Relatively prime integer proof

[rideabike]

Homework Statement

Let p be a prime and let n≥2 be an integer. Prove that p^1/n is irrational.

Homework 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.

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?

 

[Dick]

Homework Statement

Let p be a prime and let n≥2 be an integer. Prove that p^1/n is irrational.

Homework 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.

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?

How would the proof by contradiction follow from a=pb^n?

 

[rideabike]

How would the proof by contradiction follow from a=pb^n?

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?

 

[Dick]

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?

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.

 

[rideabike]

No. It works. But can't you use the same sort of idea to show that if p divides a^n, then p divides a? a^n=a*a^(n-1). If p doesn't divide a then it must divide a^(n-1).

How do we know p doesn't divide a?

 

[Dick]

How do we know p doesn't divide a?

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.

 

[rideabike]

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.

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?

 

[Dick]

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?

Yes, that's it exactly.