Q13 (b) (i)

(b)(i) Let $\sqrt2$ be rational

i.e. $\sqrt2=\frac pq$, $p,q\in\mathbb Z^+$, $p,q$ mutually prime.

Squaring both sides:

$$2=\frac{p^2}{q^2}$$ $$2q^2=p^2. \tag{1}$$

$\therefore$ as $q$ [is] not [a] factor of $p$, $2q$ is a factor of $p$.

$p=2k$.

$$p^2=4k^2. \tag{2}$$

Sub. (1) in (2):

$$2q^2=4k^2$$ $$q^2=2k^2$$

$\therefore 2$ is a factor of $q^2$.

$\therefore 2$ is a factor of $q$.

But $p$ & $q$ are mutually prime.

Contradiction.

$\sqrt2$ is IRRATIONAL.