Suppose you have an ordered field $(\mathbb{F}, +, \cdot, \le)$. How to prove that $-0 = 0$?
One way, I have thought so far is:
$$ -0 = -(0 + 0) = -0 + (-0)$$
and therefore since $-0 + a = a$, where $a = -0$ then $-0 = 0$, by uniqueness of additive inverse.
Is this correct?
On
As suggested in the comments, one way to see this is to start from $0+0=0 \; \; (\text{eq.} \; 1)$. Then, the axiom of additive inverses gives that $0-0=0 \; \; (\text{eq.} \; 2)$. Setting $(\text{eq.} \; 1)=(\text{eq.} \; 2)$ gives $0+0=0-0$, and we obtain $0=-0$ since $0$ is the additive identity.
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therefore since $-0 +a = a$, where $a = -0$ then $-0 = 0$, by uniqueness of additive inverse.
This doesn't work because the meaning of "additive inverse" is always of the form $a + (-a) = 0$ and/or $(-a) + a = 0$. Your $(-0) + (-0) = -0$ doesn't match that since the right side is not obviously $0$ (without the result we're currently to prove).
If you've already proved that for any $a$, there is only one $z$ such that $a + z = a$, then you could fix this by referencing that instead of saying "by uniqueness of additive inverse".
Or, the other answers present a different (maybe "simpler") approach: $0 + (-0) = 0$ by the definition of additive inverse. (That is, substitute $a=0$ into $a + (-a) = 0$.) And $0 + (-0) = -0$ because $0$ (appearing at the left) is the additive identity. Therefore $-0 = 0$.
We have $0=0+(-0)=-0$. The first equality is by definition of the additive inverse, and the second equality is by definition of the additive identity (zero).