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 π¦-values of an ordered pair. The set of π₯-values is called the
domain or input, and the set of π¦-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₯, π¦. 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 π¦. In this question, the number of
three-point shots made is our π₯-value and the total number of points is our
π¦-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 π¦ as a function in terms of the input π₯. As each shot scores three points,
the total points π¦ 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 π
for the
following arrow diagram.
Each pair of values in a relation
can be written as an ordered pair π₯, π¦. The π₯-value is known as the domain
or input. The π¦-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 canβt 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 π
can therefore be
written using set notation and curly brackets. π
is equal to the ordered pairs
six, eight; 10, 12; and 11, 13. Whilst weβre not asked to in this
question, we might notice that this relation is also a function. Each of our output values π¦ is two
greater than our input values π₯. This means that the general rule
for this function is π¦ 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 π
illustrated in the figure below?
Any relation is a set of ordered
pairs of the form π₯, π¦. The set of π₯-values is known as
the domain or input, and the π¦-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 π¦-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) π
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) π
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. Weβll 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 π¦ is equal to negative π₯.
Our next question involves
identifying a relation from a coordinate grid.
Write the relation π
for the
following diagram.
Any relation π
needs to be written
as a set of ordered pairs of the form π₯, π¦. In the relation, the π₯-values
correspond to the domain or input and the π¦-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 π¦-axis to one.
Our second point has coordinates
negative one, two. The next pair is zero, one. Any point that lies on the π¦-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 π
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 π
is a relation from π₯
to π¦, where π exists in π₯ and π exists in π¦, which of the following equations
correctly expresses relation π
? Is it (A) π equals π plus one,
(B) π equals two π plus two, (C) π equals two π minus two, (D) π is equal to
two π plus two, or (E) π is equal to two π minus two?
Any relation π
contains a set of
ordered pairs of the form π₯, π¦. In the diagram shown, we have three
ordered pairs: negative one, zero; four, 10; and five, 12. We can therefore say that the
relation π
is the set of these three ordered pairs. We are asked to find the correct
equation that matches any value in π₯ π to a value in π¦ π. The easiest way to do this is to
substitute our values into each of the equations. Letβs begin with the ordered pair
negative one, zero.
We will let π equal negative one
and π 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 is equal to four and
π 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 π
is π is equal to two π plus two.
Given that π₯ is the set of numbers
20, one, and three and π
is a relation on π₯, where ππ
π means that π plus two
π is equal to an even number for each π that exists in π₯ and π that exists in
π₯, determine the relation π
.
We are told in the question that
set π₯ contains three numbers: 20, one, and three. Our values of π and π are
contained in set π₯. Therefore, they must each be one of
these three numbers. We are also told that π
is a
relation on π₯. Any relation is a set of ordered
pairs π₯, π¦. Each of our pairs must satisfy that
π plus two π is an even number, where π is our π₯-value and π is our
π¦-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 plus two π 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 π 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 plus two π to be
even, then π must be even. The only one of the three values in
set π₯ that is even is 20. This means that π must be equal to
20. π 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 π
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 π¦-values is the range. A relation can often be written as
a function which links our input values π₯ to our output values π¦.