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