Lesson Explainer: Representing Relations | Nagwa Lesson Explainer: Representing Relations | Nagwa

Lesson Explainer: Representing Relations Mathematics • Third Year of Preparatory School

In this explainer, we will learn how to represent a relation using a mapping diagram or a graph, knowing the domain and range of a given relation.

Set theory is a branch of mathematics that involves the study of collections, called sets, of objects that are called elements of that set. An important tool within this area is that of the relation. Let’s begin by defining the meaning of this term.

Definition: Relation

A relation is a rule that associates the elements of one set with the elements of another. Relations can be represented as mapping diagrams, ordered pairs, tables of values, equations, or graphs.

Simply put, a relation produces one or more outputs for every input. For instance, consider a collection of students in a school. There exists a relation between each student and, for example, the number of siblings they have. If student A has 1 sibling, student B has 3 siblings, student C has 0 siblings, and student D has 1 sibling, this information can be represented in a list of ordered pairs as (,1),(,3),(,0),(,1)studentAstudentBstudentCstudentD and as a mapping (or arrow) diagram as follows:

We can see that elements in the input set are located on the left of the relation, while elements in the output set are located on the right.

In our first example, let’s demonstrate how to form ordered pairs to represent a relation given a table of values.

Example 1: Writing the Given Information as a Set of Ordered Pairs

In basketball, each shot made from outside the 3-point line scores 3 points. The table shows this relationship. List this information as ordered pairs (3-point shots made, total number of points).

Answer

Remember: a relation produces output numbers for every input number. This means that the given table represents a relation; it takes the input number, 3-point shots made, and produces an output number, total number of points.

Hence, the information in the table can be represented as a group of ordered pairs, where the first value in each pair is the input and the second value is the output. Let’s begin by looking at the first column.

The number of shots made is 0 and the total number of points is 0. The ordered pair is (0,0).

Similarly, looking at the second column, we see that when the number of shots made is 1, the total number of points is 3.

The second ordered pair is (1,3).

Repeating this process for the final two columns, we see that when the number of shots made is 2, the total number of points is 6, and when the number of shots made is 3, the total number of points is 9.

The corresponding ordered pairs are (2,6) and (3,9).

Hence, the ordered pairs are (0,0), (1,3), (2,6), and (3,9).

In our previous example, we demonstrated how to represent a relation using a list of ordered pairs. We might also observe that we could have described the relation in words. The total number of points scored is found by multiplying the number of 3-point shots made by 3: totalnumberofpointsnumberof3-pointshotsmade=3Γ—().

In our next example, we will demonstrate how to write a list of ordered pairs given an arrow diagram.

Example 2: Writing the Ordered Pairs Shown by a Relationship Given in an Arrow Diagram

Write the relation 𝑅 for the following arrow diagram:

Answer

The diagram shown is an arrow, or mapping, diagram. It links elements in an input set 𝑋 with elements in an output set π‘Œ. We can define a relation for the arrow diagram by writing it as a list of ordered pairs in a set, 𝑅.

The first input number is 6, and the arrow points to the output number 8. Hence, the first ordered pair is (6,8). Similarly, the second input number, 10, has an arrow leading to the number 12, so the second ordered pair is (10,12). The third input number, 11, has an arrow pointing to the output number 13, so the third ordered pair is (11,13). Since the final output number does not have a corresponding input number, we do not include it in an ordered pair.

Hence, 𝑅={(6,8),(10,12),(11,13)}.

We have seen one type of mapping diagrams, an arrow diagram. There are a number of alternative ways to represent relations on mapping diagrams, such as using arrows on a number line. In our next example, we will see what this looks like.

Example 3: Writing the Ordered Pairs Shown by a Relationship Given in an Arrow Diagram

Which of the following correctly expresses the relation 𝑅 illustrated in the figure below?

  1. 𝑅={(βˆ’18,18),(βˆ’9,9),(0,0),(9,βˆ’9),(18,βˆ’18)}
  2. 𝑅=ο¬ο€Όβˆ’18,βˆ’118,ο€Όβˆ’9,βˆ’19,(0,0),ο€Ό9,19,ο€Ό18,118
  3. 𝑅={(βˆ’18,18),(βˆ’9,9)}
  4. 𝑅={(βˆ’18,18),(βˆ’9,9),(9,βˆ’9),(18,βˆ’18)}
  5. 𝑅={βˆ’18,βˆ’9,0,9,18}

Answer

In order to represent a mapping diagram as a list of ordered pairs, we must carefully look at all possible inputs to the relation and follow the arrow to their corresponding output. Then, each ordered pair will have the general form (,).inputoutput

Let’s work from left to right. The first arrow starts at βˆ’18 and points to 18. Hence, the input is βˆ’18, the output is 18, and, therefore, the corresponding ordered pair is (βˆ’18,18).

The next input number is βˆ’9, and the arrow points to the corresponding output of 9. This means the ordered pair is (βˆ’9,9).

The next relation looks a little strange but is dealt with in the same way.

The input is 0 and the output is 0. The ordered pair is (0,0).

In a similar way, the final two ordered pairs are (9,βˆ’9) and (18,βˆ’18).

Hence, the correct answer is choice A, 𝑅={(βˆ’18,18),(βˆ’9,9),(0,0),(9,βˆ’9),(18,βˆ’18)}.

It is fairly intuitive that if we can represent a relation using a list of ordered pairs, we can also represent a relation on a coordinate plane. Let’s demonstrate this in our next example.

Example 4: Writing the Ordered Pairs Shown by a Relationship Given in a Cartesian Diagram

Write the relation 𝑅 for the following diagram.

Answer

Remember: we can represent a mapping diagram as a list of ordered pairs, each of which will have the general form (,).inputoutput

It follows that, given points on a coordinate plane, the ordered pairs will be of the form (π‘₯,𝑦)=(,).inputoutput

Hence, we can write the relation 𝑅 for the given diagram by first listing the ordered pair for each point.

Working from left to right and reading the π‘₯-coordinate first, we see the first pair is given by (βˆ’4,1).

Similarly, the second pair is given by (βˆ’1,2).

Then, the final three pairs are (0,1), (2,3), and (3,1).

Hence, the relation 𝑅 is given by 𝑅={(βˆ’4,1),(βˆ’1,2),(0,1),(2,3),(3,1)}.

We have already demonstrated that a relation can be represented in a number of ways, including algebraically. In our next example, we will consider how to form an equation to describe a relation given as a mapping diagram.

Example 5: Forming an Equation from an Arrow Diagram

Given that 𝑅 is a relation from 𝑋 to π‘Œ, where π‘Žβˆˆπ‘‹ and π‘βˆˆπ‘Œ, which of the following equations correctly expresses relation 𝑅?

  1. 𝑏=π‘Ž+1
  2. 𝑏=2π‘Žβˆ’2
  3. 𝑏=2π‘Ž+2
  4. π‘Ž=2π‘βˆ’2
  5. π‘Ž=2𝑏+2

Answer

The given mapping diagram takes input numbers from set 𝑋 and maps them onto output numbers from set π‘Œ. For instance, an input of βˆ’1 gives an output of 0, while an input of 4 gives an output of 10.

In order to find the correct equation that expresses the relation 𝑅, we will substitute these values into each equation. Be careful; we are told that π‘Žβˆˆπ‘‹ and π‘βˆˆπ‘Œ, so π‘Ž is the input (or independent variable) and 𝑏 is the output (or dependent variable).

Let’s begin with the equation 𝑏=π‘Ž+1. Let π‘Ž=βˆ’1: 𝑏=βˆ’1+1=0.

Since an input of βˆ’1 gives an output of 0, this equation holds for the first pair of values in our diagram. Let’s repeat this process with π‘Ž=4: 𝑏=4+1=5.

Since the input of 4 corresponds to an output of 10, this equation does not hold for all pairs of values on our diagram.

Let’s try the next equation, 𝑏=2π‘Žβˆ’2. When π‘Ž=βˆ’1, 𝑏=2Γ—(βˆ’1)βˆ’2=βˆ’4.

This does not satisfy the relation given on the mapping diagram, so this equation cannot be correct.

We will now try the equation 𝑏=2π‘Ž+2, giving our results in a table.

π‘Žβˆ’145
𝑏2Γ—(βˆ’1)+2=02Γ—4+2=102Γ—5+2=12

Since each input and output pair matches the mapping diagram, this is the correct equation. It is worth noting that since π‘Ž is the input and not the output, we would have disregarded the remaining two options.

The correct answer is choice C, 𝑏=2π‘Ž+2.

In our previous example, we demonstrated how to link a mapping diagram with an algebraic rule. We can also use an algebraic rule to find elements of a set. For instance, consider the relation from 𝑋 to π‘Œ, 𝑦=3π‘₯βˆ’1, for each π‘₯βˆˆπ‘‹ and π‘¦βˆˆπ‘Œ. This relation takes elements from set 𝑋 and generates elements in set π‘Œ. Suppose 𝑋={βˆ’1,3,4}. Then, the first element in set π‘Œ is given by substituting π‘₯=βˆ’1 into the equation 𝑦=3π‘₯βˆ’1: 𝑦=3Γ—(βˆ’1)βˆ’1=βˆ’4.

The second element is found by substituting π‘₯=3 into the same equation: 𝑦=3Γ—3βˆ’1=8.

Finally, the third element is found by substituting π‘₯=4: 𝑦=3Γ—4βˆ’1=11.

Hence, the elements of set π‘Œ are βˆ’4, 8, and 11; π‘Œ={βˆ’4,8,11}.

In our next example, we will demonstrate how to use our understanding of relations to solve problems given a rule.

Example 6: Identifying All the Ordered Pairs for a Given Relationship

Given that 𝑋={20,1,3} and 𝑅 is a relation on 𝑋, where π‘Žπ‘…π‘ means that π‘Ž+2𝑏 is equal to an even number for each π‘Žβˆˆπ‘‹ and π‘βˆˆπ‘‹, determine the relation 𝑅.

Answer

In this example, we are given a rule that links elements from set 𝑋 with other elements from the same set. The rule tells us that the relation consists of values of π‘Ž and 𝑏 from the set {20,1,3} that satisfy π‘Ž+2𝑏 is even.

Hence, we can find the ordered pairs in this relation by choosing values of π‘Ž and 𝑏 from the set given and substituting them into the equation π‘Ž+2𝑏. In fact, we can also use our understanding of odd and even numbers to deduce the relevant relation.

Let’s begin with π‘Ž=20. Since 𝑏 is an integer chosen from {20,1,3}, the expression 2𝑏 will always be even. π‘Ž=20 is even, and the sum of two even numbers is even. Hence, π‘Ž+2𝑏 will always be even for π‘Ž=20.

The ordered pairs that satisfy this relation are (20,20), (20,1), and (20,3).

Now, let π‘Ž=1. π‘Ž is odd, and the sum of an odd and an even number is odd. Hence, π‘Ž+2𝑏 is odd when π‘Ž=1 and there are no ordered pairs that satisfy the relation with an input of 1.

Finally, let π‘Ž=3. As in the previous case, this is odd, meaning that π‘Ž+2𝑏 is odd if π‘Ž=3 and there are no ordered pairs that satisfy the relation with an input of 3.

The relation 𝑅={(20,20),(20,1),(20,3)}.

In our final example, we will use a similar process to find unknowns in a set when given the rule for a relation.

Example 7: Finding Unknowns in a Relation on Pairs between Two Sets given Its Rule

The relation 𝑆 is defined on the positive real numbers by π‘₯𝑆𝑦, if and only if 𝑦=4π‘₯. Determine the values of π‘Ž, 𝑏, 𝑐, and 𝑑, given that (π‘Ž,1), (9,𝑏), (𝑐,7), and ο€Ό1214,π‘‘οˆ belong to 𝑆.

Answer

In this example, we have been given a rule that links two sets whose respective elements are π‘₯ and 𝑦. The rule tells us that the elements in each set that form the ordered pairs given and belong to the relation satisfy the equation 𝑦=4π‘₯. In this example, (π‘Ž,1), (9,𝑏), (𝑐,7), and ο€Ό1214,π‘‘οˆ belong to the relation 𝑆. Hence, given an element in each set, we can establish the corresponding element in the other set by substituting that value into this equation and solving for the unknown.

Let’s begin with the ordered pair (π‘Ž,1). Since the general form of an ordered pair is (π‘₯,𝑦), we can substitute π‘₯=π‘Ž and 𝑦=1 into the equation 𝑦=4π‘₯: 1=4Γ—π‘Ž1=4π‘Ž.

To solve for π‘Ž, we divide by 4: π‘Ž=14.

We can repeat this process for the ordered pair (9,𝑏). This time, we substitute π‘₯=9 and 𝑦=𝑏 into the equation: 𝑏=4Γ—9𝑏=36.

To solve for 𝑏, we take the square root of both sides of this equation, remembering to take both the positive and negative square root of 36: 𝑏=±√36𝑏=Β±6.

However, we are told that the relation links pairs of positive real numbers, so we disregard βˆ’6.

Hence, 𝑏=6.

Next, for the ordered pair (𝑐,7), we let π‘₯=𝑐 and 𝑦=7: 7=4𝑐49=4𝑐𝑐=494.

Finally, for ο€Ό1214,π‘‘οˆ, we substitute π‘₯=1214 and 𝑦=𝑑 to get 𝑑=4Γ—1214𝑑=121𝑑=±√121𝑑=Β±11.

Since 𝑑 is positive, we choose 𝑑=11.

Hence, the values of π‘Ž, 𝑏, 𝑐, and 𝑑 are π‘Ž=14, 𝑏=6, 𝑐=494, and 𝑑=11.

In this explainer, we have delved into the world of relations, representing them using mapping diagrams, lists of ordered pairs, graphs, and rules. This world will only expand as we move on to learn about functions, domain, range, and the various theorems that result.

For now, we will recap the key concepts from this explainer.

Key Points

  • A relation is a rule that associates the elements of one set with the elements of another. Relations can be represented as mapping diagrams, ordered pairs, tables of values, equations, or graphs.
  • A relation produces one or more outputs for every input.

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