Is my proof of the irrationality of $\sqrt{3}$ legitimate? I read the proof of the irrationality of $\sqrt{3}$ in my textbook (Richard Hammack's Book of Proof), and I was wondering if my proof is legitimate as well. 
Prop: $\sqrt{3}$ is irrational.
Suppose by way of contradiction that $\sqrt{3}$ is rational. Hence, $\sqrt{3} = \frac{m}{n}, m \in \mathbb{Z}, n \in \mathbb{N}$. Furthermore, suppose both $m,n$ are not even, so the fraction is reduced. Then, $3n^2=m^2$. Suppose $n$ is even, so $n = 2a, a \in \mathbb{Z}$. Therefore, $3 \cdot 4a^2=m^2 \longrightarrow m^2=12a^2 \longrightarrow m^2=2(6a^2)$ and $m^2$ is even, hence $m$ is even. This contradicts our assumption that both $m,n$ were not even, and hence $\sqrt{3}$ must be irrational. $\blacksquare$
Also, I'm wondering if I also need to show the case where $m$ is even?
Edit: Thank you for all the help. I realize that I was essentially trying to make the same argument as the classic proof of the irrationality of $\sqrt{2}$, but it doesn't quite work the same. I've done some research about the prime factorization theorem and I agree that $3n^2=m^2$ being a contradiction is definitely a more elegant proof that a case-by-case approach. 

There are at least two flaws:


*

*you say $m,n$ are both not even, which in fact means neither $m$ nor $n$ are even.

*you say "so the fraction is reduced", but is $\frac{15}{25}$ reduced ?

Allowing only irreducible fractions, $$\sqrt3=\frac pq\iff p^2=3q^2.$$
So $p^2$ is a multiple of $3$, and so must $p$ be. Then $p^2$ is a multiple of $9$ and $q^2$ is a multiple of $3$. And so must $q$ be !
