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Relations and functions

  1. Dec 13, 2008 #1
    Can anyone help me with this???
    Thank you very much

    Given set A={m,b,f,a,s} and B={m,b,s}

    a) Is {<m,s>, <b,m>, <f,m>, <a,b>} a function? Is it a relation or function from A to B, A to A, B to A, B to B or none of the above?
    b) Is { } a function? Is it a relation or function from A to B, A to A, B to A, B to B or none of the above?

    Please note that { } stands for the empty set
     
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  3. Dec 13, 2008 #2

    CompuChip

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    What is the definition of a function, say, from X to Y?
     
  4. Dec 13, 2008 #3
    I assumed that a function is a relation in which every element in the domain yields exactly one element.
     
  5. Dec 13, 2008 #4

    CompuChip

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    You assumed or that is what your textbook said?
     
  6. Dec 13, 2008 #5
    Textbook said!
     
  7. Dec 13, 2008 #6

    HallsofIvy

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    Good! Now, is that the case for {<m,s>, <b,m>, <f,m>, <a,b>}? (Which elements are in the domain which in the range?)
     
  8. Dec 13, 2008 #7
    I would say yes, but not sure because s is not paired with any member. Does that make it not a function? Definition says that every element should be. I had a different answer from a book so just want to check I understood it right.
    Also, what about {} ?

    Thanks
     
  9. Dec 13, 2008 #8
    Well, more formally, a relation from X to Y is any subset of X × Y, and a function f from X to Y is a subset of X × Y such that for each x in X, there is some y in Y such that <x, y> is in f, and if <x, y> and <x, y'> are in f then y = y'.

    For a), apply the definition above. s is in A, but there's no y in B such that <s, y> is in the set given, so it's not a function.

    For b), {} is a function, but not from A to B. However, it is a function from {} to S where S is any set at all; it's a subset of {} × S and the other conditions for a function are vacuously true. However, it is a relation between any two sets, since the empty set is a subset of any set.

    Note: if f is a function from X to Y, it's easy to see that X is the set of all first entries in f, so you should be able to use this as a test.
     
  10. Dec 14, 2008 #9

    HallsofIvy

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    Absolutely not! You seem to be completely misunderstanding the definition of "function". The definition of function says that every member of the domain must be paired with at most one member of the range. For example {(x,y)| y= x2} is a function even though no x is paired with a negative number. It is also true that two different values of x give the same y value so a member of the range is paired with two different members of the domain. But it is still true that no member of the domain is paired with two different members of the range and so it is a function.

    {(x, y)| x= y2} is not a function because it contains the pairs (4, 2) and (4, -2): one member of the domain, 4, is paired with two different members of the domain. Now, look at "first" and "second" members of each pair ("domain" and "range" respectively). Does your set of pairs contain two pair in which the first members are different but the second the same? If so it is not a function.
     
  11. Dec 14, 2008 #10
    Halls, you just really confused me there; what are you trying to say?

    In a function, each member of the domain must be paired with exactly one member of the codomain such that the member of the domain is the first entry of the pair. In his case, s is in the domain, but it's not paired with anything in the range (there's no pair of the form <s, ·> in the given set). That proves that it's not a function.
     
  12. Dec 14, 2008 #11

    HallsofIvy

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    I thought it simpler to note that there are two members of the domain paired with s.
     
  13. Dec 14, 2008 #12
    Where are there two members of the domain paired with s?
     
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