What does Z to R mean?
When solving a function, There usually will be a domain like Z to R, Z to Z, etc. I know that R to R means All the numbers, but what is the meaning of other ones ( for example Z to R), does it mean Z itself. If it does, why they don't write Z alone? please help
- Awms ALv 71 decade agoFavorite Answer
"R to R" or "R -> R" (as I prefer it written) is stating that the domain is the set of real numbers and the function will give a real number for each input.
In contrast, Z refers to the integers, so when we state
"Z to R" (or "Z -> R")
we mean that the domain is the set of integers and the function will output a real number for each input.
Similarly, "Z to Z" means that the domain is the set of integers, and the range will be a subset of the integers.
For instance, we could define f : Z -> R and g : Z -> Z by
f(x) = x
g(x) = x.
Now, it's obvious that the functions "do the same thing". However, it would be technically wrong to state that f and g are the same function (for categorical reasons: f is into R and g is into Z).
- Randy PLv 71 decade ago
Functions take values from one set and "map" them to values from another set. If I write f:R->Z that means that f can take any real as input, but what it produces is integers. R is the domain, and the range is a subset of Z.
It isn't that R to R or R->R means "all the numbers". The first R means that the argument of the function is any number. The second R means that the result is a real number. Usually that notation is used if the result could be any real number.
And similarly f:Z->R means that f takes integers and maps them to real numbers. An example of such a function might be
f(n) = 1/n
The first line says "f is a function defined on the integers which produces a real number as result". The second line, which is also part of the definition of f, says "the value of f for any particular integer n is defined to be 1/n". You can see that f(n) is not in general an integer, so I wouldn't define f as f:Z->Z because it doesn't produce integer-valued outputs.