Originally Posted by

**jstarks44444** Hey all, any help with the following proof would be appreciated:

The real number n-th root of 2 is irrational.

A similar proof, which says that the square root of 2 is irrational, is proved in the following way by **contradiction**:

Assume sqroot(2) = m/n for some m,n in Z (integers)

Since it is rational, you can assume m and n have no common factors.

2 = m^2 / n^2 implies m / n = 2n / m

This means n divides m, but that means sqroot(2) is an integer, which is a contradiction.

At our disposal, we have the fact that:

-The real numbers sqroot(2) is irrational

-If r in the Naturals is not a perfect square, then sqroot(r) is irrational

-Let m and n be nonzero integers. Then (m/n)*sqroot(2) is irrational

Thanks for the help!