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Basic Set Theory Proof

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that if A is a subset of B then A/D is a subset of B/D.


    2. Relevant equations



    3. The attempt at a solution

    Consider element x of A. Since A is a subset of B then for all x element of A, x is an element of B. Consider element x of A/D. If x is an element of D then x is not a member of A and thus it does not matter if x is an element of B. If x is not an element of D and is an element of A than x is also in B because x is an element of A. Thus, A/D is a subset of B.

    Not even sure if this much is correct. How do I prove this basic "subtracting a set from both subsets" identity??
     
  2. jcsd
  3. Oct 24, 2011 #2
    For one, you should not say "If x is an element of D then x is not a member of A ...", you should say "If x is an element of D then x is not a member of A/D", likewise for B/D.

    But you take a lengthy confusing approach at this point. it would be more concise to say "Consider element x of A/D.." (here comes my part): x is an element of A/D implies x is in A and x is not in D....
     
  4. Oct 24, 2011 #3
    Okay, I'll change that. But how do I finish the proof?
     
  5. Oct 24, 2011 #4
    Pretty much like you did, I think, just with that shorter way of presenting it. Talk about why it implies x in is in B/D instead of considering B and D separately.
     
  6. Oct 24, 2011 #5
    Oh! I think I get it. So the proof is: Consider x element of A/D. This means x is an element of A and not of D. This means x is a member of B, because all members of A are members of B, and since we already know that x is not an element of D we can combine these two facts and say x is an element of B/D.
     
  7. Oct 24, 2011 #6
    It seems like you missed the point of considering x is in A/D instead of considering x is D and x is not in D.

    You begin well here: "Consider element x of A. Since A is a subset of B then for all x element of A, x is an element of B. Consider element x of A/D." but then you begin taking cases of x in D or not in D. Don take cases. Just go straight into x is in A/D implies... whatever it implies. Leading into something about x is in B/D, right?
     
  8. Oct 24, 2011 #7
    Um...I don't get it. :( What does A/D imply other than x is in A and not in D?
     
  9. Oct 24, 2011 #8
    Oh, sorry, you're right. You're answer is good, I was thrown off by all the words. (we use all symbols in my class) :) Yay set theory *waves flag*
     
  10. Oct 24, 2011 #9
    Perfect! Thanks! :D
     
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