Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSets4 Marks
Question
For any two sets A and B, prove the following: $\text{A}-\text{(A}-\text{B})=\text{A}\cap\text{B}$
✓
Answer
For any sets A and B we have by De-morgan's laws $\text{(A}\cup\text{B)}' = \text{A}'\cap\text{B}',\text{(A}\cap\text{B)}' = \text{A}'\cup\text{B}'$ Also $\text{LHS}= \text{A} - \text{(A} - \text{B)}$ $=\text{A}\cap\text{(A}- \text{B)}'$ $=\text{A}\cap\text{(A}\cap\text{B}')'$ $=\text{A}\cap\text{(A}\cap\text{(B})')'$ [By De-morgan's law] $=\text{A}\cap\text{(A}'\cup\text{B})$ $[\because\text{(B}')'=\text{B}]$ $=\text{(A}\cap\text{A}')\cup\text{(A}\cap\text{B})$ $=\oint\cup\text{(A}\cap\text{B})$ $[\because\text{A}\cap\text{A}'=\oint]$ $=\text{A}\cap\text{B}$ $[\because\oint\cup$ x = x, for and set x$]$ $= \text{RHS}$ $\therefore$ LHS = RHS Proved.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.