• May 7th 2010, 11:08 PM
l flipboi l
Hello,

(A-C)∩(C-B) = 0.

Do I start off by....

suppose: ~[(A-C)∩(C-B) = 0.] ?

Also, does this (0 ∩ B') equal the empty set?

Thanks...

PS. ' indicates the complement.
• May 7th 2010, 11:24 PM
tonio
Why not start by the mere definition? Suppose $x\in\left(A-C\right)\cap\left(C-B\right)\Longrightarrow x\in A-C\,\,\,and\,\,\,also\,\,\,x\in C-B$ ...from here the way to get a contradiction is straightforward and thus no such element can exist...

Tonio
• May 7th 2010, 11:37 PM
l flipboi l
Thats exactly what I did, i'm not sure if I did it correctly though.

This is what I got :

(A ∩ 0) ∩ B'

(0 ∩ B')

0

is that correct?
• May 8th 2010, 12:52 AM
tonio
Quote:

Originally Posted by l flipboi l
Thats exactly what I did, i'm not sure if I did it correctly though.

This is what I got :

(A ∩ 0) ∩ B'

(0 ∩ B')

0

is that correct?

I've no idea what you had in mind, even assuming that $0=\emptyset$...how did you come up with the above? Besides this, $X\cap \emptyset =\emptyset$ for any set X, so from

the first line you get the empty set...(Wondering)

Tonio
• May 8th 2010, 12:56 AM
l flipboi l
Quote:

Originally Posted by tonio
I've no idea what you had in mind, even assuming that $0=\emptyset$...how did you come up with the above? Besides this, $X\cap \emptyset =\emptyset$ for any set X, so from

the first line you get the empty set...(Wondering)

Tonio

I excluded alot of the work, but I used the commutative and associative property to get c and c' to cancel out.

which brought me to my answer
• May 8th 2010, 01:05 AM
tonio
Quote:

Originally Posted by l flipboi l
I excluded alot of the work, but I used the commutative and associative property to get c and c' to cancel out.

which brought me to my answer

Good, but then why did you say you "did exactly the same" as I did?!!

Tonio
• May 8th 2010, 01:29 AM
l flipboi l
Thanks for the help! I meant, I set it up like how you did.