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Wikipedia says this:

In the strip 0 < Re(s) < 1 the zeta function satisfies the functional equation

$$ \zeta(s)=2^{s}\pi^{s-1}\ \sin\left({\frac{\pi s}{2}}\right)\ \Gamma (1-s)\ \zeta (1-s). $$

So (ignoring the trivial zeroes) if $\zeta(s)=0$ then $\zeta (1-s) = 0$ since none of the other parts of the equation can equal $0$. So if $\zeta (s) = 0$, then

$\zeta(s)=\zeta (1-s)$

$s = 1 - s$

$s = 1/2$

Is taking the inverse incorrect here because $\zeta (s)$ is multi-valued?

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    $\zeta(s)$ is not multi-valued - you mean to say it's not one-to-one. And yes that is the reason the argument is invalid. A much simpler version of the invalid argument goes as follows: $(-1)^2=1^2$ therefore $-1=1$. This logical error is far, far more basic and elementary than things like zeta functions, analytic continuation, functional equations and nontrivial zeros. – 2017-02-04

1 Answers 1

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(thanks to the properties of the $\Gamma$ function)

The functional equation really means that $$\Xi(s) = (s+1/2)(s-1/2)\pi^{-(s+1/2)/2} \Gamma((s+1/2)/2) \zeta(s+1/2)$$ is an even entire function : $\Xi(s) = \Xi(-s)$. Of course, there are many even functions having some zeros off the imaginary axis, so it tell us nothing on the Riemann hypothesis, except (together with $\Xi(\overline{s}) = \overline{\Xi(s)}$) that the (non-trivial) zeros come in pair $s,-s$ and that $\Xi(it)$ is real for real $t$, thus it has one zero at every sign change.