# Notation $f^{-1}$ for the inverse vs. notation $f^n$ for $n$-fold application

We often use to denote invertible function of $f$, as $f^{-1}$. In applied mathematics, This is the general rule. But very rigorous concepts of mathematics, like axiomatic number systems, it is just a definition to define iteration of negative numbers.

Can anyone tell me that why we have already assumed that invertible function of $f$ denoted as $f^{-1}$.

In iteration concept, we use the notation $f^{-1}$ and tell that 'we have already learnt about this notation' and without any reasoning we proceeds. Actually we did't learn why this notation used. It is clear that addtion, multiplication and exponentiation are defined by help of iteration concept So probably we haven't use any type of help of addtion, multiplication and exponentiation concepts. My question is : Then which concept we have used to define the notation of $f^{-1}$...

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It's not clear what you're asking? What do you mean by the iteration concept? – Jim Apr 5 '13 at 15:34
@Jim Probably that $f^3(x)$ stands for $f(f(f(x)))$ and not for $(f(x))^3$. Incidentylla, the inverse $f^{-1}$ of a bijection is the same as iterating $f$ negative one times (i.e. if we apply $f$ once more, we have done nothing at all). – Hagen von Eitzen Apr 5 '13 at 15:37
Extending what @HagenvonEitzen writes, note that there are two different operations on functions whose iterated application is often denoted with an exponent: multiplication and composition. The notation $f^{-1}$ can mean either the multiplicative inverse or the functional inverse. Context can sometimes make clear which one you mean, although usually without any other clues the composition sense has priority. – Sammy Black Apr 5 '13 at 15:41
Well $(F(A,A),\circ )$ is a monoid so if you take the invertible elements of $(F(A,A),\circ )$ you get a group... – xavierm02 Apr 5 '13 at 15:44

I guess by the iteration concept, you mean $$f^n = \underbrace{f \circ f\circ\ldots\circ f}_{n\text{ times}}.$$ For general functions $f : X \to X$, this definition only works for $n\in\mathbb{N}$ (with $f^0 = \operatorname{id}$). So the notation $f^{-1}$ is not covered by this definition.
For invertible functions $f : X\to X$, the inverse function is denoted by $f^{-1}$. The reason for this notation is that it perfectly fits the iteration concept, as the defining property $f^{-1} \circ f = f\circ f^{-1} = \operatorname{id}$ can be stated as $$f^{-1}\circ f^1 = f^1\circ f^{-1} = f^0.$$ ("First applying $f$ -1 times, and then applying it 1 times is the same as applying $f$ zero times.")
This is taken a step further by the definition $$f^{-n} = (f^{-1})^n.$$ Now for any $a,b\in\mathbb{Z}$, the exponentiation rule $$f^{a+b} = f^a\circ f^b$$ is true.