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adding

Last edited: August 8, 2025

Operation 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, 2025

The 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, 2025

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: \(0=0’\), \(\blacksquare\).

additive inverse is unique in a vector space

Last edited: August 8, 2025

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}

ADHD

Last edited: August 8, 2025