Number systems

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# 1.3 Further exercises

## Exercise 4

Solve the following linear equations.

• (a)   5x + 8 = −2

• (b)  x + 2 = −5

• (c)  2x −1 = 5

• (d)  5x + 4 = 3

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

### 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 x2 = 3.

### Solution

Suppose there is a rational number x such that x2 = 3. Then we can write x = p/q, where p and q are positive integers whose greatest common factor is 1.

Then the equation x2 = 3 becomes

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

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

But if p = 3k + 1, then p2 = 9k2 + 6k + 1 is not divisible by 3, and if p = 3k + 2, then

is not divisible by 3. So, since 3q2 is divisible by 3, we conclude that p = 3k.

Hence (3k)2 = 3q2, so q2 = 3k2.

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 x2 = 3.

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