Question #1963

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

Expert's answer

Assume that t = 3√2 is rational and can be expressed as a/b, a fraction in lowest terms, so t^{2} = 18 = a^{2}/b^{2}, a^{2} = 18 b^{2}. We see that a^{2 }is even, so a must be even too. Denote a = 2c, a^{2} = 4c^{2} and 4c^{2} = 18b^{2}, or 2c^{2} = 9b^{2}, 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.

It's a contradiction, since we assumed that a/b = 2c/2r should be in lowest terms, so 3√2 must be irrrational.

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