definition of vector space needs no commutativity

In the definition of vector space (http://planetmath.org/VectorSpace) oneusually lists the needed properties of the vectoral additionand 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 provedto be a consequence of the other seven axioms.  The proof canbe based on the fact that in defining the group (http://planetmath.org/Group),it suffices to postulate only the existence of a right identityelement and the right inverses of the elements (see the article“redundancy of two-sidedness in definition of group (http://planetmath.org/RedundancyOfTwoSidednessInDefinitionOfGroup)”).

Now, suppose the validity of the seven other axioms (http://planetmath.org/VectorSpace), but not necessarily the above commutative law ofaddition.  We will show that the commutative law is in force.

We need the identity  $(-1)v=-v$  which is easily justified(we have $\overrightarrow{0}=0v=(1+(-1))v=\mathrm{\dots }$).  Then we cancalculate as follows:

$v+u$$=(v+u)+\overrightarrow{0}=(v+u)+[-(u+v)+(u+v)]$

$=[(v+u)+(-(u+v))]+(u+v)=[(v+u)+(-1)(u+v)]+(u+v)$

$=[(v+u)+((-1)u+(-1)v)]+(u+v)=[((v+u)+(-u))+(-v)]+(u+v)$

$=[(v+(u+(-u)))+(-v)]+(u+v)=[(v+\overrightarrow{0})+(-v)]+(u+v)$

$=[v+(-v)]+(u+v)=\overrightarrow{0}+(u+v)$

$=u+v$

Q.E.D.

This proof by Y. Chemiavsky and A. Mouftakhov isfound in the 2012 March issue of The American MathematicalMonthly.