Someone posted this:

So we have that if sqrt(n) is rational then sqrt(n) is an integer, henceIf 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.

if sqrt(n) is not an integer it is irrational

RonL