discrete mathematics - Is this a valid proof for $sqrt5$ being irrational?

'Prove by contradiction that $\sqrt5$ is irrational.'

Proof: Assume that $\sqrt5$ is rational i.e. $\sqrt5 = p/q,$ where $p,q \in \Bbb Z$.

Then $\sqrt5q = p$

Now for p to be an integer, $\sqrt5q$ must be an integer, i.e $q=\sqrt5$, $2\sqrt5$, $3\sqrt5$, ...

$\implies$ q must be irrational for p to be an integer.

$\implies$ $p,q \notin \Bbb Z$.

Contradiction. Therefore $\sqrt5$ is irrational. #

Answer

Your proof is not correct because you have to prove that $q= k\sqrt{5}$ is irrational for every integer $k$ by proving that $\sqrt{5}$ is irrational.

Note that if we multiply a non-zero integer and an irrational the product will be an irrational number.(You can prove this for practise)

So you assume what you want to prove.

Here it is a valid proof which i hope it will help you:

Assume that $\sqrt{5}=\frac{m}{n}$ where $(m,n)=1$.

We assume that $(m,n)=1$ because if its not then we can cancel a priori every common factor of the numerator and denominator until we remain with a fraction $\frac{s}{l}$ in its lowest terms namely $(s,l)=1$.

Thus

$$5n^2=m^2 \Rightarrow 5|m^2 \Rightarrow 5|m$$ because $5$ is a prime number.

Also because $5|m$ we have that $m^2=25s^2$

Thus

$$5n^2=25m^2 \Rightarrow n^2=5m^2 \Rightarrow 5|n$$

Now we have a contradiction because $5|n$ and $5|m$ and we assumed that $(n,m)=1$

So $\sqrt{5}$ is an irrational.