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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? –  000 Dec 2 '12 at 6:24
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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 $-x$? –  Marc van Leeuwen Dec 2 '12 at 6:39
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Indeed! See my reply to Limitless. –  Dan Christensen Dec 2 '12 at 6:41

1 Answer 1

up vote 2 down vote accepted

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? –  Dan Christensen Dec 2 '12 at 6:38
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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 $R$?)? –  000 Dec 2 '12 at 6:48
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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. –  Dan Christensen Dec 2 '12 at 7:05
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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.) –  Dan Christensen Dec 2 '12 at 15:23
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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. –  anonymous Dec 2 '12 at 16:30

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