This is a proof by contradiction. Firstly, assume sqrt(2) is rational, i.e can be
represented as the irreducible fraction **m/n** where m and n are integers.
We have sqrt(2)=m/n. Squaring, and multiplying both sides by n^{2},
we get m^{2} = 2*n^{2}.

This tells us that m^{2} is even. Now the only way to get an even square is to
have its root also even, because even*even=even and odd*odd=odd. So **m** must also be even.
This means that we can write **m** = 2*k where **k** is another integer.

So now we can rewrite m^{2} = 2*n^{2} as
(2*k)^{2} = 2*n^{2} = 4 * k^{2}. Halving both sides of this, we get
n^{2} = 2 * k^{2}.

This tells us that n^{2} is even. So **n** must also be even by the
same reasoning as given above. So we can write n = 2 * **j**.

So if m is even and n is even, then m/n is not an irreducible fraction. And this argumentation can go on for ever. So the assumption that sqrt(2) is rational must be wrong, thus sqrt(2) is irrational. Q.E.D.

Now go visit my blog please, or look at other interesting maths stuff :-)

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