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Proofs in mathematics

  1. Jul 16, 2008 #1
    Hey
    Im trying to study abstract algebra, set theory and group theory, on my own. I have trouble understanding how to construct mathematical proofs though. Some of the things the excercises tells me to prove, seems so intuitively clear and obvious that I don't know what's left to prove. For example, prove that

    [tex]
    A\cup B = B\cup A
    [/tex]

    where A and B are two sets, or

    [tex]
    A\cap(B\cup C)=(A\cap B)\cup (A\cap C)
    [/tex]

    I have no idea how to start. Can someone give me a hint on these? And maybe a hint on general proof making in mathematics?
     
  2. jcsd
  3. Jul 16, 2008 #2
    Well, for starters, when you are trying to prove that the the union of A and B is equal to the union of B and A it is necessary to show that the union of A and B is contained in the union of B and A. Then you need to prove the converse. When I do proofs of this sort, I start by drawing some pretty Venn diagrams. It helps me visualize what I am trying to prove.
     
  4. Jul 16, 2008 #3

    Hurkyl

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    The first axiom of elementary set-theory is the Axiom of extensionality:
    S=T if and only if ([itex]x \in S[/itex] if and only if [itex]x \in T[/itex]).​
    In fact, this is the only axiom of Zermelo set theory that says anything about equality. So, if you want to prove two sets are equal, essentially the only method available is to apply this axiom.

    Once you learn more theorems (such as the two theorems you posted), you will learn more ways to prove sets are equal.
     
  5. Jul 16, 2008 #4
    Your book probably defines equality of sets. That definition may or may not be the same as the one Hurkyl gave.

    What's the definition given?
     
  6. Jul 17, 2008 #5
    Thanks for the answers.

    I think the definition the book gives of equality between sets, is that for two sets A and B to be equal, A must be a subset of B and B must be a subset of A.
     
  7. Jul 17, 2008 #6

    HallsofIvy

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    Yes, that's exactly what Hurkyl gave.

    And to prove "A is a subset of B" you start "let x be a member of A" and then, using the fact that x satisfies whatever the definition of A is, show that it must also satisfy the definition of B and so "x is a member of B".

    Here, since you are asked to prove [itex]A\cup B = B\cup A[/itex] you need to prove that [itex]A\cup B[/itex] is a subset of [itex]B\cup A[/itex] and then prove that [itex]B\cup A[/itex] is a subset of [itex]A\cup B[/itex].

    To prove that [itex]A\cup B[/itex] is a subset of [itex]B\cup A[/itex], start "let x be a member of [itex]A\cup B[/itex]. I said above "using the fact that x satisfies whatever the definition of A is". Here the set is defined as a union so we really need to use the definition of "union". x is in [itex]A\cup B[/itex] if and only if x is in A or x is in B. Since that is an "or", break this into two cases:
    (i) x is in A. In that case x is in [itex]B\cup A[/itex] because ...
    (ii) x is in B. In that case x is in [itex]B\cup A[/itex] because ...
     
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