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Is the following a valid proof that $\sqrt2$ is irrational?

I've seen the proof to this in Baby Rudin, but I'm trying to figure out exactly how much "self expression" (I don't know what else to call it) is allowed in proofs. The following proof I managed to arrive at readily makes more sense to me than the proof in Rudin. Is it correct and would it be considered good enough?

$\sqrt2$ is rational if $\exists\frac{m}{n}\in \mathbb{Q},m,n \in \mathbb{Z}$ and $n \neq 0$.

Suppose $\sqrt2 \in \mathbb{Q}$, then $\frac{m}{n} = \sqrt2, n\neq0$

$\frac{m^{2}}{n^{2}}=2\\m^{2}=2n^{2}$

if $m = 0$, then $0 = 2n^{2}$, which is true only if ${n=0}$

But $n$ can't be 0. So $\sqrt{2}$ is $\notin \mathbb{Q}$.

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    $\begingroup$ You haven't dealt with the case where $m \neq 0$, so the proof is not complete. $\endgroup$ – Ethan Alwaise Apr 6 '16 at 19:19
  • $\begingroup$ The m=0 and n=0 doesn't make sense. You also have to pick m and n so that they have been 'reduced', that is all the common factors have been taken out so then you will arrive at a contradiction that m2 is even, so n must also be even. $\endgroup$ – user247608 Apr 6 '16 at 19:28
  • $\begingroup$ does $2$ devide $2 n^2$? $\endgroup$ – Max Apr 6 '16 at 19:28
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    $\begingroup$ Why is this being downvoted? $\endgroup$ – tilper Apr 6 '16 at 19:30
  • $\begingroup$ @EthanAlwaise Thanks. I see what you mean. While trying to possibly complete the "proof", I also realised that that this can't be a valid proof becoz it would even suggest that $\sqrt4$ is not rational by my original line of thinking. $\endgroup$ – Sadio Apr 6 '16 at 19:47
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First of all it is useful to make explicit that since $\frac{m}{n}\in \mathbb{Q},m,n \in \mathbb{Z}$ then $m,n$ have to be coprimes. Thus, if $m\neq 0$ then $m^2=2n^2$ implies that $m,n$ are both even, which is absurd (since they were by assumption coprimes).

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  • $\begingroup$ Thanks for answering. But I was trying to see if I could at all arrive on my own at a valid proof that was different to the one in Baby Rudin and perhaps didn't rely at all on m and n being mutually prime. But as Ethan pointed out in his comment, my attempt at a proof was flawed. (The proof in Baby Rudin is similar to your answer). I'm still up voting your answer. $\endgroup$ – Sadio Apr 6 '16 at 20:14

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