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In ring $(R,+,*)$, the minus sign is often given as a unary operator for the additive inverse such that:

$\forall x\in R (-x\in R)$

$\forall x\in R(x+(-x)=0 \wedge (-x)+x=0)$

If we have $-x\in R$, can we prove (or assume) that $x\in R$?

EDIT: Although it is really Limitless's subsequent comment that I am accepting, I have indicated acceptance of his/her answer.

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    Indeed! See my reply to Limitless.2012-12-02

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We can prove it via the ring axioms. Namely, if $a \in R$ where $R$ is a ring, $a^{-1}\in R$ under one ring operation. In this case, we have that $-x \in (R,+,\cdot)$. Since the inverse of $-x$ under addition is $x$ (i.e. $-x+x=0$), we have that $x \in (R,+,\cdot)$.

See ring axioms for more. Specifically,

[. . .]$(R, +)$ is an abelian group with identity element $0$, meaning that for all $a$ and $b$ in $R$, the following axioms hold: for each $a$ in $R$ there exists $−a$ in $R$ such that $a + (−a) = (−a) + a = 0$ ($−a$ is the inverse element of $a$)[. . .]

See the comments for the full story!

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    I can think of one reasonable interpretation of the question: if $R$ is a subring of some larger ring $S$ and we have some element $x\in S$ and we know $-x\in R$, do we then also know $x\in R$? (For an explicit example, one might consider $R=\mathbb{Q}$, $S=\mathbb{R}$.) Then the usual proof using the ring axioms works fine.2012-12-02