Lesson 1: Relations

Video Lesson

Competency

Dear learner,

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

• Define and explain the concept of a relation.
• Represent Relations using ordered pairs, tables, mappings, and graphs.
• Write domain and range of a relation in interval notation, inequalities and set notation.
• Solve real-life problems using mathematical relations.

Key Terms

• Relations
• Domain and Range
• Graph of a relation

Brainstorming  Activities

Consider two sets A = {cat, rabbit} and B = {meat, milk}. Based on these sets answer the following questions.

i. Find all the subsets of A × 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

A × B={(cat, meat), (cat, milk), (rabbit, meat), (rabbit, milk)}. Hence the following sets are the subsets of A × B.

S₁ = {(cat, meat), (cat, milk), (rabbit, meat)}

S₂ = {(cat, meat), (cat, milk), (rabbit, milk)}

S₃ = {(cat, meat), (rabbit, meat), (rabbit, milk)}

S₄ = {(cat, milk), (rabbit, meat), (rabbit, milk)}

S₅ = {(cat, meat), (cat, milk), (rabbit, meat), (rabbit, milk)}.

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:

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

Example 1:

$$\begin{gathered} \text{1. Given two sets}A=\{1,3,5,7\}\text{and}B=\{6,8\}\text{and}R\text{be the relation}“\text{less than}“\text{from}A\text{to}B. \\ R=\left\{\right(1,6),(1,8),(3,6),(3,8),(5,6),(5,8),(7,8\left)\right\} \end{gathered}$$

$$\begin{gathered} \text{2. Let}A=\{1,2,3,4,5\}\text{and}B=\{a,b,c\}.\text{The following are relations from}A\text{to}B: \\ i.R_1=\left\{\right(1,a\left)\right\} \\ \mathrm{ii}.R_2=\left\{\right(2,b),(3,b),(4,c),(5,a\left)\right\} \\ \mathrm{iii}.R_3=\left\{\right(1,a),(2,b),(3,c\left)\right\} \end{gathered}$$

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.

$\mathrm{Dom}\left(R\right)=\{x\in \mathbb{A}:(x,y)\in R\text{for some}y\in \mathbb{B}\}$

• Range of R: the set of all the second components of R.  

$\mathrm{Ran}\left(R\right)=\{y\in \mathbb{B}:(x,y)\in R\text{for some}x\in \mathbb{A}\}$

 Example 2

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 the domain and the 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. Thus, we have

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

b). i. Domain = {school 1, school 2, school 3, school 4, school 5}.

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

Example 3

$\text{Let}R\text{be a relation defined by}R=\left\{\right(x,y):y=2x^2-1\}\text{for}x\in \{-2,-1,-0.1,0,1,2\}.\text{Then find domain and range of}R.$

Solution:

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

$$\begin{gathered} \mathrm{When}x=–2,y=2{(–2)}^2–1=8–1=7 \\ \mathrm{When}x=–1,y=2{(–1)}^2–1=2–1=1 \\ \mathrm{When}x=–0.1,y=2{(–0.1)}^2–1=0.02–1=–0.98 \\ \mathrm{When}x=0,y=2{\left(0\right)}^2–1=0–1=–1 \\ \mathrm{When}x=1,y=2{\left(1\right)}^2–1=2–1=1 \\ \mathrm{When}x=2,y=2{\left(2\right)}^2–1=8–1=7. \end{gathered}$$

$\text{Then,}R=\left\{\right(–2,7),(–1,1),(–0.1,–0.98),(0,–1),(1,1),(2,7)\}.$

Domain of R = { – 2 , – 1 , – 0.1 , 0 , 1 , 2 } and 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 4

$$\begin{gathered} \text{Let R be a relation defined by}R=\left\{\right(1,2),(3,4),(4,5\left)\right\}\text{, then determine domain} \\ \text{and range of R} \end{gathered}$$

Solution  

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

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

b) Correspondence Between Domain and Range

Example 5

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)}.

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

 C) Graph

Example 6

  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 of R= {-2, -1, 1, 2, 3} and the Range of R = {-1, 0, 1, 3}.

D) Equations

Example 7

$$\begin{gathered} \text{A relation R is defined by}=\left\{\text{(}x,y\text{)}:x\in \mathbb{R},y\in \mathbb{R},\text{and}y=3x+1\right\} \\ \text{Find domain and range of R?} \end{gathered}$$

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

$\begin{array}{c}y\in ℝ\text{, then}x=\frac{y–1}{3}\in ℝ\end{array}$.

Therefore, Domain = the set of all  real numbers and Range =  the set of real numbers.

E) Inequalities (Region)

Example 8

$\text{Draw the graph of the relation and determine domain and range of}R=\left\{\right(x,y):y\le x+1,y\le 1–x,\text{and}y>–4\}$.

Solution

Figure 1.3. Inequality graph representing a relation

Intersection point

$\Rightarrow x+1=1–x\Rightarrow 2x=0\Rightarrow x=0.\text{Hence,}y=0+1=1$.

$\text{Therefore, the intersection point is}(0,1)$

Intersection of the lines y = x + 1 and y = -4

⟹x+1=−4 ⟹ x=−4−1 ⟹ x=−5.

⟹ x+1=−4 ⟹ x=−4−1 ⟹ x=−5.

Therefore, the intersection point is (−5,−4).

$\text{Intersection of the lines}y=1–x\text{and}y=–4$

⟹ 1-x=−4 ⟹ x=−4+1 ⟹ x=5.

Therefore, the intersection point is (5,−4).

Hence, the domain of R=(−5,5) and the range of R=(−4,1].