Let N be a Natural Number such that square root of N is not an integer. Prove that then square root of N is even irrational.

*The problem gives you a hint stating that:

Assume square root of N is in the irrational numbers. Then the Set X := {x is in the real numbers : x multiplied by square root N is in the Natural Numbers} is nonempty. Show that if x is in X and x' := (x multiplied by square root of N) minus (x[square root of N]) then x' is in X and x' < x. thus square root of 2, 3, 5,... are irrational.

Thanks