Question

For the function f: Z->Z where f(n) = n+1, does an inverse exist and, if so, what is it?

Answer 1

The inverse for the function f: Z->Z where f(n) = n+1 is the function g(n) = n-1. This inverse function effectively reverses the operation of the original function.

Step-by-step explanation:

For the function f: Z→Z where f(n) = n+1, an inverse does exist. The inverse function would essentially "undo" the action performed by the original function. In mathematics, the inverse of an operation reverses the effect of that operation. Considering this, the inverse function of f(n) = n+1 would be g(n) = n-1 because applying g after f would return us to the original number. This can be communicated as f(g(n)) = g(f(n)) = n, where f() and g() are inverse operations of each other. To check, if we have a number n, applying f gives us n+1 and then applying g gives (n+1)-1 which simplifies back to n.