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Surjective function

  1. Apr 24, 2010 #1
    Consider the function f: Z -> Z, where f(x) 4x+1 for each x is an element in Z, here the range of F = { ... -8, -5, -2, 1, 4, 7...} is a proper subset of Z, so f is not an onto (surjective) function.

    When one examines 3x + 1 = 8, we know x = 7/3, so there is no x in the domain Z with f(x) = 8

    But if g: Q -> Q, where g(x) = 3x+1 for x is an element in Q; and h: R -> R, where h(x) = 3x+1 for x is an element in R, both g and h are surjective function.

    What I want to ask whether my understanding true or false:

    1. We consider g is a valid surjective function because with x = 7/3, g(x) = 8, we can write 8/1, and so we consider it as a rational number

    and
    2. We consider h is surjective because 7/3 is a real number (we can alswo rewrite 7/3 as demcial...)


    Thank you
     
  2. jcsd
  3. Apr 24, 2010 #2
    Yes, that's correct. This is easily seen from the form of [tex]g^{-1}[/tex] and [tex]h^{-1}[/tex].
     
  4. Apr 24, 2010 #3
    Just curious. You're not wrong to refer to 7/3 as real number, but most people would call it a rational number and reserve the term real number for those numbers that cannot be expressed as a fraction. Is there a reason for the way you're using this terminology?
     
  5. Apr 25, 2010 #4
    Hi SW. Thanks. In the given, it says "h: R -> R, where h(x) = 3x+1 for x is an element in R"

    Yeah I got the same gut feeling about these Z,Q, R, lol....

    and thank you gigasoft
     
  6. Apr 25, 2010 #5
    No, actually that is not correct; at least not completely, but you caught the essential idea.

    [itex]g:\mathbb{Q} \rightarrow \mathbb{Q}[/itex] is a surjective function because, for any rational b, there is a rational a, such that b = g(a) and this cannot proven by just one example, and the same goes for h and the real numbers.
     
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