Question

Prove that root 10 is irrational

Answer 1

Proof by contradiction means that you make a (false) assumption and from that assumption, you arrive at something that you know is not correct, therefore the original assumption must be incorrect. Assume that √10 is rational. Therefore √10 = a/b where a and b are coprime integers. Then: √10 = a/b 10 = a^2/b^2 10b^2 = a^2 2*(5b^2) = a^2 Since a^2 is a multiple of 2, a must also be a multiple of 2 (if you square an even number, you get an even number, but if you square an odd number, you get an odd number). If a is an even number, it can be expressed as a = 2c. Plugging that back into the equation we had, we get: 2*(5b^2) = a^2 2*(5b^2) = (2c)^2 2*(5b^2) = 4c^2 5b^2 = 2c^2 For the square of a number (b) multiplied by 5 to be even, the number (b) must end with 0, 2, 4, 6, or 8. That means that b is even. We have come to the conclusion that both a and b are even, but this contradicts our original statement of how we said that a and b are coprime. Therefore our original assumption that √10 is rational is false and it must be irrational.