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Cos(m) Irrational
By Yatir Halevi on Friday, September 27, 2002 - 02:34 pm:

How do you prove that cos(m) is irrational for all natural m?

Yatir

By Julian Pulman on Sunday, September 29, 2002 - 05:27 pm:

Okay, I think I worked a proof - it's largely inspired by Theorem 49 in "An Introduction to the Theorem of Numbers". I've got a few sheets of working here, so I'll try to shorten it a little - apologies for any loss of clarity in doing so.

Define f(x)= [(m - x)2n(m2-(m-x)2)n-1]/(n-1)!
A quick check shows that
0<f(x)<m4n-2/(n-1)! for 0<x<m ...(1)

Now define G(x)=f(x)-f''(x)+f(4)(x)-...-f(4n-2)(x), where f(k) is the kth derivative of f.
Þf(x)=G''(x)+G(x)
Multiply both sides by sin x, and integrate the lefthand side
d/dx [G'(x)sinx-G(x)cosx]=G''(x)sinx + G(x)sinx=f(x)sinx
Therefore, ò0 mf(x)sinx dx=G'(m)sinm-G(m)cosm+G(0) ...(2)

(i) f is a polynomial in (m-x)2, so G'(m)=0.
(ii) Clearly from (i) and f's definition we can show that f(2k)(m) is an integer Þ G(m) is an integer.
(iii) Inspection with some common sense reveals G(0) to also be an integer.

Therefore, if cosm is rational - assume cosm=p/q - then qò0 mf(x)sinx dx is an integer
Recall (1), and note that sin x is always less than or equal to 1. Because of this, multiply the RHS of (1) by 1 and integrate between the limits specified by (2). Take the modulus and we're left with:

|qò0 mf(x)sinx dx|<dm.(m4n-2/(n-1)!)
Let n get arbitrarily large, then dm.(m4n-2/(n-1)!) < 1, and the LHS cannot be an integer. Contradiction, our assumtion that cosm was rational for m¹0 was incorrect.

Julian

By Yatir Halevi on Tuesday, October 01, 2002 - 11:59 am:

Yes, thanks Julian.
I just wonder how people even think of trying function like f(x)...?

Yatir