This will be a proof by contradiction. 1. * Assume that the √5 is rational (can be expressed in the form , where a and b are integers. 2. * Then √5 can be written as an irreducible fraction (the fraction is reduced as much as possible) a / b such that a and b are coprime integers and (a / b)2 = 5. 3. * It follows that a2 / b2 = 5 and a2 = 5 b2. 4. * Therefore a2 is divisible by 5 because it is equal to 5 b2 which is obviously divisible by 5. 5. * It follows that a must be divisible by 5. 6. * Because a is divisible by 5, there exists an integer k that fulfills: a = 5k. 7. * We insert the last equation of (3) in (6): 5b2 = (5k)2 is equivalent to 5b2 = 25k2 is equivalent to b2 = 5k2. 8. * Because 5k2 is divisible by 5 it follows that b2 is also divisible