Is $\sqrt[2]{(2/7)}$ irrational?

Question

I have to show that the $\sqrt(2/7)$ is irrational. Here is my work.

Answer 1

You're skipping some steps.

You say that $7m^2$ is even, then you assume $m$ is even. This is true. You should say why though, you explain a little more why $n$ is even.

Also, how is the end a contradiction? You found that the gcd is even, when did you ever assume that it wasn't?

Answer 2

You haven't actually reached a contradiction unless you've asserted at the outset that $(m,n) = 1$. But once you assert that you can stop with this line: $2n^2 = 7m^2$. If $n$ is even, $m$ must be odd and the LHS is even, the RHS isn't. Contradiction. But if $n$ is odd and $m$ is even the LHS is an odd multiple of $2$ while the RHS is an even multiple of $2$, again leading to a contradiction.

Edited to patch a hole in the proof that I just noticed after notification of the upvote.