# prove cube root 6 is irrational

•We shall start by assuming (cube root 6) as rational. In other words, we need to find integers x and y such that

(cube root 6) = x/y.

• Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, y(cube root 6) = x.

•Cubing both side, we get, 6 y^3 = x^3.

•Thus, x^3 is divisible by 6, and by theorem we can say that x is also divisible by 6.

•Hence, x = 6z for some integer z.

•Substituting x, we get, 6 y^3 = 216 z^3 i.e. y^3 = 36 z^3; which means y^3 is divisible by 6, and so y will also be divisible by 6.

•Now, from theorem, x and y will have 6 as a common factor. But, it is opposite to fact that x and y are co-prime.

•Hence, we can conclude (cube root 6) is irrational.