# Proofs of Irrationality Correct?

1. Sep 3, 2006

### courtrigrad

Last edited: Sep 3, 2006
2. Sep 3, 2006

### shmoe

I'm just going to comment on 2b, though you could have put this in your other thread, no?

q^2*n=p^2 does NOT imply n divides p like you claim, n is not necessarily prime.

You didn't bother to show such an r and m even exist, and it doesn't really matter since you didn't use them anyways except to state their existence. It looks like you just cut and pasted some of my suggestions without understanding or even using them?

the purpose of pulling out this squarefree thing r from n was to let you isolate a prime with an odd power appearing on one side but not the other. See, this is where 2*q^2=p^2 goes wrong, you will have 2 appearing to an odd power on the left (possibly 1) and an even power on the right (possibly 0). You *know* unique factorization even if you haven't proven it at this point right? Something should be fishy about a prime appearing to different exponents on either side. This is what you exploit in the sqrt(2) case, this is what you can exploit in the sqrt(n) case when n is not a perfect square.

3. Sep 4, 2006

### shmoe

What you are now claiming in that link, n=r*m^(2k+1) with r squarefree, is no longer possible in general. I don't see why you think m would have to divide p either.

If you are going to try to invoke unique factorization, aka the fundamental theorem of arithmetic (not algebra), then you can just compare the exponents of primes in n*q^2=p^2. Since n is not a perfect square you should know that it has a prime appearing to an odd power (you MUST prove this if you haven't established it already). Again, you should only use unique factorization if this is something you, your book, or your notes have proven already.

4. Sep 4, 2006

### courtrigrad

How would you prove that $$\sqrt[p]{n}$$ is irrational? Assume that it is rational. I said it was equal to $$\frac{a}{b}$$ where $$a,b$$ are positive integers. Then we get $$b^{p}\times n = a^{p}$$. From this last equation, we see that $$n$$ divides $$a$$. Therefore there exists some integer $$k$$ such that $$a = nk$$. Substituting gives us $$b^{p}\times n = n^{p}k^{p}$$ or $$b^{p} = n^{p-1}k^{p}$$. Can we conclude that $$n$$ divides $$b$$?

By the way, this is from a calculus book not an analysis book (Courant and John)

Thanks

5. Sep 4, 2006

### shmoe

I would settle the square case first before trying to tackle this more general version.

This doesn't follow. This gives n divides a^p, you can't then conclude n divides a since n isn't necessarily prime here.

The title of the book doesn't really matter.

6. Sep 5, 2006

### courtrigrad

ok, I got the $$\sqrt{n}$$ case. What about the $$\sqrt[p]{n}$$ case? $$p$$ cant be a perfect pth power. I have to show that because of unique factorization, you cant have an odd and even number of prime factors on two different sides of the equality sign.

So a perfect pth power has an even number of prime factors?

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