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

1. Jul 7, 2008

### foxjwill

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 7, 2008

### 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.

3. Jul 7, 2008

4. Jul 7, 2008

### 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)

Last edited: Jul 8, 2008
5. Jul 7, 2008

### 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.

Last edited: Jul 8, 2008
6. Jul 8, 2008

### 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!

7. Jul 8, 2008

### 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.

8. Jul 8, 2008

### 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.

9. Jul 8, 2008

### 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.

10. Jul 8, 2008

### 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.

Last edited: Jul 8, 2008
11. Jul 8, 2008

### matt grime

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

12. Jul 8, 2008

### foxjwill

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