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Hint- Here, we will be subtracting the elements of set A’ from the universal set U.

Given, Universal set as ${\text{U}} = \left\{ {1,2,3,4,5,6,7,8,9} \right\}$

Sets ${\text{A}} = \left\{ {1,2,3,4} \right\}$, ${\text{B}} = \left\{ {2,4,6,8} \right\}$,${\text{C}} = \left\{ {3,4,5,6} \right\}$

To prove: Set ${\text{A'}} = \left\{ {5,6,7,8,9} \right\}$

As we know that set A’ (also called complement of set A) contains all the elements in the universal set which are not there in set A.

i.e., ${\text{A’}} = {\text{U}} - {\text{A}} = \left\{ {1,2,3,4,5,6,7,8,9} \right\} - \left\{ {1,2,3,4} \right\} = \left\{ {5,6,7,8,9} \right\}$

Hence proved that set ${\text{A’}} = \left\{ {5,6,7,8,9} \right\}$.

Note- In these types of problems, the complement of any set can be easily computed by subtracting all the elements in the set whose complement is required from the elements of the universal set. Also, here the sets B and C are given which are not even used to get the answer.

Given, Universal set as ${\text{U}} = \left\{ {1,2,3,4,5,6,7,8,9} \right\}$

Sets ${\text{A}} = \left\{ {1,2,3,4} \right\}$, ${\text{B}} = \left\{ {2,4,6,8} \right\}$,${\text{C}} = \left\{ {3,4,5,6} \right\}$

To prove: Set ${\text{A'}} = \left\{ {5,6,7,8,9} \right\}$

As we know that set A’ (also called complement of set A) contains all the elements in the universal set which are not there in set A.

i.e., ${\text{A’}} = {\text{U}} - {\text{A}} = \left\{ {1,2,3,4,5,6,7,8,9} \right\} - \left\{ {1,2,3,4} \right\} = \left\{ {5,6,7,8,9} \right\}$

Hence proved that set ${\text{A’}} = \left\{ {5,6,7,8,9} \right\}$.

Note- In these types of problems, the complement of any set can be easily computed by subtracting all the elements in the set whose complement is required from the elements of the universal set. Also, here the sets B and C are given which are not even used to get the answer.

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