The power of ten increases for each column. So to be consistent, the units column should be 10^{0}. Putting 10^{0} into a calculator gives 10^{0} = 1. So the units column is 1 = 10^{0}.2.2.1 Try some yourselfActivity 16Find the following powers by hand, as estimates for calculator work.(a) 10^{7}(b) 10^{8}(c) 3^{4}(d) (^{−}2)^{2}(e) (^{−}2)^{3}(a) 10^{7} = 10 000 000(b) 10^{8} = 100 000 000(c) 3^{4} = 3 × 3 × 3 × 3 = 27 × 3 = 81(d) (^{−}2)^{2} = ^{−}2 × ^{−}2 = 4(e) (^{−}2)^{3} = 4 × ^{−}2 = ^{−}8Activity 17Use your calculator to find the following and use your answers to the previous question as a check. Give your answers correct to 3 significant figures.(a) 11^{7}(b) 10.8^{8}(c) 3.14^{4}(d) (^{−}2.2)^{2}(e) (^{−}2.01)^{3}(a) 11^{7} = 19 487 171 19 500 000 (3 s.f.)(b) 10.8^{8} = 185 093 021 185 000 000 (3 s.f.)(c) 3.14^{4} = 97.21171216 97.2 (3 s.f.)(d) (^{−}2.2)^{2} = 4.84(e) ^{−}(2.01)^{3} = ^{−}8.120601 ^{−}8.12 (3 s.f.)Activity 18(a) How many zeros are there in 10^{12} when written out in full? Work out a rule for finding the number of zeros in any positive whole number power of 10.(b) Your answers to parts (d) and (e) of Question 1 may have suggested a rule for powers of negative numbers, depending on whether the power is odd or even. What is the rule?(a) There are 12 zeros in 10^{12}. The number of zeros for a positive power of 10 is the same as the power.(b) An odd power of a negative number is negative. An even power of a negative number is positive.Activity 19 Hamsters of a particular kind, left to breed in a suitable environment, double their population every month. If a friend of yours, who is thinking of breeding these hamsters, started with one pair (male and female), what population might she expect in a year's time (assuming she keeps them all)?After 0 months (i.e. at the start) there are 2 hamsters = 2After 1 month there are 2 × 2 hamsters, i.e. 2^{2} = 4After 2 months there are 2 × 2 × 2 hamsters, i.e. 2^{3} = 8So after 12 months there are 2^{13} = 8192 hamsters.Activity 20 A ploughed field has been left abandoned. It has one weed in the middle. The weed is an annual plant and its flower produces about 1000 seeds. If a tenth of these seeds germinate to produce one flower each in a year's time, estimate the possible number of dandelion flowers in the field if it is abandoned for five years. (You can assume that no other factors affect the growth of the weeds during this time). of 1000 = 100So 100 seeds per flower germinate each year.At 0 years there is 1 weed.After 1 year there are 1 × 100 = 100 weeds (which can be written as 100^{1}).After 2 years there are100 × 100 = 100^{2} = 10 000 weeds.So after 5 years an estimate is100^{5} = 10 000 000 000 weeds (or 10 billion).Activity 21 Cells in a laboratory culture divide into two every day, given sufficient nutrient. If there were five cells on day one, how many would there be on day ten?On day 1 there are 5 cellsOn day 2 there are 5 × 2 = 10 cellsOn day 3 there are 5 × 2 × 2 = 5 × 2^{2} = 20 cellsOn day 4 there are 5 × 2 × 2 × 2 = 5 × 2^{3} = 40 cells.So on day 10 there are 5 × 2^{9} = 5 × 512 = 2560 cells.2.3 Multiplying powersPowers of ten can be used to investigate what happens when two powers of the same number are multiplied together. For example, consider multiplying 10 by 100:A billion is a thousand million. In terms of powers this is:so a billion is 10^{9} or 1000 000 000.(An American billion is a thousand million. In UK a billion used to be a million million, but now the American billion is the standard usage.)Note: The number of zeros in the whole number power of ten is the same as the power.In the above calculations, 10^{1} × 10^{2} = 10^{3} and 10^{3} × 10^{6} = 10^{9}, and the way these have been worked out shows that, when you multiply numbers expressed as powers of ten, you add the powers.Powers of 10 were used because they are easy to work with, but you can just as easily show that multiplying powers of the same number always means adding the powers. For example:so 7^{2} × 7^{3} = 7^{2+3} = 7^{5}. Check this on your calculator!To multiply two powers of the same number, add the powers.2.3.1 Try some yourselfActivity 22 Write the following as one number to a single power:(a) 2^{3} × 2^{4}(b) 3^{2} × 3^{4}(c) 4^{2} × 4^{3} × 4^{4}(a) 2^{3} × 2^{4} = 2^{(3 + 4)} = 2^{7}(b) 3^{2} × 3^{4} = 3^{(2 + 4)} = 3^{6}(c) 4^{2} × 4^{3} × 4^{4} = 4^{(2 + 3 + 4)} = 4^{9}Activity 23 Without using your calculator, find the following, as estimates for calculation work in the next question.(a) (^{−}1)^{5}(b) 3^{4}(a) (^{−}1)^{5} = ^{−}1 (any odd power of ^{−}1 is ^{−}1)(b) 3^{4} = 3 × 3 × 3 × 3 = 81Activity 24 Use your calculator to find the following.(a) (^{−}1.3)^{5}(b) 3.2^{4}(Use your answers to the previous question as rough checks.) Round your answers to four decimal places.(a) (^{−}1.3)^{5}^{−}3.7129(b) 3.2^{4} = 104.85762.4 Dividing powersNow consider division in the same way. For example:So 10^{3} ÷ 10^{2} = 10^{1} and 10^{6} ÷ 10^{2} = 10^{4}, which suggests that dividing the powers of ten means subtracting the powers. You can see that the two tens underneath have cancelled with two of the tens on top in each case, which shows why the number of tens underneath is subtracted from the number on top. This argument applies equally well to any number. For example:Work out 7^{5} ÷ 7^{3} on your calculator to check this.To divide two powers of the same number, subtract the powers.2.4.1 Try some yourselfActivity 25 Write the following as a number to a single power:(a) 2^{6} ÷ 2^{2}(b) 10^{10} ÷ 10^{7}(c) 7^{8} ÷ 7^{4}(a) 2^{6} ÷ 2^{2} = 2^{(6 − 2)} = 2^{4}(b) 10^{10} ÷ 10^{7} = 10^{(10 − 7)} = 10^{3}(c) 7^{8} ÷ 7^{4} = 7^{(8 − 4)} = 7^{4}Activity 26 Express as powers of ten(a) a million(b) a thousand(c) a million divided by a thousand(a) a million = 1 000 000 = 10^{6}(b) a thousand = 1 000 = 10^{3}(c) 10^{6} divided by 10^{3} = 10^{6–3} = 10^{3} = a thousand2.5 The power zeroNow, what happens if you divide a number by itself? You should get the answer 1. For exampleso 10^{0} must equal 1. This must be true for any number (except for 0, because there is a problem in defining 0^{0}). So, for example, 7^{0} = 1, 123^{0} = 1, 1000^{0} = 1. It even works for negative and decimal numbers. Check a few numbers on your calculator to see that raising to the power 0 gives 1.Any non-zero number raised to the power 0 is 1.2.5.1 Try some yourselfActivity 27 What are the following?(a) 1^{0}(b) 0^{1}(c) 2^{0}(d) 0^{2}(a) 1^{0} = 1(b) 0^{1} = 0(c) 2^{0} = 1(d) 0^{2} = 0 × 0 = 02.6 Negative powersNow look at what happens when the power is negative. What does 10^{−3} mean? What is the result of the following calculation?100 ÷ 100 000What you are actually being asked to find is:But look at the calculation again. Using the rule for the division of powers of numbers gives:10^{2} ÷ 10^{5} = 10^{2−5} = 10^{−3}So 10^{−3} = 0.001. But you can also write this result as:This means that 10^{−3} can be thought of as 1 divided by 10^{3}.This result is true for all negative powers, not just powers of ten. For example:One over any number is called the reciprocal of the number.For example, the reciprocal of 10 is = 10^{−1} and the reciprocal of 100 is = = 10^{−2}A number raised to a negative power is the reciprocal of the number raised to the corresponding positive power.So 10^{−5} = = 0.00001.Example 8The decimal place value table columns are headed tenths, hundredths, thousandths etc. Write these as powers of ten. a tenth = =10^{−1} a hundredth = = 10^{−2} a thousandth = = 10^{−3}So the headings of the place value table are all powers of ten. To the left of the units they are positive powers, the units column is 10^{0}, and to the right the column headings are negative powers of ten.Notice thatThe negative power indicates the position of the decimal point — how many times it has moved to the left from 10^{0} = 1.10^{−1} = 0.1 (move one to the left); 10^{−2} = 0.01 (move 2 to the left) etc.2.6.1 Try some yourselfActivity 28 Find each of the following by hand, giving your answers both as a power of ten and as a decimal number. You will use these answers as a check on your calculator work in the next question.(a) 10^{−2}(b) 10^{2} × 10^{3}(c) 10^{7} ÷ 10^{4}(d) 10^{4} ÷ 10^{7}(e) 2^{−2}(a) 10^{−2} = 0.01(b) 10^{2} × 10^{3} = 10^{2 + 3} = 10^{5} = 100 000(c) 10^{7} ÷ 10^{4} = 10^{7 − 4} = 10^{3} = 1000(d) 10^{4} ÷ 10^{7} = 10^{4 − 7} = 10^{–3} = = 0.001(e) 2^{−2} = = (=0.25)Activity 29 Evaluate the following, using your calculator and your answers from Question 1 as estimates.(a) 10.3^{−2} (to four significant figures)(b) 10.1^{2} × 10.11^{3} (to four significant figures)(c) 10.112^{7} ÷ 10.21^{4} (to one decimal place)(d) 10.12^{4} ÷ 10.351^{7} (to three significant figures)(e) 2.2^{−2}Using Question 1 as a rough check, the calculator gives:(a) 10.3^{−2} = 0.009 426 (to 4 s.f.)(b) 10.1^{2} × 10.11^{3} = 105 400 (to 4 s.f.)(c) 10.112^{7} ÷ 10.21^{4} = 994.8 (to 1 d.p.)(d) 10.12^{4} ÷ 10.351^{7} = 0.000824 (to 3 s.f.)(e) 2.2^{−2} 0.2066It is a good idea to get in the habit of estimating your answers, even if not specifically asked to do so.Activity 30 Find each of the following by hand.(a) 1024^{0}(b) 1024^{1}(c) 5^{−1}(d) 10^{−4}(a) 1024^{0} = 1(b) 1024^{1} = 1024(c) (d) Activity 31 How many zeros are there after the decimal point in the number 10^{−6}? How many zeros are there after the decimal point for any given negative power of 10?10^{−6} = 0.000 001; there are five zeros after the decimal point. For any given negative power of 10, say 10^{−(NUMBER)} there is one fewer zero than NUMBER. This works for 10^{−1}, too, since 10^{−1} = 0.1, which has no zeros after the decimal point, i.e. one fewer than 1.Activity 32 Try to answer the following questions(a) Explain why a negative power of a number is one divided by the corresponding positive power of that number. (Hint: remember that a number to the power zero is 1.)(b) Use the power button on your calculator to find the following.(i) (ii) (a) Take an example, which makes the explanation easier than describing it in the abstract. Consider 10^{−6}. Now ^{−}6 is just the same as 0 − 6, so 10^{−6} = 10^{0−6}. Using the rule for dividing powers, 10^{0−6} = 10^{0} ÷ 10^{6}. But 10^{0} = 1. So 10^{−6} = 1 ÷ 10^{6}. This argument would apply equally well to any number, since any number to the power 0 is 1. (Alternatively you may have started from the other end, for example, by showing that = = 10^{0−6} = 10^{−6}.)(b)3 Scientific notation3.1 Expressing numbers in scientific notationEarlier you looked at place values for numbers, and why they were called powers of ten.

Place value

10 000

1000

100

10

1

0.1

0.01

0.001

0.0001

0.000 01

Power of ten

10^{4}

10^{3}

10^{2}

10^{1}

10^{0}

10^{−1}

10^{−2}

10^{−3}

10^{−4}

10^{−5}

Using this notation, very large numbers or small decimal numbers can be expressed in a particularly neat form. This is how scientific and graphics calculators display numbers which will not fit on the screen. This is illustrated by the following examples: 10^{7} = 10 000 000 so 25 000 000 = 2.5 × 10^{7} 10^{−6} = 0.000 001 so 0.000 0025 = 2.5 × 10^{−6}A number expressed in this form is said to be in scientific notation. Any decimal number can be converted to scientific notation, which isA calculator automatically uses scientific notation when the answer to a calculation is very large or very small. For example, return to our tale of the king and the bags of rice. In week 52 the woman would receive 2^{52} bags. Calculate this on your calculator. You should get 4.503 599 627 × 10^{15}. (This number may be displayed as 4.503 599 627 E 15.) To get a feel for how big this number is, i.e. for how many bags of rice the king would need to provide a year later, look at some other numbers in scientific notation: Great Britain has an area of 229 850 km^{2}.In scientific notation, 229 850 = 2.2985 × 100 000 = 2.2985 × 10^{5}. The distance from the Earth to the Sun is 149 600 000 km.In scientific notation, 149 600 000 = 1.496 × 100 000 000 = 1.496 × 10^{8}.Consider now the bags of rice calculation. 10^{15} is 1000 000 000 000 000. So the number of bags of rice is 4503 599 627 000 000. This is a very large number of bags of rice! The king would need to give the woman over four thousand million million bags of rice in week 52. No wonder his accountant was concerned.As mentioned above, scientific notation is also used for very small numbers. Try finding on your calculator. You should find that the answer on your calculator is 3.125 × 10^{−7}. (This number may be displayed as 3.125 E −7.) This time the index is negative. This is the same as:With practice you may not need the intermediate steps. You may be able to just move the decimal point the appropriate number of places (multiplying by 10^{−3} means moving the decimal point 3 places to the left). 5.2 × 10^{−3} = 0.00523.1.1 Try some yourselfActivity 33 Express each of the following numbers in scientific notation.(a) Light travels 9460 700 000 000 km in a year.(b) The average distance from the centre of the Earth to the centre of the Moon is 384 400 km.(c) Saturn is 1427 000 000 km from the Sun.(d) The mass of the Earth is 5976 000 000 000 000 000 000 000 kg.(e) The Mediterranean Sea has an area of 2504 000 km^{2}.(f) Annapurna 1 is 8075 m high.(g) The diameter of the Sun is 1392 000 km. Use metres in your answer.(a) Light travels at 9.4607 × 10^{12} km in a year.(b) The distance from the centre of the Earth to the centre of the Moon is 3.844 × 10^{5} km.(c) Saturn is 1.427 × 10^{9} km from the Sun.(d) The mass of the Earth is 5.976 × 10^{24} kg.(e) The Mediterranean Sea has an area of 2.504 × 10^{6} km^{2}.(f) Annapurna 1 is 8.075 × 10^{3} m high.(g) 1392 000 km is 1392 000 000 m (multiplying by 1000). In scientific notation this is 1.392 × 10^{9} m.(Alternatively: 1392 000 km is 1.392 × 10^{6} km = 1.392 × 10^{6} × 10^{3} m = 1.392 × 10^{9} m.)Activity 34 The following numbers are given in scientific notation. Write each out as a single decimal number.(a) Ben Nevis is 1.343 × 10^{3} m high.(b) Mercury is 5.8 × 10^{7} km from the Sun.(c) The distance from the Earth to the nearest star is about 4.3 × 10^{13} km.(d) The Earth has a circumference of 4.0078 × 10^{4} km at the equator.(e) The span of the Golden Gate Bridge is 1.28 × 10^{3} m.(f) The Arctic Ocean has an area of 1.3986 × 10^{7} km^{2}.(a) Ben Nevis is 1343 m high.(b) Mercury is 58 000 000 km from the Sun.(c) The distance from the Earth to the nearest star is about 43 000 000 000 000 km.(d) The Earth has a circumference of 40 078 km at the equator.(e) The span of the Golden Gate Bridge is 1280 m.(f) The Arctic Ocean has an area of 13 986 000 km^{2}.Activity 35 Express each of the following in scientific notation.(a) 0.53(b) 0.0075(c) 0.000 004(d) 0.000 020 01(a) 0.53 = 5.3 × 10^{−1}(b) 0.0075 = 7.5 × 10^{−3}(c) 0.000 004 = 4 × 10^{−6}(d) 0.000 020 01 = 2.001 × 10^{−5}Activity 36 Express each of the following in conventional decimal notation.(a) 2.03 × 10^{−4}(b) 1.4 × 10^{−3}(c) 5.67 × 10^{−5}(a) 2.03 × 10^{−4} = 0.000 203(b) 1.4 × 10^{−3} = 0.0014(c) 5.67 × 10^{−5} = 0.000 05673.2 Using scientific notationScientific notation can be very useful when estimating the answers to calculations involving very large and/or small decimal numbers.Example 9A lottery winner won £7851 000. He put the money straight into a deposit account which earns 7.5% interest per annum (i.e. each year). If he wanted to live off this interest, how much per day would it be?The amount is £7851 000 × ÷ 365.Estimate first to provide a check for the calculator work.7851 000 = 7.851 × 10^{6} 8 × 10^{6} = 7.5 × 10^{−2} 8 × 10^{−2} 365 = 3.65 × 10^{2} 4 × 10^{2}So the estimate becomes: (8 × 10^{6}) × (8 × 10^{−2}) ÷ (4 × 10^{2}).Now separate the digits from the powers of 10 to give: (8 × 8 ÷ 4) × (10^{6} × 10^{−2} ÷ 10^{2}).Since 8 × 8 ÷ 4 = 16 and 10^{6} × 10^{−2} ÷ 10^{2} = 10^{6+−2−2} = 10^{2}, the estimate is 16 × 10^{2} = 1600 (£1600 a day!).On a calculator, 7851 000 × 0.075 ÷ 365 gives £1613 rounded to the nearest pound, which is quite close to the estimate.3.2.1 Try some yourself Activity 37 Use the method outlined in Example 9 to estimate each of the following, and then use your calculator to evaluate each, giving your answers to six significant figures.(a) 2521 ÷ 38(b) 17.85 × 286.3(c) 1452 ÷ 0.0072(d) 0.0053 × 0.0078 ÷ 0.6821(e) 0.000 923 × 0.007 67.(a) Estimate: 2521 = 2.521 × 10^{3} 3 × 10^{3} 38 = 3.8 × 10^{1} 4 × 10^{1}.So 2521 ÷ 38 = × 10^{3−1}= 0.75 × 10^{2} = 75.Calculate: 2521 ÷ 38 = 66.3421 rounded to six significant figures.(b) Estimate: 17.85 = 1.785 × 10^{1} 2 × 10^{1} 286.3 = 2.863 × 10^{2} 3 × 10^{2}.So 17.85 × 286.3 2 × 10^{1} × 3 × 10^{2}= 2 × 3 × 10^{1} × 10^{2} = 6 × 10^{3} = 6000.Calculate: 17.85 × 286.3 = 5110.46 rounded to six significant figures.(c) Estimate: 1452 = 1.452 × 10^{3} 1.4 × 10^{3} 0.0072 = 7.2 × 10^{-3} 7 × 10^{−3}.(Notice that in this case it is easier to round to 1.4 than to 1 or 1.5, anticipating the next step, since it is easier to find 1.4 ÷ 7 than 1 ÷ 7 or 1.5 ÷ 7.)So 1452 ÷ 0.0072 = × 10^{3–−3}= 0.2 × 10^{3+3} = 0.2 × 10^{6} = 2 × 10^{−1} × 10^{6}= 2 × 10^{−1+6} = 2 × 10^{5} = 200 000.Calculate: 1452 ÷ 0.0072 = 201 667 rounded to six significant figures.(d) Estimate: 0.0053 = 5.3 × 10^{−3} 5 × 10^{−3}0.0078 = 7.8 × 10^{−3} 8 × 10^{−3}0.6821 = 6.821 × 10^{−1} 7 × 10^{−1}.SoCalculate: 0.0053 × 0.0078 ÷ 0.6821= 6.06069 × 10^{−5} = 0.000 060 6069 rounded to six significant figures.(Notice that the calculator gives the answer in scientific notation.)(e) 0.000 923 × 0.007 67 = 9.23 × 10^{−4} × 7.67 × 10^{−3}Estimate:Calculate: 0.000 923 × 0.007 67 = 7.079 41 × 10^{−6}.Activity 38A light year is the distance light travels through space in one year. The centre of the Milky Way galaxy is 2.6 × 10^{4} light years away from Earth. One light year is 9.46 × 10^{12} km. How far away is the centre of the Milky Way in kilometers? Estimate your answer first and then use your calculator to do the calculation, giving your answer in scientific notation.Estimate: The centre is roughly 3 × 10^{4} light years away, and each light year is roughly 9 × 10^{12} km, so the distance to the centre is approximately 3 × 10^{4} × 9 × 10^{12} = 27 × 10^{16} = 2.7 × 10^{17}.Calculate: 2.6 × 10^{4} × 9.46 × 10^{12} = 2.4596 × 10^{17}.4 Open Mark quizNow try the quiz and see if there are any areas you need to work on. ConclusionThis free course provided an introduction to studying Mathematics. 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