# Proof of the irrationality of sqrt(2)

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Suppose V 2 is rational, then we can express it as
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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.