Why is it so difficult to prove the irrationality of the zeta function for $N$?

306 Views Asked by At

I know that no one has yet been able to prove that the zeta function is irrational at every point $N$. but my question is why is it so difficult to prove the irrationality of the zeta function for $N$?

For example i find a good argument why the zeta function should be irrational (not a proof i know):

Assume that the zeta function at every point is rational you could write it like that

$$\frac{p_n}{q_n}= \zeta(n)= 1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5^n}+\ldots$$

Now divide the sum on the left into $2$ sums with the proviso that one side shares a factor with $q_n$ and the other does not.

\begin{align*}\frac{p_n}{q_n} &= (\text{not Factor with }q_n) + \left[\frac{1}{q_n^n}+\frac{1}{(2q_n)^n}+\frac{1}{(3q_n)^n}+\frac{1}{(4q_n)^n}+\ldots\right] \\ &= (\text{not Factor with }q_n) + \frac{1}{q_n^n}\cdot\left[1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5^n}+\ldots\right] \\ &= (\text{not Factor with }q_n)+ \frac{1}{q_n^n}\cdot\zeta(n) \\ &= (\text{not Factor with }q_n) + \frac{p_n}{q_n^{n+1}} \end{align*}

Then reshaping creates the requirement that both sides have a factor with $q_n$, which is a contradiction!

What are your considerations for this "proof"? And does anyone have an idea how to fully prove it with this method?

1

There are 1 best solutions below

0
On

It's always hard to say when an idea that hasn't produced a proof could lie on the road to a proof. That said, in my mind, your argument doesn't get us any closer.

The key issue: the term you call "(not Factor with $q_n$)". You don't give any suggestion of how we can prove anything about it. Specifically, how do we know it isn't the rational number $\frac{p_n}{q_n}-\frac{p_n}{q_n^{n+1}}$? Proving that looks just as hard as proving that $\zeta(n)$ is irrational.