Proof that p^(1/n) is not in Q

[foxjwill]

Homework Statement

Prove that if p is prime and r is a natural number, then p^{1/n} \not\in \mathbb{Q}.

Can someone check the validity of my proof? I have a strong feeling that it's invalid since the primality of p is never used.

Homework Equations

The Attempt at a Solution

Assume that p^{1/n}\in \mathbb{Q}. Then for a,b\in \mathbb{Z} such that a and b are coprime, p^{1/n}={a \over b}, and therefore p={a^n \over b^n}. So a^n must be a multiple of b^n which implies that a is a multiple of b. But by definition, a is not a multiple of b. Contradiction. Q.E.D.

 

[Dick]

Pick a number that isn't prime. (36)^(1/2)=6/1. Sure, 36=6^2/1^2. 6 and 1 are coprime. Where does your invalid proof fall apart? Now get in there and use that p is prime.

 

[durt]

See if you can modify the last four steps here: http://en.wikipedia.org/wiki/Square_root_of_2#Proof_by_unique_factorization.

 

[foxjwill]

Assuming that a and b are coprime, p={a^n \over b^n}. This implies that a^n=p b^n. Since a^n and b^n are also coprime, a^n can be represented as the product of primes and b^n can be represented as the product of a different set of primes. But since there are no prime factors common to both a and b, none will magically appear in the equation to cancel out and leave p.

I'm not sure whether this version is rigorous enough. (or even valid)

 

[Dick]

Actually, that's pretty much it. You've shown p divides a^n. Can you show that means p divides a? If so, at least how many factors of p divide a^n? Can you show that means p divides b^n and hence b? Then you are done.

 

[foxjwill]

Ok, so I reformatted my proof more rigorously in the format of the article durt linked to. Here it is:1) Assume that p^{1/n} is rational. Then \exists a,b \in \mathbb{Z} such that a is coprime to b and p^{1/n}={a \over b}.
2) It follows that p = {a^n \over b^n}.
3) By the unique factorization theorem, \exists x,y\in \mathbb{Z}^+_0 and odd integers r and s such that a = p^x r and b = p^y s
4) Therefore, a^n = p^{nx}r^n and b^n=p^{ny}s^n.
5) Inserting back into 3), p^{nx}r^n=p p^{ny}s^n, so p^{nx}r^n=p^{ny+1}s^n.
6) This states that an integer with an even power of p equals an integer with an odd power of p. But this contradicts the prime factorization theorem, completing the proof.
Q.E.D.

edit: it's amazing what you can do with symbols!

 

[Dick]

It's amazing what you can obscure with symbols as well. Ouch. Now you've created an obscure monster. If p divides a then it can't divide b, right? So y is zero. Can you simplify that. Like, a lot? It's making my head hurt.

 

[foxjwill]

1) Assume that p^{1/n} is rational. Then \exists a,b \in \mathbb{Z} such that a is coprime to b and p^{1/n}={a \over b}.
2) It follows that p = {a^n \over b^n}.
3) By the unique factorization theorem, \exists x\in \mathbb{Z}^+_0 and odd integers r, s not multiples of p such that a = p^x r and b = s
4) Therefore, a^n = p^{nx}r^n and b^n=s^n.
5) Inserting back into 3), p^{nx}r^n=ps^n, so p^{nx}r^n=ps^n.
6) This states that an integer has two different prime factorizations. This contradicts the prime factorization theorem, completing the proof.
Q.E.D.

How about that? I think the proof on the wikipedia article made the same complications as I did.

 

[matt grime]

I'm still with Dick on the unnecessary nature of a lot of that.

We have pa^n=b^n, so p divides b^n, hence b. We may factor that out and deduce

a^n=p^{n-1}b' for some b', thus p divides a, contradicting the assumption that a and b were coprime.Alternately, one can simple say that in pa^n=b^n, the sum of the indices of the primes in a decomposition of the LHS is one more than a multiple of n, and on the right is a multiple of n, which contradicts the uniqueness of prime decomposition.

 

[foxjwill]

Ok. That makes sense. Thanks.

Here's the (hopefully final) rewrite:
1) Assume that p^{1/n} is rational. Then \exists a,b \in \mathbb{Z} such that a is coprime to b and p^{1/n}={a \over b}.
2) It follows that p = {a^n \over b^n}, and that a^n=pb^n.
3) So, by the properties of exponents along with the unique factorization theorem, p divides both a^n and a.
3) Factoring out p from (2), we have a^n=p^{n-1}b' for some b'\in \mathbb{Z} not divisible by p.
4) Therefore p divides a.
5) But this contradicts the assumption that a and b are coprime.
6) Therefore p^{1/n}\not\in \mathbb{Q}.
Q.E.D.

 

[matt grime]

You cannot conclude that b' is not divisible by p.

 

[foxjwill]

You cannot conclude that b' is not divisible by p.

oh, right. Now that I think about, that makes sense.