Proof of square root being irrational

• Sep 14th 2007, 11:58 AM
ml692787
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
• Sep 14th 2007, 01:34 PM
CaptainBlack
Quote:

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:

Quote:

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
• Sep 14th 2007, 02:07 PM
ml692787
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.