Finding an inverse in a min function

Question

Let $M$ and $N$ are two integers with $m < n$. Let $A = \{m, m+1, ... , n\}$ and let $\min$ be the function that returns the smaller of its two arguments.

So the function's zero is $m$ because you get it back out no matter what numbers you put in, $n$ is the identity because you always get back the number that you input in with it.

Where I'm lost is finding if this has an inverse. The book states that if $x y = y x = u$ holds then the element $y$ in $C$ is called an inverse. How would I find what the inverse in the function is?

Answer 1

I interpret your question as "We define $f:A\times A\to A$ by $f(a,b)=\min(a,b)$. Does $f$ have an inverse?"

No, since more than one value in the domain maps to a single value in the range: $f(m,m+1)=f(m,m+2)=m$.

Answer 2

No, the first respondent gave a perfect explanation.

The $\min$ function takes as an argument a set, $S$, which comprises multiple values (say, $n$ not-necessarily-distinct values, $n$ could be 2). Its output is a single value, $m$, the minimum. There is no way to reconstruct the input set $S$ from the minimal value $m$ contained in $S$. There is no way to even determine the number of elements that are in $S$. Unless $S$ contains only 1 element, there is no inverse, but in such a case all you have is an identity function.

Think of it as a data compression. If you could reconstruct a whole file from a single byte describing the minimal byte value in the file, youd be able to store and recover unlimited amounts of data into the most trivial of spaces.