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}
Therefore:
\begin{equation} 0+0’ = 0 = 0’+0 = 0' \end{equation}
defn. of identity.
Hence: \(0=0’\), \(\blacksquare\).