Math problems with sets. Check my work?

[Scrapster]

Can you guys look over these problems and see if I'm on the right track?

1st problem: SOLVED

Let S = [3] x [3] (the cartesian product of {1,2,3} with itself). Let T be the set of ordered pairs (x,y) &quot;belongs to&quot; Z x Z (Z = all integers) such that 0 < (or equal) 3x + y - 4 < (or equal) 8. Prove that S is a subset of T. Does equality hold?

What I got: It's easy to make ordered pairs of [3] x [3] (1,1 1,2 1,3 2,1...)
and it's easy to prove that the equality holds for these ordered pairs. But I feel like I'm missing something else. The &quot;Does the equality hold&quot; question is a little ambiguous to me. Not sure if they are asking for a range where the equality doesn't hold for certain integers, like for x > 3 and y > 9 the equality doesn't hold. Or do I need to make a false range with integer pairs that do NOT make the the equality true?

2nd problem: Still need help!

For what conditions on sets A and B does A - B = B - A hold?

What I got: The only scenario that makes sense is if both sets A and B are equal to each other. Can anyone think of anything else?

 

[gittyup]

Sorry, but I get headache just looking at....

 

[DesignDawg]

7.

Ricky
DesignDawg

 

[Scrapster]

Any other helpful words of wisdom?

 

[Scrapster]

UPDATE: Ok. My instructor said that &quot;does the equality hold&quot; question is just asking you if S=T, which it doesn't. Easy enough.

Ok. Now for the second problem, my solution stays the same that A and B need to be the same sets in order for A-B=B-A. Can anyone think of other scenarios where this is true?

 

[Scrapster]

Anyone want to check me on #2?

 

[Cerebus451]

Darnit. Typed in a response that went into the bit bucket.

You are correct on #2.

Proof:
Suppose set S = A - B. If S is not the empty set, then S contains an element x. By definition, x is in set A, but not in set B. However, from the original equality, S also equals B - A. This means that element x is in B, but not in A. This obviously cannot be true, so therefore S must be the empty set, and A = B.