Question

If $A=1,3,5,..........17$ and $B=2,4,6,......18$ and $N$ ( the set of natural numbers) is the universal set, then show that $A^{\prime }\cup \left(\right(A\cup B)\cap B^{\prime })=N$.

Answer 1

$LetN=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}\because A=\{1,3,5,7,9,........,17\}B=\{2,4,6,8,........18\}{A}^{{}^{\prime }}=\{2,4,6,8,.......18\}{B}^{{}^{\prime }}=\{1,3,5,7,.......17\}\because A=B^{{}^{\prime }}\&B=A^{{}^{\prime }}(A\cup B)=\{1,2,3,4,5,6,............18\}(A\cup B)\cap B^{{}^{\prime }}=\{1,3,5,7,..........17\}{A}^{{}^{\prime }}\cup \left[\right(A\cup B)\cap B^{{}^{\prime }}]=\{1,2,3,4,5,6,..........18\}\therefore A^{{}^{\prime }}\cup \left[\right(A\cup B)\cap B^{{}^{\prime }}]=N$