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Union and intersection proof

by chocolatelover
Tags: intersection, proof, union
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chocolatelover
#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
#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
#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
#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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