Let f : $\mathbb{R}$ $\to$ $\mathbb{R}$ via $ f(x) = \frac{x}{1+x^2}$. Is $f$ injective?

My attempt:

I am unable to find a counter example to prove that it is not injective.

Suppose $ f(a) = f(b)$ for some $a,b \in \mathbb{R}$.

$ \frac{a}{1+a^2}=\frac{b}{1+b^2}$

$a\ +\ ab^2\ =\ b\ +ba^2$

$ ab^2-b\ \ =ba^2-a$

$ b\left(ab-1\right)=a\left(ab-1\right)$

$ \left(b-a\right)\left(ab-1\right)=0$

Would this imply that $ a = b$? In questions where I cannot easily figure out a counter example to prove a function is not injective, what should I do? Should I try to prove that it is injective and then reach a contradiction?

  • 3
    $\begingroup$ No, you got the correct answer: when $ab=1$. So $f(a)=f(1/a)$ $\endgroup$
    – gary
    Jul 16, 2017 at 0:27
  • $\begingroup$ You 've practically found the counteraxample : to have $(ab-1)=0$ we need $a=1/b$. So, take for example $a=3$ and $b=1/3$ and check that $f(3)=f(1/3)$. $\endgroup$
    – leonbloy
    Jul 16, 2017 at 3:47

4 Answers 4


It cannot be injective since $f(x) = \frac{1}{x+\frac{1}{x}}$ implies $f(x)=f\left(\frac{1}{x}\right)$ for any $x\in(0,1)$, for instance.


Following your own computations, assuming continuity throughout the answer, the condition $(a-b)(ab-1)=0$ means equality is reached either when $a=b$ ( tautological) , or when $b=1/a$. Double-checking ( assume $a \neq 0$, which anyway does not satisfy $ab=1$):

$$ f(a):= \frac {a}{1+a^2}; f(1/a):= \frac {1/a}{1+(1/a)^2}=\frac {1/a}{(a^2+1)/a^2}=\frac {a}{1+a^2} $$ ( after cancelling the a's). Notice that for a function $f: \mathbb R \rightarrow \mathbb R $ to be injective , it must be monotone, EDIT : for surjectivity, we must have $Lim f(x)_{x \rightarrow \infty} =\infty $ and $Lim f(x)_{x \rightarrow - \infty} = - \infty$ , or change the order of the infinities.

  • $\begingroup$ $\tanh(x)$ is injective, the limits do not need to be infinite. $\endgroup$
    – zwim
    Jul 16, 2017 at 0:49
  • $\begingroup$ @Zwim :Ah, yes, I was thinking surjective. Let me edit. $\endgroup$
    – gary
    Jul 16, 2017 at 0:51
  • $\begingroup$ "Notice that for a function ... to be injective, it must be monotone" -- not true. This statement is true for continuous functions, but not in general. $\endgroup$
    – zipirovich
    Jul 16, 2017 at 3:10
  • $\begingroup$ @gary: Likewise, the part about surjectivity also requires continuity.altogether. $\endgroup$ Jul 16, 2017 at 9:16
  • $\begingroup$ @MeniRosenfeld: What do you mean, that the condition does not apply to discontinuous functions? I guess in order to apply the IVT? $\endgroup$
    – gary
    Jul 16, 2017 at 17:51

A function which is continuous and injective should be strictly monotonic

[else we can invoke the intermediate value theorem before and after any local extremum and find duplicated values]

Here since $1+x^2\neq 0$ then $f$ is continuous over $\mathbb R$. Yet $f(0)=0$ and $\lim\limits_{x\to+\infty}f(x)=0$.

Since $f(x)>0$ for $x>0$ then $f$ cannot be strictly monotonic and furthermore injective.


The continuous function $f$ defined on $\Bbb R$ satisfies
$f(0) = 0$
$f(1) = 1/2$

By the intermediate value theorem, $\tag 1 f(\alpha) = 1/4$ for some $0 \lt \alpha \lt 1$.

$f(1) = 1/2$
$f(10) = 10/101$

By the intermediate value theorem, $\tag 2 f(\beta) = 1/4$ for some $1 \lt \beta \lt 10$

By (1) and (2) the function can't be injective.

If you are not familiar with the intermediate value theorem, sketch a rough graph. You might convince yourself that, say, $f(x) = 1/4$ has two solutions.
Solving for $x$,

$\frac{x}{1+x^2} = 1/4 \text { iff }$
$4 x = 1+x^2 \text { iff }$
$x^2 - 4x + 1 = 0$

Plugging into the quadratic formula, you get

$\alpha = 2 - \sqrt 3$
$\beta = 2 + \sqrt 3$

Observe that $\beta = \alpha^{-1}$.


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