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Proofs on sets help

  1. Sep 9, 2007 #1
    Hello all,

    I'm having a hard time trying to prove a few things. I'm looking for a little help because I cannot seem to grasp the concept of proofs and what constitutes a valid proof and if my proof is wrong, correcting it.

    I have a proof done and if anyone could "critique" it I would be very grateful.

    Prove: (A [tex]\cup[/tex] B) X C = (A X C) [tex]\cup[/tex] (B X C)

    Proof:
    Let x [tex]\in[/tex] (A [tex]\cup[/tex] B) X C
    Then x is of the type (y,z) where y [tex]\in[/tex] A and z [tex]\in[/tex] C
    Then y [tex]\in[/tex] A or y [tex]\in[/tex] B
    Since z [tex]\in[/tex] C, (y,z) [tex]\in[/tex] A X C or
    Since z [tex]\in[/tex] C, (y,z) [tex]\in[/tex] B X C
    Then (y,z) [tex]\in[/tex] (A X C) [tex]\cup[/tex] (B X C)
    Therefore (A [tex]\cup[/tex] B) X C = (A X C) [tex]\cup[/tex] (B X C)

    Thanks for your time,

    Ryan
     
  2. jcsd
  3. Sep 9, 2007 #2

    EnumaElish

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    What is X? does it stand for [itex]\cap[/itex]?
     
  4. Sep 9, 2007 #3
    I would assume that it represents the cartesian product.
     
  5. Sep 9, 2007 #4

    EnumaElish

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    "Duh!"

    Let x in (A U B) X C
    Then x is of the type (y,z) where y in A or B and z in C.

    Otherwise your logic is correct.
     
    Last edited: Sep 9, 2007
  6. Sep 9, 2007 #5
    Thank you guys very much for your responses, I am sure I'll have a couple more here tomorrow...

    Thanks again!
     
  7. Sep 10, 2007 #6

    HallsofIvy

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    Technical point: it would be better to say IF [itex]x\in (A\Cup B)\cross C[/itex]. "let x ..." runs into trouble if the set is empty!

    More important point: you have proved that [itex](A\Cup B) X C \subset (A X C)\Cup (B X C)[/itex], not that they are equal you still have to prove that "if x is in [itex](A X C)\Cup (B X C)[/itex], then it is in [itex](A\Cup B) X C[\itex].
     
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