How do you prove that square root of 2 is irrational? - Answers

It is proved by contradiction (reductio ad absurdum), a powerful type of proof in mathematics. Assume that the square root of 2 is rational and is equal to a/b where a and b are integers. Squaring both sides gives: 2=a2/b2 a2=2b2 Since a2 is even, it implies that a is even So, replacing a by 2k where k is an integer, we have: (2k)2=2b2 b2=2k2 Since b2 is even, it implies that b is even, which is in contradiction to our first statement: a/b is in lowest terms. Thus, the square root of 2 is irrational. (Q.E.D) Note: It can be proved that if a2 is even, then a is also even. Proof: Odd numbers are of the form 2n+1, where n is an integer. When an odd number is squared, we have: (2n+1)2 = 4n2+4n+1 = 2(2n2+2n) +1 = 2y+1 where y = 2n2+2n y is also an integer; so, 2y+1 is an odd number, which in turn means that the square of any odd number is also odd. Therefore, if the square of a number is even, the number cannot be odd; it has to be even. (Q.E.D)