I need to prove that the function $f:(-1,1)\rightarrow\mathbb{R}$ defined by $f(x) = \frac{x}{x^2 - 1}$ is surjective.

My work.

$b = \frac{a}{a^2 - 1} \iff b(a^2 -1)=a \iff ba^2 - b - a=0$

From here I did a few cases:

Case 1) $b=0$. Then $a=0$.

Using the quadratic formula: $a = \frac{1\pm \sqrt{1+4b^2}}{2b}$.

Case 2) $b \gt0$. Then $a = \frac{1+ \sqrt{1+4b^2}}{2b}$ $\notin (-1,1)$ $\forall$ b $\in R$.

Case 3) $b \gt0$. Then $a = \frac{1- \sqrt{1+4b^2}}{2b}$ $\in (-1,1)$ $\forall$ b $\in R$.

Using Case 1 and Case 3, f is subjective. Is this correct? I cannot use Calculus.

  • $\begingroup$ Maybe a more analytic argument is easier. You can observe the limit $x\to\pm 1$, and then argue with the intermediate value theorem. $\endgroup$
    – Cornman
    Mar 23 at 19:57

2 Answers 2


It is correct. But you should justify the assertions that $\frac{1+\sqrt{1+4b^2}}{2b}\notin(-1,1)$ and that $\frac{1-\sqrt{1+4b^2}}{2b}\in(-1,1)$. This follows easily from the fact that$$\frac{1+\sqrt{1+4b^2}}{2b}\times\frac{1-\sqrt{1+4b^2}}{2b}=-1.$$

  • $\begingroup$ That is what I meant by "Using Case 1 and Case 3," what would be a more rigorous way to state that? Do you mean I should ignore case 1? $\endgroup$ Mar 23 at 20:04
  • $\begingroup$ I misread what you wrote. What you did is correct indeed. Do you want me to delete my answer? $\endgroup$ Mar 23 at 20:07
  • $\begingroup$ No, it is still insightful. How do you think I should state "Using Case 1 and Case 3" where it is a little more elegant? $\endgroup$ Mar 23 at 20:09
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    $\begingroup$ I've edited my answer. Concerning your question, I would say that you should take $a=0$ if $b=0$ and $a=\frac{1-\sqrt{1+4b^2}}{2b}$ otherwise. $\endgroup$ Mar 23 at 20:14
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    $\begingroup$ @blacknapkins7 Suppose that $b>0$. Then $1+\sqrt{1+4b^2}>\sqrt{4b^2}=2b$, and therefore $\frac{1+\sqrt{1+4b^2}}{2b}>1$. Since the product of those two numbers is equal to $-1$, the other number must belong to $(-1,0)\subset(-1,1)$. A similar argument works if $b<0$. $\endgroup$ Mar 27 at 12:37

You can also just argue that $f$ is continuous on $(-1,1)$ and that $$ \lim_{x \to -1^+} f(x)=+\infty, \quad \lim_{x\to 1^-} f(x) = -\infty. $$

The intermediate value theorem (or a mild generalisation of it) takes care of the rest.


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