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# 5.3: One-to-One Functions

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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: One-to-One (Injection)

A function $${f}:{A}\to{B}$$ is said to be one-to-one if

$f(x_1) = f(x_2) \Rightarrow x_1=x_2$

for all elements $$x_1,x_2\in A$$. A one-to-one function is also called an injection, and we call a function injective if it is one-to-one. A function that is not one-to-one is referred to as many-to-one.

The contrapositive of this definition is: A function $${f}:{A}\to{B}$$ is  one-to-one  if$x_1\neq x_2 \Rightarrow f(x_1)\neq f(x_2)$

Any function is either one-to-one or many-to-one. A function cannot be one-to-many because no element can have multiple images. The difference between one-to-one and many-to-one functions is whether there exist distinct elements that share the same image. There are no repeated images in a one-to-one function.

## Definition: Identity Function

The identity function on any nonempty set $$A$$ ${I_A}:{A}\to{A}, \qquad I_A(x)=x,$ maps any element back to itself.

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

Example $$\PageIndex{1}\label{eg:oneonefcn-01}$$

The function $$h : {A}\to{A}$$ defined by $$h(x)=c$$ for some fixed element $$c\in A$$, is an example of a constant function. It is a function with only one image. This is the exact opposite of an identity function. It is clearly not  one-to-one unless $$|A|=1$$.

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

To prove $$f:A \rightarrow B$$ 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.

Example $$\PageIndex{2}$$

Prove the function $$f : {\mathbb{R}}\to{\mathbb{R}}$$ 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.

Hands-on exercise $$\PageIndex{1}\label{he:oneonefcn-01}$$

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

Hands-on exercise $$\PageIndex{2}\label{he:oneonefcn-02}$$

Determine whether the function $$h : {[2,\infty)}\to{\mathbb{R}}$$ defined by $$h(x)=\sqrt{x-2}$$ 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 $$f'(x)<0$$ for all $$x\in(a,b)$$.

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

Example $$\PageIndex{4}\label{eg:oneonefcn-04}$$

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

Hands-on exercise $$\PageIndex{3}\label{he:oneonefcn-03}$$

Use both methods to show that the function $$k:{(0,\infty)}\to{\mathbb{R}}$$ 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.

Example $$\PageIndex{5}\label{eg:oneonefcn-05}$$

Prove the function $$h : {\mathbb{R}}\to{\mathbb{R}}$$ 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.

Example $$\PageIndex{6}\label{eg:oneonefcn-06}$$

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

hands-on exercise $$\PageIndex{4}\label{he:oneonefcn-04}$$

Show that the function $$h : {\mathbb{Z}}\to{\mathbb{N}}$$ defined by $h(n) = \cases{ 2n+1 & if n\geq0, \cr -2n & if n < 0, \cr}$ is one-to-one.

Example $$\PageIndex{7}\label{eg:oneonefcn-7}$$

Let $$A$$ be the set of all married individuals from a monogamous community who are neither divorced nor widowed. Then the function $$s:{A}\to{A}$$ 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

Exercise $$\PageIndex{1}\label{ex:oneonefcn-01}$$

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

(a) $$f : {\mathbb{R}}\to{\mathbb{R}}$$, $$f(x)=x^3-2x^2+1$$.

(b) $$g : {[\,2,\infty)}\to{\mathbb{R}}$$, $$f(x)=x^3-2x^2+1$$.

Solution

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

Exercise $$\PageIndex{2}\label{ex:oneonefcn-02}$$

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

$$p :{\mathbb{R}}\to{\mathbb{R}}$$, $$p(x)=|1-3x|$$.

Exercise $$\PageIndex{3}\label{ex:oneonefcn-03}$$

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

$$q :{\mathbb{R}}\to{\mathbb{R}}$$, $$q(x)=x^4$$.

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.

Exercise $$\PageIndex{4}\label{ex:oneonefcn-04}$$

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

$$f :{\mathbb{R}}\to{\mathbb{R}}$$, $$f(x)=6-5x$$.

Exercise $$\PageIndex{5}\label{ex:oneonefcn-05}$$

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

1. $$f : {\mathbb{Z}}\to{\mathbb{Z}}$$; $$f(n)=n^3+1$$
2. $$g : {\mathbb{Q}}\to{\mathbb{Q}}$$; $$g(x)=n^2$$
3. $${k} : {\mathbb{R}}\to{\mathbb{R}}$$; $$k(x)=5^x$$
Solution

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

Exercise $$\PageIndex{6}\label{ex:oneonefcn-06}$$

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

1. $$p :{\mathscr{P}(\{1,2,3,\ldots,n\})}\to{\{0,1,2,\ldots,n\}}$$; $$p(S)=|S|$$
2. $$q :{\mathscr{P}(\{1,2,3,\ldots,n\})}\to{\mathscr{P}(\{1,2,3,\ldots,n\})}$$; $$q(S)=\overline{S}$$

Exercise $$\PageIndex{7}\label{ex:oneonefcn-07}$$

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

1. $${f_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d\}}$$; $$f_1(1)=b$$, $$f_1(2)=c$$, $$f_1(3)=a$$, $$f_1(4)=a$$, $$f_1(5)=c$$
2. $${f_2}:{\{1,2,3,4\}}\to{\{a,b,c,d,e\}}$$; $$f_2(1)=c$$, $$f_2(2)=b$$, $$f_2(3)=a$$, $$f_2(4)=d$$
Solution

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

Exercise $$\PageIndex{8}\label{ex:oneonefcn-08}$$

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

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

Exercise $$\PageIndex{9}\label{ex:oneonefcn-09}$$

List all the one-to-one functions from $$\{1,2\}$$ to $$\{a,b,c,d\}$$.

Hint

List the images of each function.

Solution

There are twelve one-to-one functions from $$\{1,2\}$$ to $$\{a,b,c,d\}$$. The images of 1 and 2 under them are listed below. $\begin{array}{|c||*{12}{c|}} \hline & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 & f_7 & f_8 & f_9 & f_{10} & f_{11} & f_{12} \\ \hline\hline 1 & a & a & a & b & b & b & c & c & c & d & d & d \\ \hline 2 & b & c & d & a & c & d & a & b & d & a & b & c \\ \hline \end{array}$

Exercise $$\PageIndex{10}\label{ex:oneonefcn-10}$$

Is it possible to find a one-to-one function from $$\{1,2,3,4\}$$ to $$\{1,2\}$$? Explain.

Exercise $$\PageIndex{11}\label{ex:oneonefcn-11}$$

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

1. $$f : {\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$h(n)\equiv 3n$$ (mod 10).
2. $$g : {\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$g(n)\equiv 5n$$ (mod 10).
3. $$h : {\mathbb{Z}_{36}}\to{\mathbb{Z}_{36}}$$; $$h(n)\equiv 3n$$ (mod 36).
Solution

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

Exercise $$\PageIndex{12}\label{ex:oneonefcn-12}$$

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

$$k : {\mathbb{R}}\to{\mathbb{R}}$$ defined by $$k(x)=3x^2-5$$ is one-to-one.

Exercise $$\PageIndex{13}\label{ex:oneonefcn-13}$$

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

$$f: {\mathbb{R}}\to{\mathbb{R}}$$ defined by $$f(x)=57$$ is one-to-one.

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.

Exercise $$\PageIndex{14}\label{ex:oneonefcn-14}$$

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