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Sets and Proofs

  1. Aug 28, 2006 #1
    Lets say you are given a bunch of statements and you need to ask some questions to prove them:

    (a) How do you show that a set is a subset of another set.
    I said to show that [itex] x\in A [/itex] and [itex] x\in B [/tex]. What else can you do to show what [itex] A\subset B [/itex]? Could you assume from the following: If [itex] A\cup B = B\cup A [/itex] then [itex] A\subset B [/itex]? (sorry, not experienced in set theory).

    (b) If [itex] a [/itex] and [itex] b [/itex] are real nonnegative real numbers, then [itex] a^{2}+b^{2} \leq (a+b)^{2} [/itex]. Is this the Cauchy-Schwarz inequality? Basically, the questions that I ask in this case, is how can I prove that [itex] a^{2}+b^{2} \leq (a+b)^{2} [/itex] or [itex] (a+b)^{2}\geq a^{2}+b^{2} [/itex] and work from this (forward or backward)?

    Thanks
     
    Last edited: Aug 28, 2006
  2. jcsd
  3. Aug 28, 2006 #2
    (a) Start by assuming x is a member of set A. Then show that it must be in set B. This will prove that A is a subset of B.

    Also, if you can show [tex] A \cap B = A [/tex] then that works too.

    (b) Multiply out the [tex] (a + b)^{2} [/tex] and notice the extra term. What can you say about the sign of this term given what you've assumed about a and b?
     
  4. Aug 28, 2006 #3

    0rthodontist

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    a. What you show is that if x is in A, THEN x is in B. Just showing that there is some x that is in A and also in B is not enough. You should convince yourself of this.

    b. What is (a + b)^2 also equal to?

    By the way, you should be careful about the [tex] \subset [/tex] symbol. Depending on the context [tex] A\subset B [/tex] can mean that A is a subset of B and not equal to B. It may be better to say [tex]A \subseteq B[/tex]
     
    Last edited: Aug 28, 2006
  5. Aug 28, 2006 #4
    Not actually trying to prove statements. Just trying to ask the right questions to develop the proof.
     
  6. Aug 30, 2006 #5
    for the first question if you don't mind some quantifiers to clear up what you need to prove:
    [tex]\forall x(x\in A \rightarrow x\in B)[/tex]
     
  7. Aug 30, 2006 #6
    I think the quantifier is superfluous in this statement, although I'm no logician.
     
    Last edited: Aug 30, 2006
  8. Aug 31, 2006 #7
    A is a subset B means every element x in A is also in B. So to show A is a subset of B, you have to show every element in A is also in B. You start by assuming there is some x(it's arbitrary) in A, then show that x is also in B. Since the x was arbitrary, it holds for all the x's in A, so that's why it works. More precisely, loop gravity's post sums up what it means for A to be a subset of B.
     
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