This is the question I have:
Show that if m is a natural number and is not the nth power of an integer, then m^(1/n) is irrational.
I'm pretty stumped. What I wanted to do was suppose that m^1/n is rational and derive a contradiction that m is equal to some integer to the nth power (contradicting the hypothesis), but I'm not sure how to do that. I tried and soon got stuck.
Suppose m^1/n is rational, then m^1/n = a/b for some a,b in the integers in lowest terms. Hence m = a^n/b^n which implies that mb^n = a^n. So m|a^n which means mi=a^n , for some integer i. But then this just takes me in a circle... that integer i is just b^n.
I'm not looking for a full solution, just something that will drive me in the right direction. Thanks for the help.
Then m^1/n = a/b, a,b is interger, b is not zero ,mean (a/b)^n=m,and m is interger >=0(Nature number), then a/b is interger, but first we said "m is not the nth power of an integer' so ,that is inconsistent.
PS: Think u should say n is positive interger first.