Prove that √2 is irrational

Prove by contradiction: Assume negation to be true i.e. √2 is rational Then √2 can be written in the form a/b where a and b are integers with no common factor (the fraction cannot be simplified) => a/b = √2 => a = b√2 => a^2 = 2b^2 => a^2 is even, so 2 is a factor of a. Therefore let a = 2k, where k is a whole number greater than zero => (2k)^2 = 2b^2 from above => 4k^2 = 2b^2 => b = 2k^2 => b is even, so 2 is a factor of b. Therefore a and b have a common factor of 2. This contradicts our original assumption that a and b have no common factor. Therefore our assumption that √2 is rational is false Therefore √2 is irrational.