Use a contradiction argument to prove that 3√2 is irrational
Assume that t = 3√2 is rational and can be expressed as a/b, a fraction in lowest terms, so t2 = 18 = a2/b2, a2 = 18 b2. We see that a2 is even, so a must be even too. Denote a = 2c, a2 = 4c2 and 4c2 = 18b2, or 2c2 = 9b2, therefore b must be even too, b = 2r. It's a contradiction, since we assumed that a/b = 2c/2r should be in lowest terms, so 3√2 must be irrrational.