Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me

Do I have Proving is true?

Question -> Prove:

(C-A)$\cup$(B-A)=(C$\cap$B)-A

My Proving:

(C-A)$\cup$(B-A)=((C$\cap$A')$\cup$ (B$\cap$A')

=(C$\cup$B)$\cap$A'

=(C$\cup$B)-A'

Do you think my Proving is correct?

share|improve this question
    
Did you mean $(C\setminus A)\cup(B\setminus A)=(C\cup B)\setminus A$, with a union on the righthand side? If you did, your argument is correct. –  Brian M. Scott Mar 6 '13 at 16:02
    
Echoing Brian's comment, the right hand side of your first equation has $\cap$ instead of $\cup$. –  copper.hat Mar 6 '13 at 16:04
    
I mean, this is exactly (C-A) ∪ (B-A) = (C ∩ B)-A –  Software Mar 6 '13 at 16:06
    
No, it needs to be $(C-A)\cup (B-A)=(C\cup B)-A$ –  Aang Mar 6 '13 at 16:07
    
Is the wrong question? –  Software Mar 6 '13 at 16:09

2 Answers 2

up vote 2 down vote accepted

In fact, $(C-A)\cup (B-A)=(C \cap A')\cup (B \cap A')$ which is $ (C \cup B) \cap A'$ and it equals to $(C \cup B)-A$ not $(C \cup B)-A'$

share|improve this answer
    
Yes exactly Thanks for your very complete Answer :) –  Software Mar 6 '13 at 16:15
    
+1 Helpful...and I'll keep my look-out, per your request! –  amWhy Mar 6 '13 at 16:55
    
very elegant answer –  Adi Dani Mar 6 '13 at 19:02
    
By the way: good mornin', Babak! –  amWhy Mar 7 '13 at 6:17
    
We had snow yesterday. Quite a bit! Kind of a slow night tonight again. But that's okay. You had a nice day again yesterday! –  amWhy Mar 7 '13 at 6:27

hint:

(C−A)∪(B−A)=(C∩A ′ )∪(B∩A ′ )

(C∪B)∩A ′ = (C∪B)−A not (C∪B)−A ′

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.