- #1

uart

Science Advisor

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## Main Question or Discussion Point

**Looking for "Easy" proof of Pi Irrational**

Hi, I just got to this forum after searching for an easy proof that Pi is irrational. The thread I found (google) was this one HERE. I wanted to reply, but since it is now “archived” I thought it would be better to post a new thread. Sorry if this topic has already been done to death or anything.

I’ve been trying to come up with a simpler proof (of Pi irrational) than the ones I’ve been able to find so far on the net. Most of the proofs I’ve found are similar to the following LINK which I can follow but find unappealing (as in I would have trouble repeating as the steps are not intuitive for me).

I’ve just come up with (what I think is) a proof that I like a lot better. If it is correct then it has almost certainly been done before. Basically I just want to know if this would be considered a valid proof (or at least if it could be massaged into one).

I was trying to get a proof of Pi^2 irrational out of the series 1 + 1/4 + 1/9 + ... 1/n^2 + ... = Pi^2 / 6. I know that this cant be a self contained proof unless it includes a proof of the above series limit, but I thought that if I used the series as a starting point then it might lead to a fairly easy proof from that point on.

The idea I was working on was that if I tried to write

**a (PI^2) / 6 = b**then somehow there would be too many prime factors in the Pi^2/6 (series) denominator for integer

**a**to be bounded. At first I couldn’t make it work but then I remembered that

**Sum{n=1..Infinity : 1/n^2} = Product{k=1..Infinity : (p_k)^2 / ((p_k)^2 - 1)}**, where

**p_k**is the kth prime number, and that seemed to give a pretty straight forward proof. So here it is.

Assume

**Pi^2 / 6 = a / b**for some finite integers

**a**and

**b**

Then

**a = b ( 1 + 1/4 + 1/9 + 1/16 + ... )**

or

**a = b 2^2 / (2^2 - 1) 3^2 / (3^2 - 1) 5^2 / (5^2 - 1) ..... (p_k^2 / (p_k^2 - 1)) ...**

But since

**p^2 - 1 = (p-1) (p+1)**, both of which are even factors for primes p > 2, then it follows that none of the denominator product terms, up to and including the (p_k^2 - 1) term, can cancel either of the two p_k prime factors in the numerator (for p_k > 2). It may be the case that the denominators of further product terms do eventually have prime factors which cancel this double

**p_k**numerator prime factor, but these further product terms will themselves introduce yet larger numerator prime factors and so on. So the largest prime factor of

**a**is unbounded and therefore

**a**is unbounded.

Is this a valid proof? And if not then can it be fixed up?

Thanks :)

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