If U = $\left\{ 1,2,3,4,5,6,7,8,9 \right\}$ , A = $\left\{ 1,2,3,4 \right\}$ , B = $\left\{ 2,4,6,8 \right\}$ and C = $\left\{ 1,4,5,6 \right\}$ , then find
(a). A’
(b). B’
(c). C’
(d). (B’)’
Answer
680.1k+ views
Hint: First we will understand the meaning of complement of a set, and with the help of that we will find all the values that have been asked and there are some properties of complement that we will also use.
Complete step-by-step answer:
Let’s start our solution by what the universal set is.
Universal set: The set containing all objects or elements and of which all other sets are subsets.
Complement of a set: Complement of a set A, denoted by A’, is the set of all elements that belongs to the universal set but does not belong to set A.
Here we have U as the universal set and A, B, C as the subsets of U.
Now we get A’ as per the above definition as:
U – A
Therefore, by taking elements not present in A, we get
A’ = $\left\{ 5,6,7,8,9 \right\}$
Similarly we get the set B’ as:
U – B
Therefore, by taking elements not present in B, we get
B’ = $\left\{ 1,3,5,7,9 \right\}$
Similarly we get the set C’ as:
U – C
Therefore, by taking elements not present in C, we get
C’ = $\left\{ 2,3,7,8,9 \right\}$
Now there is one property of complement of a set,
$\left( A' \right)'=A$
Now we will use this property for solving part (d):
(B’)’ = B = $\left\{ 2,4,6,8 \right\}$
Hence we have found all the sets that have been asked.
Note: One should be careful of the fact that when we subtract two sets the common elements of the set which is being subtract is removed and the remaining elements is the final answer, hence this point must be kept in mind to avoid any mistake.
Complete step-by-step answer:
Let’s start our solution by what the universal set is.
Universal set: The set containing all objects or elements and of which all other sets are subsets.
Complement of a set: Complement of a set A, denoted by A’, is the set of all elements that belongs to the universal set but does not belong to set A.
Here we have U as the universal set and A, B, C as the subsets of U.
Now we get A’ as per the above definition as:
U – A
Therefore, by taking elements not present in A, we get
A’ = $\left\{ 5,6,7,8,9 \right\}$
Similarly we get the set B’ as:
U – B
Therefore, by taking elements not present in B, we get
B’ = $\left\{ 1,3,5,7,9 \right\}$
Similarly we get the set C’ as:
U – C
Therefore, by taking elements not present in C, we get
C’ = $\left\{ 2,3,7,8,9 \right\}$
Now there is one property of complement of a set,
$\left( A' \right)'=A$
Now we will use this property for solving part (d):
(B’)’ = B = $\left\{ 2,4,6,8 \right\}$
Hence we have found all the sets that have been asked.
Note: One should be careful of the fact that when we subtract two sets the common elements of the set which is being subtract is removed and the remaining elements is the final answer, hence this point must be kept in mind to avoid any mistake.
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