A proof that the square root of 2 is not a fraction

© Stu Savory, 2004.

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 n2, we get m2 = 2*n2.

This tells us that m2 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 m2 = 2*n2 as (2*k)2 = 2*n2 = 4 * k2. Halving both sides of this, we get n2 = 2 * k2.

This tells us that n2 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.

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