Irrational number $\sqrt2$ proof I have seen the proof that  $\sqrt2$ is irrational
However I'm a bit confused. The proof goes something like this. first assume $m$ and $n$ are integers such that $$ \frac{m}{n}  = \sqrt{2} $$ $$ \frac{m^{2}}{n^{2}}  = 2 $$ $$m^{2}  = 2 n^{2}$$  $\implies m^{2} $ is divisible by $2 \implies m$ is divisible by 2. Let $2q = m$. Then $$m^{2}  = 4 q^{2} = 2n^{2}$$ $$ n^{2}  = 2 q^{2}$$ This means that $n^{2}$ is divisible by $2 \implies n$ is divisible by $2$...... Hence $\sqrt{2}$ is irrational. $$$$If $ m^{2} $ is divisible by $2$ how do we know that $m$ is also divisible by $2$? If you say it's a prime number then I can argue that I didn't need to do all these steps. I could have just said that its factors don't multiply to $2$ hence $\sqrt{2}$ is irrational
Suppose that $m$ is odd, then $m=2k+1$, and thus
$$m^2=(2k+1)^2=2(2k^2+k)+1$$
and thus $m^2$ is odd too. That prove that if $m^2$ is even then $m$ is even too.

You can generalize this result : If a prime number $p\mid m^s$ for $s\in\mathbb N$ then $p\mid m$. Indeed, let $m=p_1^{\alpha_1}...p_n^{\alpha_n}$ the decomposition of $m$ in prime factor. Then $$m^s=p_1^{s\alpha_1}...p_n^{s\alpha_n}$$
and thus all prime that divide $m^s$ also divide $m$.
Your entire proof is based on the assumption that m and n are integers. Till this assumption holds true. There is no fallacy in the argument. 

But, Lets say m and n can be anything. That is, we don't assume anything for m andn. Then, your statement 


  m^2 is divisible by 2 => m is divisible by 2


no longer holds true. for eg: m=k*sqrt(2).
 Then m is not divisible by 2. But m^2=2*k^2 is divisible by m.

Hence, by contradiction sqrt(2) is irrartional.