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I'm doing some very low level proofs in abstract algebra and I have derived the fact that (on a particular ring) $0 = -1$ where $0$ is the additive identity of the ring in question and $-1$ is the inverse of its multiplicative identity.

I wish to prove that this implies $0 = 1$ (which I can use to prove that my ring is the trivial ring) but I am struggling to do so. Intuitively we can add $1$ to both sides and perform the following derivation

\begin{align*} 0 = -1 &\implies 0 + 1 = (-1) + 1 \tag{???} \\ &\implies 1 = (-1) + 1 \tag{Identity} \\ &\implies 1 = 0 \tag{Inverse} \\ \end{align*}

But I'm not really sure how to justify adding $1$ to both sides. I believe the definition of equality on a ring has been inherited from set theory, however I can't figure out how one might prove this using the axioms of set theory.

How might I go about proving that $a = b \implies a+x = b+x$? Is this something that is proven or is it derived from some definition somewhere?

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  • $\begingroup$ I'm not sure I understood the question. Are you just trying to show that $0 = -1$ implies that $0 = 1$? $\endgroup$ – Ben Grossmann Sep 25 '17 at 1:05
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    $\begingroup$ Part of the definition of the definition of a ring is that the elements form an abelian group under $+$. If you go back to the definition of a group (or the definition of a binary operation), you'll be able to show that $a = b \implies a + x = b + x$. $\endgroup$ – Ben Grossmann Sep 25 '17 at 1:07
  • $\begingroup$ "I'm doing some very low level proofs in abstract algebra and I have derived the fact that 0=−1 where 0 is the additive identity of the ring in question and −1 is the inverse of its multiplicative identity." Then you've done something wrong. $\endgroup$ – fleablood Sep 25 '17 at 1:08
  • $\begingroup$ In a ring $(R,+,\cdot)$, we know (by definition) that $(R,+)$ is an abelian group (we don't really need the abelian part though, just that $R$ is a group w.r.t $+$), so $\forall~x\in R~\exists -x\in R~:~x+(-x)=0$ and hence $a=b\iff a+x=b+x~\forall~a,b,x\in R$ $\endgroup$ – Prasun Biswas Sep 25 '17 at 1:10
  • $\begingroup$ :"How might I go about proving that a=b⟹a+x=b+x? Is this something that is proven or is it derived from some definition somewhere? " Is basic substitution of equality. If a IS b then anything you do to a is the same thing as doing it to b because a IS b. SO $a = b \implies a + x = b + x$. That's a given. $\endgroup$ – fleablood Sep 25 '17 at 1:11
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It's just because addition (and all binary operations, for that matter) are functions.

If $a=b$, then the ordered pair $(a,x)=(b,x)\in R\times R$.

By the definition of a function, $a+x=+(a, x)=+(b, x)=b+x$

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  • $\begingroup$ I feel that some other answers offered (in comments too) are just exceedingly indirect or overcomplicated descriptions of this simple fact. $\endgroup$ – rschwieb Sep 25 '17 at 3:06
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Most traditional set theories imply the basic concepts of function and relation: a set is in most of the axiomatizations simply a relation. In particular 2 relations are usually considered fundamental: "is an element of" and "is equal to".

Ring axioms take for granted a certain set theory axiomatization with which they define the ring set and add to that 2 total binary functions (or 2 tertiary relations +(x,y,z) and *(x,y,z) whose meaning is self-explainatory I guess). But this isn't really relevant if I understand what you're doing.

Being the 2 ring actions just standard functions (i.e. non multi-valued) then:

(a + x = b && a + x = c) implies (b = c)

(a + x = b && b = c) implies (a + x = c)

(a + b = x && a + c = x) implies (b = c)

Etc... including of course your "both sides" theorem given the proper substitutions.

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