Let $U=\{1,2,3,4,5,6\}, A=\{2,3\}$ and $B=\{3,4,5\}$
Find $A^{\prime}, B^{\prime}, A^{\prime} \cap B^{\prime}, A \cup B$ and hence show that $(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
Let $U=\{1,2,3,4,5,6\}, A=\{2,3\}$ and $B=\{3,4,5\}$
Find $A^{\prime}, B^{\prime}, A^{\prime} \cap B^{\prime}, A \cup B$ and hence show that $(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
Clearly $A ^{\prime}=\{1,4,5,6\}, B ^{\prime}=\{1,2,6\} .$ Hence $A ^{\prime} \cap B ^{\prime}=\{1,6\}$
Also $A \cup B = \{ 2,3,4,5\} ,$ so that ${(A \cup B)^\prime } = \{ 1,6\} $
$( A \cup B )^{\prime}=\{1,6\}= A ^{\prime} \cap B ^{\prime}$
It can be shown that the above result is true in general. If $A$ and $B$ are any two subsets of the universal set $U,$ then
${(A \cup B)^\prime } = {A^\prime } \cap {B^\prime }$. Similarly, ${(A \cup B)^\prime } = {A^\prime } \cap {B^\prime }.$ These two results are stated in words as follows:
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