Thread: Proving something irrational

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.

Originally Posted by superevilcube

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.

Suppose otherwise, that is suppose that is rational, then there exist integers , such that and:



or:



Now consider the prime factorisation of and remember that and share no prime factors (in particular consider the multiplicity of the prime factors of ).

CB

Originally Posted by superevilcube

This is the question I have:

What I wanted to do was suppose that m^1/n is rational
and derive a contradiction

Yes U were right.Let's go on.

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.

proved.


PS: Think u should say n is positive interger first.