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Using the properties of sets, a show that`Acup B-AcapB=(A-B) cup (B-A)` |
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Answer» `R.H.S. = (A-B) uu (B-A)` `=(AnnbarB) uu (B nn barA)` (As `A-B = AnnbarB`) `=((AnnbarB) uu B) nn ((AnnbarB) uu barA)` `=((AuuB)nn(BuubarB)) nn ((barA uu A) nn (barAuubarB))` As, `(barA uu A) = U`(Universal set) so, our expression becomes, `=((AuuB)nnU) nn (U nn (barAuubarB))` `= (AuuB) nn (barAuubarB)` `= (AuuB) nn (bar(A nn B))` (As `(barAuubarB) = bar(A nn B)`) Again using `A-B = AnnbarB`, `= (A uu B) - (AnnB) = L.H.S.` |
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