1. ## 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.

2. 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 $\displaystyle m^{1/n}$ is rational, then there exist integers $\displaystyle a$ , $\displaystyle b$ such that $\displaystyle \gcd(a,b)=1$ and:

$\displaystyle m^{1/n}=\frac{a}{b}$

or:

$\displaystyle b^n m = a^n$

Now consider the prime factorisation of $\displaystyle m$ and remember that $\displaystyle a$ and $\displaystyle b$ share no prime factors (in particular consider the multiplicity of the prime factors of $\displaystyle m$ ).

CB

3. Originally Posted by superevilcube
This is the question I have:

What I wanted to do was suppose that m^1/n is rational