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{\large {\bf  }}
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\noindent
   {\bf Theorem.  The square root of 2 is irrational.}
   
 \noindent
{\it Proof:} Suppose not. If the square root of 2 is rational, 
$$
\sqrt{2}  =\frac{a}{b}
$$
 for some integers $ a$ and $b$. 
 
 First, note that this implies that
$$
\sqrt{2}  =\frac{x}{y}
$$
  for some integers $x$ and $y$ that are not both even.
That's because if $\sqrt{2}  =\frac{a}{b}$ and both $a$ and $b$ are even, then
it is also true that $a$ and $b$ are divisible by 2, so 

$$
\sqrt{2}  =\frac{a/2}{b/2}
$$
with both $a/2$ and $b/2$ being integers. If  $a/2$ and $b/2$ are both
even, we can repeat the division by 2, and we can keep doing
this till either the numerator or denominator is odd (or both are).
So suppose $\sqrt{2}  =\frac{x}{y}$ where $x$ and $y$ are not both even. We
will show that this implies that both $x$ and $y$ are even, a
contradiction.

First, square both sides so  
$$2 = x^2/y^2.$$ 
Then  
$$2y^2 = x^2.$$

 This
implies that $x^2$ is even. But then $x$ must be even too, since if $x$
is odd, $x(x-1)$ is even because it multiplies an odd number ($x$)
times an even one ($x-1$); and from thiswe know  that  $x(x-1)+x$ is odd because it adds
an even number ($x(x-1)$) to an odd number ($x$).

If $x$ is even, we can write it as 
$$
x = 2m
$$
 for some  integer $m$. But
then 
$$2y^2 = (2m)^2 = 4m^2,$$
  so 
  $$
  y^2 = 2m^2.$$
  
   But then $y^2$ is
even, so by our earlier argument, $y$ is even too. But that means
both $x$ and $y$ are even, which contradicts our starting
assumption! So it must be that  $sqrt{2} \neq  \frac{a}{b}$; there are no  integers $a$
and $ b$ such that  $\sqrt{2} = \frac{a}{b}$, so $\sqrt{2}$ is irrational.   {\it Quod erat demonstrandum.}

Is there a proof that $\sqrt{2}$ is irrational that does not use contradiction? I don't know. 
 
  
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