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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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    Have I adequately addressed your question?2012-12-02
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    This is a weird question. If you don't suppose $x\in R$ from the outset, what do you mean by $-x$?2012-12-02
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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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    You said, "The inverse of −x under addition is x." I don't see how you can arrive at that conclusion that from the ring axioms. Those axioms relating to inverses quantify over known elements of $R$. Perhaps there should be some partition of $R$ into positive and non-positive sets as sometimes done for the field (order) axioms?2012-12-02
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    @DanChristensen, what are you representing with $-x$? Is it not the inverse of the element $x$ under addition? Well, if we take the inverse of the inverse of the element $x$ under addition, we'd arrive back at $x$. I may be misunderstanding your question. Regarding your recent comment, what can $-x$ mean if we do not already know $x \in R$?2012-12-02
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    Yes, that is what I have always done as well. Revisiting the axioms, however, I don't see how this practice can be formally justified.2012-12-02
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    @DanChristensen, with all due respect (as I rather like this question), are you sure that this even makes sense mathematically? That is, are you sure that this is not a notational breakdown? More concisely, if we do not know that $x \in R$, isn't $-x \in R$ meaningless (the use of $-$ is only defined on elements of $R$?)?2012-12-02
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    A strict interpretation of the axioms usually given for RT certainly leads to that conclusion. So, I guess the answer to my question is no; we cannot prove or assume $x\in R$. On its own, $-x\in R$ is meaningless.2012-12-02
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    I accept with the understanding that it is really your subsequent comment that I am accepting. (There should be a way to accept comments as answers.)2012-12-02
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    @DanChristensen, I've added to my answer in order to clarify so that no one misunderstands (hopefully).2012-12-02
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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