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"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

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    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

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    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

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    No problem, I actually find this fun. *Nerd Alert*.
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