Suppose f1 and f2 are two operations on A, that is functions from A² to A.

My question is : how to express in general and formally the fact that each operation is the inverse of the other?

Is it correct to say :

f1 and f2 are inverse operations on A iff for all a, b, c belonging to A

(1) if the image of (a,b) under f1 is c, then the image of (c,b) under f2 is a


(2) if the image of (a,b) under f2 is c, then the image of (c,b) under f1 is a.

Remark. In the case of multiplication and division, maybe I would have to substract the set {0} and say " inverse operations on R - {0} "?

Is there an easier way to express the same fact using the concept of " inverse element" ? If it is the case, how to relate the " inverse element" definition and the general definiition of inverse operations?

  • 1
    $\begingroup$ Yes, there is an easier way! “Division” is not really a separate operation from multiplication, it’s just multiplication by the inverse element. $\endgroup$ – Alex Sanger Apr 20 at 16:15

This may or may not helpful: I suppose you can say multiplication is a binary function $\cdot:X^2\to X$, and then say division is the partial binary function $/:X^2\to X$ defined by $x/y=z$ iff $y\cdot z=x$ AND $y\cdot w\ne x$ for all $w\ne z\in X$, else $x/y$ is undefined.

For example: Let $X=\mathbb{N}=\{0,1,\ldots\}$. Then $6/2=3$ since $2\cdot 3=6$ and $2\cdot w\ne 6$ for all $w\ne 3\in\mathbb{N}$.

Here's some nice familiar properties:

For all $n\in\mathbb{N}$, $n/0$ is undefined: If $n\ne0$, then $0\cdot z=0\ne n$ for all $z\in\mathbb{N}$. If $n=0$, then $0\cdot 0=0$ but $0\cdot 1=0$, so $n/0$ is undefined. I'll leave you to prove that for all $n\ne0\in\mathbb{N}$, $0/n=0$.


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