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Questions Regarding Definition of One-to-One and Onto Functions?

  1. Oct 20, 2012 #1
    Hi, I was just having a little trouble of understanding what it... is saying, well first I'll state what my book says the definition is:

    A function T:D* [itex]\subseteq[/itex] R2 → R2 is called one-to-one if for each (u,v) and (u',v') in D*, T(u,v)=T(u',v') implies that u = u' and v = v'

    A function T:D* [itex]\subseteq[/itex] R2 → R2 is called onto D if for every point (x,y) [itex]\in[/itex] D there is a point (u,v) in D* such that T(u,v) = (x,y)

    So just a couple of questions (they might be stupid... sorry):

    1. Now correct me if I'm wrong but the function T, is supposed to bring a domain in D* to D right? So isn't there a problem that there are two points, (u,v), and (u',v') that goes to the same point?

    2. Isn't the definition for a function to be onto, is a definition for all functions? Because that's what a function does right? f(x) = y, you put x in, and it gives you "y"
     
  2. jcsd
  3. Oct 20, 2012 #2
    No, there is no problem...but THAT function won't be 1-1 then
    No. For example, the function [itex]\,f:\Bbb R\to\Bbb R\,\,\,,\,\,f(x)=x^2\,[/itex] isn't onto as no negative number is the square or no real number.

    DonAntonio
     
  4. Oct 20, 2012 #3
    Hmm I think I see... i still have one more question regarding the first definition, it seems to talk about (u,v) being different from (u',v') but then at the end of the definition it says they are equal?

    If you look in D* and T is one-to-one and find a (u1,v1) and then a (u3,v3) aren't they for sure different? When the definition generalizes it to just a u and v and u' and v' doesn't it say that all u's and v's are the same?
     
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