LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

COMPLEMENTARY SET

__Complementary Set__

**Given the universal set U, the complementary set of the set X
is the set containing only those members of U that do not belong to the set X
it is denoted by X’ or X ^{c} which can be written as “complement
of X”. **

**We take all the members of set U which does not belong to the
set X, thus x ****ϵ**** X’
if x ****∉**** X and x ****ϵ**** U**

__Example – 1__

**Let the universal set U is the member of the natural whole number**

**If, the universal set U = { 1, 2, 3, 4, 5 } and X
= { 3, 4, 5 } , here X’ = { 1, 2 }**

__Example – 2__

**If, X = { x / x is the consonant }**

**Then, X’ = { x / x is vowel }**

** **

__Example – 3__

**If, U = { 11, 12, 13, 14, 15, ……….., 19, 20 }, A = { x /
5< x <10 }, and B = { x / 100 < x² < 225 } then find A’ &
B’**

**Ans.) Writing both the set in tabular form A = { 6, 7,
8, 9 }, B = { 11, 12, 13, 14 }**

**A’ = the set of elements of U that do not belong to A = { 6,
7, 8, 9 }, but we can observe here no member of U is not available to A, so
A’ = ****φ ( we can say, it is
disjoint set)**

**B’ = the set of elements of U that do not belong to B = { 11,
12, 13, 14 }, we can find here B’ = { 15, 16, 17, 18, 19, 20 }.**

** **

__Example – 4__

**If, U = { 11, 12, 13, 14, 15, ……….., 19, 20 }, A = { x /
5< x <10 } and B = { x / 100 < x² < 225 } then find A’ ****⋃**** B’**

**Ans.) Writing both the set in tabular form A = { 6, 7,
8, 9 }, B = { 11, 12, 13, 14 }**

**A’ = the set of elements of U that do not belong to A = { 6,
7, 8, 9 }, but we can observe here no member of U is available to A, so A’ =
****φ ( we can say, it is
disjoint set)**

**B’ = the set of elements of U that do not belong to B = { 11,
12, 13, 14 }, we can find here B’ = { 15, 16, 17, 18, 19, 20 } .**

**So, A’ ****⋃**** B’ = ****φ U { 15, 16,
17, 18, 19, 20 } = { 15, 16, 17, 18, 19, 20 }**

** **

__Example – 5__

**If , U = { 11, 12, 13, 14, 15, ……….., 19, 20 }, A = { x /
5< x <10 } and B = { x / 100 < x² < 225 } then find A’ ****⋂**** B’**

**Ans.) Writing both the set in tabular form A = { 6, 7,
8, 9 }, B = { 11, 12, 13, 14 }**

**A’ = the set of elements of U that do not belong to A = { 6,
7, 8, 9 } , but we can observe here no member of U is available to A, so A’ =
****φ ( we can say, it is
disjoint set)**

**B’ = the set of elements of U that do not belong to B = { 11,
12, 13, 14 }, we can find here B’ = { 15, 16, 17, 18, 19, 20 }.**

**So, A’ ****⋂**** B’ = ****φ ****⋂**** { 11, 12, 13, 14 } = ****φ**

__Example – 6__

**If , U = { 11, 12, 13, 14, 15, ……….., 19, 20 }, A = { x /
5< x <10 } and B = { x / 100 < x² < 225 } then find (A ****⋃**** B)’**

**Ans.) Writing both the set in tabular form A = { 6, 7,
8, 9 }, B = { 11, 12, 13, 14 }**

**A ****⋃**** B = { 6, 7, 8, 9, 11, 12, 13, 14 }**

**(A ****⋃**** B)’ = the set of elements of U that do not
belong to A ****⋃**** B = { 6, 7, 8, 9, 11, 12, 13, 14 }, but
we can observe here no member of U is available to A ****⋃**** B, so (A
****⋃**** B)’ = ****φ
( we can say, it is disjoint set)**

__Example – 7__

**If, U = { 11, 12, 13, 14, 15, ……….., 19, 20 }, A = { x /
5< x <10 }, and B = { x / 100 < x² < 225 } then find (A ****⋂**** B)’**

**Ans.) Writing both the set in tabular form A = { 6, 7,
8, 9 }, B = { 11, 12, 13, 14 }**

**A ****⋂**** B = { 6, 7, 8, 9} ****⋂**** {11, 12,
13, 14 } = ****φ ( it is
disjoint set)**

**(A ****⋂**** B)’ = the set of elements of U that do not
belong to A ****⋂**** B = ****φ, but we can
observe here no member of U is available to A ****⋂**** B,
so (A ****⋂**** B)’ = ****φ
( we can say, it is disjoint set)**