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Suppose $S$ is a ring. So $S$ obeys the axiom of additive associativity.

If $a,b,c \in S$ then does this mean:

$$ a + b + c = a + (b + c) = (a + b) + c $$ ?

Specifically does this mean $a + b + c = a + (b + c)$?

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The point of the axiom is so the following

$$ a + (b + c) = (a + b) + c $$

holds true always. Then because of that fact, it doesn't matter in which order we compute the sum, which then allows us to simply write $a + b + c$, which then equals both of the sides in the first equation. Similarly for multiplication (which is associative by assumption in a ring).

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Formally, associativity of addition is the assertion that $(a+b)+c=a+(b+c)$ for all $a$, $b$, and $c$.

Formally, $a+b+c$ has no meaning. (Well, I am being a bit imprecise. If we had a universal convention that we do operations from left to right, then $a+b+c$ would mean $(a+b)+c$. However, the convention, though common, is not universal.)

However, because of associativity, we can safely use $a+b+c$ as shorthand for $(a+b)+c$ or $a+(b+c)$.

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  • $\begingroup$ It has a common meaning, especially in the monadic representation of semigroups. $\endgroup$ – Martin Brandenburg Mar 17 '13 at 22:20
  • $\begingroup$ @André Nicolas Thanks a lot. I understand now. $\endgroup$ – mvr950 Mar 17 '13 at 22:45
  • $\begingroup$ You are welcome. Basically, for greater readability, we leave out parentheses whenever we can safely get away with it. The resulting expression is technically not "legal," but (in the presence of associativity) not legal is OK, it will not create ambiguity. $\endgroup$ – André Nicolas Mar 18 '13 at 0:10

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