# 1.3 Further exercises

## Exercise 4

Solve the following linear equations.

(a) 5

*x*+ 8 = −2(b)

*x*+ 2 = −5(c) 2

*x*−1 = 5(d) 5

*x*+ 4 = 3

State in each case whether the solution belongs to , , and/or .

### Answer

### Solution

(a)

*x*= −2, which belongs to (and hence to and ).(b) , which belongs to .

(c)

*x*= 3, which belongs to (and hence to , and ).(d) , which belongs to (and hence to ).

*Remark* Each linear equation with coefficients in has a solution in (because the equation *ax* + *b* = 0 where *a*, *b* and *a* ≠ 0 has solution *x* = − *b*/*a*, which involves only additive inverses and division of rational numbers). Similarly each linear equation with coefficients in has a solution in . The same is not true of , as part (d) illustrates.

## Exercise 5

Show that there is no rational number *x* such that *x*^{2} = 3.

### Answer

### Solution

Suppose there is a rational number *x* such that *x*^{2} = 3. Then we can write *x* = *p*/*q*, where *p* and *q* are positive integers whose greatest common factor is 1.

Then the equation *x*^{2} = 3 becomes

Now *p* is either divisible by 3 or has remainder 1 or 2 on division by 3;

that is, *p* = 3*k* or 3*k* + 1 or 3*k* + 2 for some integer *k*.

But if *p* = 3*k* + 1, then *p*^{2} = 9*k*^{2} + 6*k* + 1 is not divisible by 3, and if *p* = 3*k* + 2, then

is not divisible by 3. So, since 3*q*^{2} is divisible by 3, we conclude that *p* = 3*k*.

Hence (3*k*)^{2} = 3*q*^{2}, so *q*^{2} = 3*k*^{2}.

But then the same argument applies to *q* to show that *q* must also be divisible by 3.

Hence 3 is a common factor of *p* and *q*.

This is a contradiction, so we conclude that the assumption must have been false. Hence there is no rational number *x* such that *x*^{2} = 3.