# Thread: Proving something irrational

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

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

or:

$b^n m = a^n$

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

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
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.