Lesson Objective:

Dear learner,

By the end of this lesson you will be able to:

• Define Relations
• Write Domain and Range of a Relation in Interval Notation, Inequalities and set Notation

Key Terms:

• Relations,
• Domain and Range
• Representing Relations

Brainstorming  Activities:

Consider Two Sets A = {cat, rabbit} and B = {meat, milk}based on this sets answer the following questions?

i. Find all the subsets of $A \times B$ containing at least three members?

ii. Find the set of the first components and the set of the second components of all members of the set in (i)?

Solution:

i). $A \times B$ = $\ \{(a,b): a \in \mathbb{A}, b \in \mathbb{B}\}$ = {(cat,meat), (cat,milk), (rabbit,meat), (rabbit,milk)} is the set containing ordered pairs which were organised from the elements of two sets:

let

• $S_ {1}$ = {(cat,meat), (cat,milk), (rabbit,meat)}
• $S_{2}$ = {(cat,meat), (cat,milk), (rabbit,milk)}
• $S_ {3}$ = {(cat,meat), (rabbit, meat), (rabbit,milk)}
• $S_{4}$ = {(cat,milk), (rabbit,meat), (rabbit,milk)} and
• $S _{5}$ = {(cat,meat), (cat,milk), (rabbit,meat), (rabbit,milk)} be the subsets of $A \times B$ with three or more elements. we can relate each ordered pair with ” a eats b” and obtain a relation so that we say a relation is the set of ordered pairs.

ii). In all subsets in number i:

• The set containing all first components is = {cat, rabbit}
• The set containing all second components is = {meat, milk}

Definition:

Given two sets A and B, any set  $\mathbb{R} =\ \{(x,y):x \in \mathbb{A} and y \in \mathbb{B}\}$ , is called a relation from set A to set B

Example 1:

$\text{1. Given two sets}~A={1,3,5,7}~\text{and}~B={6,8}$ and R be the relation “less than” from A to B. Then,

$\begin{aligned}
R={(1,6),(1,8),(3,6),(3,8),(5,6),(5,8),(7,8)}
\end{aligned}$.

$\text{2. Let}~A={1,2,3,4,5}~\text{and}~B={a,b,c}.~\text{The following are relations from A to B}$

$i. ~R_{1}=\{(1,a)\} $
$~ii. ~R_{2}=\{(2,b),(3,b),(4,c),(5,a)\} $
$~iii. ~R_{3}=\{(1,a),(2,b),(3,c)\}$

$\text{Note: A relation is the set of ordered pairs}$

Domain and Range of Relations

Given a relation R from set A to B;

• Domain of R:  the set of all the first components of R.
• Dom(R) = $\ \{x \in \mathbb{A}: (x,y) \in \mathbb{R}$ for some $y \in \mathbb{B}\}$
• Range of R: the set of all the second components of R  
• Ran(R) = $\ \{y \in \mathbb{B}: (x,y) \in \mathbb{R}$ for some$ x \in \mathbb{A}\}$

 Example: 2

1. In a Certain City There are 5 Secondary Schools. the Number of Mathematics  Teachers Taught in each School are Listed in the  Table as Shown below:

1. Find The relation Defined by The given Table ?
2. Determine Domain and Range of The Relation?

Solution:

A). In the table school names are the first components while number of mathematics teachers are the second components of the relation, then

R = {(school 1, 23), (school 2, 18), (school 3, 27), (school 4, 19), (school 5, 31)}

B). i. Domain (R) = {school 1, school 2, school 3, school 4, school 5}

ii. Range (R) = {18, 19, 23, 27, 31}

2. Let R is a relation defined by $R =\ \{(x,y):y =2 x^{2} – 1\}$  for $x \in \{-2, -1, -0.1, 0,1,2\}$ then Find domain and range of R?

Solution:

Let the value of x be any real number and y is determined by substituting x in the formula.

The equation is $ y = 2x^2 – 1 $ as follows:

$\text{ When} ~ x = -2 , y = 2(-2)^2 – 1 = 8 – 1 = 7 $
$\text{ When} ~ x = -1 , y = 2(-1)^2 – 1 = 2 – 1 = 1 $
$\text{ When} ~x = -0.1 , y = 2(-0.1)^2 – 1 = 0.02 – 1 = -0.98$
$\text{ When} ~x = 0 , y = 2(0)^2 – 1 = 0 – 1 = -1 $
$\text{ When} ~ x = 1 , y = 2(1)^2 – 1 = 2 – 1 = 1 $
$\text{ When}~ x = 2, y = 2(2)^2 – 1 = 8 – 1 = 7. $

Then, the range $R = {(-2, 7), (-1, 1), (-0.1, -0.98), (0, -1), (1, 1), (2, 7)}$.

$\text{Domain of R} = {-2, -1, -0.1, 0, 1, 2}~\text{and}~ \text{Range of R} = {-1, -0.98, 1, 7}.$

Representation of Relations

   A relation is represented by either of :

• Set of ordered pairs,
• Correspondence between domain and range,
• Graph,
• Equations,
• An inequality or combination of any of these.

A) Set of Ordered Pairs:

 Example 3:

Let R be a relation defined by $R = \ \{(1, 2), (3, 4), (4, 5)\}$, then determine domain

                    and range of R

Solution:  

Domain = {1, 3, 4} is the set of first components and

                        Range = {2, 4, 5} is the set of the second components

B) Correspondence Between Domain and Range:

Example 4:

A and B are two given sets and the relation from set A to B is given by using the diagram below, determine the relation R and find domain and range ?

Figure 1.1 Diagram representing a relation

Solution:

From the given diagram, the relation as a set of ordered pairs is given by:

 R = {(-8,26),(-6,10), (5,15)} Whereas  

Domain = {-8, -6, 5} and Range = {10, 15, 26}

 C) Graph

Example 5:

  Find the domain and Range of the relation given by the graph below?

Figure 1.2 Table representing relation

Solution :

R is represented as the set of ordered pairs of x and y.

R = {(-2, 1), (-1, 0), (1,-1), (1, 1), (2, 3),(3, 1)}

And

Domain ( R) = {-2, -1, 1, 2, 3} Range ( R) = {-1, 0, 1, 3}

D) Equations

Example 6:

A relation R is defined by  $R=\ \{(x,y): x\in \mathbb{R}, y \in \mathbb{R}, and ; y = 3x +1\}$. Find domain and range of R?

Solution:

R is defined as an infinite set of ordered pairs. The first coordinate can be any of the set of real numbers while the second coordinate becomes a real number for which it is determined by the first coordinate. as $y \in \mathbb{R}$ then $ x = \frac{y-1}{3} \in \mathbb{R}$

• Therefore Domain = the set of all  real numbers
• Range =  the set of real numbers

E) Inequalities ( Region):

Example 7:

draw the graph of the relation and determine domain and range of $R =\ \{(x,y): y \le x+1 , y \le 1-x , and y > -4}$

Solution:

Figure 1.3 inequality graph representing a relation

Intersection point:

$y = x + 1, \quad y = 1 – x \quad \text{and} \quad y = -4.$

$\text{Intersection of the lines}~ y = x + 1 ~\text {and} ~y = 1 – x $:

$\implies x+1=1-x \implies 2x=0$

$\implies x=0.$ Hence, $y=0+1=1.$

Therefore, the intersection point is $(0,1).$

$\text{Intersection of the lines}~ y = x + 1 ~\text {and} ~y = -4 $:

$\implies x+1=-4 \implies x=-4-1\implies x=-5.$

Therefore, the intersection point is $(-5,-4).$

$\text{Intersection of the lines}~ y = 1-x ~\text {and} ~y = -4 $:

$\implies 1-x=-4 \implies x=4+1\implies x=5.$

Therefore, the intersection point is $(5,-4).$

Hence, the domain of $R = (-5, 5) ~\text{and the range of} ~R= (-4, 1].$