### Video Transcript

In this video, we will learn how to
determine whether a function is injective or one-to-one. We recall that the definition of a
function requires each element of its domain to be associated with exactly one
element of its range. For a function to be injective, it
must also satisfy the statement with the roles of its domain and range reversed. We will begin by considering a
formal definition of this.

A function is injective or
one-to-one if each element of the range of the function corresponds to exactly one
element of the domain. Together with the requirement for
it to be a function, we can say that there is a one-to-one correspondence between
each element of the domain and a unique element in the range of an injective
function. This is why we also refer to an
injective function as a one-to-one function. In the picture drawn, two
functions, π and π, are described by their mapping diagrams. The function π has two elements
four and five in its range. While the range element four has
exactly one arrow pointed to it corresponding to the domain element three, the range
element five has two different arrows corresponding to the domain elements one and
two.

Since each element of the range
should correspond to exactly one element of the domain for an injective function, we
can say that π is not injective. On the other hand, each element of
the range of function π has exactly one arrow pointed to it, corresponding to
exactly one element of the domain. We can therefore conclude that
function π is injective. We will now consider an example
where we use mapping diagrams for functions to determine their injectivity.

True or false: If π and π are
both one-to-one functions, then π plus π must be a one-to-one function.

In order to prove that this
statement is true, we must prove it is true for all possible functions in this
case. However, in order to prove a
statement is false, we can do so via a counterexample. We begin by recalling that a
function is one-to-one if each element of the range of the function corresponds to
exactly one element of the domain. In mapping diagrams, this means
that each element of the range has exactly one arrow pointing to it.

Letβs consider the functions π and
π represented by the mapping diagrams shown. We see that both functions π and
π are injective or one-to-one because each element of the range has exactly one
arrow from the domain pointed at it. When we add π and π as required,
we obtain the following diagram. Noting that all three lines in the
range give the same number, we can redraw the mapping diagram. Since the range element 10 has
three arrows pointing to it, this range element corresponds to more than one of the
domain elements. We can therefore conclude the
function π plus π is not injective. Using a counterexample, we have
shown that the statement βif π and π are both one-to-one functions, then π plus
π must be a one-to-one functionβ is false.

We can easily recognize injective
functions from their mapping diagrams, but how do we recognize them from their
graphs in the π₯π¦-plane?

We know that the range of a
function is given by the portion of the vertical axis used by its graph, while the
domain is given by the portion of the horizontal axis used by its graph. Since an injective function must
associate each element of its range with a unique element in its domain, there must
be exactly one point on the horizontal axis within the domain associated with each
point of the function within its range on the vertical axis. We can test for this condition by
drawing a horizontal line for each range element and checking how many times the
graph intersects with the line. If there is more than one
intersection between the horizontal line and the functionβs graph, then that range
element is associated with more than one domain element. In this case, the function would
not be injective. Letβs consider two examples as
shown.

For function π in the first
diagram, we are able to draw a horizontal line for the range element two, which
intersects the graph of π more than once. More precisely, the range element
two corresponds to the three distinct domain elements negative two, zero, and
three. There are three points of
intersection of the horizontal line and the function at negative two, two; zero,
two; and three, two. We can therefore conclude that
function π is not injective. On the other hand, we can see that
any horizontal line drawn across the graph π in the second diagram intersects with
the graph at exactly one point. By sliding a ruler horizontally up
and down the graph, we can verify that there is no horizontal line that intersects
with the graph more than once. This means that each range element
of π corresponds to exactly one domain element. And we can conclude that function
π is injective or one-to-one.

This process is called the
horizontal line test. A function is injective or
one-to-one if each horizontal line intersects the graph of a function at most
once. Conversely, a function is not
injective or one-to-one if there is a horizontal line that crosses its graph more
than once. This is similar to the vertical
line test that is used to verify the definition of a function. In place of vertical lines, we use
horizontal lines to verify whether a function satisfies the definition of injective
or one-to-one.

We will now consider an example
where we use the horizontal line test to determine whether or not a function is
injective.

Which curve among those shown in
the graph below is a one-to-one function?

In order to answer this question,
we recall the horizontal line test, which states that a function is one-to-one or
injective if and only if its graph intersects with each horizontal line at most
once. Letβs begin with the red graph. Our goal is to draw, if possible, a
horizontal line that crosses the graph more than once, thereby indicating that the
function is not injective. One way of doing this would be to
draw the horizontal line with equation π¦ is equal to negative 10. We see that this horizontal line
crosses the red graph twice. We can therefore conclude that the
function represented by the red graph is not one-to-one.

We can repeat this process for the
green graph by drawing a horizontal line with equation π¦ equals 10. This horizontal line crosses the
green graph three times. We can therefore also conclude that
the function represented by the green graph is not one-to-one. We can also repeat this process for
the yellow graph by drawing a horizontal line with equation π¦ is equal to two. This horizontal line crosses the
yellow graph twice. So by the horizontal line test, the
function represented by the yellow graph is not one-to-one. As we have proved that the red,
green, and yellow graphs do not represent one-to-one functions, this suggests that
the blue graph does represent a one-to-one function. By firstly clearing the other
horizontal lines we have drawn, letβs consider this function in isolation.

We begin by drawing several
horizontal lines, in this case, at π¦ equals 20, 30, 40, and 50. Each of these horizontal lines
intersects our blue graph once and only once. We can check that this is true for
the whole graph by sliding a horizontal ruler up and down the graph. Each horizontal line that could be
drawn on the graph crosses the blue graph at most once. We can therefore conclude by the
horizontal line test that the function represented by the blue graph is one-to-one
or injective. This is the only one of the four
curves that represents a one-to-one function.

In this example, we used the
horizontal line test to determine whether a given graph represents an injective or
one-to-one function. If we are given the algebraic
expressions for functions rather than their graphs, then the horizontal line test is
not the ideal method to apply because it requires an accurate graph. Letβs now consider what we can do
in such circumstances. We will begin by considering the
horizontal line test, which has been applied to the function π in the graph
shown. Since the horizontal line
intersects the graph at three distinct points, there are three values π₯ sub one, π₯
sub two, and π₯ sub three for a single value π¦ zero such that π of π₯ one equals
π¦ zero, π of π₯ two equals π¦ zero, and π of π₯ three also equals π¦ zero.

Whilst it may be true, as in this
case, that there are more than two π₯-values corresponding to a single π¦-value, we
only need to find two such points since if two π₯-values correspond to the same
π¦-value, then the given function is not one-to-one or injective. This can be formally stated as
follows. Given an algebraic expression for a
function π, if there are two different π₯-values from the domain π₯ sub one and π₯
sub two such that π of π₯ sub one equals π of π₯ sub two, then π is not
injective. On the other hand, π is injective
if the conditions that π of π₯ sub one is equal to π of π₯ sub two and that π₯ one
and π₯ two belong to the domain imply that π₯ sub one is equal to π₯ sub two.

In our final example, we will use
this definition to identify injective functions from their algebraic
expressions.

Which of the following is a
one-to-one function? Is it (A) π of π₯ equals the
absolute value of π₯, (B) π of π₯ equals π₯ squared, (C) π of π₯ equals five, or
(D) π of π₯ equals π₯ plus two?

We begin by recalling that π of π₯
is not a one-to-one function if there are two different π₯-values from the domain,
π₯ sub one and π₯ sub two, satisfying π of π₯ sub one is equal to π of π₯ sub
two. Letβs consider each possible option
with respect to this condition. Option (A) is the absolute value
function. And we know that two numbers of the
same size but with opposite signs have the same absolute value. For example, if we let π₯ sub one
equal negative two and π₯ sub two equal two, then π of π₯ sub one equals the
absolute value of negative two which is equal to two and π of π₯ sub two, the
absolute value of two, is also equal to two. Since π of negative two is equal
to π of two, then the function the absolute value of π₯ is not one-to-one.

We can repeat this process for
option (B), where π of π₯ is the square function. As with the absolute value
function, two numbers of the same size but with opposite signs have the same squared
value. If we once again let π₯ sub one be
negative two and π₯ sub two be two, then π of π₯ sub one and π of π₯ sub two are
both equal to four. The square of negative two and the
square of two are both equal to four. Once again, as π of π₯ sub one is
equal to π of π₯ sub two, the function is not one-to-one.

Letβs now consider the third
option, π of π₯ equals five. This is a constant function. Whichever number we input to a
constant function, the output will always be the same. In this case, it doesnβt matter
what values we choose for π₯ sub one and π₯ sub two. π of π₯ sub one and π of π₯ sub
two will always be equal to five. This means that π of π₯ equals
five is not a one-to-one function.

Letβs now consider option (D) π of
π₯ is equal to π₯ plus two. In order to prove that this
function is one-to-one, we recall that a function is one-to-one if the conditions
that π of π₯ sub one is equal to π of π₯ sub two and that π₯ sub one and π₯ sub
two belong to the domain of π imply that π₯ sub one is equal to π₯ sub two. If π of π₯ sub one is equal to π
of π₯ sub two for the function π of π₯ equals π₯ plus two, then π₯ sub one plus two
must be equal to π₯ sub two plus two. Subtracting two from both sides of
this equation, we see that π₯ sub one is equal to π₯ sub two. This tells us that π of π₯ sub one
equals π of π₯ sub two is only possible when π₯ sub one is equal to π₯ sub two.

We can therefore conclude that π
of π₯ equals π₯ plus two is one-to-one. The only one of the functions from
the list that is one-to-one or injective is option (D).

Whilst it is outside of the scope
of this video, this leads us to the fact that all linear functions are one-to-one if
the coefficient of the variable is nonzero.

We will now recap the key points
from this video. A function is injective or
one-to-one if each element of the range of the function corresponds to exactly one
element of the domain. The horizontal line test states
that a function is injective or one-to-one if and only if each horizontal line
intersects with the graph of a function at most once. A function π is one-to-one if the
conditions π of π₯ sub one equals π of π₯ sub two and π₯ one and π₯ two belong to
the domain of π imply that π₯ sub one is equal to π₯ sub two. Finally, π is not one-to-one if we
can find distinct π₯ sub one and π₯ sub two in the domain satisfying π of π₯ sub
one is equal to π of π₯ sub two.