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Mathematics LibreTexts

5.3: One-to-One Functions

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    23271
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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.