Assume for the sake of contradiction \exists\ 0, 0’ both being additive identities in vector space V. Therefore:
\begin{equation} 0+0’ = 0’ +0 \end{equation}
commutativity. Therefore:
\begin{equation} 0+0’ = 0 = 0’+0 = 0' \end{equation}
defn. of identity. Hence: 0=0’, \blacksquare.