# CLASS-11RELATION & FUNCTION - EQUALITY OF ORDERED PAIRS

Equality of Ordered Pairs

Two ordered pairs of numbers are defined to be equal when both the first components are equal and

9       16

also their second components are equal. Thus, (- 3, 4) = (- -----, -----)

3        4

but  (4, 5) ≠ (5, 4),

(7, 9) ≠ (7, 5),  (6, 11) ≠ (3, 11)

If A = {a, b}, then all possible ordered pairs are (a, a) (b, b) (a, b) (b, a).

If A = {a, b, c} and B = {1, 2, 3, 4}. Then all possible ordered pairs such that in each ordered pair the first component is an element of set A and second component is an element of set B are obtained by pairing each element from set A with each element from set B. This can be done by the following scheme –

(a, 1), (a, 2), (a, 3), (a, 4), (b, 1), (b, 2), (b, 3), (b, 4), (c, 1), (c, 2), (c, 3), (c, 4)

Example.1) Express {(x, y) : x² + y² = 25, where x, y ∈ W} as a set of ordered pairs.

Ans.) Let us find possible values of whole numbers x, y which make x² + y² = 25.

x = 0, y = 5 => x² + y² = 0² + 5² = 25

x = 3, y = 4 => x² + y² = 3² + 4² = 25

x = 4, y = 3 => x² + y² = 4² + 3² = 25

x = 5, y = 0, => x² + y² = 5² + 0² = 25

so, the required set of ordered pairs is {(0, 5), (3, 4), (4, 3), (5, 0)}

Example.2) The number of elements in the set {(a, b) : 2a² + 3b² = 35, a, b ∈ Z}, where Z is the set  of all integers is –

(a) 2             (b) 4                     (c) 8              (d) 12

Ans.)      Given set is {(a, b) : 2a² + 3b² = 35, a, b ∈ Z

By trial, we see that 2(±2)² + 3 (±3)² = 35, and 2(±4)² + 3(±1)² = 35

So, 8 elements of the sets are –

{(2, 3), (2, -3), (-2, 3), (-2, -3), (4, 1), (4, -1), (-4, 1), (-4, -1)}