5.3: One-to-One Functions

Harris Kwong
State University of New York at Fredonia via OpenSUNY

We distinguish two special families of functions: one-to-one functions and onto functions. We shall discuss one-to-one functions in this section. Onto functions were introduced in section 5.2 and will be developed more in section 5.4.

One-to-One (Injective)

Recall that under a function each value in the domain has a unique image in the range. For a one-to-one function, we add the requirement that each image in the range has a unique pre-image in the domain.

Definition: Identity Function

It is clear that all identity functions are one-to-one.

For domains with a small number of elements, one can use inspection on the images to determine if the function is one-to-one. This becomes impossible if the domain contains a larger number of elements. In practice, it is easier to use the contrapositive of the definition to test whether a function is one-to-one: \[f(x_1) = f(x_2) \Rightarrow x_1 = x_2\]

To prove a function is One-to-One

Assume $f(x_1)=f(x_2)$
Show it must be true that $x_1=x_2$
Conclude: we have shown if $f(x_1) = f(x_2)$ then $x_1 = x_2$, therefore $f$ is one-to-one, by definition of one-to-one.

Prove the function $ f : <\mathbb>\to<\mathbb>$ defined by $f(x)=3x+2$ is one-to-one.

Solution

Assume $f(x_1)=f(x_2)$, which means $3x_1+2 = 3x_2+2.$
Thus $3x_1=3x_2$
so $x_1=x_2$.
We have shown if $f(x_1)=f(x_2)$ then $x_1=x_2$. Therefore $f$ is one-to-one, by definition of one-to-one.

Prove the function $ g : <\mathbb>\to<\mathbb>$ defined by $g(x)=5-7x$ is one-to-one.

Determine whether the function $ h : <[2,\infty)>\to<\mathbb>$ defined by $h(x)=\sqrt$ is one-to-one.

Interestingly, sometimes we can use calculus to determine if a real function is one-to-one. A real function $f$ is increasing if \[x_1 < x_2 \Rightarrow f(x_1) < f(x_2),\] and decreasing if \[x_1 < x_2 \Rightarrow f(x_1) >f(x_2).\] Obviously, both increasing and decreasing functions are one-to-one. From calculus, we know that

A function is increasing over an open interval $(a,b)$ if $f'(x)>0$ for all $x\in(a,b)$.
A function is decreasing over an open interval $(a,b)$ if \(f'(x)

Therefore, if the derivative of a function is always positive, or always negative, then the function must be one-to-one.

The function $p : <\mathbb>\to<\mathbb>$ defined by \[p(x) = 2x^3-5\] is one-to-one, because $p'(x)=6x^2>0$ for any $x\in\mathbb^*$. Likewise, the function $q:<\big(-\frac<\pi>,\frac<\pi>\big)>\to<\mathbb>$ defined by \[q(x) = \tan x\] is also one-to-one, because $q'(x) = \sec^2x > 0$ for any $x\in \big(-\frac<\pi>,\frac<\pi>\big)$.

Use both methods to show that the function $k:<(0,\infty)>\to<\mathbb>$ defined by $k(x) = \ln x$ is one-to-one.

To prove a function is NOT one-to-one

To prove $f:A \rightarrow B$ is NOT one-to-one:

Exhibit one case (a counterexample) where $x_1\neq x_2$ and $f(x_1)=f(x_2).$
Conclude: we have shown there is a case where $x_1\neq x_2$ and $f(x_1)=f(x_2)$, therefore $f$ is NOT one-to-one.

Prove the function $ h : <\mathbb>\to<\mathbb>$ given by $h(x)=x^2$ is not one-to-one.

Solution

Consider $a=3$ and $b=-3$. Clearly $a \neq b$. However, $h(3)=9$ and $h(-3)=9$ so $h(a)=h(b).$
we have shown there is a case where $a \neq b$ and $h(a)=h(b)$, therefore $h$ is NOT one-to-one.

The function $ f : <\mathbb>\to<\mathbb>$ defined by \[f(n) = \cases< \frac & if $n$ is even \cr \frac & if $n$ is odd \cr>\] is not one-to-one, because, for example, $f(0)=f(-1)=0$. The function $ g : <\mathbb>\to<\mathbb>$ defined by \[g(n) = 2n\] is one-to-one, because if $g(n_1)=g(n_2)$, then $2n_1=2n_2$ implies that $n_1=n_2$.

Let $A$ be the set of all married individuals from a monogamous community who are neither divorced nor widowed. Then the function $s:\to$ defined by \[s(x) = \mbox < spouse of >x\] is one-to-one. The reason is, it is impossible to have $x_1\neq x_2$ and yet $s(x_1)=s(x_2)$.

Summary and Review

A function $f$ is said to be one-to-one if $f(x_1) = f(x_2) \Rightarrow x_1=x_2$.
No two images of a one-to-one function are the same.
Know how to write a proof to show a function is one-to-one.
To show that a function $f$ is not one-to-one, all we need is to find two different $x$-values that produce the same image; that is, find $x_1\neq x_2$ such that $f(x_1)=f(x_2)$.

Exercises

Which of the following functions are one-to-one? Explain.

Solution

(a) No. For example, $f(0)=f(2)=1$
(b) Yes, since $g'(x)=3x^2-4x=x(3x-4)>0$ for $x>2$.

Decide if this function is one-to-one or not. Then prove your conclusion.

Solution

No. For example, $2 \neq -2$, but $q(2)=16$ and $q(-2)=16$.
We have shown a case where $x_1 \neq x_2$ and $q(x_1)=q(x_2)$, so $q$ is NOT one-to-one.

Decide if this function is one-to-one or not. Then prove your conclusion.

Determine which of the following are one-to-one functions.

(a) One-to-one (b) Not one-to-one (c) One-to-one

Determine which of the following are one-to-one functions and explain your answer.

Determine which of the following functions are one-to-one.

$:>\to>$; $f_1(1)=b$, $f_1(2)=c$, $f_1(3)=a$, $f_1(4)=a$, $f_1(5)=c$
$:>\to>$; $f_2(1)=c$, $f_2(2)=b$, $f_2(3)=a$, $f_2(4)=d$

(a) Not one-to-one (b) One-to-one

Determine which of the following functions are one-to-one.

$:>\to>$; $g_1(1)=b$, $g_1(2)=b$, $g_1(3)=b$, $g_1(4)=a$, $g_1(5)=d$
$:>\to>$; $g_2(1)=d$, $g_2(2)=b$, $g_2(3)=e$, $g_2(4)=a$, $g_2(5)=c$

List all the one-to-one functions from $\$ to $\$.

Hint

List the images of each function.

Solution

There are twelve one-to-one functions from $\$ to $\$. The images of 1 and 2 under them are listed below. \[\begin<|c||*> \hline & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 & f_7 & f_8 & f_9 & f_ & f_ & f_ \\ \hline\hline 1 & a & a & a & b & b & b & c & c & c & d & d & d \\ \hline 2 & b & c & d & a & c & d & a & b & d & a & b & c \\ \hline \end\]

Is it possible to find a one-to-one function from $\$ to $\$? Explain.

Determine which of the following functions are one-to-one.

$ f : <\mathbb_>\to<\mathbb_>$; $h(n)\equiv 3n$ (mod 10).
$ g : <\mathbb_>\to<\mathbb_>$; $g(n)\equiv 5n$ (mod 10).
$ h : <\mathbb_>\to<\mathbb_>$; $h(n)\equiv 3n$ (mod 36).

(a) One-to-one (b) Not one-to-one (c) Not one-to-one

Decide if this function is one-to-one or not. Then prove your conclusion.

Solution

No. For example, $8 \neq 17$, but $f(8)=57$ and $f(17)=57$.
We have shown a case where $x_1 \neq x_2$ and $f(x_1)=f(x_2)$, so $f$ is NOT one-to-one.

Give an example of a one-to-one function $f$ from $\mathbb$ to $\mathbb$ that is not the identity function.

This page titled 5.3: One-to-One Functions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong (OpenSUNY) .

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- 5.2: Deﬁnition of Functions
- 5.4: Onto Functions and Images/Preimages of Sets

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5.3: One-to-One Functions

One-to-One (Injective)

Definition: Identity Function

To prove a function is One-to-One

To prove a function is NOT one-to-one

Summary and Review

Exercises

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