What is the maximum number of additive inverse can a number have?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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 $a'=a$.