Lesson Video: Representing Relations | Nagwa Lesson Video: Representing Relations | Nagwa

Lesson Video: Representing Relations Mathematics • 8th Grade

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

17:35

Video Transcript

In this video, we will learn how to represent a relation using a mapping diagram or graph, knowing the domain and range of a given relation. We will begin by defining what we mean by a relation in mathematics.

A relation is a relationship between a set of values. This relation is between the π‘₯-values and 𝑦-values of an ordered pair. The set of π‘₯-values is called the domain or input, and the set of 𝑦-values is called the range or output. Relations can be displayed as a table, a mapping, or a graph. We will look at all three types in this video. A relation produces one or more output number for every valid input number. Our answers are written as the ordered pair or coordinate π‘₯, 𝑦. We will now look at a variety of questions on relations in different contexts.

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

An ordered pair is written in the form π‘₯ comma 𝑦. In this question, the number of three-point shots made is our π‘₯-value and the total number of points is our 𝑦-value. We need to look at each column of the table one at a time.

When there were no three-point shots made, there were no total points. Therefore, our first ordered pair is zero, zero. One three-point shot being made resulted in three points. Therefore, our second ordered pair is one, three. There were a total of six points from two three-point shots being made, giving us an ordered pair two, six. Our final column in the table gives us the ordered pair three, nine. The four ordered pairs from the table are zero, zero; one, three; two, six; and three, nine.

In this question, we could write the output 𝑦 as a function in terms of the input π‘₯. As each shot scores three points, the total points 𝑦 is equal to three multiplied by π‘₯ or three π‘₯. Whilst it is true that all functions are relations, not all relations are functions.

Our next question involves a mapping or arrow diagram.

Write the relation 𝑅 for the following arrow diagram.

Each pair of values in a relation can be written as an ordered pair π‘₯, 𝑦. The π‘₯-value is known as the domain or input. The 𝑦-values are the range or output. We can see from the arrow diagram or mapping that the input of six gives an output of eight. Therefore, our first ordered pair is six, eight. The input of 10 gives an output of 12. So this is our second ordered pair: 10, 12.

The third and final ordered pair is 11, 13. We recall that a relation produces one or more output values for every input value. When considering a mapping or arrow diagram, every value in the domain must have a corresponding value in the range. We can’t have any numbers that are not matched in the domain. However, we can in the range. In this case, 28 is not matched to any of the input values.

The relation 𝑅 can therefore be written using set notation and curly brackets. 𝑅 is equal to the ordered pairs six, eight; 10, 12; and 11, 13. Whilst we’re not asked to in this question, we might notice that this relation is also a function. Each of our output values 𝑦 is two greater than our input values π‘₯. This means that the general rule for this function is 𝑦 is equal to π‘₯ plus two. The output value will always be two more than the input value.

Our next question involves a more complicated mapping diagram.

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

Any relation is a set of ordered pairs of the form π‘₯, 𝑦. The set of π‘₯-values is known as the domain or input, and the 𝑦-values are the range or output. In terms of a mapping diagram like this, the π‘₯-values will be where the arrows start and the 𝑦-values will be where the arrows finish. We could go straight to our diagram and list the correct ordered pairs. Alternatively, we could look at the five options and eliminate some immediately.

Option (A) 𝑅 is equal to the set of negative 18, negative nine, zero, nine, and 18. This is just a set of values and not a set of ordered pairs. Therefore, option (A) cannot be correct. Option (B) 𝑅 is equal to negative 18, 18 and negative nine, nine. As there are five arrows on the figure, we know there must be five ordered pairs. This means that option (B) cannot be correct. Option (C) has four ordered pairs: negative 18, 18; negative nine, nine; nine, negative nine; and 18, negative 18. This means that this option is also incorrect.

Both options (D) and (E) contain five ordered pairs. However, option (E) contains fractional values: negative one eighteenth, negative one-ninth, one-ninth, and one eighteenth. As these values do not appear on the figure, this answer is also incorrect. By elimination, we have found that option (D) must be the correct answer β€” negative 18, 18; negative nine, nine; zero, zero; nine, negative nine; and 18, negative 18. We’ll now check that these are indeed the five ordered pairs on the figure. The arrow that starts furthest to the left starts at negative 18 and finishes at 18. This means it has an input of negative 18 and an output of 18 and is therefore the ordered pair negative 18, 18.

The second arrow along goes from negative nine to nine. This is therefore another ordered pair. The third arrow starts at zero but also does a loop and goes back to zero. This corresponds to the ordered pair zero, zero. Our next arrow has an input of nine and an output of negative nine. This is the ordered pair nine, negative nine. Finally, we have an arrow that starts at 18 and finishes at negative 18. The five ordered pairs are negative 18, 18; negative nine, nine; zero, zero; nine, negative nine; and 18, negative 18. This confirms that option (D) is correct.

We might notice a pattern here between our input and output values. If we multiply or divide each of the input values by negative one, we get the output values. This means that this mapping corresponds to the function 𝑦 is equal to negative π‘₯.

Our next question involves identifying a relation from a coordinate grid.

Write the relation 𝑅 for the following diagram.

Any relation 𝑅 needs to be written as a set of ordered pairs of the form π‘₯, 𝑦. In the relation, the π‘₯-values correspond to the domain or input and the 𝑦-values are the range or output. In this question, there are five points marked on the diagram. Therefore, we need five ordered pairs. These can be written in any order, but we will start furthest to the left, the lowest π‘₯-value. The first ordered pair is therefore equal to negative four, one. We go along the π‘₯-axis to negative four and up the 𝑦-axis to one.

Our second point has coordinates negative one, two. The next pair is zero, one. Any point that lies on the 𝑦-axis will have an π‘₯-coordinate of zero. Our fourth point has coordinates two, three. The fifth and final ordered pair is three, one. We can therefore conclude that the relation 𝑅 is the set of values negative four, one; negative one, two; zero, one; two, three; and three, one.

Our final two questions involve matching equations to the given relation.

Given that 𝑅 is a relation from π‘₯ to 𝑦, where π‘Ž exists in π‘₯ and 𝑏 exists in 𝑦, which of the following equations correctly expresses relation 𝑅? Is it (A) 𝑏 equals π‘Ž plus one, (B) 𝑏 equals two π‘Ž plus two, (C) 𝑏 equals two π‘Ž minus two, (D) π‘Ž is equal to two 𝑏 plus two, or (E) π‘Ž is equal to two 𝑏 minus two?

Any relation 𝑅 contains a set of ordered pairs of the form π‘₯, 𝑦. In the diagram shown, we have three ordered pairs: negative one, zero; four, 10; and five, 12. We can therefore say that the relation 𝑅 is the set of these three ordered pairs. We are asked to find the correct equation that matches any value in π‘₯ π‘Ž to a value in 𝑦 𝑏. The easiest way to do this is to substitute our values into each of the equations. Let’s begin with the ordered pair negative one, zero.

We will let π‘Ž equal negative one and 𝑏 equal zero. Zero is equal to negative one plus one. This means that equation (A) does work for the first ordered pair. Likewise, two multiplied by negative one plus two is also equal to zero. This means that equation (B) also works for the first ordered pair. In option (C), two multiplied by negative one minus two is equal to negative four and not zero. This means that equation (C) is not the correct answer. This is also true of options (D) and (E) as negative one is not equal to two multiplied by zero plus two or two multiplied by zero minus two. We can therefore rule out both of these options.

We will now consider the second ordered pair four, 10 for equation (A) and equation (B). This time, π‘Ž is equal to four and 𝑏 is equal to 10. 10 is not equal to four plus one. This means that equation (A) is also incorrect. 10 is equal to two multiplied by four plus two. This means that equation (B) holds for the first and second ordered pairs. We can now move on to the third ordered pair five, 12. Two multiplied by five plus two is equal to 12. As equation (B) holds for all three ordered pairs, this is the correct answer. The equation that correctly expresses relation 𝑅 is 𝑏 is equal to two π‘Ž plus two.

Given that π‘₯ is the set of numbers 20, one, and three and 𝑅 is a relation on π‘₯, where π‘Žπ‘…π‘ means that π‘Ž plus two 𝑏 is equal to an even number for each π‘Ž that exists in π‘₯ and 𝑏 that exists in π‘₯, determine the relation 𝑅.

We are told in the question that set π‘₯ contains three numbers: 20, one, and three. Our values of π‘Ž and 𝑏 are contained in set π‘₯. Therefore, they must each be one of these three numbers. We are also told that 𝑅 is a relation on π‘₯. Any relation is a set of ordered pairs π‘₯, 𝑦. Each of our pairs must satisfy that π‘Ž plus two 𝑏 is an even number, where π‘Ž is our π‘₯-value and 𝑏 is our 𝑦-value. We could write down the nine possible ordered pairs as shown: one, one; one, three; one, 20; three, one; and so on. We could then substitute each of these into the expression π‘Ž plus two 𝑏 to identify which are even and which are odd.

This could be quite time-consuming, so we will look for a quicker method based on the equation in this question. Multiplying any integer by two gives an even number. This means that two 𝑏 will always be even. We recall that adding an even number to another even number gives an even answer, whereas adding an odd number and an even number gives an odd answer. In order for π‘Ž plus two 𝑏 to be even, then π‘Ž must be even. The only one of the three values in set π‘₯ that is even is 20. This means that π‘Ž must be equal to 20. 𝑏 on the other hand could take any one of the three values as multiplying any of them by two would give an even answer. The relation 𝑅 is therefore the set of three ordered pairs: 20, one; 20, three; and 20, 20.

We will now summarize the key points from this video. We saw questions in this video where a relation was displayed as a table, a mapping, and a graph. In each case, our relation was written as a set of ordered pairs. This is demonstrated as shown. The set of π‘₯-values in the ordered pairs is the domain, and the set of 𝑦-values is the range. A relation can often be written as a function which links our input values π‘₯ to our output values 𝑦.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy