If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, and = {2, 3, 5, 7} verify that:

Question:

If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, and = {2, 3, 5, 7} verify that:

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

(ii) $(A \cap C)^{\prime}=\left(A^{\prime} \cup B^{\prime}\right)$

 

Solution:

(i) $A^{U} B=\{2,3,4,5,6,7,8\}$

$\left(\mathrm{A}^{U} \mathrm{~B}\right)^{\prime}=\{1,9\}$

$\mathrm{A}^{\prime}=\{1,3,5,7,9\}$

$\mathrm{B}^{\prime}=\{1,4,6,8,9\}$

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

$\Rightarrow\left(\mathrm{A}^{\mathrm{U}} \mathrm{B}\right)^{\prime}=\mathrm{A}^{\prime} \mathrm{n}_{\mathrm{B}}^{\prime}$

Hence proved

(ii) $A^{\cap} B=\{2\}$

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

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

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

Hence proved

These are also known as De Morgan’s theorem  

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