Q12 (a)

Criteria	Marks
Provides correct proof	3
Shows that $23$ is a factor of a relevant integer, or equivalent merit	2
Attempts a proof by contradiction or writes $\sqrt{23}=\frac pq$ for integers $p,q$, or equivalent merit	1

Sample answer:

Assume the contradiction, that $\sqrt{23}$ is rational.

Let $\sqrt{23}=\frac pq$ where $p,q\in\mathbb Z$ and $q\ne0$.

Also assume that $p$ and $q$ are coprime.

$\left(\sqrt{23}\right)^2=\left(\frac pq\right)^2$
$23=\frac{p^2}{q^2}$
$23q^2=p^2$

$23$ divides LHS, so $23$ divides $p^2$. Since $23$ is prime, $23$ also divides $p$.

So $p=23m$, for some $m\in\mathbb Z$.

$\therefore23q^2=(23m)^2$
$q^2=23m^2$

So similarly $23$ also divides $q$.

But now $p$ and $q$ have a common factor of $23$, contradicting the assumption. Therefore, the assumption is false and $\sqrt{23}$ is irrational.