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In writing proofs, I'll sometimes end up with two separate but similar results, such as $[a] + [b] = [a + b]$ and $[a] \times [b] = [a \times b]$.

Out of curiosity, is there some standard notation to denote two operations at once, besides the $\pm$ sign? For example, maybe there's some symbol like $\ast$ that represents addition and multiplication?

Cisplatin
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    Maybe say "the map $x\mapsto [x]$ is a ring homomorphism"? My idea is not to find a compact notation, but a sentence that communicates effectively. – Stephen Montgomery-Smith Dec 14 '13 at 18:54
  • I would think that $\pm$ communicates more effectively than writing two equations, one with $+$ and one with $-$; it also shows the meaning of the results more elegantly. I'm just trying to find other notations like this, that are both compact and effective. – Cisplatin Dec 14 '13 at 18:57
  • Yes, but I think the $\pm$ sign is effective communication. If there were an equivalent $*$ to mean $\times$ or $+$, I don't think it would be effective communication. – Stephen Montgomery-Smith Dec 14 '13 at 19:00
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    The symbol $\pm$ has no business here,it's not a binary operation, nor is it short for two binary operations. The string of symbols $a\pm b =c$ is short hand notation for the statement $a+b=c\lor a-b=c$, $\pm$ is not a binary operation at all. – Git Gud Dec 14 '13 at 19:03
  • There isn't a widely-known symbol to denote both $+$ and $\times$ (if there is one at all) so if you were to use a symbol it would just make things less clear because not many people would recognize it. – andraiamatrix Dec 14 '13 at 19:06
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    You can abbreviate by saying something like: let $\in {+,\times}$, for all $a,b$ the following holds: $[a][b]=[a*b]$. Clearly $*$ can take multiple meaning, the same way $+$ in $[a] + [b] = [a + b]$ has multiple meanings. That's a problem that arises from the way you set up question. – Git Gud Dec 14 '13 at 19:07
  • @GitGud Isn't it in some cases? For example, the quadratic equation – Cisplatin Dec 14 '13 at 19:08
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    @Sim If you're talking about $$ax^2+bx+c=0\iff x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$$ then no. As you know - and as I have pointed out - the RHS of the equivalence is short for $$x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\lor x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}.$$ – Git Gud Dec 14 '13 at 19:10

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