The symmetric difference of A and B, denoted by A ⊕ B, is the set containing those elements in either A or B, but not in both A and B. Show that A ⊕ B = (A − B) ∪ (B − A).
i guess (A − B) is the elements in A but not in B and (B − A) is the elements in B but not A
ouh.. thank you.. unkle
i am not familiar with this notation , but it kinda makes sense \(A\cup B\) is the elements in A and B, but if an element is in both sets it only counts once
so, u think we need to draw that answer ?
i believe this is trying to define the "exclusive or". Such as, we can either be in Kansas or Texas; but there is no way we can be in both places at the same time
let x be in the set A(+)B by definition, x is either in A, but not B. OR x is in B, but not A not really sure if thats a valid showing tho. I was never good at set proofing in this way
the way you usually prove that two sets are equal is to show each one is contained in the other i.e. pick an arbitrary element \(x\) of \(A\otimes B\) and show it is contained in \((A-B)\cup (B-A)\) and vice versa
\[\large A\oplus B=(A\cup B)\setminus(A\cap B)=(A\cup B)\cap(A\cap B)^c \] \[\large =(A\cup B)\cap(A^c\cup B^c)=[(A\cup B)\cap A^c]\cup[(A\cup B)\cap B^c] \] \[\large =(A\cap A^c)\cup(B\cap A^c)\cup(A\cap B^c)\cup(B\cap B^c) \] \[\large =\emptyset\cup(B\setminus A)\cup(A\setminus B)\cup\emptyset= (A\setminus B)\cup(B\setminus A) \]
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