CLASS-7
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 Xc 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)