By no means an answer but an extended comment. I should also preface I am an amateur mathematician with no formal education. Following Wikipedia's entry for the Riemann zeta function we have the following functional equation:
$$
\zeta(s)=2^{s}\pi^{s-1}\sin\left(\pi s/2\right)\Gamma(1-s)\zeta(1-s)\label{a}\tag{1}
$$
Suppose $s=3/2.$ Via direct substitutions I can rewrite LHS side of $\ref{a}$ explicitly as
$$
2^{3/2}\pi^{3/2-1}\sin\left(3\pi/4\right)\Gamma(1-3/2)\zeta(1-3/2);\label{b}\tag{2}
$$
which after regrouping and gathering like terms is equal to
$$
2^{3/2+1/2}\pi^{3/2-1+1/2}\zeta(-1/2);\label{c}\tag{3}
$$
which is reducible to
$$
4\pi\zeta(-1/2).\label{d}\tag{4}
$$
And so
$$
\zeta(3/2)=4\pi\zeta(-1/2), \label{e}\tag{5}
$$
Observe:
Case 1: Note $4\pi$ is an irrational number. If $\zeta(-1/2)$ is rational then RHS side of $\ref{e}$ is an irrational number and so to by equality is $\zeta(3/2).$
Case 2: Again note $4\pi$ is an irrational number. If $\zeta(3/2)$ is a rational number then the RHS of $\ref{e}$ is a rational number. If $\zeta(-1/2)$ is an rational number then the RHS of $\ref{e}$ is irrational; which is a contradiction. Subsequently $\zeta(-1/2)$ is an irrational number.
From cases 1 and 2 it seems at least one of $\zeta(3/2)$ or $\zeta(-1/2)$ is an irrational number. In particular they cannot both be rational. More strongly it would seem of the infinite number of pairs
$$
\{\zeta(j/2),\zeta(1-j/2)\}_{j\in\mathbb{N}}
$$
at least one number from each of the infinite pairs is irrational.
