Here I am thinking of using $-(x+y)$ and show that it equals $-(y+x)$.
$-(x+y)=-x-y$ by distributivity
=$-x-y+0=...$ Here I don't know how to continue, could someone suggest?
2026-03-31 05:24:40.1774934680
If commutativity of vector space is omitted, can we still use other axioms to prove the commutativity?
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Yes this is possible even for a (left) module $M$ over a ring $R$ with a multiplicative identity $1$: $$(1+1)(x+y)=x+y+x+y$$ by the left distributive law. But also $$(1+1)(x+y)=2(x+y)=2x+2y=x+x+y+y.$$Cancelling $x$ and $y$ at both sides yields $x+y=y+x$, that is, $M$ must be an abelian group.