# Homework Help: Relatively prime integer proof

1. Dec 2, 2012

### rideabike

1. The problem statement, all variables and given/known data
Let p be a prime and let n≥2 be an integer. Prove that p1/n is irrational.

2. Relevant equations
We know that for integers a>1 and b such that gcd(a,b)=1, a does not divide b^n for any n≥
1.

3. The attempt at a solution
To prove irrationality, assume p^(1/n)=a/b for integers a and b≠0.
This is equivalent to an=pbn

If we've assumed a and b have been reduced to lowest terms, gcd(a,b)=1.
Then the proof by contradiction would follow directly if it were just a=pbn
But what do I do since it's an?

2. Dec 2, 2012

3. Dec 2, 2012

### rideabike

Since gcd(a,b)=1 and a does not divide b^n for any n≥1, by Euclid's Lemma a divides p, which is a contradiction since p is prime.

Or does this not work because a is not necessarily greater than 1?

4. Dec 2, 2012

### Dick

a isn't necessarily prime. You can't use Euclid's lemma that way. Sorry about my previous post. I deleted it. I was reading too fast.

Last edited: Dec 2, 2012
5. Dec 2, 2012

### rideabike

How do we know p doesn't divide a?

6. Dec 2, 2012

### Dick

You can show p divides a. If p divides a^n (which it does) then p divides a. Would you agree with that? Sorry about the confusion of my previous post. I modified it.

7. Dec 2, 2012

### rideabike

Okay, I see that now, it's recursive.
Then you could replace a with pk for some integer k.
Then (pk)^n=pb^n
p^(n-1)k^n=b^n

Could the same logic as before be used to show p divides b, hence a and b have a common divisor, producing a contradiction?

8. Dec 2, 2012

### Dick

Yes, that's it exactly.