Proving that $f\colon \mathbb{R} \to \mathbb{R}$ $\frac{x}{1+x^2}$ is not an injection I needed help proving that this function is not an injection. I know that I can try a few $x$ values and see if I get the same output but I wanted a more rigorous proof using the definitions.
This is what I tried.
\begin{align}
f(m)=f(p)
&\implies \frac{m}{1+m^2} = \frac{p}{1+p^2} \\
&\implies m+mp^2 = p + pm^2 \\
&\implies m(1+p^2) = p(1+m^2)
\end{align}
I don't really know where to go from here, how would I derive a contradiction showing that this is not injective?
 A: Note that for $x \ne 0$, we have $$\frac{1/x}{1 + (1/x)^2} = \frac{(1/x)(x^2)}{(1 + (1/x)^2)(x^2)} = \frac{x}{x^2 + 1},$$ so if $f(x) = x/(1+x^2)$, then $f(1/x) = f(x)$ for all $x \ne 0$.
An easier way to have seen this particular property of $f$ is to have written $$f(x) = \frac{x}{1+x^2} = \frac{x/x}{1/x + x^2/x} = \frac{1}{x + \frac{1}{x}},$$ and now this relationship between $f(x)$ and $f(1/x)$ is laid bare.
A: Hint let $f(x)=C$ for some constant $C$ chosen wisely. You then get $x=(1+x^2)C$ which is a quadratic with two solutions if you picked $C$ smartly. 
A: I would like to clear some of the smoke surrounding heropup's very slick–but somewhat mysterious–answer.  The following solution is very elementary in the sense that it doesn't use anything more advanced than high school algebra (e.g. we don't make reference to the extreme value theorem or any other result from calculus), and in the sense that it doesn't try to do anything clever.  The argument is a bit tedious because I am trying not to do anything that is even a little bit clever–I am just following my nose in the most naive way possible (which is what I would expect from most students).
Suppose that there $f(x) = f(y)$ for some $x,y\in\mathbb{R}$ (where these two values are not assumed to be distinct).  Then we have
$$ \frac{x}{1+x^2} = \frac{y}{1+y^2}
\implies x(1+y^2) = y(1+x^2)
\implies x y^2 - (1+x^2) + x = 0.$$
This equation is quadratic in $y$, so, using the quadratic formula, we obtain
\begin{align} y
&= \frac{(1+x^2) \pm \sqrt{(1+x^2)^2 - 4x^2}}{2x} \\
&= \frac{(1+x^2) \pm \sqrt{1-2x^2+x^4}}{2x} \\
&= \frac{(1+x^2) \pm \sqrt{(1-x^2)^2}}{2x} \\
&= \frac{(1+x^2) \pm (1-x^2)}{2x}.
\end{align}
Therefore, simplifying the $\pm$, we conclude that
$$ y \in \left\{ \frac{(1+x^2) + (1-x^2)}{2x}, \frac{(1+x^2) - (1-x^2)}{2x} \right\}
= \left\{ \frac{1}{x}, x\right\}.
$$
As long as $x \ne 0$, we obtain two possible values of $y$:  either $y=x$ (which is uninteresting, as our goal is to show that the function $f$ is not injective), or $y=\frac{1}{x}$, which is interesting as it shows that it is possible for $f(x)$ to be equal to $f(y)$ even if $x\ne y$ (demonstrating a lack of injectivity).  Indeed, what we have now shown is exactly heropup's solution.  If we take $y=\frac{1}{x}$, then we have
$$
f(y)
= f(1/x)
= \frac{1/x}{1+(1/x)^2}
= \frac{x}{1+x^2}
= f(x)
$$
for all $x\ne 0$.
A: Actually, your attempt would have gotten you the answer if you had pushed a little further:
$m(1+p^2) = p(1+m^2)$ yields $(p-m)(mp-1) = 0$, which you should automatically have found because you already know that one equality case is $p=m$ by symmetry and so can factor out $(p-m)$. This immediately tells you that you should try $m = 1/p$, which indeed works.
A: Note that $f(0)=0$ and $\lim_{x \to \infty} f(x)=0$.  There is a unique local maximum at $x=1$ with value $1/2$, so this is an absolute maximum on $[0, +\infty)$.  Intuitively, the graph "begins" at $(0,0)$, goes up to $(1,1/2)$, then descends toward the $x$-axis in the long run. Such a function cannot be one-to-one for continuity reasons.  (This is a little flabby, but can be made rigorous.)
