This is proved by a reductio in absurdum, i. e. by assuming that the opposite is true, and showing that this leads to a logical contradiction.

Let the square root of two, assumed to be a fraction, be = p/q, where p is the numerator and q the denominator of the fraction. The fraction is assumed to have been reduced to its simplest form, so that there are no common factors between p and q. (For instance 3/12 would already have been simplified to 1/4.)

Then p²/q² = 2 by definition, or p² = 2q².

This means that p² is an even number (a multiple of 2). In that case p itself must be an even number, for if it were an odd number, then p² would also be an odd number. So let us write p = 2r, where r is an integer.

We now have 4r² = 2q², or 2r² = q². So by the same argument that led us to conclude that p was an even number, q must be an even number. But this means that both p and q are divisible by 2, which contradicts our initial assumption.

So the square root of 2 cannot be a fraction.