adding
Last edited: August 8, 2025Operation that adds elements in a set
constituents
- A set \(V\)
- Each non-necessarily-distinct elements \(u,v \in V\)
requirements
addition on a set \(V\) is defined by a function that assigned an element named \(u+v \in V\) (its closed), \(\forall u,v\in V\)
additional information
See also addition in \(\mathbb{F}^n\)
additive identity
Last edited: August 8, 2025The additive identity allows another number to retain its identity after adding. That is: there exists an element \(0\) such that \(v+0=v\) for whatever structure \(v\) and addition \(+\) you are working with.
additive identity is unique in a vector space
Last edited: August 8, 2025Assume 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: \(0=0’\), \(\blacksquare\).
additive inverse is unique in a vector space
Last edited: August 8, 2025Take 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}