the decimal point. You can think of all numbers as being decimals if you think of whole numbers as having an invisible decimal point after the units digit. You usually only make the decimal point visible if there are digits after it. Activity 10 Using decimal times tables Allow 3 minutes Write out the first ten members of the 1.5 times table. You will know that you have them correct if the last one is 15.0.
5 This week’s quiz Check what you’ve learned this week by doing the end-of-week quiz. Week 1 practice quiz Open the quiz in a new tab or window and come back here when you’ve finished. 6 Summary of Week 1 Now that you have completed Week 1 you have learned how early counting leads to early addition, which then leads to subtraction, multiplication and division. Careful development of these skills helps children to achieve a sense of how numbers work, providing a firm foundation for future mathematics learning. You have explored varied methods, both mental and written methods, for adding, subtracting, multiplying and dividing. Finally you have looked at place value and decimal numbers. You should now be able to: briefly describe how number sense develops in young children demonstrate a variety of mental, informal and formal written methods of basic arithmetic understand how place value underpins our number system in base ten, for both whole numbers and decimals. Next week you will learn about the conceptual basis of fractions and percentages, the pitfalls learners often meet when studying fractions in particular and effective approaches to the teaching of fractions and percentages. You can now go to Week 2.
Week 2: Teaching and learning: percentages and fractions 1 Equivalence as a concept applied to fractions, decimals and percentages In mathematics the term ‘equivalence’ is used to mean ‘has the same value’ or ‘can be substituted for’. Fractions, decimals and percentages each have two main uses: Fractions, decimals and percentages are numbers, for example, $\frac{1}{2}$, 3.2, 78% are all numbers and have a position on a number line Fractions, decimals and percentages can be used to denote a proportion of a quantity, for example we can find $\frac{1}{2}$ of £6, 3.2 of 5 kilograms or 78% of 250 people. Confident learners may be able to move between the number meaning and the proportion meaning of fractions, decimals and percentages but many learners find this confusing. Consider the following well-known equivalences: $\frac{1}{2}=0.5=50%$ If you think of these as numbers then their equivalence means that they all occupy the same position on a number line. Their equivalence in this case means ‘has the same value’. If you think of these as proportions then their equivalence means that finding $\frac{1}{2}$ of a given quantity is the same as finding 0.5 of that same given quantity or 50% of that same given quantity. Their equivalence in this case means ‘can be substituted for’.
1.1 Diagrammatic representations of equivalence The first video shows a mathematical cake which is shaped like a rectangle and has been cut into ten slices. Learners can draw their own mathematical cake on paper. It can have a varied length but it is useful to ensure that the width is 10 cm, which means each slice is 1 cm wide. Diagrams can be very useful aids to teaching concepts in mathematics. Video 1 Mathematical cake: a diagrammatic representation of the equivalence of fractions, decimals and percentages INSTRUCTOR: Mathematical cakes, a diagrammatic representation of the equivalence of fractions, decimals, and percentages. Here is a mathematical cake, which is rectangular in shape and split into 10 equal slices. We'll consider the percentages first the whole cake is 100%. so one slice must be 100% divided by 10, or 10%. No slices is 0%, one slice is 10%, two slices are 20%, three slices of 30%, and so on and one whole cake is 100%. Next the decimals, the whole cake is 1.0. So each slice must be 1.0 divided by 10, which is 0.1. No slices is 0.0, one slice is 0.1. two slices are 0.2, three slices are 0.3, and so on. And the whole cake is 1.0. Finally, the fractions, the whole cake is one whole, so each slice must be 1/10. No slices is 0, one slice is 1/10, two slices are 2/10, three slices are 3/10, and so on. And the whole cake is one whole. Some of these fractions have equivalent simpler fractions. 5/10 is half of the cake. And that can be seen to be equivalent to 0.5 and 50%. 3/10 is 0.3 and 30%. and 60% is equivalent to 0.6 and 6/10, which can be simplified to 3/5. You can easily read across to find equivalences from the diagram. The fractions, decimals, and percentages shown on the diagram are some of the important equivalences to know. Supposing you wanted to find out 1/4 from this diagram, how would you find where 1/4 is and work out its equivalences in decimals and percentages? What would you do about 3/4? Activity 1 Finding equivalences to one-quarter and three-quarters Allow 5 minutes Look at the mathematical cake diagram which was shown in the video. Use it to find equivalences to $\frac{1}{4}$ and $\frac{3}{4}$. The mathematical cake diagram can be used to demonstrate the equivalences of percentages, decimals and fractions in tenths. One-quarter can be found by finding the mid-way point between 0 and one-half. Three quarters would be found mid-way between one-half and one-whole of the cake. One-quarter is 25% and 0.25. Three-quarters is 75% and 0.75. The fraction wall Figure 1 shows a fraction wall divided into strips of the same length which can be split into equal parts. The fraction wall can be used to identify equivalent fractions. Activity 2 Finding equivalent fractions Allow 5 minutes Use the fraction wall to find as many fractions as you can that are equivalent to $\frac{6}{8}$. Use the fraction wall to find as many fractions as you can that are equivalent to $\frac{1}{3}$.
1.2 Challenges to understanding the equivalence between fractions You have been introduced to two diagrams which can help people learn about the equivalence of fractions, decimals and percentages through visualisation. Activity 3 Thinking of other useful diagrammatic representations Allow 5 minutes Can you think of any other diagrams which you have used, or seen in use, which help learners to understand fractions, decimals and percentages? The concept of equivalence is important in mathematics. Can you think of other areas of mathematics where equivalence is used? Equivalence is an important concept in mathematics. Algebra is a particular area of mathematics where equivalence is important. Equivalent algebraic expressions are different ways of denoting the same relationship.
2 Approaches to teaching fractions In this section you will meet again the ideas of equivalence in fractions. You will also be introduced to some different approaches to teaching the calculation of fractions, including the use of diagrams. Activity 4 Where are fractions used in mathematics? Allow 5 minutes The topic of fractions is a substantial one in the school curriculum. List as many of the different learning activities as you can which relate to the topic of fractions. Did you think of any of the following? finding fractions of amounts finding equivalent fractions ordering a list of fractions from smallest to greatest or vice versa adding and subtracting fractions multiplying and dividing fractions and using fractions within other topics: solving problems using proportion algebraic equations which involve fractions calculating probabilities
2.1 What are fractions? Fractions are numbers made of two integers (positive and negative whole numbers), one acting as the numerator and one acting as the denominator. (Fractions typically refer to quantities which are smaller than 1. In this case the numerator is smaller than the denominator. Fractions with a numerator larger than the denominator have a value greater than 1. They can also be written as mixed numbers; the whole number part and the fraction part). Consider the fraction illustrated in the diagram below which shows $\frac{3}{4}$ of a rectangle shaded. The rectangle has been divided into 4 equal parts and 3 of these parts have been shaded. Three- quarters of any quantity can be found using the same process. $\frac{1}{4}$ is found (by dividing by 4 to calculate four equal parts), and then multiplied by 3 to give $\frac{3}{4}$. This is referred to as a part-whole relationship (Figure 2). However the number $\frac{3}{4}$ can also be taken to mean that 3 has been divided into 4 equal parts and in this case the value of the fraction is equal to the numerator divided by the denominator. This interpretation of the fraction is used when converting fractions to their equivalent decimals. If you input $3÷4$ into a calculator you will get 0.75, which is the decimal equivalent to $\frac{3}{4}$. This takes the meaning of a quotient (the result of dividing one number by another). Essentially a fraction such as $\frac{3}{4}$ can have two meanings (the whole-part relation and the quotient) and this can cause difficulty in understanding for some students. One way to address this difficulty is to tell students, for example, that $\frac{3}{4}$ means 3 lots of $\frac{1}{4}$ of something, and it can also mean $3÷4$. This is not the only such case in mathematics, where symbolic notation can take on different meanings and which causes difficulties in students’ understanding.
2.2 Ordering fractions The understanding of equivalence among fractions can also cause a challenge because numbers which effectively have different labels can mean the same quantity of proportion (depending on how they are being used). As an example $\frac{4}{6}$ and $\frac{10}{15}$ are equivalent fractions. They can both be shown to be equivalent to the simplest form of $\frac{2}{3}$ by ‘cancelling down’ i.e. dividing the numerator and denominator in each fraction by the same common factor. Since the value of the fraction is determined by the relation between the numerator and the denominator it can be difficult to compare the sizes of the two fractions. The usual method when comparing the size of two fractions is to find equivalences which have the same denominator. Activity 5 Which fraction is larger? Allow 10 minutes Find the larger of these two fractions: $\frac{2}{3}$ or $\frac{5}{8}$. $\frac{2}{3}$ and $\frac{5}{8}$ which is larger? Both the denominators 3 and 8 are factors of 24 (the least common multiple or LCM) so you can find equivalent fractions whose denominators are 24 $\frac{2}{3}$ = $\frac{16}{24}$ multiplying numerator and denominator by 8 $\frac{5}{8}$ = $\frac{15}{24}$ multiplying numerator and denominator by 3 So $\frac{2}{3}$ is the larger fraction
2.3 Adding and subtracting fractions Learners are often told that the denominator in the fraction denotes what kind of a fraction it is (halves, thirds, etc.) and that it is only possible to add or subtract the same types of fraction, i.e. the denominator has to be the same number. This is where the difficulty lies as learners struggle to find out which number should be the common denominator and what needs to be done to change each fraction so that they do have the same denominator. Several rules and procedures exist to help learners to add and subtract fractions but, since the conceptual basis is poorly understood, learners typically do not remember the rules and procedures. This often leads to fractions being taught again and again with little success. It is worth looking at other methods which build on and develop learners’ conceptual understanding of fractions. The next section gives an approach to teaching the adding of fractions which is based on learners’ prior knowledge of finding fractions of a quantity. Activity 6 Shading a grid Allow 10 minutes Try the following activity. Draw a 6 by 4 grid. Next shade the grid according to this schedule (with no shading overlapping): Shade $\frac{1}{2}$ in red Shade $\frac{1}{4}$ in blue Shade $\frac{1}{8}$ in green Shade $\frac{1}{24}$ in yellow Shade $\frac{1}{12}$ in grey The whole rectangles should have been shaded once the activity is completed. Did your grid look like this, or at least have the same number of parts per colour? Activity 7 Use the grid to add fractions Allow 5 minutes I hope it is clear that each box in the rectangle is $\frac{1}{24}$ since the grid has been split into 24 equal parts. This should help you (and your students) to calculate the following additions of fractions based on the shaded grid. $\begin{array}{cc}\frac{1}{2}+\frac{1}{8}& =\\ \frac{1}{4}+\frac{1}{6}+\frac{1}{12}& =\\ \frac{1}{8}+\frac{1}{24}& =\end{array}$ Did you get these answers? $\begin{array}{cc}\frac{1}{2}+\frac{1}{8}& =\frac{5}{8}\\ \frac{1}{4}+\frac{1}{6}+\frac{1}{12}& =\frac{12}{24}=\frac{1}{2}\\ \frac{1}{8}+\frac{1}{24}& =\frac{4}{24}=\frac{1}{6}\end{array}$ With some experience of adding fractions using this grid shading methods students should begin to see that they are looking for a common denominator which would have been the number of parts in the grid. Subtraction can be carried out in a similar way by thinking about shading parts on the grid and then rubbing some of them out.
2.4 Multiplying fractions The procedure for multiplying fractions is straightforward but the conceptual understanding is not so easy to develop. $\frac{a}{b}×\frac{c}{d}=\frac{a×c}{b×d}$ Watch video 2 on multiplying fractions which demonstrates three different approaches. Video 2 Multiplying fractions using three different approaches INSTRUCTOR: Multiplying fractions using three different approaches. Suppose we calculate 3/4 times 2/3. Using the procedure of multiplying the two numerators and multiplying the two denominators, we get 6/12, and that cancels to 1/2. Or I could have cancelled numbers top and bottom before multiplying. I can see I've got a 3 on the top and a 3 on the bottom, so I can cancel those 3's. And I've got a multiple of 2 on the top and a multiple 2 on the bottom. 2's into 2 go once, and 2's into 4 go twice. And that leaves me with 1/2. These procedures seem straightforward, but they don't really build the conceptual understanding which underpins the multiplication of fractions. So let's look at other approaches, which can help build that understanding. The method I'm going to demonstrate now, it helps to remember that in mathematics the word of is interchangeable with multiply. So 3/4 times 2/3 can be thought of as 3/4 of 2/3. At present, it would be difficult to calculate 3/4 of 2/3. I need to change 2/3 into an equivalent fraction, which is divisible by 4. If I had a numerator of 4, that would be helpful because I can find 3/4 of 4. If I multiply the numerator and denominator in 2/3 by 2, that will give me 4/6, so that will be 3/4 of 4/6. Now 3/4 of 4 is 3, so 3/4 of 4/6 is 3/6, and that's equivalent to a 1/2. I could display 3/4 times 2/3 as 3/4 of 2/3 on a grid. And a four by three grid would make sense in this case. First, I'm going to shade 3/4 of the grid, so I will shade three out of the four columns. Then, I'm going to shade 2/3, so I will shade two out of the three rows. The overlap is 3/4 of 2/3, and that's 6 out of 12 parts, and that's equivalent to a 1/2.
2.5 Dividing fractions The standard procedure for dividing by a fraction is one of those where learners can get incredibly confused. ‘Turn it upside-down and multiply’ is the common mantra but which fraction (if there are two) and why does it work? Division is the most difficult of the arithmetic functions, even for whole numbers. It can help to go back to remembering what division means. The grouping method of division is most useful when dealing with fraction division. Watch Video 3 for some useful approaches to teaching division with fractions. Video 3 Dividing with fractions INSTRUCTOR: I'm going to look at the problem 2 divided by 2/3. What does this mean? One understanding of division is that the problem is asking, how many 2/3's are there in 2? I can work this out by counting up in 2/3's and seeing where I get 2. So if I've got one 2/3 and then add on another 2/3 to that, I'd have 4/3. And if I add on another one, I'll get 6/3. Now, 6/3 is 2. So the answer to the question 2 divided by 2/3 is 3. This question was easily solved by counting up in 2/3's, but some questions are not so easy. Consider this question, 1 and 3/4 divided by 1/2, which we could say as, how many 1/2's are there in 1 and 3/4? If I count up in 1/2's, I'd get 1/2, 1, 1 and 1/2, 2, and so forth. So I can see that 1 and 3/4 is in between 1 and 1/2 and 2. So I expect my answer to be a bit bigger than 3. I'm going to look at this by drawing a diagram. I'm going to draw 1 and 3/4 circles. So that's one circle and then 3/4 circles. OK. And I'm going to split those up into halves. So the whole circle will have 1/2 in it and another 1/2. And then this 3/4 circle will be 1/2 here, and then we've got a quarter of a circle here. So I can see from this that I've got three halves, and I've also got the quarter. Now, that quarter is a quarter of a circle. But we're actually looking at 1/2's. So when we're looking at that, we're actually looking at a half of a half. This quarter is a half of a half. So we've actually got one, two, three halves, 3/2, and then a half of a half. So 1 and 3/4 divided by 1/2 is 3 and 1/2. I can also choose to do these calculations using the well-known method of turn it upside down and multiply. So it's a nice, quick, and easy way to do it. But you do have to remember the procedure and how it works. And it's OK if only the procedure is done properly. And what a lot of learners do is think that it's both the numbers that have to be turned upside down, which obviously is the problem here. So it's actually the second number which is turned upside down and multiplied. If we look at this one here, so 2 divided by 2/3 is the same as 2 multiplied by 3/2. OK. So 2 lots of 3/2 is 6/2, which is 3. And in this case, 1 and 3/4 divided by 1/2 is the same as 1 and 3/4 times 2/1, 2/1 being 2. That basically means you're doubling it, so that's 3 and 1/2. What you're doing when you turn it upside down and multiply is you are multiplying by the reciprocal of the second number. In other words, the number which is 1 divided by it. So 2/3 and 3/2, they actually are reciprocals. Because if you multiply them together, you would get 1. And similarly here, 1/2 and 2/1 are reciprocals, because if you multiply them together, you get 1. So this is what's happening here. You're multiplying by the reciprocal, which is, as it were, equivalent to dividing by the original number. The ‘turn it upside-down and multiply’ method works because turning a fraction upside-down produces its multiplicative inverse. Fractions are unlike the integers in that they have a multiplicative inverse which can be used in fraction division because multiplying is also the inverse of dividing. In the next section you will meet the mathematical cakes model again as a way to help learners to find percentages of amounts.
3 Approaches to teaching percentages Percentages are typically easier to handle than fractions as they are effectively fractions with a denominator of 100 (hence ‘per cent’ which means ‘out of one hundred’). This makes it very easy to compare the size of percentages. But remember that, for example, 1% of a large amount could well be bigger than 10% of a small amount.
3.1 Calculating percentages of amounts When calculating a percentage of an amount, many learners reach for the method of putting the percentage over one hundred and multiplying by the amount over one. As an example, 35% of 60 is calculated below using this method. $\begin{array}{cc}35%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}60& =\frac{35}{100}×\frac{60}{1}\\ & =\frac{35×60}{100×1}\\ & =\frac{2100}{100}=21\end{array}$ This method works if you are good at arithmetic and can remember the procedure! However, it is more intuitive to use some common knowledge: that 10% is equivalent to one-tenth, 1% is equivalent to one-hundredth, 50% is equivalent to one-half, and 25% is equivalent to one-quarter. Other percentages can be worked out from these. The mathematical cake, which you met in Section 1 on equivalence between fractions, decimals and percentages, also proves useful in helping younger learners to understand the concept of percentages of amounts. It can be used to find multiples of 10% of amounts by imagining the cake cut into ten slices as before. Watch Video 4 to find out about a useful approach to working out percentages of an amount. Video 4 Using the mathematical cake diagram to find percentages of an amount INSTRUCTOR: Here is a mathematical cake. Mathematical cakes are split into 10 equal slices. The whole cake is 100%. So each slice is 100% divided by 10, which is 10%. Here is a percentage question. Suppose the whole cake costs 3 pounds. How much would 70% of the cake cost? Well, 70% is seven slices of cake. If we work out how much one slice, or 10%, of the cake costs, then we can find the cost of 70% of the cake. 100% of the cake cost 3 pounds. So one slice costs 3 pounds divided by 10. That's 300 pennies divided by 10, or 30 pence. We can write this as naught pounds 30. One slice cost naught pounds 30, two slices cost naught pounds 60, three slices cost naught pounds 90, four slices costs 1 pound 20, and so on, all the way up the cake. Finally, 10 slices cost 3 pounds. Now we can read off the answer to 70% of 3 pounds as 2 pounds 10. Work out these other percentages using the mathematical cake-- 40% of 3 pounds. Read across, and that's 1 pound 20. 80% of 3 pounds-- read across, and that's 2 pounds 40. 50% of 3 pounds-- read across, and that's 1 pound 50. Suppose the cake weighs 500 grammes. We can use the mathematical cake method to work out multiples of 10% of 500 grammes. 10% of 500 grammes is the same as 1/10 of 500 grammes, or 50 grammes. We can label the right-hand side of the cake going up in 50 grammes. No slices weigh naught grammes. One slice weighs 50 grammes, two slices weigh 100 grammes, and so on all the way up the cake. Finally, 10 slices weigh 500 grammes. Then we can read off the answer to the following questions-- 20% of 500 grammes, 90% of 500 grammes, 30% of 500 grammes. Do these questions for yourself and then click on reveal answers to check your answers. $\begin{array}{c}20%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}500\mathrm{g}=100g\\ 90%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}500\mathrm{g}=450g\\ 30%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}500\mathrm{g}=150g\end{array}$ Once learners have been introduced to the mathematical cake diagram, they will certainly be familiar with finding 10% of an amount and multiples of 10% of an amount. It is also intuitive to find 25% and 50%. For example, 35% of 60 can be found by doing the following: $10%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}60=6$ $30%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}60=18$ (multiplying 10% of 60 by 3) $5%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}60=3$ (finding half of 10% of 60) $35%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}60=21$ (adding the results of finding 30% and 5%) Activity 9 Find percentages of 60 Allow 10 minutes A useful exercise to do with your learners is to ask them what percentages of 60 they can find, encouraging them to modify the methods they have previously used for finding simple percentages of amounts. Try this one yourself (Figure 4). There are many possible questions and answers that you could have posed in the diagram. Examples could be 50% of 60, 30% of 60, or 25% of 60. Extend learners’ understanding by posing questions such as: 15% of 60, how could you find that? What about 76% of 60? Or 120% of 60? Ask learners to pose ‘What percentage of 60?’ questions themselves. So far, if your learners have been using the methods shown above, they will have found percentages of amounts by making reference to the equivalent fractions. Decimal equivalents can also be used when calculating percentages of amounts and this is a more helpful form when using a calculator. Referring back to the mathematical cake diagram in Section 1.1, remember that $10%=0.1,\text{\hspace{0.17em}}20%=0.2,$ etc. To convert from percentages to decimals, the percentage has been divided by 100. This follows from an earlier observation about fractions, which stated an alternative meaning of the numerator and denominator, i.e. numerator divided by denominator. In other words: $x%=\frac{x}{100}=x÷100$ For example, if calculating 76% of 60, the learner keys into the calculator: $76÷100×60=$ However, it becomes easy to convert 76% straight away to 0.76 and then multiply by 60. It is worth raising a caution here about using the percentage button on a calculator. Since the percentage button functions in different ways on different makes of calculator , it is best not to use it. Instead, encourage learners to convert the percentage to a decimal and then multiply by the amount.
3.2 Percentage change: using multiplying factors Consider the following problem. A customer goes into a restaurant and pays for a meal. Because they had the ‘early bird’ menu, their bill is discounted by 25%. There is also a 10% service charge added to their bill. What difference does it make to the customer's bill if the waiter deducts the 25% and then adds the 10% service charge, or alternatively, adds the service charge and then makes the deduction? Activity 10 Does the order matter? Allow 15 minutes Work out the solution to the above problem in both ways and see what the difference is. Don’t forget that the second percentage change is enacted on the result of the first percentage change, i.e. having deducted 25% from the bill, the 10% service charge is worked out on the resulting discounted bill. You could choose to calculate the solution of the problem using different amounts for the bill. Or you could prove it once and for all using a general case, which requires you to use algebra, by using a letter to represent the amount of the bill. One form of the solution uses algebra to find the general case: Let the cost of the meal be c. For 25% deduction followed by 10% increase: $\begin{array}{c}25%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=\frac{1}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{find 25% of the cost of the meal}\right)\\ c-\frac{1}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=\frac{3}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{subtract 25% from the cost of the meal}\right)\\ 10%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{o}\mathrm{f}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=\frac{1}{10}×\frac{3}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=\frac{3}{40}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{find 10% of the amended bill so far}\right)\\ \frac{3}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c+\frac{3}{40}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=\frac{33}{40}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{finall cost of the meal}\right)\end{array}$ For 10% increase followed by 25% deduction: $\begin{array}{c}10%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=\frac{1}{10}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{find}10\mathrm{% of the cost of the meal}\right)\\ c+\frac{1}{10}c=\frac{11}{10}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{add 10% to the cost of the meal}\right)\\ 25%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{o}\mathrm{f}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{11}{10}c=\frac{1}{4}X\frac{11}{10}c=\frac{11}{40}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{find}25\mathrm{% of the amended bill so far}\right)\\ \frac{11}{10}c-\frac{11}{40}c=\frac{44}{40}c-\frac{11}{40}c=\frac{33}{40}c\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{find cost of the meal}\right)\end{array}$ Were you surprised at the result? It does not matter which way round the final bill is calculated because a 25% deduction followed by adding on 10% works out the same as adding 10% first followed by deducting 25%. To see why this is, consider the use of multiplying factors for each of the percentage changes. A 10% increase means that, having started with 100% of the amount and adding an extra 10%, you end up with 110% of the amount. $\mathrm{That is equivanet to 1.1 as a decimal or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\frac{1}{10}\text{\hspace{0.17em}}\mathrm{as a fraction}$ A 25% decrease means that, having started with 100% of the amount and subtracting 25%, you end up with 75% of the amount. $\mathrm{That is equivanet to 0.75 as a decimal or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{4}\text{\hspace{0.17em}}\mathrm{as a fraction}$ Using decimals for the multiplying factors (although you can also use fractions as the multiplying factors), an elegant solution to the problem can be found as follows: Subtracting 25% then adding 10% gives c $×0.75×1.1$ Adding 10% then subtracting 25% gives c$×1.1×0.75$ Both multiplications come to 0.825c because it does not matter which way round you multiply (this is known as commutativity). In short, a percentage change can be calculated by using the appropriate multiplying factor, which is calculated by considering what has happened to the original 100%. The use of multiplying factors for percentage change does make the calculations much simpler. However, it may be a difficult concept for younger learners to grasp. If this is the case, it is better to allow learners to work the long way round. In mathematics it is always best to work with the conceptual understanding that learners have demonstrated and to build on this carefully. Learners can do, understand and succeed when they have grasped the conceptual underpinning for the mathematics they are doing. This was summarised by Richard Skemp (1976)as relational understanding. Otherwise, if there is no understanding, learners will resort to trying to learn rules which are easily forgotten. (Skemp referred to the learning of rules and procedures without a conceptual underpinning as instrumental understanding.)
3.3 Finding the original amount when given the result of a percentage change It is necessary to understand what percentage change is before this type of problem can be understood. Finding the original amount from the result of percentage change requires some understanding of what happens when algebra is manipulated. However, first consider two easier examples where intuitive thinking can be sufficient. Activity 11 25% more sunscreen Allow 10 minutes A ‘special offer’ bottle of sunscreen promises 25% more than usual and holds 500 ml. How much does the bottle of sunscreen normally hold? Here is one way to arrive at the solution: 25% more means one-quarter more than the usual amount in the bottle of sunscreen. This means that the special offer holds: $125%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{5}{4}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of the usual amount}$ You need to find out how much 100% or 1 whole of 500 ml. $\mathrm{divide}\text{\hspace{0.17em}}500\text{\hspace{0.17em}}mls\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{by}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{to find}\text{\hspace{0.17em}}\text{\hspace{0.17em}}25%\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\mathrm{or}\frac{1}{4}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{of the amount in the normal size bottle}$ This gives 100 ml. Then multiply by 4 to get the whole of the amount in the normal size bottle, i.e. 400 ml. You can see that the solution was found by dividing by 5 and multiplying by 4. Remembering that a fraction can also denote numerator divided by denominator, this is equivalent to multiplying by the fraction $\frac{4}{5}$ $\frac{4}{5}\text{\hspace{0.17em}}\mathrm{is the inverse of the fraction}\frac{5}{4}\text{\hspace{0.17em}}\mathrm{which is the multiplying factor for an increase in 25%}$ This means that one is the inverted (upside-down) version of the other and, if they are multiplied together, you get 1. It also works by using the equivalent decimals 0.8 and 1.25, which can be checked on a calculator. Activity 12 A coat is reduced in a sale Allow 10 minutes A coat is reduced in a sale by 10% and the sale price is £81. What was the original price of the coat? Have a go at this problem yourself. The sale price is 90% of the original price and you need to find 100% of the original price. 90% of the original price = £81 10% of the original price = £9 (dividing by 9) 100% of the original price is £90 (multiplying by 10)
3.4 Using multiplying factors to find an original amount given the result of a percentage change Now consider the same problem using algebra. The sale price of a coat is £81 and this is the result of a 10% discount. Let p be the original price of the coat. 10% discount leaves 90% or 0.9 of the price You know that 0.9 p = 81 Divide both sides by 0.9 to get p $p=\frac{81}{0.9}=90$ So, the original cost of the coat is £90. This method can be used with much harder numbers and it is a simpler and more elegant way of reaching the solution. However, learners need to understand the method and why it works. This is known as procedural understanding. If learners do not have this procedural understanding then the method becomes a rule to be learned and there are too many such rules in mathematics if your approach is to learn rules rather than develop conceptual understanding.
4 This week’s quiz Check what you’ve learned this week by doing the end-of-week quiz. Week 2 practice quiz Open the quiz in a new tab or window and come back here when you’ve finished 5 Summary of Week 2 After completing Week 2 you have met some mathematical diagrams which can be used to represent fractions, percentages and decimals. You will have seen how such diagrams can be effective in helping to: demonstrate equivalence in fractions, decimals and percentages show how to calculate fractions and percentages of quantities support methods for addition, subtraction, multiplication and division of fractions. An important message from Week 2 is that it is important to build on learners’ understanding of how mathematics works; conceptual understanding, and of how mathematical methods work; procedural understanding. Next week you will build on the learning from week two when you learn about proportional reasoning, one of the important ideas from mathematics. You can now go to Week 3.
Week 3: Developing understanding of proportion and ratio 1 Developing proportional reasoning and an appreciation of multiplicative structure Proportional reasoning involves thinking about relationships and making comparisons between two or more quantities using multiplicative reasoning. It is a complex way of thinking which requires learners to think about numbers and values in relative, rather than absolute, terms. Proportional reasoning is a hugely important part of school mathematics as it underpins many other topics. For instance, it is instrumental in working with fractions, decimals and percentages, as well as similarity and enlargements. Proportional reasoning also assists learners when working on other topics throughout the mathematics curriculum, from simple probability to problems involving trigonometry. In addition to being relevant to several topics in the mathematics curriculum, proportional reasoning also has many practical applications. It is used to calculate best buys, to work with scale drawings and maps, to adjust recipes or create particular concentrations of mixtures and solutions and to perform currency conversions.
1.3 Developing learners’ proportional reasoning Teaching basic algorithms for solving problems involving ratio and proportion will not develop learners’ proportional reasoning on their own. Learners need their conceptual understanding of proportion to be flexible so that they can apply it to the many varied situations in which proportional reasoning is required. Working on repetitive exercises can support learners in remembering methods for solving particular types of ratio and proportion questions. But it does not develop their proportional reasoning or prepare them for unfamiliar problems. Instead, learners need to be exposed to a variety of different problems and approaches. You will consider a range of problems for learners involving ratio and proportion in Section 2 of this week.
1.4 Using concrete resources It is important to give learners of all ages a variety of concrete experiences connected with proportional reasoning to encourage them to compare, make conjectures, find relationships and generalise their learning. Young learners may start with weighing and measuring objects in the classroom, and discussing comparisons between amounts. Activity 6 Sharing money When asked to share out £24 between two cousins in the ratio of 3:1, learners can use counters to represent the pounds (Figure 7). How can you use counters to demonstrate the solution to this problem? Write responses in the box below. Grouping counters in groups of 1 and 3 is one way to work on this problem. By grouping counters in this way, learners share out all the counters in the ratio of 1:3. They can then look at how many of each colour there are altogether. Noticing that there are 6 groups of 3 and 6 groups of 1 would result in finding a total of 18 blue counters and 6 yellow counters, representing the £18 and £6 the cousins will each get. Another way to approach this problem using counters is to create boxes (on paper or on mini-whiteboards) to represent the ratio. In this case, since the ratio is 3:1, you will need 3 boxes for the first cousin and 1 box for the second cousin. The counters, which represent pounds, are then shared out equally between the 4 boxes. The first cousin is given 3 boxes, so receiving £18, and the second cousin is given 1 box, so receiving £6 (Figure 8). This approach can also be used with pens on mini-whiteboards. Learners need to create the boxes to represent the ratio and then put a mark in every box until they get to the total, in this case 24 (Figure 9). This imagery can then support learners in moving towards a more abstract concept of sharing amounts into a ratio. In other words, it can help them to understand that in a ratio of 3:1 there are 4 equal parts in total which the £24 needs to be divided equally between. Note: When moving on to ratio problems involving fractions and decimals, using manipulative resources can become problematic. It is sensible to move onto more abstract methods with your learners before introducing these kinds of problems. <b>Teaching idea: Cooking with ratios</b> Why not incorporate some creative cooking into a maths lesson or some maths into a cooking lesson? To make 12 rice crispy cakes you need: 60 g unsalted butter 3 tablespoons golden syrup 1 x 100 g bar of milk or dark chocolate 90 g Rice Krispies©. Before making the delicious treats, learners could first work out how much of each ingredient they will need to make, say, 6, 24, 36 or more challenging amounts such as 30 or 50. Decorating biscuits with icing is an even simpler recipe which could be used with younger learners.
1.5 Moving to abstract thinking and using pictorial representations Proportional reasoning does not always involve physical objects, so learners need to be able to abstract their thinking about ratio and proportion. When moving from concrete resources to more abstract thinking, it can be helpful to encourage learners to draw diagrams to support their thinking. Activity 7 An example of a proportionality problem Allow 5 minutes If it takes Sakina 30 minutes to read 45 words, how many words could she read in 60 minutes? Minutes can be compared to minutes. Here 60 minutes is twice 30 minutes. This multiplicative relationship can then be applied to the relationship between pages. This is known as a scalar comparison. A pictorial representation of this approach is shown below. Alternatively, pages can be compared to minutes. This is a proportional comparison. Here, 45 is 1$\frac{1}{2}$ times 30, and this relationship can then be related to the 60 minutes. A pictorial representation of this approach is shown below. Note that both approaches give the same result. In the next section you will look at problem solving with proportional reasoning and discuss misconceptions that often arise.
2 Problem solving using proportional reasoning, including common misconceptions Learners in the middle years often struggle with proportion problems. Research consistently highlights learners’ difficulties with proportion and proportion-related tasks and application. There are many common misconceptions that arise within the topic of ratio and proportion and it is important to be aware of these in order to plan to address these in your teaching.
2.1 Problem solving using proportional reasoning Proportional reasoning questions cover a huge variety of real-life applications, from best buys to map scales. Maps and scales Maps and scale drawings use proportionality. In order for a map to be useful, it needs a scale which represents how much smaller it is than the place it represents. This means that journey distances can be accurately calculated. Activity 8 Example problem: model boat Allow 5 minutes A scale model of a sailboat is made to a scale of 1:20 (Figure 13). If the model boat is 17 cm wide, how wide is the actual boat? If the boat has a sail of height 5 m, how high is the sail on the model? The actual boat is 20 times larger, so it will be 340 cm or 3.4 m wide. The model boat is 20 times smaller, so the sail will be 0.25 m or 25 cm high Teaching idea: School or home map Learners can create scale maps of their school or home by taking measurements using trundle wheels, measuring tapes and metre rulers. Then, using an appropriate scale, they can produce accurate scale drawings or maps. Alternatively, learners can calculate how far away different countries are by measuring distances on scale maps in a school atlas (Figure 14). Best buys Having proportional reasoning helps to ensure you get the best deals when shopping, particularly when it comes to offers and multi-buys. Often deals are misleading and larger ‘value’ packs can work out to be more expensive. It is important to be able to compare prices properly in order to make smart choices. Activity 9 Example problem: washing-up liquid Allow 5 minutes Two shops, Fast-Save and Super-Mart, sell the same brand of washing-up liquid in the same size bottles (Figure 15). At which shop is the washing-up liquid the best value for money? You need to be able to compare the two prices, so you need to find the price at each shop for a common number of bottles. One approach to this problem is to find a common multiple of 5 and 6. In this case, 30 is a sensible choice as it is the lowest common multiple of 5 and 6. The price for 30 bottles at Fast-Save is £17.10 (6 $×$ £2.85). The price for 30 bottles at Super-Mart is £19.50 (5 $×$ £3.90). An alternative approach would be to find the unit price for 1 bottle of washing-up liquid at each shop. This would involve dividing the cost by 5 and 6, respectively. Exchange rates A common use for proportionality is currency exchange rates. When you change your money at the Post Office or travel agent ‘commission free’, they will use a different exchange rate when they sell you foreign currency to when they buy it, thus making a profit. Activity 10 Example problem: Japanese yen Allow 5 minutes Zoë is going on holiday from England to Japan. Before leaving, she changes 1500 pounds sterling (GBP) into Japanese yen (JPY). The exchange rate is 1 GBP = 177.415 JPY. Calculate the amount of foreign currency Zoë takes on holiday. This question is quite straightforward since you are given the unit rate for £1 in yen. All that is required here is to multiply the number of pounds by the rate, giving 266 122.50 yen. If the rate was given in another form, for instance: 20 GBP = 3548.30 JPY, another step would be needed to calculate the unit rate. Mixtures and solutions Proportionality is used when producing mixtures and solutions. For example, to make concrete you need 1 part cement, 3 parts sand, and 3 parts aggregate. This is a ratio of 1:3:3, or it can be said that the proportion of cement is 1/7. Activity 11 Example problem: orange drink Allow 10 minutes A bottle contains 750 ml of concentrated orange squash. It is enough to make fifteen 250 ml glasses of diluted orange drink. How much water is needed to make 10 litres of this drink? This question tackles proportion in a real-life context. It is a multistep problem as, first, the total amount of diluted drink needs to be calculated before then comparing ratios. This problem comes from the Nrich website. Suggested approaches and solutions can be found via the link in the ‘Further reading’ section. Teaching idea: A giant’s hand investigation One way to engage learners in working on ratio and proportion is to leave a ‘giant’s handprint’ on the classroom wall for them to find (Figure 16). The learners are then asked to investigate the question ‘How big must the giant be?’ Note: Week 8 of this course looks at how you might conduct statistical investigations with learners. You might find it useful to read that section before carrying out this investigation in your classroom. Activity 12 Reflecting Allow 10 minutes What do the learners need to know? What do they need to find out? How does this problem relate to proportion? The learners will need to measure the ‘giant’s hand’ and use proportion to calculate the size of the giant. The idea behind this activity is that learners will investigate the proportions found in the human body. One way to do this is to collect data from the whole class, with each learner measuring their hand size (length, width or both) and height. This is a hands-on, practical activity which will help learners to get a sense of what proportionality means. Whole-class discussions looking at the class data set may draw out some conclusions about the proportional relationship between hand size and height (or other measures: foot size, arm span, etc.). These relationships can then be related to the giant’s hand measurements to calculate the height (and foot size, arm span, etc.) of the giant. Different learners may come to slightly different conclusions, so encourage them to explain their answers. This will help to develop their proportional reasoning. Note: in exploring the proportions in the human body, you might like to discuss the golden ratio with your learners. This idea was briefly discussed in the Introduction to this week. Some teaching ideas based around investigating the golden ratio can be found on the Nrich website.
2.2 Common misconceptions with proportional reasoning problems In this section, you are offered several problem-solving tasks relating to proportional reasoning. As you read through each problem, consider how you would approach each task yourself and think about the issues, and possible misconceptions, that might arise for learners as they work on each problem. Proportional problem 1: Soup Here is a recipe for vegetable soup (Figure 17). Vegetable soup (serves 2) 200 g chopped mixed vegetables (e.g. onion and carrot) 300 g potato 700 ml vegetable stock 1 tablespoon olive oil. Zoë makes the soup using 600 g of mixed vegetables. What weight of potato should she use? Activity 13 Reflecting Allow 10 minutes Look at this learner’s incorrect response below. Zoe needs 200 g of chopped vegetables in the original recipe. Now she needs 600 g of chopped vegetables. 200 g + 400 g = 600 g She needed to add 400 g of chopped veg, so she must need to add 400 g of potato. So, it must be 300 g + 400 g = 700 g of potatoes. How can you explain the thinking behind this approach? This is a common approach to this type of problem, known as the constant difference strategy. The learner seems to be thinking in absolute terms. They have identified that the difference between the amount of veg and potato in the original recipe is 400 g, so have kept this difference constant by adding 400 g to the new amount of veg, rather than thinking about the relationship between the ingredients. This is an example of additive reasoning in place of multiplicative reasoning, which was discussed in Section 1 of week 3. It is challenging for learners to identify when they should be thinking proportionally rather than additionally. Using diagrammatic representations of ratios and having discussions related to the context of the problem can support learners in moving from additive to multiplicative reasoning. Proportional problem 2: Sharing the workload It takes 3 workers 8 days to complete the foundations of a new building (Figure 22). How long would it take if there was an extra worker? Assume that all the workers will work at the same rate. Activity 14 Reflecting Allow 10 minutes Look at the learner’s incorrect response below. How could you explain to a learner that this approach is incorrect? There are 8 days of work completed by 3 workers. So, 8 ÷ 3 = 2 2/3 days of work per person. Now there are 4 workers, so: 4 $×$ 2 2/3 = 10 2/3 days. This requires some logical thinking, relating the answer back to the original problem. If it takes 3 workers 8 days to complete the job, would it take 4 workers more time to complete the same job? No, it would take them less time, so this answer cannot be correct. If learners have been working on ‘sharing in ratio’ problems, they may go into automatic mode and try to use the same approach with these types of problems, causing them to make mistakes. When working on problems with ratio and proportion, the context of the problem is important. $3\text{\hspace{0.17em}}\text{\hspace{0.17em}}workers×8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{days}=24\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{days of work in total}\text{\hspace{0.17em}}$ $\mathrm{An extra worker}=4\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{workers in total}\text{\hspace{0.17em}}$ $4\text{\hspace{0.17em}}\text{\hspace{0.17em}}workers×?\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{days}=24\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{days of work in total}\text{\hspace{0.17em}}$ $24÷4=6\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{days}$ It will take 4 workers 6 days to complete the same work. Proportional problem 3: Paint A particular shade of green is made by mixing 3 litres of blue paint with 5 litres of yellow paint (Figure 23). How much yellow paint is needed to make this shade of green if you only have 2 litres of blue paint? Activity 15 Reflecting Allow 5 minutes Look at the learner’s incorrect response below. How can you explain the thinking behind this approach? There are 3 litres of blue and 5 litres of yellow paint. So, there is a total of 8 litres of paint needed. There are 2 litres of blue paint. So, to make the 8 litres, 6 litres of yellow paint are needed. This strategy is known as the constant sum strategy. This can be easily tackled using examples for which this strategy is impossible. For instance, if the question had asked ‘How many litres of yellow paint is needed for 12 litres of blue paint?’, the total sum could not have been consistent. In this particular example, discussions about proportionality and the need for consistency in order to preserve the particular shade of green, may support learners. Proportional problem 4: Printing press A printing press takes exactly 12 minutes to print 14 dictionaries. How many dictionaries can it print in 30 minutes? Activity 16 Reflecting Allow 5 minutes Look at the learner’s incorrect response below. How can you explain the thinking behind this approach? The printing press can print 28 dictionaries. This learner appears to have multiplied by 2, despite 12 not being half of 30. This is a common mistake made by learners, who have previously found that doubling has been an effective strategy. This is known as the magic doubling method, when learners use a doubling strategy but it is inappropriate to do so. Similarly, learners often make mistakes by using multiplication but not by the correct factor. It is important to offer learners a wide range of varied examples and to have discussions about strategies, identifying that some approaches will work for particular problems but not for others. As mentioned in the first part of this week, teaching learners methods for solving ratio problems is not enough to develop their proportional reasoning. Learners need to be able to make decisions about when to use different strategies. Proportional problem 5: Bookshop Martha buys 4 books for £11. In this bookshop, all books are the same price. Billy buys 9 books from the same shop. How much do they cost in total? Activity 17 Reflecting Allow 5 minutes Look at the learner’s incorrect response below. How can you explain the thinking behind this approach?
Books Price
4 £11
8 £22
9 £23
This learner has used an incomplete application of the building-up method. They have attempted to use a building-up method, by doubling the original amount to get the price for 8 books and then they have made a mistake in the final step. In the left-hand column, the learner doubled the number of books and then added a quarter of the original number of books. On the right-hand side, this has been translated to ‘double the price and add 1’. As the teacher, unpicking this mistake means you can see that there is some understanding to build upon. Bookshop
Books Price
4 £11
8 £22
1 £2.75
9 £24.75
Adding the extra step into the working could help to avoid the mistake made above.
2.3 Teaching to address common misconceptions In a topic such as this, there are many common misconceptions. It is important not to avoid these issues but instead to build opportunities to discuss these misunderstandings as part of your teaching. Through identifying mistakes and discussing why answers are incorrect, learners will develop both their proportional reasoning and their problem-solving skills.
3 Connecting ratio and proportional reasoning According to the Oxford English Dictionary (OED) Online, ratio is defined as ‘the quantitative relation between two amounts showing the number of times one value contains or is contained within the other’. For example, the ratio of male to female professors is 4 to 1. In contrast, proportion is described by the OED as a part, share, or number considered in comparative relation to a whole. For example, the proportion of greenhouse gases in the atmosphere is rising. In other words, a ratio is used to describe a comparison between two (or more) objects, amounts or values, and a proportion is used to describe one object, amount or value, using a comparison to a whole. For example if the ratio of boys to girls in the class is 2 to 3, or 2:3, the proportion of boys is 2/5. Since the proportion of boys in the class is expressed as a fraction of a whole, it is not necessary to know the proportion of girls. If there are 30 children in the class and 2/5 of them are boys, it can be inferred that 3/5, or 18, of them are girls. In contrast, when expressed as a ratio, the ‘2’ is meaningless without the comparison to the ‘3’. Proportional thinking and reasoning is knowing the multiplicative relationship between the base ratio and the proportional situation to which it is applied. Understanding ratio and proportion is more than merely being able to perform appropriate calculations, being able to apply rules and formulae or manipulating numbers and symbols in proportion equations.
3.1 Using simple ratios to make comparisons First, look at Figures 20–22 and then do the activity based on them. Activity 18 Writing ratios Allow 5 minutes Write the ratios for each of the images shown above in the simplest form. What is the ratio of red rods to yellow rods? What is the ratio of the lengths of the green and white rods? What is the ratio of orange to white cubes? 8:12 which simplifies to 2:3. 10:4 which simplifies to 5:2. 1:3. Using manipulatives When first introducing ratio and proportion, it is important to have a range of concrete resources and manipulative materials available for learners to use. For example, multilink cubes, counters, coins and Cuisenaire rods. (Figure 23) Cuisenaire rods are a great resource for comparing ratios as the rods can all be related in size. For example, the red rod is twice as long as the white rod, so if the white rod is assigned the value of 3 for a particular question, the red rod will represent 6. Rods are a great tool for exploring multiplicative relationships. Note: learners need to be given some time to familiarise themselves with the rods and the relationships between the different rods before they can be useful in the classroom. If you do not have access to a set of Cuisenaire rods, there is a virtual version of them available on the Nrich website, although this does not provide the same concrete resource that the physical rods do. Depending on the age of the learners and the teaching setting, using edible resources for work on ratio and proportion can be a concrete resource while providing a meaningful real-life context for learners (Figure 24). The ability to cut up certain foods, such as cakes or pizzas, into halves, quarters and so on, means that fractional ratios can be explored in a practical, hands-on way.
3.2 Connecting ratio and proportion Throughout this week you have looked at various applications of proportional reasoning. Finally, you will look at how notation for ratio and proportion are linked. Recapping what was discussed at the start of this section, a ratio represents a relationship between two quantities. A proportion represents one quantity in relation to the whole. For example, a tropical drink uses 150 ml of orange juice and 100 ml of mango juice. The ratio of orange to mango, in the simplest form, is 3:2, as for every 3 parts of orange juice, 2 parts of mango juice is needed. Expressed as a fraction, the proportion of orange juice in the drink is $\frac{3}{5}$ as there are 5 parts in total and 3 of these are orange juice. Activity 19 Matching ratio and proportion Allow 10 minutes For each of the sets of counters displayed, write the ratio and proportion in the box below. Ratio of red to blue is 3:1 The proportion of red is $\frac{3}{4}$ Ratio of red to blue is 1:2 The proportion of red is $\frac{1}{3}$ Ratio of red to blue is 4:5 The proportion of red is $\frac{4}{9}$ Ratio of red to blue is 2:3 The proportion of red is $\frac{2}{5}$ Ratio of red to blue is 1:1 The proportion of red is $\frac{1}{2}$ Ratio of red to blue is 1:3 The proportion of red is $\frac{1}{4}$
3.3 Using ratio tables One way to support learners in developing their own mental strategies for solving proportion problems is through the use of ratio tables. Ratio tables are a way to symbolise the problem and can support learners in finding strategies for solution. They encourage approaches such as halving, doubling, and multiplying by 10. Activity 20 Reflecting Allow 10 minutes Two learners have used ratio tables to work on the problem shown below. They have both taken slightly different approaches. Have a look at their workings below. Can you identify each of their strategies? Problem: Seedling plants come in boxes of 35 plants. How many plants would be in 16 boxes? Seedling plants – Sophie’s strategy
 Boxes 1 2 4 8 16 Plants 35 70 140 280 560
Seedling plants – Alejandro’s strategy
 Boxes 1 10 2 6 16 Plants 35 350 70 210 560
Sophie’s strategy Repeated doubling to get to 16 boxes. This works because 16 is a power of 2. If the question was asking for 15 boxes, she could use the same strategy but then subtract one lot of 35. Alejandro’s strategy Multiply by 10. Separately also multiply original amount by 2, then multiply by 3 to get 6 lots. Add 10 lots and 6 lots to get 16. Alejandro has used a building-up strategy. He did not just use a scalar multiplier. Instead, he had to find different parts and sum these. This can be known as the addition and scaling method. In the next example, Alejandro was given the unit amount (that 1 box contained 35 seedlings). This meant that he only needed to use multiplication. In the second example below, the unit cost is not given, so some division, as well as multiplication, is required.
Activity 21 Proportion problem Allow 10 minutes Think about how you might approach this problem before reading the learner’s response below. Problem: Mangoes are 2 for £3. How many could you buy for £7.50? Mangoes – Yasmin’s strategy
 Mangoes 2 4 1 5 Price £3 £6 £1.50 £7.50
This problem is slightly different from the previous examples because it is not as clear how to get from £3 to £7.50 as it was to get from 1 to 16. Because £7.50 is not a multiple of 3, it cannot be made using either Sophie’s or Alejandro’s strategy. Yasmin’s strategy: Double to get £6. Need £1.50 more so adding another £3 will be too much. Half the original amount to get £1.50. Add £1.50 to £6 (1 lot to 4 lots) to get £7.50.
3.4 Continuing to develop learners’ proportional reasoning This week discussed how learners might approach different problems relating to proportion and ratio. You have considered how misconceptions could be addressed and the types of resources that might help learners to develop their understanding. Ratio and proportion underpin many key mathematical topics and have applications in STEM subjects as well as real-life contexts. It is important that learners are given ample opportunity to practise, problem solve and discuss proportional reasoning.
4 This week’s quiz Check what you’ve learned this week by doing the end-of-week quiz. Week 3 practice quiz Open the quiz in a new tab or window and come back here when you’ve finished 5 Summary of Week 3 In Week 3 the focus has been on proportional reasoning and its link to multiplicative relationships. You have seen how the use of concrete objects can support early understanding of proportional reasoning, progressing to pictorial representations and, later, abstract mathematical notation as learners’ thinking becomes more sophisticated. You have worked on a range of proportional reasoning problems, looked at examples of common misconceptions and considered how these could be addressed. You should now be able to: define what is meant by proportional reasoning recognise when a learner is using additive reasoning or when they are using multiplicative reasoning use effective approaches for teaching proportional representations have an awareness of the common misconceptions which learners meet when working through proportional reasoning problems. Next week you will study the first of two weeks on algebra, looking in particular about how the foundation of algebraic understanding is introduced and developed. You can now go to Week 4.
Week 4: Introducing algebra 1 Mathematical sentences and algebraic expressions It is tempting to define algebra as the strand of mathematics that uses letters for numbers. However, the roots of algebraic thinking lie in paying attention to quantities and their structural relationships. One definition that respects this broader view holds that algebraic thinking can be can characterised by: Indeterminacy: the problem involves unknown quantities. Denotation: the unknown quantities are represented, for example by words or symbols. Analyticity: the unknown quantities are operated on as if they were known numbers (Radford, 2014). In the next section you will be asked to have a go at an algebraic task. At the end of the section you will consider how it meets these three requirements.
 1 P = (a + b + 3) – 3 + (a + b + 3) – 3 A On the green triangle it’s only a and b because the 3 is missing. On that one it’s 3 and b and a but then you’ve got to take 3 away again. 2 (a + b) + 3 + b+ a – 3 B The perimeter is 3 add side b then it’s the little bit down there, then it’s add side b and side a. That bit is what’s left over when you put the 3 next to a. 3 P = 3 + b + (a – 3) + b + a C To make the perimeter – for each whole triangle it’s a plus b plus 3, but it’s missing 3 from each one. 4 (a + b + 3) + b + (a – 3) D Count all the sides, it’s a plus b plus 3, doubled, and then you miss out 3 twice. 5 2 (a + b + 3) – 2 x 3 E That side is the same as the missing side so you have all of one triangle. For the other triangle you have side b and just a bit down side a, a less 3.
All of the expressions are equivalent to 2a + 2b but they represent different thinking processes. You should have matched 1C, 2A, 3B, 4E, 5D.
The next section examines the idea that you can express mathematical reasoning in words or in algebraic expressions. The mathematical reasoning that you are doing in this task is giving instructions for calculations without actually doing the calculation.
2 Finding the general in the particular Early algebra involves moving between the particular and the general. When we talk about moving to the general in mathematics (also called generalising), we are concerned with finding features of a situation, thinking about how they may change and extending our reasoning to cover those changes. This has many similarities with our aims for learning – we want our learners to generalise beyond today’s lesson. You have seen that algebraic thinking involves working with unknown quantities, so you can think of this movement in two ways. First, teachers and learners can choose different representations for the unknown quantities. Some representations are closely linked to their own particular classroom experiences, while others draw on generally accepted conventions for symbolic notation. Second, the indeterminate quantity being represented can be thought of as having a specific (but unknown) value or it can be thought of as a variable, capable of taking a range of values. Mathematical statements about a specific unknown are likely to lead to finding its value; whereas statements about a variable say something more general, since they must hold for any of its possible values. This section looks at these two movements, starting with representations.
2.1 Representations The work of Jerome Bruner (1966) has been influential in early algebra. He identified three modes of representation for mathematical objects: the enactive, the iconic and the symbolic, which move broadly from the concrete to the abstract. Learning a new concept is supported by meeting each of these modes and translating between them. In the enactive mode, the concept is represented through learners acting on concrete, physical objects. For algebra, the concept of an unknown quantity can be represented by a bag or tin containing an unknown number of counters. The teacher can ask questions such as 'What happens if three counters are placed in a tin? What if one counter is taken out? What if there are ten tins with the same number of counters in each?' A variable can be represented as a changing length, for example learners can measure the height of a sunflower. In the iconic mode, a picture illustrates the unknown quantity (Figure 3). The picture takes its meaning from the learner’s previous enactive experiences. In the symbolic mode, the concept is represented in an abstract or a conventional way. Bruner considers words (such as ‘five’, ‘add’ or ‘tin’) to be symbolic in the same way as conventional letters and signs are (e.g. 5, + or t). Concrete–pictorial–abstract Some teaching traditions describe a learning progression through concrete–pictorial–abstract and develop their materials accordingly. There are often intermediate stages in the final translation towards symbols/abstract representations. For example, Figure 3 shows two different pictorial/iconic representations of the tin activity. One use photographs of objects and arrows that represent the actions of adding and taking away counters. In the other, the learner has drawn their picture of a tin and combined it with familiar symbols such as +, 3 and =. Bruner’s three modes offer a useful way of planning for progression, but their order does not have to be interpreted too strictly. Teachers have found that even young children can reason algebraically about a quantity that they have not handled, when the meaning is carefully established through discussion. Equally, what feels concrete or abstract to a learner can change with experience. Words such as the ‘price of a book’ are considered as symbolic for younger learners, but they can feel iconic once their language has developed. For example, learners often choose to use symbols such as b (book) or p (price) which remind them of the words they are using for a quantity. A learner studying A-level mathematics will be able to use x, y or any other letter for the price and will eventually have enough experience of manipulating symbols with pencil and paper that this will start to feel like a concrete activity on which she can base further learning. Making sense of symbols Research in early algebra suggests that learners need to establish meaning in two ways. First, they need to understand how calculations and arithmetical relationships between quantities are expressed using the familiar symbols for numbers and operations. This is why teachers emphasise forming mathematical sentences in words and in number symbols as well as calculating answers. Second, learners need to make sense of working with an algebraic symbol to represent an indeterminate quantity. These are both necessary stages, which lay the foundations for manipulating symbols out of a context. If learners move too quickly to symbolic representation they risk interpreting the symbols incorrectly, introducing misconceptions that use their prior experiences from number work to deal with the troublesome letter. Activity 6 Reflecting Allow 10 minutes Look at this question and the range of responses. Which would you expect to see from your learners (try it with them)? What is the misconception that underlies each answer? When you have finished the task watch the video. $\begin{array}{c}\mathrm{If}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4,\text{\hspace{0.17em}}\mathrm{what is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2a+1?\\ A\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}25\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}B\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}C\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}3}a\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}D\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}E\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}17\end{array}$ Video 2 Discussing the misconceptions behind wrong answers INSTRUCTOR: We are interested in learner responses to the question-- if a is 4, what is 2a plus 1? We know that 2a plus 1 means 2 times a, add 1. The most common learner response is likely to be 9, in which case, they're working out 2 times 4-- 4 instead of a-- then plus 1. And the answer is 9. And that's the most common response. But you're also likely to get a range of incorrect responses. One of these is the answer 3a. In this case, the learner's probably looking at 2a plus 1 and ignoring the letter. So they see 2a plus 1 and write the answer 3. In some cases, they then add the a back in, again, just to do something with it. So that's likely to be an incorrect response you'd get, and we call it letter not used or letter ignored. This is also the case with the incorrect response 7. Here, we have 2. The a is replaced by 4. And we have plus 1. That's read as an instruction to add up the numbers 2, 4, and 1, which gives 7. Sometimes learners, aware that this is algebra, will add an a in. Those three answers are the ones that you're most likely to find. Sometimes you'll also find an answer such as 25. Here, the learner looks at the 2a next to each other and thinks about this as a two-digit number. So 2 with a 4 is 24, giving the answer 24 plus 1 is 25. This is quite a reasonable view, as it is how we write two-digit numbers. But in algebra, 2 followed by a has to be recognised as 2 times a. Finally, you might find you get strange numbers, such as 17. What probably happened here is a confusion between 2a and a squared. 2a is a plus a or 2 times a. But confusing the addition and the multiplication can give us a times a. When a equals 4, that is 4 times 4 plus 1 equals 17. So those were a range of correct and incorrect responses to the question-- if a is 4, what is 2a plus 1? You have the correct response-- 9. You have responses such as 3a and 7, where the letter is ignored. And the calculation is done on the numbers available, perhaps with the letter added in afterwards. And you have a sensible, but incorrect interpretation of 2a as a two-digit number. In your class of, say, 30 learners. Most will give the correct answer. But you will have a few who show these misconceptions. If you see these responses, you will know that they are based on some reasoning, but it includes a misunderstanding of algebraic conventions.
2.3 Comparing algebraic expressions Learners find it easier to follow word problems that are posed in a way that matches a typical sentence structure, such as the THOAN problems above. However, even a small variation on these will start to disrupt the order. For example, it is much harder to work on a problem like this one (which comes from an examination for 11-year-olds): Younus takes three large letters to the Post Office. He pays with a £20 note. He gets change of £14.80. How much did each letter cost to post? Earlier you have seen there is a progression in moving from working with a quantity that represents a specific unknown to one that can take a range of values. You will return to the ideas of variables and generality from a different perspective in Week 5, when you will look at introducing functions. In the middle years, learners will move between these perceptions of number as they tackle different problems. Some learners strongly prefer to think of an unknown quantity as having a specific value. They experience what researchers call ‘a compulsion to calculate’. You can test this preference by asking diagnostic question such as these: $\begin{array}{c}\mathrm{If}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2m=12,\text{\hspace{0.17em}}\mathrm{what is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2m-1?\\ \mathrm{If}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2m=52,\text{\hspace{0.17em}}\mathrm{what is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2m-1?\\ \mathrm{If}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m+n=15,\text{\hspace{0.17em}}\mathrm{what is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}m+n+2?\end{array}$ Activity 9 Reflecting Allow 5 minutes How did you answer these questions? Did you choose specific values for m and n? Do you see now that you did not need to? The answers are: 11, 51 and 17. The first question involves such a familiar number bond that you are very likely to have thought of: m as 6. The second was chosen to have the same structure but vary the number involved so that it becomes easier for you to pay attention to the comparison of 2m and 2m – 1. Some of your learners will not do this. Instead they will work out m = 26 and then calculate $26×2-1=51$. The third question again has a similar structure but here m and n can take a range of values, so they are not specific unknowns. Again, some learners will choose values, for example 10 and 5, and then work out 10 + 5 + 2 = 17. Some will show awareness that these are not the only possible values by trying another pair, for example 7 and 8. Another way is to use tasks that allow learners to treat quantities as variables while providing some support by being embedded in a number context. The strategies discussed above lead learners to a correct answer but they reveal the ‘compulsion to calculate’. These learners have difficulty in dealing with unknowns by comparing algebraic expressions, and this will prevent their progression to more complex algebra. One way of working on this difficulty is for teachers to ‘productively linger’ with algebraic problems – asking learners to explain their thinking and listen to others. Obtaining the answer is not enough: learners should use two different methods and be able to compare them. Activity 10 Symbol sums Allow 10 minutes This task involves working with several unknowns and there are several different ways to approach it. Have a go at it yourself. Each symbol in the grid in Figure 5 has a value. The totals of the rows and columns are shown. Can you find what each symbol is worth? Activity 11 Thinking points Allow 5 minutes Where would you encourage learners to start? How long would you leave learners to work independently before you ask them to listen to other people’s strategies? What kind of verbal explanations would you expect from learners? How do you want them to record their thinking? Is the task finished when the first learner works out the values? How could you ‘productively linger’ over this task to encourage algebraic thinking? You can find a similar activity at https://nrich.maths.org/1053 You could start the problem in several ways: The first row gives you the sum of heart and smiley. Compare it with the fourth column to find the value of a cloud. Compare the second and third rows to find that heart is 1 more than smiley. The fourth row shows a cloud has to be a multiple of 3: try 3 or 6 or 9. Compare the third and fourth column to see that a diamond is 3 more than a smiley. The third row and third column show a diamond is 5 more than a lightning. Activity 12 Symbol sums continued Allow 5 minutes It is a good exercise of your own mathematical knowledge to invent similar problems and check that they can be solved. In the two grids in Figure 6, the symbols in the grid are the same but the level of mathematics is slightly harder. Can you see why?
3 The equals sign One of the strong findings of early algebra research is that learners in the middle years need to change their perception of the equals sign. This sign is recognised from an early age (Figure 7), and it is usually understood as standing for an instruction such as ‘makes’ or ‘work it out now’. This is called the operational understanding of the equals sign and it works well for number problems. Text books, worksheets and even national examinations use the equals sign in this way: $15+5=$ $\mathrm{Area}=\mathrm{length}×\mathrm{breadth}$ To make progress in algebraic thinking, learners need to develop a new relational understanding of the equals sign. In algebra, the equals sign only makes sense when it is used as a way of comparing two expressions and saying that they have the same value as each other. All of the following equations can be understood as statements that express a relationship between two quantities: $10×4=40$ $10×4=20×2$ $5\text{\hspace{0.17em}}\mathrm{m}=500\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{cm}$ $2/3=4/6$ $2\left(\mathrm{y}+6\right)=2\mathrm{y}+12$ Two other relational symbols are the inequality signs < and >. Tasks that require learners to decide which of these symbols to use can develop a relational understanding of the equals sign. Activity 13 Relationships between expressions Allow 5 minutes Which is bigger: 3ab or 2a + b? Choose values of a and b and decide. Record your results using the signs <, > and =. You could write: If we choose $a=1,b=2$ then $3×1-2=1$ and $2×1+2=4$, so $1<4$. If $a=1,b=4$ then $3×1-4=-1$ and $2×1+4=4$ so $3a-b. Or $a=2,b=1$ then $3×2-1=5$ and $2×2+1=5$ Activity 14 Reflect Allow 5 minutes You might have noticed the different ways in which the three responses to the previous question were written. How would you ask your learners to record their results? Does this ensure that they have practice in selecting and writing the correct signs, in comparing two expressions?
 a b 3a – b 2a + b Write your inequality or equation here 1 2 1 4 3a – b < 2a + b 10 3 6 3
Using the equals or inequality signs between numbers or symbolic expressions is correct mathematics. Learners should not write =, < or > on its own. Using a table is often helpful for keeping track of several calculations.
 a b 3a – b 2a + b Write your inequality or equation here 1 2 1 4 3a – b < 2a + b 10 3 27 23 3a – b > 2a + b 3a – b 2a + b 15 15 3a – b > 2a + b
3.1 Relational understandings Treating the equals sign as a relation is important for two aspects of algebraic thinking. The first is maintaining equal value. The two expressions on either side of the equals sign are equivalent; their values balance. Anything that is done to one side of the equation must be done to the other side, in order to keep that balance. This aspect is seen in the common metaphor of an equation as a balance scale (Figure 8). Sometimes this aspect is referred to as sameness. However, looking at the balance makes it obvious that there is not the same expression on both sides. Instead, they have equal value. What is the equation represented by this balance? What happens if you take 1 from both sides of this balance? If you take the number represented by x from both sides? The second benefit of the relational understanding of the equals sign is that it allows substitution. If two expressions are equal in value then one can be substituted for the other. This is useful in simplifying expressions and solving equations. You met substitution in the symbol grid activities. Watch the video below to see how. The equals sign as substitution is used twice in this process: the first time it is used to replace an indeterminate quantity (expressed symbolically as a sum) with a number that has equal value. The second time the recall of number facts is used to determine a number that could be substituted for symbolic quantities, and a new equation can be written. Video 5 The equals sign – substitution Substitution or ‘do the same to both sides’ It is useful for teachers to be aware of the substitution aspect of the equals sign as learners will use this in their speech. They are likely to draw on the substitution when they know the number facts involved (here it is 9 + 6 that equals 15). When they do not know the number facts, they will use the principle of ‘maintaining equal value’ and ‘do the same to both sides’ in order to work out 15 – 9 = 6. Both aspects need to be recognised and taught.
3.2 Always, sometimes, never true? As previously discussed, algebraic expressions are like mathematical phrases or sentences. Equations are like mathematical statements. A mathematician will always ask whether or not an equation is true. When equations contain no unknown quantities, use calculation to check if they are true: $3+4-2=6$ is an equation but it is not true. $3+22=24+1$ is an equation that is true. When equations include unknown quantities, they may be true for only one value of the unknown: $2n-5=17$ is only true when n is 11. If n = 10 then 2n – 5 = 17 is false. They may be true for several values: ${n}^{2}+1=10$ is true when n = 3 and when n = –3. They may be true for all values of the unknown: $2\left(n+3\right)=2n+6$ In formal mathematics you can use the sign ≡ to show that these expressions are ‘identically equal’: $2\left(n+3\right)\equiv 2n+6$ Sometimes an equation is given as a definition or a formula for working something out. We can assume that someone else has done the work of showing it is always true. For example: ${a}^{3}=a×a×a$ (definition). $\mathrm{Area of a triangle}=\mathrm{half base}×\mathrm{height}$ (formula a learner could prove). $\mathrm{E}={\mathrm{m}\mathrm{c}}^{2}$ (formula that a learner could not prove!). Activity 15 Always, sometimes or never true Allow 10 minutes Decide whether these equations are true for all values of the variable, or only some values or none. Trying values will help you see that there are no values of b or m that make their equations true: they are never true. Using algebraic reasoning will help us see that e = 1, t = 1.5, c = 0, q = 10 are specific values that make their respective equations true. Also, some equations are true for a range of values: k < 15, p either greater than 2 or less than -2, and x greater than 1 or less than 0. A final one, that is more subtle because it involves two variables, is that p + 12 = f + 12 is true only when p = f. All these equations are sometimes true. Finally the equations 2n+3 = 3+2n and 16s2 = (4s)2 are true for all values of n and s. The first is because addition is commutative, and the second because squaring applies to the whole bracket. They are always true.
4 This week’s quiz It’s now time to complete the Week 4 badge quiz. It’s similar to previous quizzes but this time, instead of answering five questions, there will be 15. Week 4 compulsory badge quiz Remember, this quiz counts towards your badge. If you’re not successful the first time, you can attempt the quiz again in 24 hours. Open the quiz in a new tab or window and come back here when you’ve finished. 5 Summary of Week 4 In Week 4 you met the idea that algebra is more than simply using letters for numbers. Algebraic thinking involves an appreciation of structure and generality. It allows us to work with unknown quantities as if they were numbers, using words or symbols to represent those numbers. You have seen how algebraic thinking can be supported by the progression in the use of concrete, pictorial and abstract representations. You should now be able to: describe algebraic expressions and algebraic terms support learners to move between working with numbers and working with symbols deal carefully with the equals sign and be aware that it is often used to mean ‘makes the answer of’ but that in algebra it has the meaning of equivalence (i.e. both expressions on the right and left of the equal sign have the same value). Next week you will build on what you learned in week four to begin to think about functional reasoning and specifically how unknown quantities can vary. You are now halfway through the course. The Open University would really appreciate your feedback and suggestions for future improvement in our optional end-of-course survey, which you will also have an opportunity to complete at the end of Week 8. Participation will be completely confidential and we will not pass on your details to others. You can now go to Week 5.
2.1 Term-to-term and position-to-term rules For three of the sequences in Figure 6 you probably thought about how to get from one term to the next, and continued that pattern. You could have done this by adding 5, doubling, or drawing another triangle. These are all ways of describing the change between any two consecutive terms, so this is called finding a term-to-term rule. It would be much more difficult to identify a term-to-term rule for the other two sequences in Figure 6. Instead, you need position-to-term rules. This type of rule connects the position number (1, 2, 3, etc.) to the associated term. A table is a useful representation to show the relationship between position number and the term.
 Position 1one 2two 3three 4four 5five 6six n Term(letter sequence) O T T F F S ? Term (fraction sequence) 1 1/4 1/9 1/16 1/25 1/36 1/n2
For the letter sequence, the position-to-term rule is ‘the term is the first letter of the position number as an English word’. For the fraction sequence, ‘the term is the fraction 1 divided by the position squared’. You can write this symbolically as $\mathrm{nth term is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{{n}^{2}}$ Term-to-term rules describe the pattern you see by moving along a row of the table. Position-to-term rules describe the pattern you see when moving down from the position row. The rule must describe a general pattern that holds for every term of the sequence. Learners often prefer looking for term-to-term rules. This is for two reasons. First, they only need to look along the changes in one variable – along the terms of the sequence. Second, term-to-term rules involve simpler operations. However, a term-to-term rule is not helpful if you want to find a term that is far away from the ones that are known. The position-to-term rule is more powerful. Activity 4 Position-to-term rules Allow 10 minutes Imagine finding the 37th term in each of the five sequences in Figure 6. Which one would be easiest? Which one would be hardest? Write the position-to-term rules for the first two sequences.
 Sequence 37th term Position-to-term rule 5, 10, 15, 20, 25 185 Multiply position number by 5.nth term is 5n 3, 6, 12, 24, 48 3 × 236= 206 158 430 208; most calculators will not display all of these digits. Raise 2 to the power of one less than the position number. Multiply the answer by 3.nth term is 3 x 2n-1 1, 1/4, 1/9, 1/16, 1/25, ... 1/ 372 nth term is the fraction 1 divided by the position squared.nth term is1/n2 O, T, T, F, F T take the first letter of the position number
There is another reason to emphasise position-to-term rules. Sequences are an early example of functions. If we relate this to the function machine: the position number is the input, the term is the output and there is a rule that connects them.
When learners start to work with sequences, they work with specific values of the position numbers: 1, 2, 3, etc. These are not indeterminate quantities, so they are thinking numerically. The algebraic thinking only starts when learners consider what the rule is for a general input number, which is usually denoted by the symbol n. The corresponding term is called the nth term. It is relatively easy to understand how the position number can change, since it follows the natural numbers. So, although, teachers may never mention the word ‘function’ when teaching sequences, by emphasising position-to-term rules they are nevertheless preparing the underlying idea of a function as a relationship between two variable quantities.
2.2 Generalising with sequences Sequences also give learners a context in which to generalise, particularly when they are defined using diagrams or shapes. If you look at sequence resources you will find two main approaches to generalising activities. We can describe these as looking for changes versus looking for structure. Many textbooks written in the 2000s focus on the first approach, while more recent research (Kieran et al., 2016) suggests that learners also need to look for structure. The next two activities in this section exemplify each approach. In the first approach, the intention is for learners to observe a pattern of change across a sequence of examples by asking ‘What is the same and what is different?’ Activity 5 Looking for changes Allow 10 minutes Each term of the sequence in Figure 7 is a house. Draw the next two terms. What is the same and what is different about the houses? What would the 26th house look like? How many matches are needed for each of the houses? Complete the table below to make a number sequence. Then describe a method for finding how many matches are needed for house n.
 Position 1 2 3 4 10 26 Number of matches
Did you continue the pattern by looking at the changing shapes? Or by thinking about numbers – how many matches you would need? You will explore each of these ways of finding the nth term. (This is the position-to-term rule.)
Changing shapes Learners will come up with several different ways of looking at the way these ‘houses’ grow. As you saw in Week 4, having multiple solution methods can give rise to fruitful algebraic reasoning if there is time planned in the lesson for learners to compare their methods, correct errors and explain why they are the same. One very common method of describing the changing shapes is to focus on the 3 horizontal rows of matches. Another method is to note that each house has the same beginning and the same end. The middle section grows each time by adding 3 matches (1 each to the roof, the floor and the middle). We will now follow this second method to show its details. The method is described in pictures, and then words and symbols. This sequence can be used for any methods. Pictures (see Figure 8) Words The number of matches for the 10th shape will be 5 (for the beginning) plus 2 (for the end) plus 3 for each of term added on after the first. You add one less group of 3 than the shape number. Number symbols It can be helpful for learners to see the calculations written as a sequence. Shape 1: 5 + 2 or 7 Shape 2: 5 + 3 + 2 or 7 + 3 Shape 3: 5 + 3 + 3 + 2 or 7 + 3 + 3 Shape 4: 5 + 3 + 3 + 3 + 2 or 7 + 3 + 3 + 3 Now that we have got past 3 × 3, it is worth moving to multiplication instead of repeated addition. Shape 5: 5 + 4 × 3 + 2 or 7 + 4 x 3 Words and algebraic symbols Looking at the calculations shows a pattern. A method for finding the nth term is (starting with the part that changes) to calculate the number that is one less than n, multiply that by 3 and add it to 7. In symbols this is (n-1) × 3 + 7. We can simplify this by expanding the brackets: 3n – 3 + 7 3n + 4. So, the nth term of the sequence is 3n + 4.
2.3 Changing numbers If, instead of looking at the shapes, you focused on the number of matches, you probably noticed that the terms went up the same amount each time, in threes. Sequences that have a constant difference have two names (teachers need to know these but learners do not). They are called linear or arithmetic sequences. For linear sequences, it is possible to work only with the numbers and ignore the shapes, and some teachers prefer to work like this. They rely on learners remembering a procedure: Remember that the 3 times table increases in threes. Add a row in the table for the multiples of 3. This row has the rule 3n. Work out what the change is between 3n and the sequence you want. Adapt the rule to include this change Let’s demonstrate this procedure. Step 1 and 2
 Position 1 2 3 4 10 26 Number of matches 7 10 13 16 3n 3 6 9 12 30 78
Step 3 Looking at the third and the second rows (e.g. look at 3 and 7, or 6 and 10), the difference between them is + 4, so we write the rule as number of matches = 3n + 4 This helps complete the table for the 10th and 26th terms. Should teachers work with the shapes or the numbers? The advantage of working only with number patterns is speed. It provides a relatively quick method for finding a position-to-term rule for a given linear sequence. On the other hand, it is relatively common for learners to misapply this procedure and give the answer 3n – 4. Another disadvantage is the reduced opportunity for thinking algebraically since learners tend to focus only on the first term of the sequence (so they are not working with indeterminate quantities). In addition, simply finding a position-to-term rule is a skill that is worth a few marks in examinations, but is not used directly to support other topics. If learners can use the numerical method without writing an extra row in the table, it is a good indication of fluent thinking with symbols. If they cannot, the teacher needs to balance the effects on their learners of spending time memorising and perfecting the procedure, or of trying to make sense of change within a context.
2.4 Looking at structure in a sequence We now return to the second main approach, where the focus is on looking for structure in a sequence. This is similar to looking at shape, rather than number, but it uses only one term from the sequence. A typical term is chosen – usually the third or the fourth, since the first term is often too simple to show the structure. Activity 6 Looking for structure Allow 10 minutes An activity based on this approach will have two phases of exploration. In the first short period, learners decide on their ‘shape four’. They can be supported by asking ‘Where do you see the “threes” in shape three?’. This phase ends with collecting several examples of ‘shape fours’, displaying them for comparison and asking ‘If this is shape four, do we know how to make shape five?’ For this problem, you might allow the class to divide into two groups: those who want to extend only the ‘arms’ of the C (so that the fourth shape has 9 squares) and those who have also extended its ‘spine’ (so it has 10). Either method can be justified but they lead to different expressions. The task for the second phase is for pairs to explain in their books using a labelled diagram what ‘shape one million’ would look like, and find how many squares would be used for any shape. Isometric dotty paper is a good resource for creating more patterns to practise this approach (Figure 10). Ask ‘If these are shape 3, what is shape 4?’
2.5 Summary So far this week, you have met the idea of a function machine, with input and output, and the idea that sequences are functions where the input is the position number and the output is the term. In both of these representations, the function can be considered as a process, that is, as something that you do to the input. In Activity 1 you became more sophisticated in your treatment of the function by reversing the process. In Activity 2 you compared two functions. In Activities 3 to 6 you also saw how a function can be considered as a relationship between two quantities, that is, as something to notice and describe. This relationship stems from the structural connections that link the two quantities (which gives the position-to-term rule) and it determines how they change together. This will be the focus of the next section.
3 Extending functions: variables and graphs The idea of a function as a relationship is sometimes called covariance; because it describes how two quantities vary together. If your learners go on to study advanced mathematics, they will need to be able to shift fluently between the process perspective on function (associated with rules, function machines and sequences) and the relationship (covariance) perspective. In the middle years, the goal for most learners is to have a sound understanding of a function as a process and to develop some awareness of the representations related to functions that will prepare them to notice different kinds of covariance. In this section you will meet some of the more advanced vocabulary and concepts associated with these representations.
3.2 Increasing and decreasing graphs In the Video 2, the graph of time in minutes against time in hours was described as increasing, since whenever the $x$-variable increases, the y-variable does too. It is also possible for graphs to be decreasing. The definition for a decreasing graph is: whenever the $x$-variable increases, the y-variable decreases. Most graphs are neither increasing or decreasing. Linear graphs In the video you saw an example of a linear (straight-line) graph (Figure 11). If a linear graph is increasing then its gradient is positive, since when $x$ increases by 1, y changes by a positive number. If a linear graph is decreasing, the gradient is a negative number. What happens when the gradient of a linear graph is 0? The y-value does not change when $x$ increases and so the graph is horizontal. Activity 8 Function detective Allow 5 minutes Recall the variables from Activity 7: age / number of hours of sleep number of people / how many chocolates each person gets sharing a box of 24 chocolates your age / your elder sibling’s age. Match each pair with one of the graphs in Figure 12 below. Is each graph increasing, decreasing or neither? Graph A You should have found that graph A showed the relationship between the number of people sharing a box of 24 chocolates and how many chocolates each person gets. One way of checking this is to look for the points (1,24), (2, 12), (3, 8) that all lie on the graph. The graph is decreasing. Notice also that this graph is shown as a set of discrete points since the number of people and the number of chocolates are both discrete variables. You could have chosen either variable for the x-axis (and the other for the y-axis) because this graph is the graph of a reciprocal function. The rule is that: the number of people sharing a box of 24 chocolates x how many chocolates each person gets = 24 In symbols we could write that xy = 24 Or y = 24/x Any function where y is a constant divided by x is a reciprocal function. A reciprocal function does not have a constant gradient since the slope changes along the graph. Graph B Graph B shows your age (on the x-axis) and (on the y-axis) the age of an (imaginary) sibling who was 3 years old when you are born. So it starts at (0,3). This point is called the y-intercept: it is the point where x takes the value zero. The graph is increasing. The gradient of the graph is 1 since your sibling’s age increases by 1 for every year you gain. The graph shows a linear function since it is a straight line with a constant gradient, but it is not a proportional relationship (see Week 3) because it does not start at (0,0). It is drawn as a line graph since age can be considered as a continuous variable. Graph C This graph is neither increasing or decreasing since there are intervals (between ages 20 and 60) in which the x-value increases but the y-value stays the same. It represents age on the x-axis and recommended hours of sleep on the y-axis. It only shows some data points, not joined by a line, so the variables are being treated as discrete variables. However, in the context we are clearly meant to use the data at the given ages to infer (make predictions about) the sleep needs at other ages. It is unlikely that there is a simple mathematical function that will describe this graph. Graph D This graph is not one of those shown. It is an example of a piece-wise linear function, since the different sections are each straight lines. It shows weekly salary on the x-axis and how much income tax you pay on the y-axis (using the 2018-19 tax bands). The curriculum in England suggests that learners should be able to read information from graphs in financial contexts. Research suggests that it is necessary to discuss the context with learners before asking them to interpret graphs. Most 11-14 year olds would not be familiar with the idea that employers take tax out of your weekly pay. So far we have met the linear graph which increases regularly so we can calculate a gradient. It has equation y = mx + c where m is the gradient and c is the (starting) value of y when x is zero. These graphs typically occur for costs, for travel or for conversion graphs are linear. the piece-wise linear graph which has straight line sections. It does not have a simple equation. These graphs occur when there are different types of cost or travel combined on one graph. the reciprocal graph with equation y = k/x .This graph typically occurs when sharing a constant amount k. There is one final graph that we will meet in the next section.
3.3 The exponential function In the next two activities you will meet one last important type of function and graph: the exponential function. Activity 9 Word of mouth Allow 10 minutes Three friends share a secret. On day 2 they each tell one new person. Then every day for a week, everyone who knows the secret tells it to one new person. How many people know on day 2? How many people know on day 3? Estimate how many people will know the ‘secret’ on day 8. On day 2, 6 people know the secret. On day 3, 12 people know. This is a sequence you met before: 3, 6, 12, 24, 48. The term-to-term rule is ‘double each term to get the next’. The position-to-term rule is 3 × 2n-1. By day 8, a large number of people will know. You can use the formula or a graph to find the value (Figure 14). In this problem, the number of new people who learn the secret each day is equal to the number who know it already. This makes it an example of an exponential function. The characteristic feature of these functions is that their rate of change is proportional to their value. In more everyday words, this means that the higher the y-value, the more it grows. It is worth comparing this with a linear function, which grows at a constant rate. This feature lies behind the English phrase ‘growing exponentially’ which is used to emphasise rapid growth. It also makes exponential functions difficult to graph. The scale on the y-axis has to be large to include the points for higher values of x, but then the y-values for low values of x are not easy to read. As well as exponential growth you can have exponential decay, which gives a decreasing exponential function. Activity 10 Exploring exponentials Allow 20 minutes Exponential functions are used in many real-life situations. In biology, they describe the spread of diseases and the growth of populations. In engineering, they describe waves, and how movement dies away. In finance, they describe how money grows when it is invested at a constant rate of interest. Although exponential functions appear on the school curriculum for the middle years, learners are expected to plot and read their graphs and not to work with their equations. This has been included only for your own knowledge asa teacher. The role of teachers is to support learners in gaining a good mathematical understanding of graphical representation in a range of contexts. Learners need to be able to choose scales and plot x-values and y-values on a graph. They need to find the y-value for a given x-value and vice versa. They also need to interpret change on a graph. They need to know whether the graph is increasing or decreasing, and whether changes are constant or can vary. The linear, piece-wise linear, reciprocal and exponential graphs are all functions that illustrate different kinds of change and this is their value for teachers.
4 This week’s quiz Check what you’ve learned this week by doing the end-of-week quiz. Week 5 practice quiz Open the quiz in a new tab or window and come back here when you’ve finished 5 Summary of Week 5 In Week 5 you have looked at functional thinking and how this is introduced using the idea of the function machine which works by operating on an input to produce an output. This led to a consideration of number sequences where the input is the position in the sequence and the output is the number (or term) at that position. Finally the idea of covariance was introduced where there are two quantities which vary together according to a relationship. The two quantities can be plotted on a graph and the shape of the graph is determined by the function itself. Although the concept of the function is a sophisticated one, the approaches you met in week five show that it is possible to carefully introduce it to learners in the middle years. You should now be able to: introduce learners to the concept of functions using concrete and pictorial representations (such as the function machine) and to develop this towards more abstract representations of functions use other representations, with which the learners are familiar, to build on their understanding of functions explain to learners how two unknowns can vary together according to the rule (or function) connecting them. Next week you will learn about the teaching and learning of geometry, including how shapes are defined according to their properties and problem solving in geometry. You can now go to Week 6.
Week 6: Teaching and learning about geometry 1 How children think and reason about geometric shapes The study of geometry has its foundation in the learner’s experience of the environment in which they live. More than any other area of mathematics we learn about shape and space in an intuitive way through our experience. By the time children go to school, they have already learned much about shapes and have developed spatial perception in a natural way by interacting with their environment. It would seem appropriate that any approach to learning about the geometry of shape must build on the intuitive understanding which is gleaned from our lived experience. Nevertheless, even though children come to school with an impressive understanding of shape and space pertaining to their environment, most encounter difficulties when learning about geometry.
1.2 Defining shapes using flexible definitions In this section you will consider how more flexible definitions of shapes can be used to include other kinds of shapes. Being able to write a definition of a shape is an example of analytical thinking. As learners’ reasoning develops, they are able to define shapes in different ways and to develop definitions which are more inclusive. As an example of this, consider the rectangle and the square (Figure 5). What happens if you were to take one side of the square and pull it? The following video shows this happening. A square being ‘pulled’ into a rectangle INSTRUCTOR: Here is a square. If I pull on one of its sides, I can make this square into a rectangle. I can make the square into different rectangles whose width is the same as the side of the square, but whose length changes. You have just watched a square being pulled into a rectangle. Might a square be thought of as a special kind of rectangle? (In mathematics we call this ‘a special case’.) Activity 4 Writing definitions Allow 5 minutes Write down a definition of the rectangle. Write down a definition of the square. Activity 5 More definitions Allow 10 minutes Below are two definitions of a rectangle, with a subtle difference between them, and one definition of a square. Is it possible to include the definition of a square into either of the two definitions of a rectangle? Definition 1 of a rectangle A rectangle has four sides and four right angles. It has two longer sides of the same length, which are opposite each other, and two shorter sides of the same length which are opposite each other. Definition 2 of a rectangle A rectangle has four sides and four right angles. It has two pairs of equal sides which are opposite each other. Definition of a square A square has four sides which are the same length and four right angles. The definition of a square cannot be included in definition 1 because its definition of a rectangle stipulates that there are two sizes of equal sides: the longer sides and the shorter sides. The definition of a square can be included in definition 2 because its definition of a rectangle says that there are two pairs of equal and opposite sides. This also includes the case when these two pairs of equal sides are also equal to each other, in which case the rectangle is also a square. The understanding that some classes of shapes can be included in other classes of shapes is called the concept of inclusion. So, for example, squares can be seen to be special rectangles. However, in order to accept the concept that squares are special rectangles, a more inclusive definition of the rectangle, such as definition 2, needs to be used. Activity 6 Including some shapes as special cases of other shapes Allow 5 minutes Try to identify other inclusive relationships. For example, is there a definition of the kite which would include the rhombus? Can rectangles be thought of as special cases of the parallelogram?
1.3 The visual or physical representation of shapes If you consider a shape, such as a square or a circle, you have a mind’s-eye picture of the shape which includes the ‘squareness’, or the ‘circleness’, and this shape may be filled in (with colour) in your mind, and it may have a particular orientation. You may draw a diagram of this shape with a pencil and the outside edges are likely to have some width (or you could not see them). If you have a square or circular tile then the tile will have depth (or you could not pick it up). The diagram and the tile are representations of the shape and as such are examples of what the shape can look like. In contrast, the mathematical shape itself is entirely abstract and is described using its geometric properties. Orientation and thick pencil lines are not included in those properties. However, teachers and learners need to work with representations of shapes as if they are mathematical objects because it is all they have to work with. It is worth being aware of the differences because learners’ attention may be distracted by particular features of the physical representations of the shapes.
2 Investigating the properties of triangles and quadrilaterals In the previous section you read about the properties of some shapes and how combinations of those properties form the definitions of those shapes. Triangles and quadrilaterals are important because they form the basis of other shapes and are generally included in the school curriculum for the middle years.
2.1 Triangles Triangles are the most basic of the polygons (a polygon is a straight-sided shape). They have three sides, the smallest number of sides it is possible to have. Triangles also make a rigid structure, as any engineer will know. You will know the names of the triangles which have further properties in addition to having three sides (Figure 6). In addition, the isosceles right-angled triangle has a right angle, two equal sides, and two equal angles of 45 each. Acute-angled triangles have three angles all smaller than a right angle. Obtuse-angled triangles have one angle that is bigger than a right angle.
3 Developing problem-solving skills in geometry So far this week, you have read about children’s geometrical thinking as it develops through the early to the middle school years. This has been linked to the description of the properties of shapes. You have looked at definitions of shapes and how these can be written in an inclusive way to include some shapes as special cases of others. You have considered triangles and how these can generate special quadrilaterals by reflecting a triangle in one of its sides or rotating the triangle using a half-turn around the midpoint of one of its sides. In this section you will think about developing learners’ problem-solving skills in geometry. Many problems in geometry are entirely abstract from the point of view that there may be no practical real-world application. However, geometry can be approached in a practical way and this makes it accessible and engaging for learners. Geometry does also have real-world applications in nature, in design and architecture for example. Three problems will be posed that you will be asked to have a go at answering. After each one you will consider how to best pose these problems for learners, including the resources needed and how to manage these in the classroom context.
3.3 Enlarging shapes using repeating tiles The activities in this section can be linked to enlargement as a transformation. Some shapes can be enlarged when several smaller versions of themselves are laid together (Figure 19). Squares, rectangles and parallelograms all come into this category. Note how many of these shapes fit together to make an enlargement which is two or three or four times as big. Triangles are also repeating tiles. Try this with different versions of a triangle. You can either make a triangle tile from card or using ICT. You may have to rotate some of the triangle tiles. Activity 16 Using repeat tiles to make a version which is twice the size of the original Allow 10 minutes Explore other shapes which can be used as repeating tiles. For example, show how the shape below can be generated from four smaller versions of itself. Repeating tiles as an activity with learners This task gives the opportunity for learners to create and manipulate cardboard tiles in order to explore which shapes are repeat tiles. Alternatively, the tiles could be produced using ICT. For example, some shapes are generated by the ‘Insert Shapes’ facility of Microsoft Word. Repeat tiles provide an approach to enlarging shapes because the final result of placing together several repeat tiles must be a correct larger version of the original repeat tile. When four repeat tiles are joined together then the length of each side of the enlargement is 2 times the length of the side of the original tile. In order to create a shape where the lengths of each side are 3 times the original tile, you need to use 9 tiles. Learners can investigate the number of tiles needed to generate enlargements of 2, 3, 4 and 5 times the original tile. This can lead to the observation that the factor of enlargement for area is the square of the factor of enlargement for lengths.
4 This week’s quiz Check what you’ve learned this week by doing the end-of-week quiz. Week 6 practice quiz Open the quiz in a new tab or window and come back here when you’ve finished 5 Summary of Week 6 In Week 6 you read how the study of geometry can be very intuitive, drawing on learners’ experience of the lived in world and as such it is a very practical area of mathematics. On the other hand teachers and their learners work with diagrams and physical objects which are themselves representations of abstract geometrical objects. There is a tension between the specific physical representation which can be seen and touched and the abstract concept in geometry (e.g. the triangle on paper represents a whole set of all possible triangles in the abstract). You should now be able to: understand that geometrical shapes have properties and that their definitions draw on these properties use varied activities to support the exploration of geometric shapes, both 2D and 3D think about how to set up the classroom to facilitate problem solving activities in geometry, including use of practical resources. Next week you will learn about measures, how these were developed through history, and how to address the appropriate use of measures within mathematics lessons. You can now go to Week 7
Week 7: Developing understanding about using measures 1 Measures and measuring In this section you will consider different types of measuring scales and read about the history behind the development of measures.
1.1 Measuring scales Quantities of things are measured using measuring tools such as a ruler or a thermometer. These measuring tools may have numerical scales and the quantity is read from the scale. Measurement scales can be: Ordinal scales, when things are put in order, e.g. 1st, 2nd, 3rd. Interval scales, such as temperature, where equal differences on the scale represent equal differences in temperature. However, there may be no true zero point. This means it is not possible to say that a temperature which is 40⁰ Fahrenheit is twice the temperature as 20⁰ Fahrenheit. Ratio scales, which satisfy the properties of measurement, in that equal intervals of the scale represent equal intervals of what is being measured and there is a non-arbitrary zero. Examples include length, area, volume, angle and age. Measurements are thought of as being rational numbers (i.e. whole numbers and fractions or decimals) even though they may represent irrational numbers such as the diagonal of a square or the circumference of a circle which involve numbers such as the square root of 2 or π (pi). This links with the idea of approximation in measurement, which will be considered later. Activity 1 Reflecting on learners’ difficulties with using measures Allow 5 minutes Have you observed learners using measuring tools such as a ruler or tape measure, weighing scales, a thermometer or clocks? What problems or misconceptions were evident? Learners may struggle with understanding that measuring (a length, a weight, a time interval, etc.) means dealing with a continuous quantity which starts with zero on the measuring scale, rather than with 1 as in counting. Research evidence indicates that when students measure lengths, they may be applying a poorly understood procedure, rather than focusing on the correspondence between the units on the ruler (seeing it erroneously as a counting device) and the length being measured (Battista, 2006). Time measurement is given in analogue form (a round-faced clock with the numbers 1 to 12 on the face where the passage of time is analogous to the turning of the hour and minute hands) and digital form (using numbers to indicate hours and minutes passed). Time has a different base (e.g. time uses base 60 and 24) from decimal (base 10).
1.2 The historic development of measures The development in the use of measures happened because people wanted to trade (in foodstuffs, amounts of cloth, etc.) and because communities needed to divide areas of land between family groups. Informal measurements began with reference to parts of the body, for example: The inch was the length from the tip of the thumb to its joint. The foot was the length of a man’s foot. The yard was the length from a man’s nose to the tip of the index finger on his outstretched arm. I know that, if I wanted to buy some cloth from a merchant, I would have sent the tallest man in my family! Hence the need for these measures to be standardised. The first time that measures were standardised in the UK appears to have been in 1215 in the Magna Carta. Some of the old imperial measures have strange names. For example, 22 yards is equal to one chain. My historian friend tells me that every community had a chain. This was literally a chain made of links, 22 yards long and used to measure the ground when setting out fields. Imperial measures of length Below is a list of imperial measures of length and how they relate to each other. 12 inches to 1 foot 3 feet to 1 yard 22 yards to 1 chain (real chains, see previous comment) 10 chains to 1 furlong (still used in horse racing) 8 furlongs to 1 mile 1760 yards to 1 mile. Imperial measures of weight Below is a list of imperial measures of weight and how they relate to each other. 16 ounces (oz) to 1 pound (lb) 14 pounds to 1 stone 8 stones to 1 hundredweight (cwt) 20 hundredweight to 1 ton. Imperial measures of volume (liquid capacity) Below is a list of imperial measures of volume and how they relate to each other. 20 fluid ounces (fl oz) to 1 pint 8 pints to 1 gallon. If you are as old as me, you will have had to learn all of these conversions between units of measure! The United Kingdom was ‘decimalised’ in 1971 which meant that the UK changed from using money in bases of 12 and 20 (12 pence in a shilling, 20 shillings in a pound) to the current system of 100 pence in a pound. This change in the measurement units for money happened fairly quickly but metrication (the change from the imperial system to the metric system), which began officially in 1973, remains unfinished. Schools teach the use of metric units but also have to teach learners about imperial measures because there are still areas of public life where these are used. Examples include distance in miles, vehicle speeds in miles per hour (mph) and beer sold in pints.
1.3 The metric system The metric system is based on grams (g), litres (l) and metres (m). The prefixes always mean the same. The common prefixes are as follows. Kilo 1000 1 kg = 1000 g 1 km = 1000 m Centi 1/100 1 cm = 1/100 metre 100 cm = 1 metre Milli 1/1000 1 mm = 1/1000 m 1000 mm = 1 m 1 millilitre = 1/1000 litre 1000 ml = 1 litre Water connects all three measures! 1 cm × 1 cm × 1 cm = 1 cubic centimetre (cm3) and it holds 1 ml of water which weighs 1 gram (g). 10 cm × 10 cm × 10 cm = 1000 cubic centimetres and it holds 1 litre of water weighing 1 kg.
1.4 Useful ‘rough’ metric to imperial equivalences Below is a list of useful ‘rule of thumb’ equivalences between metric and imperial measures. 10 cm is about 4 inches 30 cm is about 12 inches = 1 foot 8 km is about 5 miles A litre of water is a pint and three-quarters. Two and a quarter pounds of jam, weighs about a kilogram.
1.5 Everyday measures In this section you will use your everyday experience to think about appropriate use of measures. Activity 2 Having a sense of common objects and their measures Allow 10 minutes It helps to have a sense of what a metre looks like, what a kilogram feels like, and so on. Take a look in the kitchen cupboard and list items which are measured in grams, kilograms, millilitres, centilitres, litres and metres. It can help children to consider everyday items and their measures. Here are some common measures (clearly they are approximate). A packet of crisps weighs 25 grams. An apple weighs about 100 grams. A bag of sugar weighs 1 kilogram. A family car weighs 1 tonne. (Note: imperial tons and metric tonnes are similar in size.) A teaspoon holds 5 millilitres. A bottle of wine holds 75 centilitres. A bottle of fizzy drink holds 2 litres. A ruler is 30 centimetres long. The room is 2½ metres high. Activity 3 ‘The tailor’s rule of thumb’ Allow 15 minutes ‘The tailor’s rule of thumb’ was said to be used to make shirts and trousers in the past. It is quoted in the novel Gulliver’s Travels by Jonathan Swift: The seamstresses took my measure as I lay on the ground, one standing at my neck, and another at my mid-leg, with a strong cord extended, that each held by the end, while a third measured the length of the cord with a rule of an inch long. Then they measured my right thumb, and desired no more; for by a mathematical computation, that twice round the thumb is once round the wrist, and so on to the neck and waist, and by the help of my old shirt, which I displayed on the ground before them for a pattern, they fitted me exactly.(Swift, 1726, p73) Work out how the Lilliputian seamstresses calculated Gulliver’s wrist, neck and waist measurements. The mathematical computation is: ‘Twice round the thumb is once round the wrist Twice round the wrist is once round the neck Twice round the neck is once round the waist.’ By measuring round the base of the thumb, a tailor can get a rough estimate of wrist, neck and waist measurements. An interesting classroom activity is to get learners to see how accurate this is with string. However, caution is needed where learners have issues about body image.
2 Reading measures and accuracy In this section you will look at different types of measuring scales and consider appropriate degrees of accuracy in measurement.
2.1 Measuring scales Reading a length from a ruler or tape measure is straightforward because these measuring devices are generally marked in unit lengths and each unit length is subdivided. A metre ruler may be marked in centimetres, each of which are subdivided into tenths or 0.1 of a centimetre. The tape measure shown in Figure 2 has inches on one side, subdivided into quarters and sixteenths, and centimetres on the other. Note that the 10s figure for centimetres is only marked for every 10 centimetres. Reading temperature from thermometers is usually straightforward because they are marked in degrees Celsius (⁰C) and/or degrees Fahrenheit (⁰F) (Figure 3). However, although each gradation may be a unit length, they may be numbered in 20s, 10s or 5s. Reading a weight from a weighing scale may be more challenging. Some scales may be marked in kg or lb but the gradations might not be in the subunits. It may be necessary to count how many intervals a unit has been divided into and enact a division to find out what each subdivision represents. See the picture of the scales in Figure 4. Reading the amount of fuel left in the tank of a car is a case in point. Fuel gauges rarely tell you how much fuel is in the tank. Instead, they tell you what fraction of a full tank is left (Figure 5). It is worth saying that many measuring tools are digital and provide a measurement in digits rather than on a scale. Time measurement provides a common example. We continue to measure time using analogue clocks, where the passage of time is reckoned by the position on the clock face of the hour-hand and the minute-hand (Figure 6). Yet we have become used to digital clocks which provide the time as a series of digits (Figure 7).
2.2 Degrees of accuracy in measurement Depending on the context, it may not always be necessary to give an exact answer to a measurement. At other times, it is actually more acceptable to have rounded the amount to an appropriate degree of accuracy. The population of countries is often quoted in millions because the exact figure changes constantly. For example, the population of the UK is ‘about 65 million’. It can also depend on what you are going to do with the measurement. For example, when measuring a window for curtains, the width to the nearest 5 cm would be more than sufficient. But if you are measuring the same window for a replacement pane of glass, a much higher degree of accuracy would be needed. It should be understood that measurement is continuous, so lengths, weights, etc. can take any value on the number line including decimals and fractions. This is in contrast to amounts which are counted, such as the number of sweets in a jar, which can only take whole number values. However, large numbers of items to be counted may act as continuous measures, such as national population sizes. Accuracy is often quoted using the number of decimal places. If you need to round a measurement to 1 decimal place that implies to the nearest one-tenth. If you need to round a measurement to 2 decimal places that implies to the nearest one-hundredth. A number such as 3.8567 lies in between 3.8 and 3.9; numbers with 1 decimal place. You might remember the video on decimal places which explains how the digits after the decimal point denote decreasing powers of ten: tenths, hundredths, thousandths, and so on. Therefore, the only useful digit in deciding whether to round 3.8567 to 3.8 or 3.9 is the second number after the decimal point, i.e. the 5 in this example. You could call this the first unwanted number. Since that number is 5 or more, convention dictates that the number is rounded up to 3.9. Similarly, with large numbers such as millions, you need to consider the first unwanted number to decide whether to round up or down. So, 65 million people in the UK may have been rounded from, say, 65 345 211. This number lies between 65 million and 66 million but is clearly closer to 65 million. The first digit in the column below one million (i.e. in the 100 000 column) is 3, which is less than 5, and so you round down. Activity 4 Rounding numbers Allow 5 minutes Now have a go at rounding these numbers to the degree specified. 62 to the nearest 10. 14.237 to the nearest 1 decimal place. 14.237 to the nearest 2 decimal places. 465 to the nearest 100. 60 14.2 14.24 500 Finally: measurement as a human activity is not an exact science. We are bound by our own skills with measuring and the tools we use for measuring. If we are using a school ruler, for example, we can measure accurately to one decimal place but not more than that. Suppose we do a calculation such as the ones mentioned earlier for the length of the diagonal of a rectangle. We are likely to get a calculator answer with many figures after the decimal point. We cannot measure more accurately than one decimal place and so it makes sense to round the answer to 1 decimal place.
3 Teaching angle measures and angle facts Learners struggle with the idea of angle being a measurement of turn and also a measure of the gradient between two lines. In particular, learners can be distracted by the lengths of the lines which make up the angles. The angles between the two sets of lines shown in Figure 8 may be considered to be different even though the angle of turn from one line to the other is the same.
3.1 Using a protractor to measure angles The half-circle protractor is commonly used to measure angles (Figure 9). The inner number scale rises from 0 to 180, from right to left, and the outer scale goes from 0 to 180, left to right. Learners commonly make mistakes because they are not using the correct scale when measuring the angle between two lines. The point to remember when measuring angles is that you need to start at zero on one of the lines and read round on that scale to see the number on the second line. In order to read the number from the protractor, it may help to use a ruler to make the lines longer so that they meet the scales on the protractor. Make sure that one of the lines in the angle sits on the zero line. Then, using the scale which starts at zero (the inner scale in this case), read off where the second line meets the scale (24⁰ for this particular angle). If a learner gets 156⁰ for the angle, it is obvious that they have made a mistake. Why is this obvious? The angle is smaller than a right angle (acute), so the number of degrees must be less than 90. Degrees in a circle As you can see from the half-circle protractor displaying 180⁰, it is commonly accepted that there are 360⁰in a circle. No one is entirely sure where this originated, possibly with the Babylonians [2000-1600 BC]. It is certainly worth having at least one plausible explanation ready when teaching this in order to answer the question ‘Why?’
3.2 Angle sum of triangles The angles of any triangle sum to 180⁰. This can be proved in several ways and can easily be shown to convince learners as a classroom activity. Activity 5 Demonstrating the angle sum of a triangle Allow 15 minutes You will need paper, pen, ruler and scissors. Draw any triangle and mark the angles at the corners. Cut out the triangle. Tear off the corners then put the corner angles together. They should all form a straight line. If we accept that there are 360⁰ in a circle and 180⁰ on a straight line then we have shown that the three corners add up to 180⁰. This activity can be done in the classroom. If 30 learners can do this activity using 30 different triangles then they will have demonstrated the angle sum of a triangle in 30 different ways. This is not a proof, merely a demonstration that it works. It is much more fun for learners if they discover angle facts for themselves, rather than being simply told them. Think of other ways for learners to discover the angle facts using drawing and cutting, and measuring using a protractor. Activity 6 A problem using angle facts Allow 15 minutes Below is a design using 8 congruent (i.e. exactly the same size and shape) rhombuses. Calculate the two different angles in the rhombuses. Think about the number of rhombuses which meet in the middle of the design. Their sum must be 360⁰. Think about the rhombus as a quadrilateral. Its angles must sum to 360⁰. A rhombus has two pairs of equal angles. Use compasses and a protractor to recreate this design. Begin by drawing a circle and marking off 8 equally spaced points on the circumference. Use these points to draw the bottom halves of the 8 rhombuses. To find the top points for each rhombus use the compasses opened to the same radius as when you drew the circle. Draw 2 arcs (parts of circles) from each point which will intersect in pairs above the rhombus halves. These intersections are the top points of the rhombuses. Join these to the circumference to complete the rhombuses.
4 Teaching about area and volume Area is a measurement of the amount of space inside a 2D shape and is reckoned by the number of unit squares inside the shape. Common metric units of area are square centimetres (cm2), square metres (m2) or hectares (ha) for measuring land areas. A hectare is defined as 100 m × 100 m or 10 000 m2. Imperial measures of area are square inches, square feet and acres for measuring land areas.
4.1 Area It is best to introduce the topic of area by considering the inside of shapes as being divided into unit squares with area being recorded as the number of squares and the units of area (e.g. 10 cm2). In contrast, the perimeter of a shape is the total length of its boundary but this is often confused with area by learners. The next stage in learning about area is to find the area of rectangles which have sides measured as whole numbers. It is straightforward for learners to note that the rectangles contain an array of unit squares (Figure 12). This leads to the general formula for the area of a rectangle = length × width. Using the area formula for a rectangle to derive other area formulas The area formulas for parallelograms and triangles can be easily derived from the area formula of a rectangle (see Video 1). Watch this video to see how the area formula of a rectangle can be used to derive the area of a parallelogram and then a triangle. Video 1 Deriving the area of a parallelogram and the area of a triangle INSTRUCTOR: Step one, draw a rectangle. The area is a times b, which is length times width. Step two, draw a line from one corner to the opposite side to make a triangle. Step three, cut out the triangle. Step four, move the triangle to the other side to make a parallelogram. Area equals a times b, which is base times height of the parallelogram. Step one, draw a parallelogram. Area equals base times height. Step two, draw one of its diagonals. Step three, cuts along the diagonal. Check the resulting two triangles are the same by placing one over the other. The area of one triangle is 1/2 the base times the height. Since the parallelogram has been made by moving parts of the original rectangle without losing anything, the area of the parallelogram must be the same as the area of the rectangle. It is worth bearing in mind that if the area formulas are promoted too quickly, before learners have conceptualised area as being a measure of the inside of a shape, then area measurement becomes a calculation without reference to the geometric situation.
4.2 Converting square metres into square centimetres If you imagine a tile which is one square metre, then 100 square centimetre tiles would line up along its edge. You could fit 100 rows of these hundred square centimetre tiles inside the metre square. That’s 100 × 100 = 10 000 square centimetres in one square metre (Figure 13).
4.3 Volume Volume is a measurement of the amount of space inside a 3D shape and is reckoned by the number of unit cubes inside the shape. Common metric units of volume are cubic centimetres (cm3) or cc, and cubic metres (m3). Examples of imperial measures of volume are cubic inches and cubic feet. It is best to begin the topic of volume by considering the inside of 3D shapes as being divided into unit cubes with volume being recorded as the number of cubes and the units of volume (e.g. 12 cm3). Volume can be investigated practically using 1 × 1 × 1 cm multilink cubes. For example, learners can be asked to use multilink cubes to make a cuboid which is 3 cm by 4 cm by 2 cm and to work out how many cubes are needed to make it (Figure 14). Investigating different sized cuboids should lead to the formula for the volume of a cuboid as Volume = length × width × height. As with the area formulas, the volume of other more complicated 3D shapes can be derived from the formula of the volume of a cuboid.
4.4 Converting cubic centimetres into cubic metres If you imagine a box which is one cubic metre then 100 cubic centimetres would line up along its edge. You could fit 100 rows of these 100 cubic centimetres in one layer. There would be 100 layers of cubic centimetres in the cubic metre. That’s 100 × 100 × 100 = 1 000 000 cubic centimetres in one cubic metre (Figure 15).
5 This week’s quiz Check what you’ve learned this week by doing the end-of-week quiz. Week 7 practice quiz Open the quiz in a new tab or window and come back here when you’ve finished. 6 Summary of Week 7 In Week 7 you have seen how it is important to be aware of different measuring scales and the various types of measuring instruments. Taking a measurement involves human judgement which is therefore limited to a particular degree of accuracy. For example, we can usually only measure lengths correct to the nearest millimetre. You have also read a brief summary of the development of measuring systems such as the imperial system and the metric system. Giving learners a history of how measures were developed can be interesting and demonstrate how the use of measures drives from human activity. You should now be able to: check that your learners use measuring tools such as rulers, appropriately, for example making sure they put the zero on the ruler at the beginning of the line they are attempting to measure. When measuring an angle, checking that they place the protractor correctly and read from the appropriate scale help learners understand that area is a measurement (by unit squares) of the amount of space within a 2D shape and that volume is a measurement (by unit cubes) of the amount of space within a 3D shape derive the area formulas of a parallelogram and a triangle from the area formula of a rectangle. Next week you will focus on data handling, specifically statistics and probability. You can now go to Week 8
Week 8: Working with data and uncertainty 1 Collecting data and illustrating it in charts In this section you will consider different methods for collecting and recording data. In Sections 2 and 3 we will look at analysing and interpreting data.
1.1 Collecting data Data is a collective name for information recorded for statistical purposes. There are many different types of data, which we will discuss in this section. Data can be collected in several ways. These include face-to-face or telephone interviews, or asking participants to complete questionnaires in person or by post. Data can also be collected online via an internet search or through other more active data- collecting methods, such as counting the number of passengers on a bus at a specific time of day. When data is collected from an original source for a specific purpose, this is known as primary data. For example, a learner who is investigating the most popular make and model of cars in their town might count cars which pass a particular set of traffic lights. When data is not collected from an original source, this data is secondary. For example, a learner might find data on goals scored in the football Premier League last year in order to investigate whether strikers from a particular country score more goals on average. Since the data was not collected for this purpose originally and was not collected by the learner themselves, this data is secondary.
1.2 Population and sampling The population of a survey is everyone who can be questioned in relation to that survey. For example, if a learner wanted to find out the most popular meal in their school canteen, the population of the survey would be all of the students, and possibly also the teachers, in the school. A sample is a small selection of the population, for example just one class in a school. There are advantages and disadvantages to using entire populations and samples. Table 1 Advantages and disadvantages of sampling
Population Every opinion is includedRepresentative and therefore reliable Time consumingExpensive
Sample QuickMore cost effective Only some opinionsSelection method could cause bias
Activity 1 Reflecting Allow 5 minutes How could a selection method mean that the results of a sample could cause bias? When data is collected from the entire population, all opinions are taken into consideration. When data is collected from a sample of the population, it is important that the sample is representative of the population. For example: Joe wants to find out whether motorbikes or cars are more popular, so he has asked 20 people at a motor bike convention which they prefer. Comment: Joe’s data selection method is clearly flawed as he is much more likely to find that people prefer motorbikes within this sample of the population.Thus they are unlikely to be representative of the wider population. Considering potential bias when sampling data is part of having data sense. Sampling When collecting data from a sample of the population, decisions have to be made about how the sample is chosen. It is important to think about how to ensure the sample is representative of the wider population. There are many different sampling methods. Here are a few examples. Convenience sampling is the most simple and straightforward way to collect a sample of data since participants are chosen in the most convenient way and are often asked to volunteer. Statistical investigations carried out by young learners are likely to use this form of sampling, for example when investigating favourite pets earlier in this section, the learners used their own classmates for convenience. Random sampling can be done by assigning a number to each member of the population and then using a random number generator to select the required number of individuals. Systematic sampling involves choosing individuals at regular intervals. For instance, if a sample of 20 is needed from a population of 100, every 5th individual would be used in the sample. This sample will obviously vary depending on how the population is ordered. This is a convenient and simple way to collect a sample but may result in some bias as some smaller groups within the population could be omitted from the sample. A stratified sample takes into consideration different groups and their proportions within the overall population, using the same proportions in the sample. For example, if a school has more students in year 7 and fewer students in year 11, the sample would take a greater number of participants from year 7 and fewer from year 11. This is done using percentages. A quota sample does not take into consideration the proportion of each group in the population. Instead, a quota sample may require there to be 20 male and 20 female participants in the sample, regardless of whether there are many more of either gender in the wider population. Sampling methods are not generally covered in the middle school curriculum. But it is important for learners to be aware that their sampling choices will affect the reliability of their findings when conducting statistical investigations.
1.4 Tally charts and frequency charts Tallying is a way of recording data in groups of five. Recording frequencies in this way means totalling the number of tally marks made (figure 1). Tally charts can be used to collect and organise primary data. Look at the tally chart in Figure 2. The first column contains all of the possible data options. Note that there is an ‘other’ box in case any of the answers do not fit into the first four categories. As with questionnaire questions, it is important that every possible answer can be recorded in a data collection table such as this one. The second column is for a tally, so that results can be recorded and added to.The final column is the frequency. This is the total for each data category once all the data has been collected. It is also possible to have a frequency chart without a tally column, like the example shown below. Table 2 What is your favourite ice cream flavour?
 Flavour Frequency Chocolate 14 Vanilla 5 Strawberry 7 Other 20
Activity 3 Reflecting Allow 10 minutes Did you notice that the ‘Other’ category in the example above had the highest frequency? What conclusions could you draw from this? Sometimes when collecting data, the results can be surprising. Participants sometimes give answers you do not expect. It may be that there are popular icecream flavours that have not been listed. In some instances, it is worth introducing further categories once some initial data has been collected.
1.5 Grouping data When there are a large number of possible outcomes, it may be necessary to group the data. For example, if a learner is investigating how many DVDs each of their classmates own, the values could vary from 0 to 100 (or more). Introducing 100+ different possible outcomes into a table is going to be very time consuming and difficult to plot in any kind of graph. Instead, grouping the data in a sensible way can make data collection and analysis much more straightforward. Below you will see a grouped frequency table which could be used to investigate how many DVDs the learner’s classmates own. Table 3 How many DVDs do you own?
Number of DVDs Tally Frequency
0–5
6–10
11–20
21–40
Over 40
Activity 4 Reflecting Allow 5 minutes Did you notice that there was no overlap in the class intervals and that all possible outcomes were included? Did you notice that the class intervals in the example above are not equal? If there is no upper bound for a numerical category, it is common to use ‘over’ or ‘more than’ in order to cover all of the possible responses. Similarly, if there is no lower bound, it is common to use ‘less than’, so the first category could have been ‘less than 5’. Depending on how the data is going to be used, it might be necessary to use classes with equal widths. We will discuss this later in this week.
1.6 Two-way tables When data falls into more than one category or a learner is interested in investigating more than one variable, data cannot be organised easily in a frequency table. Two-way tables are a way of sorting data so that the frequency of each category can be seen quickly and easily. For example, a learner has asked 20 people about whether they like football and whether they like rugby. The results can be seen in the two-way table below. From the table, it can be seen that there were 9 people who liked both rugby and football and 4 people who liked neither sport. It can also be seen that there were people who liked rugby but not football, and 5 people who liked football but not rugby. Table 4 Do people who like football also like rugby?
Like football Do not like football
Like rugby 9 2
Do not like rugby 5 4
Activity 5 Interpreting a two-way table Allow 5 minutes Use the information presented in the two-way table below to answer the following questions. How many right-handed students are there in the class? How many students are there in the class in total? Table 5 Do left-handed students like art?
Left-handed Right-handed
Likes art 2 12
Do not like art 1 10
There are 22 right-handed students in the class as this is the sum of the ‘right-handed’ column. There are 25 students in total. This is the sum of the ‘left-handed’ and ‘right-handed’ columns.
1.7 Discrete and continuous data Earlier this week we discussed primary and secondary types of data. Numerical data can be further categorised into discrete or continuous data. Discrete data is numerical data that can only take certain values. The number of people on a fair ground ride, the score on a pair of dice, or a shoe size are all examples of discrete data. Continuous data is numerical data that can take any value within a given range. The heights of a group of adults, the lengths of some leaves, or a dog’s weight are examples of continuous data. Discrete data is counted and continuous data is measured. Activity 6 Are these measures discrete or continuous? Allow 5 minutes A child’s foot length A child’s shoe size The time taken to run a race The number of runners in a race The number of matches in a box The speed of a car Air temperature Time displayed on a dial watch Time displayed on a digital watch Annual salaries of teachers The discrete measures are: examples (2), (4), (5) and (9) since the number of people or matches cannot be anything other than a whole number and the time displayed on a digital watch is restricted to hours, minutes and seconds. Note that in the case of example (2), a discrete variable need not be restricted to whole numbers (shoes in the UK can be in half sizes). The continuous measures are examples(1), (3), (6), (7) and (8) since length, time, speed, temperature and time can be measured to any degree of accuracy (there is no limit to the number of decimal places that could be included in the measurement). Example (10) is rather ambiguous. Strictly speaking, it is a discrete variable, given that there is a basic unit of 1p below which measurements cannot be taken. However, 1p is so small in relation to an annual salary (even the annual salary of a teacher) that in practice it is treated like a continuous variable.
1.8 Grouping continuous data Earlier you used a grouped frequency table to organise the number of DVDs owned by school students. In a similar way, grouped frequency tables are needed for continuous data. There is a slight difference in the way the class intervals are organised for continuous data. Since continuous data can take any values within a given range, you need to allow for this in your table, as follows. Table 6 How long do you spend on your mobile phone each week?
Hours (h) spent on phone Tally Frequency
0 ≤ h < 10
10 ≤ h < 20
20 ≤ h < 30
More than 30
In this grouped frequency table we are using ‘less than’ and ‘less than or equal to’ notation. Earlier, when discussing the discrete data example of DVDs, we discussed the importance of having no overlap in the categories. For this reason, the categories were 0–5, 6–10, and so on. In this example, since time is continuous, we need to allow for values such as 10.5 hours. The first category includes 0 but does not include 10. This means that any value less than 10 hours will be included in the first row. This would include values such as 9.5 or 9.7. In the second row, 10 is included but 20 is not. This means that any value which is between 10 (including 10) and 20 (not including 20) will go here. This concept is covered towards the end of lower secondary school but learners in primary school will have seen the notation <, to mean ‘less than’ and ≤ to mean ‘less than or equal’ to. So, we can write 10 ≤ h < 20.
1.9 Stem and leaf diagrams As well as recording data in frequency tables, results can be recorded and presented in a stem and leaf diagram. This visual representation allows for the data to be grouped in classes and it shows the shape of the data. A group of students challenged their teachers to a speed-texting competition. The 21 participants (12 students and 9 teachers) were each required to send a short text message on a mobile phone. Their times are given in Figure 3 in order of size. Creating a stem and leaf diagram involves finding the ‘stems’, in this case the tens, and then recording the ‘leaves’ in order. The stem and leaf diagram for the student times is shown in Figure 4. You can see that 1|9 represents 19 and 4|1 represents 41. There are no values between 0 and 10 which is shown as there are no leaves next to the stem 0. The greatest value in this data set is 41, so there are no values next to the stem 5 either. Activity 7 Creating a stem and leaf diagram Allow 10 minutes On paper, have a go at creating an ordered stem and leaf diagram for the teacher’s data from Figure 3. Your diagram should look like Figure 5.
2 Developing data sense and forming sensible conclusions So far this week, we have discussed how data can be collected and recorded in frequency tables Later, in Section 3, you will look at how data can be presented in different graphs and charts. In this section you will look how learners can develop data sense. You will look at how data can be analysed using appropriate statistical measures and consider how sensible conclusions can be made from these statistics.
2.1 Statistical thinking and the PCAI cycle Statistical thinking involves more than just calculating averages and plotting graphs. In the English National Curriculum (2014), students are expected to ‘interpret and present data’ from 7 years of age. Students in lower secondary school, aged 11–14, are expected to ‘describe, interpret and compare’ data and to ‘describe simple mathematical relationships between two variables’. In order for learners to develop statistical thinking, they need to be encouraged to discuss the meaning behind their calculations and to interpret measures and graphs within relevant contexts. Statistics lessons must therefore involve more than just teaching techniques and methods. Outside a school setting, statistics are rarely calculated in isolation. Data is collected, analysed and interpreted with a purpose, to investigate something or to answer a research question. Learners can be introduced to the concept of a statistical investigation using the PCAI cycle (Figure 6). PCAI stands for: pose (a question), collect, analyse and interpret (the data). Example Here is an example of a simple statistical investigation with young learners. Pose: learners want to find out which type of pet is favoured by their classmates (Figure 7). The question that they will investigate is: ‘What is the most popular pet?’ This could be further refined as: ‘What is the most popular pet in our class?’ Collect: learners will ask each of their classmates what their favourite pet is, and record the results in a tally chart, similar to the one in Figure 8. Analyse: learners will add up the totals in their tally charts, to compare the popularity of each pet. They will then produce a bar chart to present this data visually (Figure 9). Interpret: learners will look at their bar charts and totals from their table to decide which is the most popular pet in their class. This is a very simple example of a statistical investigation, but the PCAI cycle can be used with learners of all ages and experiences as the question posed, data collection method, data analytics and interpretation of data can all be varied in levels of sophistication. For instance, in the UK, older students may use a large data set from the Office for National Statistics to investigate employment levels, using box plots to compare employability in different ages groups.
Mean Takes all values from the data set into consideration Can be skewed by very large or very small values in the data set
Median Very large or very small values do not affect it It can be very slow to calculate for a large data set
Mode The only average which can be used for non-numerical data There could be more than one modeThere may be no modeIt might not represent the data set
When analysing data, it is important to consider which measures of spread will be most suitable. Activity 10 Calculating measures of spread Allow 10 minutes Below are the English and Maths test scores for 10 school children. Find the mean, median, mode and range of the English scores. The Maths measures have already been calculated. Table 8 Maths and English scores
 Maths 77 78 76 78 78 76 80 79 78 80 English 67 73 101 68 66 85 69 86 101 64
Table 9 Measures of central tendency and spread
 Maths English Mean 78 Median 78 Mode 78 Range 4
Table 10 Measures of central tendency and spread
 Maths English Mean 78 78 Median 78 71 Mode 78 101 Range 4 37
3 Charts and graphs asvisual representations of data In Section 1, you looked at how data can be collected and recorded in tables. Visual representations of data, namely graphs and charts, can provide insights which are not easily observed from the data alone. In this section you will look at how data can be represented visually using several different charts and graphs, including: pictograms bar charts line graphs pie charts scatter graphs. We will also discuss when it might be appropriate to use each of these graphs and charts.
3.1 Basic charts and graphs In this section you will consider pictograms and bar charts as highly visual ways of displaying data. Pictograms Pictograms use images to represent data (Figure 17). A key is needed in order for a pictogram to make sense. Since one cupcake image represents 6 cupcakes sold, you can see that 30 cupcakes (5 × 6) were sold on Monday and 15 cupcakes (2½ × 6) were sold on Tuesday. Bar charts Bar charts are used to represent categorical data, such as favourite colours, and discrete numerical data, such as the number of children in a group of families. In a bar chart the height of each bar represents the frequency corresponding to each category (Figure 18). Activity 12 Reading data from a bar chart Allow 5 minutes How many people prefer sci-fi films? How many people like comedy and romance films? How many people were asked in this survey? Two people prefer sci-fi films. There are 21 (10 + 11) people who like comedy and romance films. A total of 36 people were asked (the sum of the frequencies). Other types of bar chart Bar charts can also be used to compare two sets of data in either dual or composite bar charts. A dual bar chart shows two sets of data side by side for comparison (Figure 19). A composite bar chart is useful when the two sets of data are connected, as the frequencies are combined. (Figure 20) For example, when recording the takings from two branches of a shop, a composite bar chart will show the total takings from both branches but also show the breakdown of sales from each shop. Line graphs Line graphs are often used to show changes or trends over time. They are produced by plotting a series of points which are joined up by a straight line. The general shape of a line graph allows some interpretation at first glance. For example, in Figure 21, it is clear that the rainfall was highest on the third day. Activity 13 Interpreting a line graph Allow 5 minutes On which day was the rainfall lowest? On which day did 7 mm of rain fall? Days 2 and 8 both had the lowest rainfall of 2 mm. A total of 7 mm of rain fell on day 4.
3.2 Pie charts In this section you will read about using pie charts, how to draw pie charts and the important point that pie charts demonstrate the proportions of each type of data rather than the frequency of each type of data. Understanding and interpreting pie charts Pie charts use different sized sectors of a circle to represent data. The pie chart below represents the first instrument played by a sample of people. Understanding this pie chart In this example, the most common instrument is the piano. This can be seen easily as the light blue part of the pie chart, representing the piano, is much larger than any of the other sectors. It is also straightforward to see that ¼ of the people in this sample had a guitar as their first instrument. You can see this because the angle within the dark blue section, representing guitar, is 90°. Using this information, you know that if the total number of people in this survey was 80, there must be 20 people who play guitar (80 divided by 4). Similarly, if you knew that 6 people play guitar, you could work out that the total number of people in the sample must be 24 (6 × 4). Using angles in pie charts In the example above, we used the fact that ¼ of the pie represented the number of people for whom guitar was their first instrument. But it would have been more difficult to give an accurate answer if the question had asked about the number of people whose first instrument was drums. Since pie charts use sectors of circles, we can relate the proportions of the circle, in degrees, to the proportion of the sample and use this to find out how much each sector represents (Figure 23). We need some more information, either the total number of participants in this survey or the value for one of the categories. There were 48 participants in this survey. We can now calculate how many people each of the sectors of the pie chart represent. Table 11 Instrument, angle and frequency
Instrument Angle   Frequency
Guitar 90° 90/360 × 48 = 12
Trumpet 15° 15/360 × 48 = 2
Drums 60° 60/360 × 48 = 8
Flute 45° 45/360 × 48 = 6
Piano 150° 150/360 × 48 = 20
To check for accuracy, it is important first to check that the angles sum to 360° and, second, that the total frequency sums to what you expect, in this case to 48.
Drawing pie charts Drawing pie charts accurately requires the same process in reverse (Figure 24). Teaching idea: Do Smarties<sup>®</sup> have the answer? In Week 3, we discussed the use of concrete materials, including edible resources when appropriate. In this classroom activity, learners are given a tube of Smarties® (or similar sweets with a range of colours), either one each or in pairs. The learners are then required to record the frequency of each colour of sweet in a tally and frequency chart. Learners can use the sweets to create their own visual representations of data. Bar charts work really well – see Figure 25. Data from each individual, or pair, can then be added to a whole- class table of results. Using an Excel spreadsheet is a good idea here. Learners can compare their frequency charts and bar graphs with the whole class results, allowing for discussions about natural variation in samples. This will lead to the understanding that larger samples will generally be more representative of the whole population. This fun investigation allows learners to use the PCAI cycle, while thinking about how they can visually represent data.
3.3 Scatter graphs So far this week, you have mainly focused on graphs and charts which are used to display single variables. In this section you will consider bivariate data, which will involve looking at data for two variables at once. One example would be to look at the sales of umbrellas per day alongside the hours of rainfall per day. Scatter graphs show the relationship between two sets of variables. By looking at a scatter graph you can see whether there is a link between two variables. When there is a link it is called correlation. Correlation Scatter graphs can have positive correlation, negative correlation, or no correlation. Positive correlation means as one variable increases, so does the other variable (Figure 26). They have a positive connection. For example: as it snows more, sales of de-icer sprays go up. Negative correlation means as one variable increases, the other variable decreases (Figure 27). They have a negative connection. For example: as the temperature increases, fewer hot chocolate products are sold. No correlation means there is no connection between the two variables (Figure 28). For example: an increase in rainfall in the UK and the sales of bananas in supermarkets. Activity 14 Expected correlation Allow 5 minutes Would you expect there to be a positive correlation, a negative correlation, or no correlation between the following pairs of variables? age and weight cold drinks sales and temperature height and intelligence In children, you might expect a correlation between age and weight but with adults, you would not expect a correlation. Yes, in hot weather, people are more likely to buy cold drinks. No, there is no relationship between height and intelligence. Interpreting correlation Correlation shows a link between two variables. Interpreting this relationship with the context of the data is an important part of the PCAI cycle (Section 2.1). When correlation is found between two variables, some conclusions can usually be drawn. For example, a negative correlation was found in one of (a), (b) and (c) in Activity 14. This suggests that there is a relationship between these two variables and that as one increases, the other decreases. Thinking about the context of this example helps to ensure that a sensible conclusion is made.
3.4 Misleading correlation It is important to use data sense when interpreting relationships based on correlation. Correlation, not causation Sometimes correlation exists between two variables but this does not automatically mean there is causation. An example of this is that the sale of ice-creams is highest when the number of accidents in swimming pools is highest. To claim that ice- cream sales cause accidents in pools, or vice versa, would be false. In reality, both factors may have been caused by a heat wave. The heat wave is a hidden or an unseen variable, also known as a confounding variable. Spurious correlation Sometimes a correlation can be found between two variables but there is no relationship or connection between them at all. This is known as spurious correlation or, in other words ‘false correlation’. For example, the number of astronauts dying in spacecraft is directly correlated to seatbelt use in cars. At first glance there appears to be a link between the two variables but there is no relationship, so the correlation is spurious. Having discussions with your learners about when correlation does and does not imply a relationship is a useful way to support the development of their statistical thinking.
4 Addressing common misconceptions with probability In this section we will discuss some of the common misconceptions that arise within the topic of probability and how you might address them in your teaching.
4.1 Equal likelihood ‘All events are equally likely.’ When discussing how likely an event is to occur, a common misunderstanding is that all events must be equally likely. In some cases, for example when throwing a dice or tossing a coin, the outcomes are all equally likely. However, there are many instances when this is not the case. Tim says, ‘Tomorrow it will either rain or not rain, so the probability that it will rain is 50:50.’ Activity 15 Reflecting Allow 5 minutes What is wrong with Tim’s logic in this instance? Tim has identified two outcomes: rain or not rain. He has therefore deduced that the probability of rain is 0.5. While he is right that rain and not rain are mutually exclusive (i.e. cannot both occur at the same time), there are many factors involved in calculating a weather forecast. It is not usually the case that the probability of rain is exactly the same as the probability of it not raining.
4.2 Gambler’s fallacy ‘Later events may be affected by or compensate for earlier ones.’ Many people hold the incorrect belief that dice or coins have memory. When playing games or placing bets, people often believe that their luck is about to change because ‘a 6 is due’ or ‘my numbers must come up as they haven’t come up yet’. In fact, when using a dice, coin, lottery machine, etc., the probability of an event occurring is irrelevant of the previous results. For example, Chloe says: ‘I have thrown a dice 12 times and not yet got a 6. The probability of getting a six on my next throw must be more than $\frac{1}{6}$.’ Comment Chloe believes that the events that have occurred already have some kind of effect on future events. In fact, while it might seem unlikely to throw a dice 12 times and not get a 6, the probability of getting a 6 on the next throw remains $\frac{1}{6}$.
4.3 Sample space considerations ‘If events are random, the results of a series of independent events are equally likely.’ For example: if you toss a coin twice, it is equally likely that you will get heads, heads as heads, tails. Comment It is important to draw a sample space diagram for multiple events as it is not necessarily the case that all outcomes are equally likely. Tossing a coin twice is a very good example of when this is not the case. When one coin is tossed, the outcome is either heads (H) or tails (T), so the probability of each outcome is $\frac{1}{2}$. When two coins are tossed, the first coin can be either H or T and the second coin can be either H or T. This results in the sample space: HH, HT, TH, TT. This means that the probability of getting two heads (HH) is $\frac{1}{4}$ and the probability of getting one head and one tail (TH or HT) is $\frac{1}{2}$. The two events are not equally likely.
4.4 Spinners The probability of landing on a particular colour on a spinner is determined by the number of sections rather than the size of the angles. When looking at spinners, it is important to look at the size of the sectors. In Figure 32 there are 3 possible outcomes but the angle for the green section is much larger than for the yellow section, so the probability of landing on green is much higher. Activity 16 True or false? Allow 5 minutes Read each of the statements below and decide whether they are true or false. Give reasons for your decision and try to explain the misconception you think each learner has formed. (a) I flipped a coin 3 times and got 3 heads. It is more likely to be tails than heads if I spin it again. (b) I tossed a coin 8 times and got tails every time. That proves that this coin must be biased as half of the time I should have got heads. (c) I roll two dice and add the results. The probability of getting a total of 6 is$\frac{1}{2}$ because there are 12 different possibilities and 6 is one of them. (d) If six dice are thrown at the same time, I am less likely to get 1, 1, 1, 1, 1, 1 than 1, 2, 3, 4, 5, 6. (e) I spin two coins. The probability of getting heads and tails is $\frac{1}{3}$ because I can get heads and heads, head and tails, or tails and tails. (f) Each spinner has two sections – one black and one white. The probability of getting black is $\frac{1}{2}$ for each spinner. Provide your answer... (a) False. The probability remains $\frac{1}{2}$ . (b) False. It could be a biased coin but the point is, this is not enough evidence to prove that it is. If it were rolled 100 times with only tails appearing, that would give a fair indication that the coin is biased. However, it is possible to get tails 8 times in a row, so it is not enough in itself to prove that the coin is biased. (c) False. First, the sample space is the totals which can be made from two dice. Totals can be made from 2 to 12 (1 is not possible with two dice). Second, the probability of each score is not equal. For instance, 7 can be made in 6 different ways: 1 + 6, 2 + 5, 3 + 4, 4 + 5, 5 + 2, 6 + 2, whereas 2 can only be made in 1 way: 1 + 1. The probability of getting a 6 is $\frac{5}{12\phantom{\rule{0ex}{0ex}}}$ . (d) False. Both outcomes are equally likely. The sample space would be all of the possible combinations. So 1, 1, 1, 1, 1, 1 is no less likely than 1, 2, 3, 4, 5, 6. Of course, if you are not worried about the order (i.e. you can have the numbers 1 to 6 in any order) then that outcome is more likely than 1, 1, 1, 1, 1, 1 as there is only one arrangement of this but the numbers 1 to 6 can be ordered in 6 × 5 × 4 × 3 × 2 × 1 ways. (e) False. We discussed this example earlier. Yes, there are three outcomes if you don’t consider the order, but there are two ways to get heads, tails – HT and TH – so the probability of getting heads, tails is $\frac{1}{4}$ . (f) False. In the first spinner in Figure 33 the probability of landing on black is $\frac{1}{2}$ as half of the spinner is shaded black. In the second spinner, the probability of landing on black is only $\frac{1}{4}$ . In the third spinner the probability of landing on black is $\frac{3}{4}$ . It is important to take the proportion into consideration, not just how many different outcomes there are.
4.5 Building misconceptions into your teaching You have considered some of the common misconceptions that arise when teaching probability in the middle years. As you teach this topic you are likely to find others. In Week 3, where you looked at common misconceptions with proportional reasoning, we discussed the importance of tackling misconceptions directly. Due to the abstract and sometimes counter-intuitive nature of probability, it will be is important to ensure that your learners have a clear understanding of the rules and logic involved. Using quotes like those in Activity 16 as starter activities to stimulate discussion is a good way to address misconceptions with your class.
5 Making the connections between theoretical and experimental probability In this section you will look at the connections between theoretical and experimental probability.
5.1 Experimental probability Have a go at this probability game and then reflect on what it demonstrates. Activity 17 Horse race Allow 10 minutes Several people can play this game (or a whole class can play). Every person chooses a number to represent their horse. They each take it in turns to roll two dice and sum their numbers and when their number comes up they shade one box for their horse. So if someone gets a 2 and a 3, their total is 5 and they shade one square on the 5 line. The race is won by the first horse to get to the finishing line (that is when all five boxes on their line are shaded). When the game is finished you should have a horizontal bar chart. If you do not have any dice, there are simulators available online, for instance: www.random.org/dice/?num=2 www.transum.org/software/SW/SnailRace/ Activity 18 Reflecting Allow 5 minutes Some questions to get you thinking about the activity. How did your horse do? Did anything surprise you about the results you recorded? Was the winning horse one you might have expected to win? Did you notice anything about horse number 1? You will calculate the theoretical probabilities of each horse moving forward in the race later, in Section 5.3.
5.2 Relative frequency The experimental probability, or relative frequency, of an event is the ratio of the number of outcomes in which a specified event occurs to the total number of trials in an actual experiment. This is different from the theoretical probability of an event, which we will discuss in Section 5.3. In the horse race example, the event is ‘getting a score of 8 on two dice’ (or any of the other totals) and the total number of trials is the number of times the two dice are thrown. For example: the horse race game was played by rolling a pair of dice 100 times. This means the total number of trials is 100. The frequencies of each total score have been recorded in the table in Figure 34. These represent the number of outcomes of each specific event (i.e. the number of times each score was rolled). The bar chart gives an overview of the distribution of the data. The relative frequency is calculated by dividing the number of times each event occurred by the total number of trials. For example, if the event is ‘scoring an 8 when rolling two dice’, the relative frequency is $\frac{6}{100}$ since a score of 8 was recorded 6 times and there were 100 rolls altogether. You can see the relative frequencies of all 12 scores, shown as fractions, in Figure 35. Note that the total of the relative frequencies must sum to 1, since all possibilities have been included. We will discuss how experimental probability (relative frequency) relates to theoretical probability in Section 5.4.
5.3 Calculating theoretical probability As well as conducting your own probability experiments, you can calculate the theoretical probability of events occurring using what you know about the event and the other possible outcomes. In order to calculate the probability of each horse moving forward, you need to understand the sample space, which includes all of the possible outcomes. Looking at one dice is simpler. There are 6 possible outcomes: 1, 2, 3, 4, 5, 6.Assuming that you are using a dice, each event is equally likely, so the theoretical probability of getting a 6 (or any other score) on a dice is: $\frac{1}{6}$ When looking at two dice, it gets a bit more complicated as you can get 1, 2, 3, 4, 5, 6 on each of the two dice. So you have to consider all of the possible combinations and the resulting scores. The sample space diagram in Figure 36 shows all of the possible combinations of scores. Note: the smallest score is 2, since it is not possible to get a score of 1 using two dice. From the sample space diagram, you can see that there are 36 possible outcomes when rolling two dice. The probability of each event is not 1 in36 though. Some events are more likely than others. For example, a score of 4 can be made in three ways: 3 then 1, 2 then 2, or 1 then 3, whereas a score of 2 can only be made in one way. Figure 37 lists the frequency of each event in the sample space. We will use this to calculate the theoretical probability. Since there are 36 possible outcomes, the theoretical probability for each event will be: $\frac{\mathrm{The number of outcomes for each event}}{\mathrm{The total number of outcomes}}=\frac{\mathrm{Number of ways to get each score}}{36}$
5.4 Comparing experimental probability with theoretical probability As sample size increases, experimental probability gets closer to theoretical probability. In the horse race example, this means that if the pair of dice are rolled 1000 times, the relative frequency will be much closer to the theoretical probability than if the dice were only rolled 10 or 100 times.
5.5 Estimating probability When you don’t know enough about a population to calculate the theoretical probability, you can estimate the probability using the relative frequency from a sample of the population. For example:a bag contains some red and some blue sweets. There are 7 red sweets and 3 blue sweets. Amrit does not know how many sweets of each colour are in the bag. She took out one sweet from the bag, recorded its colour and replaced it. Amrit did this 10 times and found that she picked a red sweet 5 times. This gave the relative frequency $\frac{5}{10}$ . At this point Amrit predicted that the probability of picking a red sweet was $\frac{5}{10}$ , or ½. Therefore, if there were 10 sweets in the bag in total, she thought there were probably 5 red sweets. In order to get a more accurate estimate of the probability, Amrit should repeat the experiment many more times. As we discussed earlier, the more times an experiment is carried out, the closer the relative frequency will be to the theoretical probability.