union and intersection proof


by chocolatelover
Tags: intersection, proof, union
chocolatelover
chocolatelover is offline
#1
Mar7-08, 11:39 AM
P: 239
1. The problem statement, all variables and given/known data
Prove (A is a union of B)/(A is an intersection of B)=(A/B) is a union of (B/A)

2. Relevant equations



3. The attempt at a solution

Could someone first help me translate all of this into plain English. I don't really understand what I need to prove. Would I start off with the contrapositive? Is the contrapositive "If (A/B) is not the union of (B/A), then A is not the union of B/(A is not the intersection of B) and it is not equal to the antecedent"? Could someone please show me where to go from here?

Thank you very much
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GregA
GregA is offline
#2
Mar7-08, 12:47 PM
P: 213
Quote Quote by chocolatelover View Post
1. The problem statement, all variables and given/known data
Prove (A is a union of B)/(A is an intersection of B)=(A/B) is a union of (B/A)

2. Relevant equations



3. The attempt at a solution

Could someone first help me translate all of this into plain English. I don't really understand what I need to prove. Would I start off with the contrapositive? Is the contrapositive "If (A/B) is not the union of (B/A), then A is not the union of B/(A is not the intersection of B) and it is not equal to the antecedent"? Could someone please show me where to go from here?

Thank you very much
I assume the question was given as:
[tex](A \cup B)/(A \cap B) = (B/A)\cup(A/B)?[/tex]
If x is an element of the set on LHS then x is in A or x is in B but x is not in both A and B
what can you say about RHS? does it imply something about x that will help you get LHS?
andytoh
andytoh is offline
#3
Mar7-08, 02:59 PM
P: 363
This is called the symmetric difference of two sets. It can be proven the the associative, distributive, and commutative laws holds with symmetric difference. Those are good exercises.

chocolatelover
chocolatelover is offline
#4
Mar7-08, 03:17 PM
P: 239

union and intersection proof


Thank you very much

I assume the question was given as:
That's correct, except A and B are switched in the second part. (A/B) U (B/A)

Would the contrapositive also prove it?

I know how to use the associative property, but I'm sure how how use the others to prove this. I know that, say, A upside B upside C=(A upside B) upside U C=A upside U (B upside U C) I'm not sure how to do that or the others for this problem. Would it be (A U B)/(A upside U B)=A U B/A upside U B?

Could some please help me on this?

Thank you


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