Question

12. Let R be a relation in N defined by (x, y) belongs to R=> x+2y = 8.
Express Rand R^-1 as the set of ondered pairs and also find
domain and range of R.​

Answer 1

Understanding the Basic Concept :-

Relation :-

A relation R is the subset of the cartesian product of X x Y, where X and Y are two non-empty sets. It is derived by stating the relationship between the first element and second element of the ordered pair of X × Y.

The set of all primary elements of the ordered pairs is called a domain of R

The set of all second elements of the ordered pairs is called a range of R.

So,

Mathematically,

R = {(a,b) : a ∊ X, b ∊ Y}

and

Domain of relation R = { a : (a, b) ∊ R }

and

Range of relation R = { b : (a, b) ∊R }

Inverse Relation

Let A, B be two sets and Let R be a relation from Set A to Set B. Then the Inverse relation of R is defined as a relation from B to A.

Mathematically,

$\sf \: R ^{ - 1} = \{ (b, a) : (a,b) \in \: R \}$

$\large\underline{\bold{Solution-}}$

We have,

$\sf \: (x, y) \in \: R \: such \: that \: x + 2y = 8$

$\sf \: 2y = 8 - x$

$\sf \: y \: = \: \dfrac{8 - x}{2} \: where \: x, \: y \in \: N$

Now,

$\sf \: For \: x = 2, \: y = \dfrac{8 - 2}{2} = 3 \in \: N$

$\therefore \: (2, 3) \in \: R$

Now,

$\sf \: For \: x = 4, \: y = \dfrac{8 - 4}{2} = 2 \in \: N$

$\therefore \: (4, 2) \in \: R$

Now,

$\sf \: For \: x = 6, \: y = \dfrac{8 - 6}{2} = 1 \in \: N$

$\therefore \: (6, 1) \in \: R$

Hence,

$\bf \: R = \{(2, 3), \: (4, 2), \: (6, 1) \}$

$\therefore \bf \: Domain \: of \: R = \{2, \: 4, \: 6 \}$

and

$\bf \: Range \: of \: R \: = \{3, \: 2, \: 1 \}$

and

$\bf \: R ^{ - 1} = \{(3, 2), \: (2, 4), \: (1, 6) \}$