Question

Use a contradiction argument to prove that 3√2 is irrational

Answer 1

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.