Lesson Video: Injective Functions Mathematics

In this video, we will learn how to determine whether a function is a one-to-one function (injective).

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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.

Letβs 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 βif 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 functionβs graph, then that range element is associated with more than one domain element. In this case, the function would not be injective. Letβs 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. Letβs 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, letβs 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. Letβs 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. Letβs 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.

Letβs 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 doesnβt 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.

Letβs 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.