Prove that √2 is irrational
Begin by assuming that √2 is rational, and can therefore be written as √2 = p\q where p and q are coprime integers.By squaring both sides, you get the result 2 = p2\q2, which rearranges to show that p2=2q2.This implies that p2 is even, and therefore p must also be even. Therefore p=2a where a is an integer.By substituting p=2a into our equation, and then rearranging, we get the result q2=2a2This implies that q2 is even, and therefore q must also be even, so we can write q=2b, where b is an integer.From this it follows that √2 = p/q = 2a/2b which shows that p and q have a common factor of 2, however, we have stated that p and q are coprime, and therefore we have a contradiction. Our original assumption must therefore be false, and therefore √2 must be irrational.
