Commutative axiom of vector spaces using other axioms.

Question

So I have following axioms. $$\begin{array} &[A_1]&(u+v)+w=u+(v+w) \\ [A_2] & u+0=0+u=u \\ [A_3] & u+(-u) =(-u)+u=0 \\ [A_4] &u+v=v+u\end{array} $$

$u,v\in V$, a vector space and $0$ is the zero element. I want to prove axiom 4 using others. Here is my proof, I wanted to know if I made unnessary steps. $$u=u \\ u+v+(-v)=(-v)+v+u \\ (u+v)+(-v)=(-v)+(v+u) \\ \implies [(u+v)+(-v)]+[-((-v)+(v+u))]=0 \\ (u+v)+(-(v+u))=0 \\ u+v=v+u$$

And here I have taken $-(-v)=v$ which can be proved seperately without using $A_4$.

Answer 1

Here is the usual proof. It uses distributivity.

$(1+1)(u+v) = (1+1)u + (1+1)v = u + u + v + v$

$(1+1)(u+v) = 1(u+v) + 1(u+v) = 1u + 1v + 1u +1v = u + v + u + v$

Therefore, $u + u + v + v = u + v + u + v$.

Now add $-u$ on the left of both expressions and $-v$ on the right to conclude $u+v=v+u$.

Note how associativity is used everywhere.

Answer 2

Be very careful about trying to prove an axiom from others in the list; we usually make it as long as it is because they're all needed.

You shouldn't be able to prove axiom 4 from axioms 1-3. After all, the multiplication of non-singular square matrices is associative with an identity element and inverses, but axiom 4 is the only one on the list it doesn't satisfy.

So your argument must have a flaw, and a comment by Shervin Sorouri has already identified it.

Answer 3

Your mistake is in assuming $-(a+b) = -a-b$, which isn't true in general.

However, if you add A5: $\lambda (x+y) =\lambda x+ \lambda y$ for $\lambda \in K$ ($K$ is the underlying field) and $x,y\in E$, and A6: $1_Kx = x$, and A7 $(\lambda +\mu)x= \lambda x+\mu x$ for $\lambda, \mu \in K$ and $x\in E$ then you can prove A4 from the rest, because then :

$0_Kx = 0_E$: indeed $0_Kx = (0_K+0_K)x= 0_Kx+0_Kx$ hence $0_Kx= 0_E$

$-x = (-1_K) x$: indeed $x+(-1)x = 1_Kx+(-1_K)x = (1_K-1_K)x= 0_Kx = 0_E$

$-(u+v) = -u-v$: follows from A5 and the previous line

$u+v= v+u$: $u+v -(v+u) = u+v -v-u = u-u = 0_E$.

As the other answers and comments mentioned, you can't prove A4 from A1, A2,A3 because that would mean any group is abelian, which is clearly not true (the group of permutations on more than $3$ elements is not abelian for instance, but also the group of affine invertible transformations of the euclidean plane, etc. there are many examples)

A5,A6 and A7 are axioms that are specific to vector spaces that make it so that indeed, you can prove A4 from the rest; but they add a specific structure to the group.