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Some numbers have no additive inverses. Can someone prove that a number can have at most 1 additive inverse?

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closed as unclear what you're asking by Jonas Meyer, mez, Claude Leibovici, Jyrki Lahtonen, 5xum Jul 29 at 6:35

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What numbers have no additive inverses? –  john w. Mar 17 '12 at 21:36
@john: $1$ in $\mathbb{N}$? –  TMM Mar 17 '12 at 21:42
The point being that the statement "some numbers have no additive inverses" is very vague. What is meant by number? What is meant by additive? Once these have been established, then a proof like below can be done. –  john w. Mar 17 '12 at 21:52

3 Answers 3

For an example of a commutative magma in which an element has more than one inverse, consider $M=\{0,a,b,c\}$ with the following addition table: $$\begin{array}{c||cccc} + & 0 & a & b & c\\ \hline 0 & 0 & a & b & c\\ a & a & b & 0 & 0\\ b & b & 0 & c & a\\ c & c & 0 & a & a \end{array}$$ Then $a$ has two additive inverses ($b$ and $c$); the reason the "usual" proof does not work any more is that $c+(a+b)\neq (c+a)+b$, since the operation is not associative.

Associativity is necessary; commutativity can be avoided if you specify that the two inverses act on "different sides" (e.g., $ab=1$ and $ca=1$). It is possible for an element to have two distinct inverses on the same side and no inverse on the other side even in the presence of associativity. For example, if $X$ is an infinite set, and $f\colon X\to X$ is a one-to-one function that is not onto, then $f$ will have many left inverses under composition (in fact, infinitely many), but no right inverses.

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Suppose a number $a$ has two additive inverses $b$ and $b^\prime$. $$ \begin{align*} b &= b + 0\\ &= b + (a + b^\prime)\\ &= (b + a) + b^\prime\\ &= 0 + b^\prime\\ &= b^\prime \end{align*} $$

Therefore, $b$ and $b^\prime$ were really the same number after all.

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+1 though you have assumed the associative property –  Henry Mar 17 '12 at 21:41
I'm not sure how this proof might go (or not) without associativity, but I gather OP isn't interested in such strange algebras. –  Austin Mohr Mar 17 '12 at 21:41

Can you specify the setting more? Are you talking about rings, groups, or fields?

A simple naive go at your question without any more specificity would be, assume $a$ and $a'$ are additive inverses of $b$, then $a=a + 0 = a+(b+a')=(a+b)+a'=0+a'=a'$ so $a'=a$.

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Like Austin said, you probably should specify what axioms we are working under. Usually associativity and commutativity come with the term "additive" –  Steven-Owen Mar 17 '12 at 21:45

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