hi all,
can anyone prove that pi is irrational ?
I hope it will not involve very high mathematics ?
thanks
cheers
Niranjan.

It goes (sort of) like this: let q be any integer.

Let A_n = qⁿ /n! ò₀^p xⁿ (p-x)ⁿ sin(x) dx

Then show, by integrating A_n by parts twice, that A_n+2 = (4n+6)q A_n+1 -(qp)² A_n. This is quite fiddly, as it is easy to get very confused with the sheer volume of algebra, but it doesn't need you to do anything particularly clever.

Now for the important bit. Suppose p is rational; then we can write p = p/q, so that both q and qp are integers.

Then if we use this q in the recurrence relation above, we see that A_n+2 is a sum of integer multiples of A_n+1 and A_n. So if these are integers, A_n+2 is also an integer. We can check the values of A₀, which is 2, and A₁, which is 4q. So, if our q exists, then A₀, A₁, and hence A₂ and all the other A's are integers.

Now we are almost there, although it may not look like it. How big is A_n? Since sin(x) ³ 0 for x=0..p, A_n is positive for all n.

However, for x between 0 and p, xⁿ £ pⁿ,

(p -x)ⁿ £ pⁿ,

and sin(x) £ 1.

So A_n is at most qⁿ /n! ò₀^pp²ⁿ dx = p(qp²)ⁿ /n!.

Can you see that as we make n larger and larger, this will inevitably get smaller and smaller? In other words, it tends to 0 as n gets big.

So, we know three things about A_n:

• it's getting smaller and smaller
• it's never zero
• it's an integer

So we have, at last, reached a contradiction. Thus our original assumption that q existed must have been false; we can't write p = p/q, i.e. it is irrational. QED.

David

Maybe looking at this archive is a good idea.

Kerwin

a much similar but slightly different variation of the proof is given here....
http://www.mathpages.com/home/kmath313.htm

love arun