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: \(0=0’\), \(\blacksquare\).

Take a vector \(v \in V\) and additive inverses \(a,b \in V\).

\begin{equation} a+0 = a \end{equation}

defn. of additive identity

\begin{equation} a+(v+b) = a \end{equation}

defn. of additive inverse

\begin{equation} (a+v)+b = a \end{equation}

associativity

\begin{equation} 0+b = a \end{equation}

defn. of additive inverse

\begin{equation} b=a\ \blacksquare \end{equation}

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