Prove:
\[ (A-B) \cup (B-A) = (A \cup B) - (A \cap B) \]
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OpenStudy (zzr0ck3r):
what is cup?
OpenStudy (swissgirl):
what did i do wrong?
OpenStudy (anonymous):
Union..
OpenStudy (swissgirl):
cup = union
cap = intersection
OpenStudy (anonymous):
Cup is Union and cap is Intersection..
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OpenStudy (anonymous):
I am writing it once again..
OpenStudy (swissgirl):
Why didnt it work :(((
OpenStudy (anonymous):
\[ (A-B) \cup (B-A) = (A \cup B) - (A \cap B)\]
OpenStudy (zzr0ck3r):
what is A/cupB
OpenStudy (zzr0ck3r):
ahh
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OpenStudy (swissgirl):
i must have done smth wrong
OpenStudy (anonymous):
Do not use \{\} brackets instead use \[\] these brackets..
OpenStudy (zzr0ck3r):
sorry phone be back in a sec
OpenStudy (anonymous):
For this:
Take any element that belongs to the set \((A-B) \cup (B-A)\)
Suppose it as x..
So,
\[x \in (A-B) \cup (B - A)\]
Or you can say:
\[x \in (A - B) \quad \color{green}{Or} \quad x \in (B - A)\]
\[(x \in A \; and \; {x \cancel{\in} B)} \quad Or \quad (x \in B \; and \; x \cancel{\in} A)\]
\[(x \in A \; Or \; { \in B)} \quad and \quad (x \cancel{\in} B \; Or \; x \cancel{\in} A)\]
\[x \in (A \cup B) \; and \; x \cancel{\in} (A \cap B)\]
\[x \in (A \cup B) - (A \cap B)\]
OpenStudy (swissgirl):
THHANNKKSSS GUYYSSSS
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OpenStudy (anonymous):
|dw:1342982392324:dw|
both (A-B)U(B-A) and (AUB)-(A^B) represent the same region.
so both must be equal