Question

If $U = \left\{ {4,8,12,16,20,24,28} \right\},A = \{ 8,16,24\} ,B = \{ 4,16,20,28\} .$
Verify that ${\left( {A \cap B} \right)^\prime } = A' \cup B'$

Answer 1

Hint: In these types of questions operations of sets are used. Mainly properties of complement of sets are involved in this question. The given question is an example of de morgan’s law : ${\left( {A \cap B} \right)^\prime } = A' \cup B'$ .These laws can also be verified by using venn diagrams.

Complete step-by-step answer:
The given sets are $U = \left\{ {4,8,12,16,20,24,28} \right\},A = \{ 8,16,24\} ,B = \{ 4,16,20,28\} .$
According to question we have to verify , ${\left( {A \cap B} \right)^\prime } = A' \cup B'$
As we know that compliment of a set is given by,
$ \Rightarrow A' = U - A$
Here we want ,$A',B',{\left( {A \cap B} \right)^\prime }$
$
   \Rightarrow A' = U - A = \{ 4,8,12,16,20,24,28\} - \{ 8,16,24\} \\
   \Rightarrow A' = \{ 4,12,20,28\} \\
    \\
   \Rightarrow B' = U - B = \{ 4,8,12,16,20,24,28\} - \{ 4,16,20,28\} \\
   \Rightarrow B' = \{ 8,12,24\} \\
    \\
   \Rightarrow \left( {A \cap B} \right) = \{ 8,16,24\} \cap \{ 4,16,20,28\} \\
   \Rightarrow \left( {A \cap B} \right) = \{ 16\} \\
    \\
   \Rightarrow {\left( {A \cap B} \right)^\prime } = U - \left( {A \cap B} \right) \\
   \Rightarrow {\left( {A \cap B} \right)^\prime } = \{ 4,8,12,16,20,24,28\} - \{ 16\} \\
    \\
$
$ \Rightarrow {\left( {A \cap B} \right)^\prime } = \{ 4,8,12,20,24,28\} .$ -----(i)
Now ,
$ \Rightarrow $ $
  A' \cup B' = \{ 4,12,20,28\} \cup \{ 8,12,24\} \\
    \\
$
$ \Rightarrow A' \cup B' = \{ 4,8,12,20,24,28\} .$ ------(ii)
From equations (i) and (ii) we get both ${\left( {A \cap B} \right)^\prime },A' \cup B'$
Hence , we can verify that ${\left( {A \cap B} \right)^\prime } = A' \cup B'$

Note : In this question we have used some properties of complement of sets . first we will get all the given sets and then we have to find compliments of those sets and other necessary operations that are used to verify the statement. Then we will get both the sets L.H.S and R.H.S and by comparing both sides we can verify the given statement.