Square root of $2$ is irrational I am studying the proof that $\sqrt 2$ is an irrational number. Now I understand most of the proof, but I lack an understanding of the main idea  which is:

We assume $\frac{m^2}{n^2} = 2$. Then both $m$ and $n$ can't be even.

I do not understand, why can't both $m$ and $n$ be even?
 A: Because if $n=2n_1$ and $m=2m_1$ then $\frac{m_1^2}{n_1^2}=2$... and we get an infinite series $n>n_1>n_2>...$ of natural numbers, which is impossible.
A: Okay, this may be a bit pendantic.  
There are a few fundamentals assumed but not stated.
1) Given that $a = k*n; a,k,n \in \mathbb Z$ and $b = k*m; b, m \in Z$ then then rational value of the ratio $\frac ab $ is the same as the ratio $\frac mn$ and we say $\frac ab = \frac nm$.  
Why?  Well... because. (If someone has a better answer feel free to leave a comment).  I feel this is the basic concept of ratio or proportion and ratio of "$n$ units to $m$ units" is axiomatically the same value regardless of the unit.  And if the "units" is a measure of $k$ value then "$n$ $k$-s are in proportion to $m$ $k$-s" as "$n$ is in proportion to $m$" is simply an axiom.
If anyone has a better or more correct way of putting this, let me know.
2) Given any two integers $a,b$ there is some common factor $k$ so that if $a\div k =n$ and $b \div k = m$ and $m,n$ have only $1$ as a common (positive integer) factor.
Why?  If $a,b $ don't have any common factor other than then $k=1$ is precisely that common factor.  If $a, b$ have a common $j$ other than $1$ then $j > 1$ and $a\div j = n$ then $a > n$ and if $b\div j = m$ then $b > n$.  If $a$ and $b$ were such that there is no such common factor, then $j$ can not be that factor so $m$ and $n$ have a common factor $j_2 > 1$ and if $m =j_2*m_2$ and $n = j_2*n_2$, we have $a > n > n_2$ and $b > m > m_2$.  If there is no such terminating common factor we can do this infinite so we can get an infinite chain of $a > n > n_2> n_3>.....$ and $b > m > m_2 > m_3> .....$.
This is clearly impossible.  Why? Because these are integers the difference between $n_i$ and $n_{i+1}$ is at least $1$ and so $a$ and $b$ must be larger than an infinite $1$s.  (i.e. $a$ and $b$ are infinite.)
From those two assumptions, we can conclude:
A) If $q = \frac ab$ is any rational number with $a,b$ are integers.  Then $a,b$ have a common factor $k$ so that $n = \frac ak; m = \frac bk$ and $n$ and $m$ have no common factor other than $1$. Thus  we can state $q =\frac nm$ where $n$ and $m$ have no factors other than $1$ in common.  
We can express any rational $q$ in such a way.
.....
So now you can start your proof:
Assume $\sqrt{2}$ is rational.  Then $\sqrt{2} = \frac ab$ for integers $a,b$ so that $a$ and $b$ have no common factor other than $1$.
Then...... < < details omitted > >... $2$ is a factor of $a$ and ..... < < details omitted > > .... $2$ is a factor of $b$.
And therefore $a$ and $b$ have a factor of $2$ in common, which contradicts that $a$ and $b$ have no factors in common other than $1$.
So $\sqrt{2}$ is not rational.
A: An another way 
Be $\sqrt{2}\in \mathbb{Q}$, and $\mathcal{Q}=\bigg\{q\in\mathbb{N}^*,\quad q\sqrt{2}\in\mathbb{N}\bigg\}\implies \mathcal{Q}\neq\varnothing$
Let $q_s:=\min \mathcal{Q} $
Let $p:=q_s\sqrt{2}-q_s$
$p=q_s\sqrt{2}-q_s<2q_s-q_s=q_s\iff \boxed{p<q_s}\quad (1)$
$p=q_s\sqrt{2}-q_s\iff\sqrt{2}p=2q_s-\sqrt{2}q_s \iff \boxed{\sqrt{2}p\in\mathbb{N}}\quad (2)$
$(1)\land(2)\implies$ we have found $p<q_s$ such that $\sqrt{2}p\in\mathbb{N}$ so we have a contradiction thus $\sqrt{2}\notin \mathbb{Q}$
A: Before you continue the argument you write the fraction $m/n$ in lowest terms. Then the numerator and denominator can't both be even.
A: we can assume that $$\gcd(m,n)=1$$ and $$2=\frac{m^2}{n^2}$$ then $$2n^2=m^2$$ thus the left-hand side is even and so $$m^2$$ this is a contradiction, both numbers $m,n$ can not be even
A: Okay thanks I understand it now after seeing that the fraction m/n in the lowest terms, in this situation both m and n cannot both be even.
Here is my proof
Prove that $√2$ is irrational
Let's assume $√2$ is rational then $2=m^2/n^2$
by writing it in the lowest terms not both $m$ and $n$ can be even
We obtain $m^2=2n^2$
Since any integer $n^2$ even or odd multiplied by $2$ is even, then $m^2$ must be even as well, hence m is also even because any square root of an even number is even.
Hence $m=2k$ for some positive integer $k$
Hence $m^2=(2k)^2$
so $4k^2=2n^2$
therefore $n^2=2k^2$ for some positive integer $k$, which indicates that $n^2$ is even.
However from the assumption $2=m^2/n^2$ we know that not both m and n can be even, therefore $m^2/n^2=2$ cannot be true  
A: In more concise terms : 
if $\sqrt{2}$ is rational number It can be written like $\sqrt{2}=\dfrac{m}{n}, \quad m\in\mathbb{Z},n\in \mathbb{Z}^*\quad\gcd(m,n)=1$
$\sqrt{2}=\dfrac{m}{n}\iff  2n^2=m^2\implies 2|m^2\iff\boxed{2|m}\;(1)\iff4|m^2\implies 4|2n^2\iff2|n^2\iff \boxed{2|n}\;(2)\qquad (1)\land(2)\implies\gcd(m,n)\geq2$
So $\sqrt{2}$ is not a rational number
A: In addition to the other answers, I like the following way of proving it: Assume there are $a\in \mathbb Z$ and $b\in \mathbb N$ such as 
$(\ast)\qquad \sqrt 2 = \frac a b $  
with $\gcd(a,b)=1$. 
We know that $\sqrt2>1$ as $1^2=1$ and $\sqrt2<2$ as $2^2=4$. 
Squaring $(\ast)$, we obtain
$(+)\qquad 2=\frac{a^2}{b^2}$
As 2 ist a natural number, the numerator in equation $(+)$ must be a multiple of the denominator. Because of $\gcd(a,b)=1$, we know that $a$ and $b$ do not have any common prime factors. Thus, $a^2$ and $b^2$ cannot have any common prime factors, either. We can deduce $b^2=1$ and thus $b=1$. As a consequence, we get $a^2=2$ or $a=\sqrt2$, but $a$ is meant to be an integer.
So $\sqrt2$ being rational would imply $\sqrt2\in\mathbb Z$ and even $\sqrt2\in\mathbb N$, as roots are not negative. So $\sqrt2$ would be natural, greater than 1 and smaller than 2. However, there is no natural number between 1 and 2. Contradiction.
A: $\sqrt{2}$ is not a rational number using $\mod 3$:
Suppose $\sqrt{2}$ is rational. note that $\sqrt{2}>1$, so w.l.o.g. I can take $a,b\in \mathbb{N}$ such that $\sqrt{2}=\frac{a}{b}$
Then $a^2=2b^2$. Note that $a^2\equiv 0,1(\mod 3)$
Case 1:
If $a^2\equiv 1(\mod 3)$, then $b^2\equiv 1(\mod 3)\Rightarrow 2b^2\equiv 2(\mod 3)$, but $a^2=2b^2$, then $a^2\equiv 2(\mod 3)\Rightarrow 1\equiv 2(\mod 3)$, which is impossible.
Case 2:
If $a^2\equiv 0(\mod 3)$, then $b^2\equiv 0(\mod 3)$, then $a=3c$, $b=3d$ where $c,d \in \mathbb{N}$.
Now we again get $c^2=2d^2$, since $9c^2=2\times9d^2$. This lead to a loop. The loop is we can not go to case 1(since that case is impossible). Then we have stay in case 2, which will lead to $3\mid c$ and $3\mid d$ and this process will continue until we end up to $1=2$ which is again impossible.
So there does not exist $a,b\in \mathbb{N}$ such that $a^2=2b^2$. Hence $\sqrt{2}$ is not a rational number$.\space\space\space\blacksquare$
A: The usual proof assumes from the beginning that $m$ and $n$ don't have a common divisor. Then finding one is a contradiction to that assumption.
However you can write the proof also without that assumption; it just gets a bit more complicated.
Assume there exist integers $m,n$ such that $\left(\frac mn\right)^2 = 2$, that is, $m^2=2n^2$. Now be $k$ the largest natural number such that $2^k|m$, and $l$ the largest natural number such that $2^l|n$. Then we can write $m=2^kM$ and $n=2^lN$ with $M$ and $N$ odd.
Therefore $m^2 = 2^{2k} M^2$ and $n^2 = 2^{2l} N^2$, where of course $M^2$ and $N^2$, as products of odd numbers, are also odd.
Thus the claim is that $2^{2k}M^2 = 2\cdot 2^{2l} N^2 = 2^{2l+1} N^2$. Since $M^2$ and $N^2$ are both odd, this implies $2^{2k} = 2^{2l+1}$, which is equivalent to $2k = 2l+1$. But $2k$ is even and $2l+1$ is odd, therefore they cannot be equal.
