Formula for $\mathrm{ch}(E \wedge E)$ gives fractional rank of $E \wedge E$?

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$\newcommand{\ch}{\mathrm{ch}}$In this answer, we have the following formula for the Chern character of a wedge of the vector bundle $E$ on $X$:

$$ \ch(E \wedge E) = \frac{1}{2} \ch(E)^2 - r_2 \cdot \ch(E)$$ where $r_2$ acts on $H^{2k}(X, \mathbb{Q})$ by multiplication by $2^k$.

My question is as follows: say $E$ has odd rank i.e. $\mathrm{rk}(E) = \ch_0(E)$ is odd. Then $\ch_0(E \wedge E) \in \frac{1}{2} \mathbb{Z}$, but I thought $\ch_0$, i.e. the rank, should always be integral. Am I going wrong somewhere?