Prove that $13\sqrt{2}$ is irrational. I am currently a beginner at proofs and I am having trouble proving this problem...
I know that the square root of $2$ is irrational because the square root of $2$ can be expressed as $\frac{p}{q}$ and once both sides are squared it is true that both $p$ and $q$ are even which is a contradiction to the assumption that they have no common factors.
I am having trouble proving that $13$ and the square root of $2$ is irrational though and any help would be greatly appreciated! Since we are not dealing with the square root of $13$, I do not know how to start since we can not set it equal to $\frac{p}{q}$.
Thank you in advance!

A much less elegant but maybe clearer proof than those involving greatest common denominator (g.c.d.):-
If $\sqrt{2}$ is rational, the we can express it as a ratio of two integers, $A$ and $B$.
So:
$$ \sqrt{2} = \frac {A}{B}  $$
where
$$  A, B \in \mathbb{N} $$
$$ \implies A = \sqrt{2}B $$
Since $ A^2 = 2B^2$ we can see that $A^2$ is even, i.e.
$$ A^2 \in \mathbb{E} $$
Each integer can be written as the product of a unique set of prime factors to various powers. So any integer that is the square of a smaller integer has each of its prime factors at least twice. So the integer square root of a number that is even will itself also be even. (And indeed if the integer square is divisible by $N$ then so will the starting integer be divisible by $N$.)
$$\implies A \in \mathbb{E} $$
Writing $A$ as some even number $2K$, we see that
$$ B^2 = \frac{A^2}{2} = \frac{(2K)^2}{2} = 2K^2 $$
$$ \implies B^2 \in \mathbb{E} $$
$$ \implies B \in \mathbb{E} $$
But from our earliest definition, we have:
$$A^2 = 2B^2$$
and, with $B^2$ resolved into pairs of prime factors, this would imply that $A^2$  has one unpaired prime factor, i.e. $2$.
Yet, from our reasoning on squared integers, we know that this cannot be so since squared integers may only have pairs of prime factors.
Hence, since our initial assertion that $\sqrt{2}$ is rational has led us to an erroneous deduction, we can conclude that $\sqrt{2}$ is not a rational number.
It is clear from the above reasoning that neither is any $\sqrt{K} $ rational, where $K \in \mathbb{N}$ and $\sqrt{K} \notin \mathbb{N}$
The latter generality covers $\sqrt{338}$ or $13 \sqrt{2}$.
It also covers the more obvious situation where $K$ is a prime number, i.e. $K$ has no factors (unless you regard $1$ as a factor) let alone an integer root.
