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Using ONLY the nine other vector space axioms and clearly justifying each step prove that (-a) + a = 0

I got as far as: Since we are allowed to use all other axioms by the existence of V5 (the inverse) this must be true.

This felt too easy though so I tried to do more but am not sure if I'm making this unnecessarily complicated.

Since we know that V is a vector space and by definition a vectors space has (–x) = (-1)x

0 = (-1)x + x By V10 there exists an identity such at 1x = x

0 = (-1) x + (1)x By V8 we can group the scalar multiples

0 = (-1 + 1)x

0 = 0x

0 = 0 So we have proved this to be true.

user131875
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    I would prefer if you referred by the axioms by more standard names (at first you used closure under scalar multiplication, V10 is the multiplicative identity, etc) but yeah you've got it. More specifically this shows that the inverse $(-a)$ described in the axioms is equal to $(-1)a.$ – Stephen Donovan May 16 '22 at 22:30
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    If you want to improve your post and get more informative answers, you can tell us what your symbols for the axioms represent. Those symbols are not standard, and we are left to guess what you might mean by them, which might not be very helpful to you. – Lee Mosher May 16 '22 at 22:55
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    "By definition $-x = (-1)x$" is almost certainly not true. This is a theorem about vector spaces that you would have to justify if you're doing a question like this. Just saying it's true because of the existence of inverses also seems a bit too short to me - you will at least have to talk about commutativity, if you've got that as an axiom. But in fact the only axioms you need are existence of right inverses, existence of a right identity, and associativity. See here. – Izaak van Dongen May 17 '22 at 14:18
  • You've given this some thought, but without any source for the exercise or list of what axioms are "the nine other", I cannot be sure your thoughts are free of a circular argument. – hardmath May 27 '22 at 00:58

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Your proof seems fine to me (although I would distinguish $0_\mathbb{K}$ and $0_V$ to be more clear, since they are different objects, and say explicitly in which set each object lives).

Personally, I would consider your "too simple" proof to be fine, and just use the group axioms.

A vector space V is an additive group, so it is outfitted with an involution (self-inverse) inverse mapping $\text{inv}: V \to V$ such that $\forall a \in V, \text{inv}(a) + a = (-a) + a = a + \text{inv}(a) = a + (-a) = 0_V$

Tristan Duquesne
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