Question

If $U=\{1,2,3,4,5,6,7,8,9\},A=\{2,4,6,8\}$ and $B=\{2,3,5,7\}$ verify that
$(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$

Answer 1

We have,

$U=\{1,2,3,4,5,6,7,8,9\}$

$A=\{2,4,6,8\}$

$B=\{2,3,5,7\}$

Since,

${(A\cup B)}^{\prime }=U-(A\cup B)$

${(A\cup B)}^{\prime }=\{1,2,3,4,5,6,7,8,9\}-(\{2,4,6,8\}\cup \{2,3,5,7\})$

${(A\cup B)}^{\prime }=\{1,2,3,4,5,6,7,8,9\}-\{2,3,4,5,6,7,8\}$

${(A\cup B)}^{\prime }=\{1,9\}$

Now,

$A^{\prime }\cap B^{\prime }=(U-A)\cap (U-B)$

$A^{\prime }\cap B^{\prime }=(\{1,2,3,4,5,6,7,8,9\}-\{2,4,6,8\})\cap (\{1,2,3,4,5,6,7,8,9\}-\{2,3,5,7\})$

$A^{\prime }\cap B^{\prime }=\{1,3,5,7,9\}\cap \{1,4,6,8,9\}$

$A^{\prime }\cap B^{\prime }=\{1,9\}$

It is clear that,

${(A\cup B)}^{\prime }=A^{\prime }\cap B^{\prime }$

Hence, proved.