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Basic Cardinal Arithmetic

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data

    prove that (a x b)[tex]^{}c[/tex] = (a[tex]^{}c[/tex] x b[tex]^{}c[/tex] where a,b,c are any cardinal numbers

    2. Relevant equations



    3. The attempt at a solution

    i know that they should first be interpreted as sets A,B,C but what functions should I use.
     
  2. jcsd
  3. Mar 16, 2009 #2

    tiny-tim

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    Hi saadsarfraz! :smile:

    (use the X2 tag just above the Reply box, instead of tex :wink:)
    AC is the set of functions from C to A …

    so pick a typical function on one side of the equation and show how to define a corresponding function on the other side :wink:
     
  4. Mar 16, 2009 #3
    Hi and thanks but im still a bit confused so is this how i shld do it:

    A^c is defined as f: C --> A and I also define B^c as g: C --> B and h can be defined as h: C --> A X B, and since assuming f and g are bijections h too is a bijection. Am I in the right direction?
     
  5. Mar 16, 2009 #4

    tiny-tim

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    No, not at all …

    AC is the set of all functions from C to A, not one function
     
  6. Mar 16, 2009 #5
    i still dont know how to do it, can you please help me in this.
     
  7. Mar 17, 2009 #6

    tiny-tim

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    Start:

    Let f:A → C be a member of AC and g:B → C be a member of BC …​

    and then construct a member h:AxB → C of (AxB)C using f and g :smile:
     
  8. Mar 17, 2009 #7
    is h going to be like this h(a,b) = (f(a),g(b) next to show that this is an injection?
     
  9. Mar 18, 2009 #8

    tiny-tim

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    Hi saadsarfraz! :smile:

    Yes, that's exactly right! :approve:

    ('cept you missed out a bracket! :wink:)

    ok, now the other way round …

    starting with an h, how do you define an f and g? :smile:
     
  10. Mar 18, 2009 #9
    I dont know how to do define an f and g starting with an h?
     
  11. Mar 18, 2009 #10

    tiny-tim

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    Hint: if h:C→ AxB is a member of (AxB)C, then define the projections hA:C→ B and hB:C→ A :wink:
     
  12. Mar 18, 2009 #11
    h_a going to be (B^c) and h_b(c) = (A^c)
     
  13. Mar 18, 2009 #12

    tiny-tim

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    mmm :frown: … i suspect you've got it …

    but what you've actually written makes no sense​
     
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