Prove that $f: (0,1)\to \mathbf{R}$ is surjective.

Question

I wanted to find an algebraic bijection from $(0,1)$ to $\mathbf{R}$ (not the overused $\tan^{-1}$ variation.) I came up with $f$$: (0,1)\to \mathbf{R}$ $$f(x)=\frac{1}{x}-\frac{1}{1-x}$$ and proved injectivity quite trivially. But I came to a screeching halt when trying to prove that it was indeed onto. I supposed $f(x_c)=c$ for some $c\in \mathbf{R}$ and tried to unravel $f$ for $x_c$ in terms of $c$. I came to $$x_c=\frac{c+2-\sqrt{c^2+4}}{2c}$$ but I have to show that $x_c\in (0,1).$ This graph settled my nerves as it shows what I want to prove (indeed, it is the graph of $f^{-1}$), but how would one argue this algebraically?

Answer 1

Leaving the $c=0$ case aside, which $f^{-1}(0)$ should be $\frac 12$.

Rationalise the numerator in different ways for $x_c$, $x_c-1$ and $x_c-\frac12$:

$$\begin{align*} x_c &= \frac{c+2-\sqrt{c^2+4}}{2c} \tag{0}\\ &= \frac{(c+2)^2-(c^2+4)}{2c\left(c+2+\sqrt{c^2+4}\right)}\\ &= \frac{4c}{2c\left(c+2+\sqrt{c^2+4}\right)}\\ &= \frac{2}{c+2+\sqrt{c^2+4}} \tag{1}\\ \end{align*}$$

Line $(1)$ shows that: if $c>0$ then $x_c > 0$.

(Slightly less obviously, this holds also for $c<0$. For all $c\ne 0$, $\sqrt{c^2+4}+c > 0$, so $x_c>0$.)

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$$\begin{align*} x_c &= \frac{c+2-\sqrt{c^2+4}}{2c}\\ &= 1+\frac{-c+2-\sqrt{c^2+4}}{2c}\\ &= 1+\frac{-4c}{2c\left(-c+2+\sqrt{c^2+4}\right)}\\ &= 1-\frac{2}{-c+2+\sqrt{c^2+4}} \tag{2}\\ \end{align*}$$

Line $(2)$ shows that: if $c<0$ (i.e. $-c > 0$), then $x_c < 1$.

(Slightly less obviously, this holds also for $c>0$. For all $c\ne 0$, $\sqrt{c^2+4}-c > 0$, so $x_c<1$.)

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$$\begin{align*} x_c &= \frac{c+2-\sqrt{c^2+4}}{2c}\\ &= \frac12 + \frac{2-\sqrt{c^2+4}}{2c}\\ &= \frac12 + \frac{-c^2}{2c\left(2+\sqrt{c^2+4}\right)}\\ &= \frac12 - \frac{c}{2\left(2+\sqrt{c^2+4}\right)} \tag{3} \end{align*}$$

Line $(3)$ shows that:

• if $c>0$, then $x_c < \frac12 < 1$;
• if $c<0$, then $x_c > \frac12 > 0$.

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To conclude, for all real $c\ne 0$,

$$0 < \frac{c+2-\sqrt{c^2+4}}{2c} < 1$$

Answer 2

Being a rational function with the denominator having no zeros in the open interval $(0,1)$ this function is differentiable there and its derivative is negative showing it is monotonically decreasing. And hence it is 1-1.

Let us evaluate it rational numbers of the form 1/n which turns out to be $$f\bigg(\frac1n\bigg) = n-1-\frac1{n-1}.$$ This shows that the function takes arbitrarily high values.

Evaluating at the rational number $(n-1)/n$ we see that the value is $\frac{n}{(n-1)}-n$ which shows it takes values highly negative without bounds. Combined with intermediate value theorem this shows the range is the whole real line.