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

2 Answers

  • Awms A
    Lv 7
    1 decade ago
    Favorite 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).

  • 1 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.

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