ICSE Class 8 Maths Sets Chapter 02 Operations and Venn Diagrams

Operations and Venn Diagrams

Operations on Sets

When we carry out the operations of addition, multiplication, etc., on two numbers, we get new numbers. Similarly, when we carry out the operations of union and intersection on two sets, we get new sets. Just as we can find the difference between two numbers, we can also find the difference between two sets.

Union of two sets

The union of the sets A and B is the set of all the elements that belong to either A or B or both. It is denoted by \(A \cup B\) (read as "A union B").

We can write, \(A \cup B = \{x : x \in A \text{ or } x \in B\}\).

Examples

(i) Let \(A = \{2, 4, 6\}\) and \(B = \{6, 8, 10\}\).

Then \(A \cup B = \{2, 4, 6\} \cup \{6, 8, 10\} = \{2, 4, 6, 8, 10\}\).

(ii) Let \(P = \{a, b, c\}\) and \(Q = \{x, y, z\}\).

Then \(P \cup Q = \{a, b, c, x, y, z\}\).

(iii) Let \(A = \{5, 6, 8\}\).

Then \(A \cup A = \{5, 6, 8\} \cup \{5, 6, 8\} = \{5, 6, 8\} = A\).

Also, \(A \cup \phi = \{5, 6, 8\} \cup \phi = \{5, 6, 8\} = A\).

(iv) Let \(A = \{a, b, c\}\) and \(B = \{a, b, c, d\}\).

Here, \(A \subseteq B\) then \(A \cup B = \{a, b, c, d\} = B\).

Note: The results we have arrived at in (iii) and (iv) are always true. In other words, they are laws of operations on sets.

Intersection of two sets

The intersection of the sets A and B is the set of all the elements which belong to both A and B. It is denoted by \(A \cap B\) (read as "A intersection B").

We can write, \(A \cap B = \{x : x \in A \text{ and } x \in B\}\).

If A and B do not have any element in common then \(A \cap B =\) a null set = \(\phi\).

Examples

(i) Let \(A = \{2, 4, 6, 8\}\) and \(B = \{4, 8, 12\}\).

Then \(A \cap B = \{2, 4, 6, 8\} \cap \{4, 8, 12\} = \{4, 8\}\).

(ii) Let \(P = \{1, 3, 5, 7, 9, 11, 13\}\) and \(Q = \{2, 4, 6, 8, 10, 12, 14\}\).

Then \(P \cap Q = \{1, 3, 5, 7, 9, 11, 13\} \cap \{2, 4, 6, 8, 10, 12, 14\}\)

= \(\phi\), because there is no element that is common to both P and Q.

Disjoint sets

Two sets A and B are called disjoint sets if they have no element in common, that is, \(A \cap B = \phi\).

Example

Let \(A = \{x : x \text{ is a positive integer}\}\) and \(B = \{x : x \text{ is a negative integer}\}\).

Then \(A = \{1, 2, 3, 4, \ldots\}\) and \(B = \{-1, -2, -3, -4, \ldots\}\).

Then \(A \cap B = \{1, 2, 3, 4, \ldots\} \cap \{-1, -2, -3, -4, \ldots\} = \phi\). So A and B are disjoint sets.

Overlapping sets

Two sets A and B are called overlapping sets if they have at least one element in common, that is \(A \cap B \neq \phi\).

Example

Let \(A = \{x : x \text{ is prime and } x < 10\}\) and \(B = \{\text{first two natural numbers}\}\).

Then \(A = \{2, 3, 5, 7\}\) and \(B = \{1, 2\}\).

So, \(A \cap B = \{2, 3, 5, 7\} \cap \{1, 2\} = \{2\}\).

\(\therefore\) \(A \cap B \neq \phi\). So, A and B are overlapping sets.

Difference of two sets

Let A and B be any two sets. Then the difference of A and B is the set of elements which belong to A but not to B. This is denoted by \(A - B\).

We can write \(A - B = \{x : x \in A \text{ and } x \notin B\}\).

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

Example

Let \(A = \{1, 2, 3, 4\}\) and \(B = \{2, 4, 6\}\).

Then \(A - B = \{1, 3\}\) and \(B - A = \{6\}\).

Complement of a set

Given the universal set U, the complement (or complementary set) of a set A is the set of those elements of U which are not elements of A. It is denoted by A' (or \(A^C\) or \(\overline{A}\)). Symbolically, \(A' = \{x : x \in U \text{ and } x \notin A\}\). In other words, A' is the difference of the universal set U and A.

\[A' = U - A\]

Examples

(i) Let \(U = \{1, 2, 3, 4, 5\}\) and \(A = \{1, 3, 5\}\).

Then \(A' = U - A = \{1, 2, 3, 4, 5\} - \{1, 3, 5\} = \{2, 4\}\).

(ii) Let \(U = \{x : x \in \mathbb{N} \text{ and } x \leq 10\}\) and \(A = \{x : x \in \mathbb{W} \text{ and } 4 \leq x \leq 6\}\).

Then, \(U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) and \(A = \{4, 5, 6\}\).

Thus, \(A' = U - A = \{1, 2, 3, 7, 8, 9, 10\}\).

Solved Examples

Example 1 Let \(A = \{\text{letters of the word CRICKET}\}\) and \(B = \{\text{letters of the word HOCKEY}\}\).

Find (i) \(A \cup B\), (ii) \(B \cup A\), (iii) \(A \cap B\), (iv) \(B \cap A\), (v) \(A - B\), (vi) \(B - A\).

Solution In the tabular form: \(A = \{C, R, I, K, E, T\}\), \(B = \{H, O, C, K, E, Y\}\)

(i) \(A \cup B = \{C, R, I, K, E, T\} \cup \{H, O, C, K, E, Y\} = \{C, R, I, K, E, T, H, O, Y\}\).

(ii) \(B \cup A = \{H, O, C, K, E, Y\} \cup \{C, R, I, K, E, T\} = \{H, O, C, K, E, Y, R, I, T\}\).

(iii) \(A \cap B = \{C, R, I, K, E, T\} \cap \{H, O, C, K, E, Y\} = \{C, K, E\}\).

(iv) \(B \cap A = \{H, O, C, K, E, Y\} \cap \{C, R, I, K, E, T\} = \{C, K, E\}\).

(v) \(A - B = \{C, R, I, K, E, T\} - \{H, O, C, K, E, Y\} = \{R, I, T\}\).

(vi) \(B - A = \{H, O, C, K, E, Y\} - \{C, R, I, K, E, T\} = \{H, O, Y\}\).

Note: From (i) and (ii) it should be clear that \(A \cup B = B \cup A\). Similarly, from (iii) and (iv), \(A \cap B = B \cap A\). These two are laws of operations on sets.

Union and intersection of sets work similarly to combining or finding common elements in real-world groups, like when you combine guest lists from two parties or find common friends among different social circles.

Example 2 If \(U = \{x : x \in \mathbb{W} \text{ and } 6 \leq x \leq 11\}\), \(A = \{6, 8, 9\}\), \(B = \{7, 8, 11\}\) and \(C = \{6\}\), find the following sets.

(i) A' (ii) B' (iii) C' (iv) \(A - B\) (v) \((B \cup C)'\) (vi) \((A \cap C)'\)

(vii) \(A - (B \cup C)\) (viii) \(A - (B \cap C)\)

Solution Here, \(U = \{x : x \in \mathbb{W} \text{ and } 6 \leq x \leq 11\} = \{6, 7, 8, 9, 10, 11\}\)

\(A = \{6, 8, 9\}\), \(B = \{7, 8, 11\}\) and \(C = \{6\}\).

(i) \(A' = U - A = \{6, 7, 8, 9, 10, 11\} - \{6, 8, 9\} = \{7, 10, 11\}\).

(ii) \(B' = U - B = \{6, 7, 8, 9, 10, 11\} - \{7, 8, 11\} = \{6, 9, 10\}\).

(iii) \(C' = U - C = \{6, 7, 8, 9, 10, 11\} - \{6\} = \{7, 8, 9, 10, 11\}\).

(iv) \(A - B = \{6, 8, 9\} - \{7, 8, 11\} = \{6, 9\}\).

(v) \(B \cup C = \{7, 8, 11\} \cup \{6\} = \{6, 7, 8, 11\}\).

\(\therefore\) \((B \cup C)' = U - (B \cup C) = \{6, 7, 8, 9, 10, 11\} - \{6, 7, 8, 11\} = \{9, 10\}\).

(vi) \(A \cap C = \{6, 8, 9\} \cap \{6\} = \{6\}\).

\(\therefore\) \((A \cap C)' = U - (A \cap C) = \{6, 7, 8, 9, 10, 11\} - \{6\} = \{7, 8, 9, 10, 11\}\).

(vii) \(A - (B \cup C) = \{6, 8, 9\} - \{6, 7, 8, 11\} = \{9\}\).

(viii) \(B \cap C = \{7, 8, 11\} \cap \{6\} = \phi\). So, \(A - (B \cap C) = \{6, 8, 9\} - \phi = \{6, 8, 9\}\).

Teacher's Note

These operations help us organize and filter information in databases and search engines, where union finds all results and intersection narrows them down to the most relevant matches.

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