Nontrivial Zeroes of Riemann Zeta Function are Symmetrical with respect to Critical Line

Theorem

That is, suppose $s_1 = \sigma_1 + i t$ is a nontrivial zero of $\map \zeta s$.

Then there exists another nontrivial zero $s_2$ of $\map \zeta s$ such that:

$s_2 = 1 - s_1$

$\ds \pi^{-s/2} \map \Gamma {\dfrac s 2} \map \zeta s = \pi^{\paren {s/2 - 1/2 } } \map \Gamma {\dfrac {1 - s} 2} \map \zeta {1 - s}$

We suppose $s_1 = \sigma_1 + i t$ is a nontrivial zero of $\map \zeta s$. Then we have:

$\ds \map \zeta {s_1}$

$=$

$\ds 0$

$\ds \pi^{-s_1/2 } \map \Gamma {\dfrac {s_1} 2} \map \zeta {s_1}$

$=$

$\ds 0$

$\ds \pi^{\paren {s_1/2 - 1/2 } } \map \Gamma {\dfrac {1 - s_1} 2} \map \zeta {1 - s_1}$

$=$

$\ds 0$

It remains to be shown that of the three terms on the right hand side, $\map \zeta {1 - s_1}$ MUST equal zero.

We can rewrite the first term in terms of the exponential function

$\ds \pi^{\paren {s_1 / 2 - 1 / 2} } = \map \exp {\map \ln {\pi^{\paren {s_1 / 2 - 1 / 2} } } }$

From the definition of the exponential function, we know $\map \exp {\map \ln {\pi^{\paren { {s_1}/2 - 1/2 } } }}$ never equals zero.

From Zeroes of Gamma Function, we know $\map \Gamma {\dfrac {1 - s_1} 2}$ never equals zero.

Therefore:

$\map \zeta {1 - s_1} = 0$

$\blacksquare$