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

Yatir

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.

A quick check shows that

0 < f(x) < m^4n-2 /(n-1)! for 0 < x < m ...(1)

Now define G(x)=f(x)-f ' ' (x)+f⁽⁴⁾ (x)-¼-f^(4n-2) (x), where f^(k) is the kth derivative of f.

Þ f(x)=G ' ' (x)+G(x)

Multiply both sides by sinx, and integrate the lefthand side

d/dx [G ' (x)sinx-G(x)cosx]=G ' ' (x)sinx + G(x)sinx=f(x)sinx

Therefore, ò₀^m f(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^m f(x)sinx dx is an integer

Recall (1), and note that sinx 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^m f(x)sinx dx| < dm.(m^4n-2/(n-1)!)

Let n get arbitrarily large, then dm.(m^4n-2/(n-1)!) < 1, and the LHS cannot be an integer. Contradiction, our assumption that cosm was rational for m ¹ 0 was incorrect.
Julian

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

Yatir