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 $G(x)=f(x)-f\text{'}\text{'}(x)+f^{(4)}(x)-\dots -f^{(4n-2)}(x)$, where $f^{(k)}$ is the $k$th derivative of $f$.

$\Rightarrow f(x)=G\text{'}\text{'}(x)+G(x)$

Multiply both sides by $\sin x$, and integrate the lefthand side

$d/\mathrm{dx}[G\text{'}(x)\sin x-G(x)\cos x]=G\text{'}\text{'}(x)\sin x+G(x)\sin x=f(x)\sin x$

Therefore, ${\int }_0^{m}f(x)\sin x\mathrm{dx}=G\text{'}(m)\sin m-G(m)\cos m+G(0)$ ...(2)

(i) $f$ is a polynomial in $(m-x{)}^2$, so $G\text{'}(m)=0$.

(ii) Clearly from (i) and $f$'s definition we can show that $f^{(2k)}(m)$ is an integer $\Rightarrow G(m)$ is an integer.

(iii) Inspection with some common sense reveals $G(0)$ to also be an integer.

Therefore, if $\cos m$ is rational - assume $\cos m=p/q$ - then $q{\int }_0^{m}f(x)\sin x\mathrm{dx}$ 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:

$|q{\int }_0^{m}f(x)\sin x\mathrm{dx}|<\mathrm{dm}.(m^{4n-2}/(n-1)!)$

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

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

Yatir