ICSE Class 8 Maths Chapter 39 Ordered Pair Cartesian Product PDF Download

Chapter 39: Ordered Pair; Cartesian Product

39.1 Ordered Pair

An ordered pair is the pair of two objects which occur in a particular order (i.e. in which the order of objects is important).

e.g. consider two objects (numbers) '6' and '8'. If written as (6, 8), they form one ordered pair and if written as (8, 6), they form another ordered pair.

Since, the order of writing the objects in both the pairs is different, therefore, (6, 8) and (8, 6) are two different ordered pairs.

1. In an ordered pair (a, b), a is called the first component and b is called the second component of the ordered pair. Conversely, if first and second components of an ordered pair are x and y respectively, then the ordered pair is (x, y).

2. The components of ordered pairs may be the same e.g. (3, 3), (a, a), (p, p), etc.

3. It is important to know that {a, b} and (a, b) are not the same. Reason: {a, b} represents a set whose elements are a and b, whereas (a, b) represents an ordered pair, whose components are a and b. Moreover, {a, b} = {b, a} as the change in order of writing the elements does not change the set, but (a, b) ≠ (b, a) as by changing the order of writing the components, the ordered pair is changed.

39.2 Equality Of Ordered Pairs

Two ordered pairs are said to be equal, if they have the same (equal) first components and the same second components.

e.g. (7, 10) = (7, 10); (5, -3) = (5, -3); Also, if (x, y) = (5, 7), then x = 5 and y = 7.

Example 1:

Use the elements of set A = {a, b, c} to form all possible ordered pairs.

Solution:

The possible ordered pairs are: (a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b) and (c, c).

Example 2:

Given A = {5, 7} and B = {4, 6, 8}, form all ordered pairs, so that in each ordered pair, (i) the first component is from A and the second component is from B. (ii) the first component is from B and the second component is from A.

Solution:

(i) Associate each element of set A with each element of set B. This can be done in the following manner:

5 - 4, 5 - 6, 5 - 8

7 - 4, 7 - 6, 7 - 8

Thus, the required ordered pairs are: (5, 4), (5, 6), (5, 8), (7, 4), (7, 6) and (7, 8).

(ii) Associate each element of set B with each element of set A, i.e.

4 - 5, 4 - 7

6 - 5, 6 - 7

8 - 5, 8 - 7

The required ordered pairs are: (4, 5), (4, 7), (6, 5), (6, 7), (8, 5) and (8, 7).

Teacher's Note

Ordered pairs are used in real-life applications like GPS coordinates (latitude, longitude), where the sequence matters - (40.7128, -74.0060) is New York, but reversing them gives a different location.

Test Yourself

1. If (x - 3, 5) = (4, 2 - y), then x - 3 = .............. and 5 = .............., so x = .............. and y = ..............

2. If (3x - 2, 5 - y) = (4, -1), then .............. = .............. and .............. = .............., so .............. = .............. and .............. = ..............

Exercise 39 (A)

1. Use elements of set P = {2, 3} to form all possible ordered pairs.

2. Use the elements of set A = {x, y, z} to form all possible ordered pairs.

3. Given A = {5, 6, 7} and B = {3, 4}. Form all possible ordered pairs, so that in each ordered pair, (i) first component is from set A and second component is from set B. (ii) first component is from B and second is from A; (iii) both the components are from A; (iv) both the components are from B. In each case, write the total number of ordered pairs formed.

4. State, true or false:

(i) {x, y} is an ordered pair, whose first component is x and second component is y.

(ii) (x, y) is an ordered pair, whose components are x and y.

(iii) (a, b) is a set, whose elements are a and b.

(iv) {5, 7} = {7, 5}

(v) (5, 7) = (7, 5)

(vi) If {x, y} = {3, 5}; then x = 3 and y = 5

(vii) If (x, y) = (3, 5); then x = 3 and y = 5

(viii) Ordered pairs (a, 3) and (5, x) are equal means, a = 5 and x = 3.

5. Given (2a - 3, 3b + 1) = (7, 7); find a and b.

6. Given (3x, -5) = (x - 2, y + 3); find x and y.

7. Given (a - 2, \frac{b}{3}) = (0, 0); find a and b.

8. If the ordered pairs (a - 3, a + 2b) and (3a - 1, 3) are equal, find the values of a and b.

39.3 Cartesian Product

If A and B are two non-empty sets, then their cartesian product is the set of all possible ordered pairs such that the first components of all the ordered pairs are from set A and the second components of all the ordered pairs are from set B.

i.e. cartesian product of sets A and B = set of all possible ordered pairs (x, y) such that x ∈ A and y ∈ B.

The cartesian product of sets A and B is represented by A × B and is read as A cross B.

Example 3:

Given A = {b, c, d} and B = {x, y}; find:

(i) the cartesian product of sets A and B, i.e. A × B.

(ii) the cartesian product of sets B and A, i.e. B × A.

(iii) A × A.

(iv) B × B.

Solution:

(i) A × B = Set of all possible ordered pairs such that their first components are from A and second components are from B.

= {(b, x), (b, y), (c, x), (c, y), (d, x), (d, y)}

(ii) B × A = Set of all possible ordered pairs (x, y) such that x ∈ B and y ∈ A.

= {(x, b), (x, c), (x, d), (y, b), (y, c), (y, d)}

(iii) A × A = {(x, y): x and y both belong to A}

= {(b, b), (b, c), (b, d), (c, b), (c, c), (c, d), (d, b), (d, c), (d, d)}

(iv) B × B = {(x, x), (x, y), (y, x), (y, y)}

1. The cartesian product A × B is not the same as the cartesian product B × A, i.e. A × B ≠ B × A. Hence, the cartesian product of two unequal sets is not commutative.

2. The product sets A × B and B × A have an equal number of ordered pairs. i.e. n(A × B) = n(B × A) = n(A) × n(B)

3. If A × B = B × A, then A = B. Conversely, if A = B, then A × B = B × A.

4. Since empty set ∅ contains no element in it, therefore for any set A, the cartesian product A × ∅ also contains no element, i.e. n(A × ∅) = O = n(∅ × A).

Example 4:

Given A = {a, b, c} and B = {x, y}, show that the product sets A × B and B × A are equivalent sets.

Solution:

A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}

B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}

Since the number of ordered pairs in A × B is 6, i.e. n(A × B) = 6.

Also, the number of ordered pairs in B × A is 6, i.e. n(B × A) = 6.

Therefore, A × B and B × A are equivalent sets.

[Two sets are said to be equivalent if they contain an equal number of elements].

Teacher's Note

Cartesian products are used in creating databases and spreadsheets, where you combine data from multiple sources - for example, pairing all possible products with all possible colors in an online store.

Test Yourself

If A = {5, 6, 7} and B = {6, 8}, then

3. A × B = ....................... 4. B × A = ....................... 5. n(A × B) = ....................... 6. n(B × A) = ....................... 7. n(A) × n(B) = ....................... 8. Is A × B = B × A ? ....................... 9. Is n(A × B) = n(B × A) = n(A) × n(B)? ................ Is this result always true?.................. 10.Are A × B and B × A equivalent sets? .......................

Exercise 39 (B)

1. If A = {1, 2} and B = {2, 3, 4}; find: (i) A × B (ii) B × A (iii) A × A (iv) B × B. Is A × B = B × A?

2. If M = {x : x ∈ N, 1 < x ≤ 4} and N = {y : y ∈ W, y < 3}; find:

(i) M × N (ii) N × M (iii) n(M × M) (iv) n(N × N) Is M × N = N × M? Is n(M × N) = n(N × M)?

3. State, true or false:

(i) If n(A) = 4 and n(B) = 5, then n(A × B) = 20.

(ii) If n(A) = m and n(B) = n; then A × B = mn.

(iii) If A = {x, y} and B = {y, x}; then A × B = {(x, y), (y, x)}

(iv) A × B and B × A are equal sets.

(v) If A = B; then A × B = B × A.

4. Given A = {5, 6, 7} and B = {6, 8, 10}; find: (i) A ∪ B (ii) (A ∪ B) × B (iii) B ∩ A (iv) A × (B ∩ A)

5. If A = {x: x ∈ W, 3 ≤ x < 6}, B = {3, 5, 7} and C = {2, 4}; find: (i) A - B (ii) (A - B) × C (iii) B - C (iv) A × (B - C)

6. If A = {5, 6, 7, 8} and B = {6, 8, 10}; find: (i) A ∪ B (ii) A ∩ B (iii) (A ∪ B) × (A ∩ B)

7. If P = {a, b, c} and Q = {b, c, d}; find (P ∩ Q) × P.

8. If A = {5, 7}, B = {7, 9} and C = {7, 9, 11}, find: (i) A × (B ∪ C) (ii) (A × B) ∪ (A × C). Is A × (B ∪ C) = (A × B) ∪ (A × C)?

9. Given A = {x ∈ W: 7 < x ≤ 10}; find: A × A.

10. Given M = {0, 1, 2} and N = {1, 2, 3}; find: (i) (N - M) × (N ∩ M) (ii) (M ∪ N) × (M - N)

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