If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Solution and Explanation
U = {a, b, c, d, e, f, g, h}
(i) A = {a, b, c}
A' = {d,e,f,g,h}
(ii) B = {d, e, f, g}
∴ B' = {a, b , c , h}
(iii) C = {a, c, e, g}
∴ C' = {b, d, f, h}
(iv) D = {f, g, h, a}
∴ D' = {b, c,d,e}
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Concepts Used:
Complement of a Set
The complement of a set is described as A’ = {x: x ∈ U and x ∉ A}
where,
A’ stands for the complement.
Complement of Sets Properties:
1. Complement Laws: The union of a set A and its complement A’ allows the universal set U of which, A and A’ are a subset.
A ∪ A’ = U
Also, the intersection of a set A and its complement A’ cause the empty set ∅.
A ∩ A’ = ∅
For Example: If U = {11, 12 , 13 , 14 , 15 } and A = {11 , 12 , 13 } then A’ = {14 , 15}. From this it can be seen that
A ∪ A’ = U = { 11 , 12 , 13 , 14 , 15}
Also, A ∩ A’ = ∅
2. Law of Double Complementation: According to the law, if we take the complement of the complemented set A’ then, we get the set A itself.
(A’)’ = A
In the previous example we can see that, if U = {11 , 12 , 13 , 14 , 15} and A = {11 , 12, 13} then A’ ={14 , 15}. Now if we consider the complement of set ‘A’ we get,
(A’)’ = {11 , 12 , 13} = A
This gives back the set A itself.
3. Law of empty set and universal set: According to this law the complement of the universal set gives us the empty set and vice-versa i.e.,
∅’ = U And U’ = ∅
This law is accessible.