TITLE: Proof that $\sqrt{2}$ is irrational
QUESTION [5 upvotes]: I am having hard time flowing a proof from Introduction to the theory of computation (3rd ed.) by Michael Sipser.   Please points out what I am missing.  My questions is "Now, at least one of $m$ and $n$ must be add odd number." Why can we make such a statement already? 
PROOF First, we assume for the purpose of later obtaining a contradiction
that $\sqrt{2}$ is rational.
Thus $\sqrt{2} = \dfrac{m}{n}$,
where $m$ and $n$ are integers. If both m and n are divisible by the same integer greater than $1$, divide both by the largest such integer. Doing so doesn’t change the value of the fraction. Now, at least one of m and n must be an odd number. We multiply both sides of the equation by n and obtain
$n\cdot \sqrt{2} = m$.
We square both sides and obtain 
$2\cdot n^2 = m^2$.
Because m2 is 2 times the integer n2, we know that m2 is even. Therefore, m, too, is even, as the square of an odd number always is odd. So we can write m = 2k for some integer k. Then, substituting 2k for m, we get
$2n^2 = (2k)^2 = 4k^2$. 
Dividing both sides by $2$, we obtain 
$n^2 = 2k^2$.
But this result shows that $n^2$ is even and hence that $n$ is even. Thus we have established that both $m$ and $n$ are even. But we had earlier reduced $m$ and $n$ so that they were not both even—a contradiction.

REPLY [3 votes]: If both m and n are divisible by the same integer greater than 1, divide both by the largest such integer.
If the they are both even then you didn't divide by the bigger integer.  Because you can still divide both by $2$.
I'd have worded it differently.  I'd have said "If they have any factors in common divide both by that factor and keep repeating until they have no factors in common".  
It means the same thing but it just makes it more clear that in the end, they will have no factors in common.  And if the have no factors in common, they can't have $2$ in common.  And the can't both be even.