I have just taught the classic proof by contradiction that $\sqrt 2$ is irrational, and one of my students came up with the following proof:
Assume that $\sqrt 2$ is rational, so $\sqrt 2=\frac{a}{b}$ where $a$ and $b$ are integers such that $\frac{a}{b}$ is irreducible.
$2=\frac{a^2}{b^2}$
$b^2=\frac{a^2}{2}$
$b^2$ is a square number, and $a^2$ is a square number, but a square number divided by 2 cannot equal a square number, so there is a contradiction.
To justify this claim:
If a number $a$ is even, then $a=2n$, so $a^2=4n^2$.
$\frac{a^2}{2}=2n^2$ and the square root of $2n^2$ is $\sqrt{2}n$, which is clearly not an integer, therefore $2n^2$ is not a square number.
If a number $a$ is odd, then $a=2n+1$, so $a^2=4n^2+4n+1$.
$\frac{a^2}{2}=2n^2+2n+\frac{1}{2}$ which is an integer add a half, so it is not an integer. Therefore it is not a square number.
So, since a square number divided by 2 is not a square number, the contradiction is reached in the line:
$b^2=\frac{a^2}{2}$
Therefore $\sqrt 2$ is irrational
Is this sound?
This step appears to be implicitly making the inference \begin{gather}n\ne0\implies\Big(\sqrt2\not\in\dfrac{\mathbb Z}n \implies \sqrt{2}n\not\in\mathbb Z\Big)\tag✓\\\text{and}\quad\color{red}{\sqrt2\not\in\dfrac{\mathbb Z}n};{\tag!}\\\text{therefore,}\quad{\big(n\ne0\implies\sqrt{2}n\not\in\mathbb Z\big)}.\end{gather} Although this inference per se is sound, by begging the question, it invalidates the proof containing it. Therefore, the proof is fallacious.