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Functions: 1-1, onto

  1. Apr 25, 2006 #1
    Given any two functions f: A --> B, g: C --> D, define a new function h: AxC ---> BxD by h(x,y) (f(x)), g(y))

    Show by counterexamples that the converse of each statement is false. What additional assumptions are needed to make the converses true?

    (a) if both f and g are 1-1, then h is 1-1
    (so show that if h is 1-1, both f and g are be 1-1)

    (b) if both f and g are onto, then h is onto
    (so show that if h is onto, then both f and g are onto)

    i am having some trouble finding counterexamples,
    for example can h(x) = x^2 - y^2, or does it have to be h(x,y) = (x^2, y^2)????

    i believe the assumptions needed to make the converses true has to do with the domains of f and g.

    but i'm pretty lost on this problem. any help is appreciated, thanks guys
     
  2. jcsd
  3. Apr 25, 2006 #2
    You can't find a counterexample because the converse of each statement is TRUE.

    By the definition of h, an element of the domain of h is a pair (a,c) where a is in A, and c is in C. The image of (a,c) under h is the pair (b,d) where b=f(a), d=g(c).

    h(x) = x^2 - y^2 does not fit the above description. x must be an ordered pair, and the output of the function, h(x), must be an ordered pair.

    If you wanted to give a specific example of h, then you need to specify what the sets A,B,C,D are. You also need to specify f and g, that is show how to obtain the image for each element in their respective domains.
     
    Last edited: Apr 25, 2006
  4. Apr 25, 2006 #3

    AKG

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    Try proving that if h is 1-1, then f and g are 1-1. Do this by trying to prove the converse, i.e. if either f or g is non 1-1, then h is not 1-1. So first try to prove that if f is not 1-1, h is not 1-1. You will almost be able to prove it, but if you're careful, you'll notice some subtle condition that must be imposed on C, the domain of g (yes I mean g, not f) in order for your proof to go through.
     
  5. Apr 25, 2006 #4

    AKG

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    No, the converse of each statement is false in general.
     
  6. Apr 25, 2006 #5
    How can you say the converse is false, but ask him to prove that it's true in your first post? (post #3)

    Perhaps I misread the OP's question, but what I interpreted is that:
    The original statement is "If f and g are both 1-1 then h is 1-1"
    The converse is "If h is 1-1 then f and g are both 1-1" , which is a true statement.

    similarly for part b, the original statement is "if f and g are both onto, then h is onto."
    The converse is "If h is onto, then both f and g are onto," again a true statement.

    regards,
    nocturnal
     
    Last edited: Apr 25, 2006
  7. Apr 25, 2006 #6

    AKG

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    Did you read what I said in my post? I said he should try to prove it. I also said that he will almost be able to, but he would have to impose some subtle condition on C in order to really prove it. This will tell him that the converse is not true in general, and will simultaneously tell him what conditions are needed to make the converse true. Also, once he knows this condition, it will be easy to come up with a counterexample.

    The original statement is "If f and g are both 1-1 then h is 1-1"
    The converse is "If h is 1-1 then f and g are both 1-1" , which is a true statement.


    No, its not. Consider the following nocturnal (by the way, if you haven't figured the answer out yet JasonJo, don't read the following):

    Suppose f has empty domain and g is not 1-1. Then h has empty domain and hence is 1-1, but it's not true that both f and g are 1-1. Now suppose f has empty codomain and g is not onto. Then h has empty codomain and hence is onto, but it's not true that both f and g are onto. This makes it clear that if h is 1-1 and f and g have nonempty domains, then f and g are 1-1. Similarly, if h is onto and f and g have nonempty codomains, then f and g are onto. Given this, it is very easy to find a simple counterexample to both converses simultaneously. Let f:{} -> {}, g:{0,1} -> {0,1} be a constant function.
     
  8. Apr 25, 2006 #7
    oops. You're right AKG. I didn't consider that subtle point you mentioned :redface:
     
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