Proof of the irrationality of sqrt(2)

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Suppose V 2   is rational, then we can express it as
 ___
V 2  =  p / q

Where p and q are the smallest positive integers that 
satisfy the above equation.

p and q cannot both be even since if they were, we 
could divide each by 2 and have a smaller p and q
to work with.


From this we can by squaring both sides get:

     2   2                            2
2 = p / q     then by multiplying by q

   2     2
2 q  =  p

                              2              2
From this we can deduce that p is even.  If p is even,
then q must be odd since p and q cannot both be even.

       2
Since p  is even, then p must be even.

                    2
If p is even, then p must be divisible by 4.

Thus, 

   2
  p
-----  is an even integer.
  2

But since

   2     2
2 q  =  p

Then, 

   2
  p        2
-----  =  q
  2

      2 
Thus q  must be even, and therefore q is even.  We had previously

determined that q must be odd.  q cannot be both even and odd, so

we conclude that no such integer exists, which invalidates our 
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premise that V 2  is rational.