Proving that $\sqrt{2}$ is irrational with a math level of a middle school student? So I have a friend, whose professor challenged the class to prove that $\sqrt{2}$ is irrational by using only middle school math level.
No one managed to do it, and neither did I (although I didn't give it much thought, as I have other stuff to do for the moment), but I am still curious as to how to do it.
Apparently, most of the class thought about using the Pythagorean theorem: $$ a^2 + b^2 = c^2$$ And taking $a=b=1$, but it didn't help them at all.
So any ideas? I know that the middle school levels vary from country to country, but let's just suppose that we are talking about YOUR country's middle school.

Hopefully, the following argument (adapted from Cauchy)) might be at the middle school level, as it uses only the notions of integer and decimal parts of a number.
We prove that if $x=\sqrt 2$ is rational, it must necessarily be an integer. As it is clear no perfect square is equal to $2$, it will prove that $x$ is irrational. 
For this, consider the set $S$ of integers $n$ such that $nx$ is an integer. If $x$ is rational, equal to, say $\dfrac rs$, this set is not empty since $sx=r$ is an integer, so $s$ belongs to $S$.
Let $q$ be the smallest positive integer in $S$. Note that if $y$ is the decimal part of $x$, $qy$ is an integer: indeed, $x=1+y$ since $1<\sqrt 2<2$, so
$$qy=q(x-1)=\underset{\text{integer}}{qx}-\underset{\text{integer}}{q}$$
We'll denote $\;q'=qy$.
Claim: $\;q'x$ is an integer.
Indeed $\; q'x=(qx-q)x=qx^2-qx=\underset{\text{integer}}{2q} -\underset{\text{integer}}{qx}$.
Now, as by definition a decimal part is  $\;<1$, we know that $\;0\le q'=qy <q$. As $q$ was chosen to be the smallest positive integer in $S$, this implies that $q'=qy$ is $0$, hence $y=0$. In other words, if $x$ is rational, it has no decimal part, which means it's an integer.
