mustang19 said:
Well how about this. Let f(x) be whatever function makes the equation work.
What equation?
The denominator is the zeta function can be rewritten as (n^1/2)(n^f(x)). Because taking a square root requires you to factor out the perfect squares in the radicand, you will end up with
Sqrt(perfect squares)*(n^f(x))
So the 1/2 just provides the list of perfect squares to be excluded from the result by the rest of the function.
It's difficult to know what in the world you are talking about. What equation are you talking about? What denominator are you talking about? Why are you talking about taking a square-root?
We have a well-defined function, [itex]\zeta(s)[/itex], which for real [itex]s > 0[/itex] can be written as [itex]\sum_n n^{-s}[/itex], but which can be analytically extended to complex values of [itex]s[/itex]. When we try to look for values of [itex]s[/itex] making [itex]\zeta(s) = 0[/itex], we find find solutions for:
[itex]s = -2, -4, -6, ...[/itex]
and for values of the form
[itex]s = +1/2 + i x[/itex]
The conjecture is: All zeros of [itex]\zeta(s)[/itex] are of this form. It's just one of those conjectures, like Fermat's last theorem, that seems to be true, but nobody has a proof. It's definitely not "just a convention", so that's the answer to your original question, isn't it? Have you changed to a different question, or are you still wondering whether it's just a convention?