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!
The following statement, which is a generalization, is true " where is rational if and only if is an integer[/tex]" this can be seen as a corollary of the fact that the only rational algebraic integers are integers themselves. The proof is simple and basically follows from the rational root theorem. So, is never rational since it's never an integer since for .