1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: "Determine if ___ is a function"

  1. Feb 3, 2008 #1
    [SOLVED] "Determine if ___ is a function"

    1. The problem statement, all variables and given/known data
    Determine whether f is a function from Z to R if
    a) f(n) = ±n.

    b) f(n) = Sqrt(n^2 + 1).

    c) f(n) = 1 / (n^2 – 4).

    3. The attempt at a solution
    Have the book right in front of me, and reading the definitions of functions, transitions, and maps, but nothing's registering in my brain.

    To me, ±n would not be a function because it would be the same as the equation x = y^2.
    However, this whole "Z to R" business has me lost. I know what Z and R represent, but I don't see how that has any meaning on whether or not f(n) is a function. What is a function? More importantly, what makes something not a function?

    Haven't attempted b) or c), as they look even worse. Trying to comprehend the concept of a) first.
  2. jcsd
  3. Feb 3, 2008 #2

    Gib Z

    User Avatar
    Homework Helper

    In simplest terms, any set of ordered values, in this case (x, f(n) ) is called a relation. A relation is called a function if for every input value, there is only one output value. Geometrically, this means a function is a function if its graph of a Cartesian Plane has no place where a vertical line cuts the graph twice, the so called "vertical line test".

    This question is only a bit harder, because of the "from Z to R" part. Basically this means that f has to be a function where only integers can be inputted, and only real numbers can be output.

    So; even if a function has points where there are 2 output values, and maybe even perhaps these 2 are complex numbers, as long as this only occurs on non-integer points, it is still a function from Z to R, if at every integer point the function gives a single Real number.

    Or, f(n) = ni +1 is a function, but not a function from Z to R, though it can be a function from Z to C.
  4. Feb 3, 2008 #3
    This input/output talk you gave me is making all the sense in the world :)

    So... we can input any integer, and get back any real number.
    a) would not be a function, as it fails the vertical line test
    b) would be a function
    c) would not be a function for n = ±2... er? I think that means it IS a function, but just has a restriction?
  5. Feb 3, 2008 #4

    Gib Z

    User Avatar
    Homework Helper

    Well it seems you are getting the hang of this, just a bit more to learn =]

    a) Correct =] It wouldn't even be a function if we changed the restriction that the function has to be from Z to R, because no matter what, even if we reduced the condition to be from C to C, which is more general because the complex numbers include Z and R, there would always be two values. That is only a function when the domain is restricted to one point. n=0

    b) That is correct as well!

    c) Only mistake :(

    A lot of things can be functions with appropriate conditions! For example, I made a) a function, all i had to do was severely limit my input value range =[ But with that range, the condition was satisfied - For every input value, only one being 0 this time, there is only 1 output value, also only one being 0 this time.

    What you must know is that Z is shorthand for the set of the integers, so in set notation,

    [tex]\mathrr{Z} = \{ 0, \pm 1, \pm 2, \pm 3 \right} ...\}[/tex]

    However, this function does not work, as you said, for n = 2 or n=-2. So it can be a function for that new set [tex]G = \{ k \in Z | k \neq \pm 2 \}[/tex] which basically reads
    "The Set G, which is equal to the set of all numbers, k, in the integers, such that k is not equal to plus or minus 2"

    This set is not equal to Z :( For them to be equal, they must have identical elements, which these don't. So If I reduced the condition to say, is c) a function from G to R, then I could say yes. But since it asked Z to R, I must say no :(
  6. Feb 3, 2008 #5

    Thank-you for your assistance!
    /cookie :3
  7. Feb 3, 2008 #6

    Gib Z

    User Avatar
    Homework Helper

    No problem, I actually find this fun. *Nerd Alert*.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook