ICSE Class 8 Maths Chapter 02 Operations on Sets Venn Diagrams PDF Download

Chapter 2: Operations on Sets; Venn Diagrams

In previous classes, you studied the union and intersection of two sets, and also the complement of a set. You also learnt how to draw simple Venn diagrams to represent sets and the various relationships between them. In this chapter, we shall strengthen these ideas. We shall also introduce another operation on sets - difference of two sets.

Operations on Sets

Union of Sets

The union of two sets A and B, written as A \(\cup\) B (read as 'A union B'), is the set consisting of all those elements which belong to either A or B or both.

Thus, A \(\cup\) B = {x | x \(\in\) A or x \(\in\) B}.

For example:

(i) If A = {a, b, c, d, e, f} and B = {a, e, i, o, u}, then
A \(\cup\) B = {a, b, c, d, e, f, i, o, u}.

(ii) If A = {1, 5, 9} and B = {0, 2, 4, 6, 8, 10}, then
A \(\cup\) B = {1, 5, 9, 0, 2, 4, 6, 8, 10}.

(iii) If A = {0, 3, 6, 9, 12, 15, 18, 21, 24} and B = {6, 12, 18, 24}, then
A \(\cup\) B = {0, 3, 6, 9, 12, 15, 18, 21, 24} = A.

Intersection of Sets

The intersection of two sets A and B, written as A \(\cap\) B (read as 'A intersection B'), is the set consisting of all those elements which belong to both A and B.

Thus, A \(\cap\) B = {x | x \(\in\) A and x \(\in\) B}.

For example:

(i) If A = {a, b, c, d, e, f} and B = {a, e, i, o, u}, then
A \(\cap\) B = {a, e}.

(ii) If A = {1, 5, 9} and B = {0, 2, 4, 6, 8, 10}, then
A \(\cap\) B = \(\emptyset\).

(iii) If A = {0, 3, 6, 9, 12, 15, 18, 21, 24} and B = {6, 12, 18, 24}, then
A \(\cap\) B = {6, 12, 18, 24} = B.

Difference of Two Sets

If A and B be two sets, then A - B is the set consisting of all those elements which belong to A but do not belong to B.

Thus, A - B = {x | x \(\in\) A and x \(\notin\) B}.

Similarly, B - A = {x | x \(\in\) B and x \(\notin\) A}.

Teacher's Note

Understanding set operations helps when organizing groups in real life, like sorting students who play different sports or belong to various clubs.

Note that A - B is the set consisting of elements of A only and B - A is the set consisting of elements of B only.

For example:

(i) If A = {a, b, c, d, e, f} and B = {a, e, i, o, u}, then
A - B = {b, c, d, f} and B - A = {i, o, u}.
Note that A - B \(\neq\) B - A.

(ii) If A = {1, 5, 9} and B = {0, 2, 4, 6, 8, 10}, then
A - B = {1, 5, 9} and B - A = {0, 2, 4, 6, 8, 10}.

(iii) If A = {0, 3, 6, 9, 12, 15, 18, 21, 24} and B = {6, 12, 18, 24}, then
A - B = {0, 3, 9, 15, 21} and B - A = \(\emptyset\).

Complement of a Set

If \(\xi\) is the universal set and A is any set, then the complement of A, denoted by A' or \(\overline{A}\) or A^c (read as 'complement of A'), is the set consisting of all those elements of \(\xi\) which do not belong to A.

Thus, A' = {x | x \(\in\) \(\xi\) and x \(\notin\) A}.

Note that A' = \(\xi\) - A.

For example:

(i) If A = {1, 3, 5, 7, 9} and \(\xi\) = {1, 2, 3, ..., 10}, then
A' = {2, 4, 6, 8, 10}.

(ii) If A = {a, b, c, d, e, f} and \(\xi\) = {letters of English alphabet}, then
A' = {last twenty letters of English alphabet}.

(iii) If A = {January, June, July} and \(\xi\) = {months of a year}, then
A' = {months of a year which do not begin with letter 'J'}.

Teacher's Note

The complement operation is useful for finding what is excluded; for example, finding which students did not participate in an event from a full class list.

Remarks

- If A is any set, then
(i) A \(\cup\) \(\emptyset\) = A, A \(\cup\) \(\xi\) = \(\xi\), A \(\cup\) A = A
(ii) A \(\cap\) \(\emptyset\) = \(\emptyset\), A \(\cap\) \(\xi\) = A, A \(\cap\) A = A.

- \(\xi\)' = \(\emptyset\), \(\emptyset\)' = \(\xi\).

- If A is any set, then
A \(\cup\) A' = \(\xi\), A \(\cap\) A' = \(\emptyset\).

- If A and B are any sets, then
(i) A \(\cup\) B = B \(\cup\) A, A \(\cap\) B = B \(\cap\) A (Commutative laws)
(ii) A \(\subseteq\) A \(\cup\) B, B \(\subseteq\) A \(\cup\) B, A \(\cup\) B \(\subseteq\) \(\xi\)
(iii) A \(\cap\) B \(\subseteq\) A, A \(\cap\) B \(\subseteq\) B
(iv) A - B = A \(\cap\) B', B - A = B \(\cap\) A'
(v) (A \(\cup\) B)' = A' \(\cap\) B', (A \(\cap\) B)' = A' \(\cup\) B' (De Morgan's laws)

- If A and B are two sets, then
(i) A and B are disjoint sets if and only if A \(\cap\) B = \(\emptyset\)
(ii) A and B are overlapping sets if and only if A \(\cap\) B \(\neq\) \(\emptyset\).

- If A \(\subseteq\) B, then
(i) A \(\cup\) B = B, A \(\cap\) B = A (ii) A - B = \(\emptyset\).

Example 1

If A = {factors of 24} and B = {factors of 36}, then find
(i) A \(\cup\) B (ii) A \(\cap\) B
(iii) A - B (iv) B - A.

Solution. The given sets in the roster form are:
A = {1, 2, 3, 4, 6, 8, 12, 24} and B = {1, 2, 3, 4, 6, 9,12, 18, 36}
(i) A \(\cup\) B = {1, 2, 3, 4, 6, 8, 12, 24, 9, 18, 36}
(ii) A \(\cap\) B = {1, 2, 3, 4, 6, 12}
(iii) A - B = {8, 24}
(iv) B - A = {9, 18, 36}.

Example 2

If A = {letters of SECUNDRABAD} and B = {letters of BENGALURU}, then find
(i) A \(\cup\) B (ii) A \(\cap\) B
(iii) A - B (iv) B - A.
Also verify that:
(a) n(A \(\cup\) B) = n(A) + n(B) - n(A \(\cap\) B)
(b) n(A - B) = n(A \(\cup\) B) - n(B)
(c) n(B - A) = n(B) - n(A \(\cap\) B)
(d) n(A - B) + n(B - A) + n(A \(\cap\) B) = n(A \(\cup\) B).

Solution. The given sets in the roster form are:
A = {S, E, C, U, N, D, R, A, B} and
B = {B, E, N, G, A, L, U, R}
(i) A \(\cup\) B = {S, E, C, U, N, D, R, A, B, G, L}.
(ii) A \(\cap\) B = {E, U, N, R, A, B}.
(iii) A - B = {S, C, D}.
(iv) B - A = {G, L}.

Verification
Here n(A) = 9, n(B) = 8, n(A \(\cup\) B) = 11, n(A \(\cap\) B) = 6,
n(A - B) = 3 and n(B - A) = 2. Therefore,
(a) n(A) + n(B) - n(A \(\cap\) B) = 9 + 8 - 6 = 11 = n(A \(\cup\) B)
(b) n(A \(\cup\) B) - n(B) = 11 - 8 = 3 = n(A - B)
(c) n(B) - n(A \(\cap\) B) = 8 - 6 = 2 = n(B - A)
(d) n(A - B) + n(B - A) + n(A \(\cap\) B) = 3 + 2 + 6 = 11 = n(A \(\cup\) B).

Example 3

If \(\xi\) = {all digits in our number system}, A = {x : x is prime}, B = {x : x is a factor of 18} and C = {multiples of 3}, then verify the following:
(i) (A \(\cup\) B)' = A' \(\cap\) B' (ii) (A \(\cap\) B)' = A' \(\cup\) B'
(iii) A - C \(\neq\) C - A (iv) A - B = A \(\cap\) B'.

Solution. Here
\(\xi\) = {0, 1, 2, ..., 9}
A = {2, 3, 5, 7},
B = {1, 2, 3, 6, 9} and
C = {0, 3, 6, 9}.
(i) A \(\cup\) B = {1, 2, 3, 5, 6, 7, 9}
\(\therefore\) (A \(\cup\) B)' = {0, 4, 8}

Also, A' = {0, 1, 4, 6, 8, 9} and B' = {0, 4, 5, 7, 8}
\(\therefore\) A' \(\cap\) B' = {0, 4, 8}.
Hence, (A \(\cup\) B)' = A' \(\cap\) B'.

(ii) A \(\cap\) B = {2, 3}
\(\therefore\) (A \(\cap\) B)' = {0, 1, 4, 5, 6, 7, 8, 9}

Also A' \(\cup\) B' = {0, 1, 4, 6, 8, 9} \(\cup\) {0, 4, 5, 7, 8}
= {0, 1, 4, 6, 8, 9, 5, 7}
Hence, (A \(\cap\) B)' = A' \(\cup\) B'.

(iii) A - C = {2, 5, 7} and C - A = {0, 3, 9}
Hence, A - C \(\neq\) C - A.

(iv) A - B = {5, 7},
A \(\cap\) B' = {2, 3, 5, 7} \(\cap\) {0, 4, 5, 7, 8} = {5, 7}
Hence, A - B = A \(\cap\) B'.

Some Basic Results About Cardinal Number

- If A and B are finites sets, then
(i) n(A \(\cup\) B) = n(A) + n(B) - n(A \(\cap\) B)
(ii) n(A - B) = n(A \(\cup\) B) - n(B) = n(A) - n(A \(\cap\) B)
(iii) n(B - A) = n(A \(\cup\) B) - n(A) = n(B) - n(A \(\cap\) B)
(iv) n(A \(\cup\) B) = n(A - B) + n(B - A) + n(A \(\cap\) B)

- If \(\xi\) (universal set) is finite set and A is any set, then
n(A) + n(A') = n(\(\xi\)).

These results are very useful in solving problems.

Example 4

If n(\(\xi\)) = 40, n(A) = 25, n(B) = 12 and n((A \(\cup\) B)') = 8, find
(i) n(A \(\cup\) B) (ii) n(A \(\cap\) B) (iii) n(A - B).

Solution. (i) We know that n(A \(\cup\) B) + n((A \(\cup\) B)') = n(\(\xi\))
[\(\because\) n(A) + n(A') = n(\(\xi\))]
\(\Rightarrow\) n(A \(\cup\) B) + 8 = 40
\(\Rightarrow\) n(A \(\cup\) B) = 40 - 8 = 32.

(ii) We know that n(A \(\cup\) B) = n(A) + n(B) - n(A \(\cap\) B)
\(\Rightarrow\) 32 = 25 + 12 - n(A \(\cap\) B)
\(\Rightarrow\) n(A \(\cap\) B) = 25 + 12 - 32 = 5.

(iii) We know that n(A - B) = n(A) - n(A \(\cap\) B)
\(\Rightarrow\) n(A - B) = 25 - 5 = 20.

Example 5

If n(\(\xi\)) = 50, n(A) = 35, n(B) = 20 and n((A \(\cap\) B)') = 40, find
(i) n(A') (ii) n(B') (iii) n(A \(\cap\) B)
(iv) n(A \(\cup\) B) (v) n((A \(\cup\) B)') (vi) n(B - A).

Solution. (i) n(A') = n(\(\xi\)) - n(A) = 50 - 35 = 15
(ii) n(B') = n(\(\xi\)) - n(B) = 50 - 20 = 30
(iii) n(A \(\cap\) B) = n(\(\xi\)) - n((A \(\cap\) B)') = 50 - 40 = 10
(iv) n(A \(\cup\) B) = n(A) + n(B) - n(A \(\cap\) B) = 35 + 20 - 10 = 45
(v) n((A \(\cup\) B)') = n(\(\xi\)) - n(A \(\cup\) B) = 50 - 45 = 5
(vi) We know that n(B - A) = n(A \(\cup\) B) - n(A)
\(\Rightarrow\) n(B - A) = 45 - 35 = 10.

Example 6

If n(A - B) = 15, n(B - A) = 10 and n(A \(\cap\) B) = 5, find
(i) n(A \(\cup\) B) (ii) n(A) (iii) n(B).

Solution. (i) We know that n(A \(\cup\) B) = n(A - B) + n(B - A) + n(A \(\cap\) B)
\(\Rightarrow\) n(A \(\cup\) B) = 15 + 10 + 5 = 30
(ii) We know that n(A - B) = n(A) - n(A \(\cap\) B)
\(\Rightarrow\) 15 = n(A) - 5 \(\Rightarrow\) n(A) = 15 + 5 = 20
(iii) We know that n(B - A) = n(B) - n(A \(\cap\) B)
\(\Rightarrow\) 10 = n(B) - 5 \(\Rightarrow\) n(B) = 10 + 5 = 15.

Teacher's Note

Cardinal numbers (the counts of elements in sets) help us solve practical problems like determining how many people have certain characteristics in a survey.

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