# definition of vector space needs no commutativity

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In the definition of \PMlinkname{vector space}{VectorSpace} one
usually lists the needed properties of the vectoral addition
and the multiplication of vectors by scalars as eight axioms,
one of them the commutative law
$$u+v = v+u.$$
The latter is however not necessary, because it may be proved
to be a consequence of the other seven axioms.\, The proof can
be based on the fact that in defining the \PMlinkname{group}{Group},
it suffices to postulate only the existence of a right identity
element and the right inverses of the elements (see the article
\PMlinkname{redundancy of two-sidedness in definition of group}
{RedundancyOfTwoSidednessInDefinitionOfGroup}'').

Now, suppose the validity of \PMlinkname{the seven other axioms}
{VectorSpace}, but not necessarily the above commutative law of
addition.\, We will show that the commutative law is in force.

We need the identity\, $(-1)v = -v$\, which is easily justified
(we have $\vec{0} = 0v = (1+(-1))v = \ldots$).\, Then we can
calculate as follows:
\begin{align*}
v+u &= (v+u)+\vec{0} = (v+u)+[-(u+v)+(u+v)]\\
&=\; [(v+u)+(-(u+v))]+(u+v) = [(v+u)+(-1)(u+v)]+(u+v)\\
&= [(v+u)+((-1)u+(-1)v)]+(u+v) = [((v+u)+(-u))+(-v)]+(u+v)\\
&= [(v+(u+(-u)))+(-v)]+(u+v) = [(v+\vec{0})+(-v)]+(u+v)\\
&= [v+(-v)]+(u+v) = \vec{0}+(u+v) \\
&=u+v
\end{align*}
Q.E.D.\\

This proof by {\sc Y. Chemiavsky} and {\sc A. Mouftakhov} is
found in the 2012 March issue of {\it The American Mathematical
Monthly}.\\
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