Thread: Proof of square root being irrational

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

Originally Posted by ml692787

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

Someone posted this:

If you know the rational root theorem we have:

let y=sqrt(n), then:

y^2-n=0

and if this has rational roots they are amoung the factors (positive or
negative) of n.

So we have sqrt(n) is an integer, and [sqrt(n)]^2=n, that is n is a
perfect square.

So we have that if sqrt(n) is rational then sqrt(n) is an integer, hence
if sqrt(n) is not an integer it is irrational


RonL

I'm pretty sure we have not covered the rational root theorem in class, and as much sense as that theorem makes, I don't think I can use that to solve the problem on a test if we have not covered it yet.