Vtyjkyuk

I am trying to show that the function $f:(-1,1) rightarrow mathbbR$, $f(x) = fracxx^2-1$ is bijective. I have the following two questions:

1) Proving it using calculus: First I show that $f$ is injective. Since $f'(x) = -fracx^2+1(x^2-1)^2 < 0$ for all $x in (-1, 1)$, then $f$ is monotonic and hence injective.

I am having trouble showing that it is surjective using the Intermediate Value Theorem (IVT). The IVT states that if $f: [a,b] rightarrow mathbbR$ is continuous and $L$ is a real number satisfying $f(a) < L < f(b)$ or $f(b) < L < f(a)$, then there exists a $c in (a,b)$ where $f(c) = L$. Clearly the function at hand is continuous on $(-1, 1)$, but to use the aforementioned theorem, we need to have a closed interval $[a,b]$ (whereas $(-1, 1)$ is not a closed interval...). I "sort of" have an idea: Since $f$ is continuous on $(-1, 1)$, $f(-1)$ isn't defined but it is basically $+infty$ and $f(1)$ is "like" $-infty$, so if $L$ is a real number $-infty< L < +infty$, i.e., $L in mathbbR$, then there exists a $c in (-1, 1)$ such that $f(c) = L$, hence $f$ is surjective. But I feel this is wrong because $f(-1)$ and $f(1)$ aren't defined and the IVT needs a closed interval.

2) How can I prove bijection without using calculus? E.g., to show injection, let $fracx_1x_1^2-1 = fracx_2x_2^2-1$ and I want to reach the conclusion that $x_1 = x_2$, but I am stuck. Also, how can I prove surjection?

You can show it is surjective by solving the quadratic equation. Given $y in Bbb R$ we want to find $x$ such that $frac xx^2-1=y$, or $yx^2-x-y=0, x=frac 12y(1pm sqrt1+4y^2)$ and we want the signs of $x$ and $y$ to be opposite, so we take the minus sign. The fact that you get a unique answer shows it is injective, so answers your second part as well.

$$=>x_1x_2^2-x_1=x_2x_1^2-x_2$$
$$=>x_1x_2(x_2-x_1)=-(x_2-x_1)$$
$$(x_1-x_2)(1+x_1x_2)=0$$
$$textSo, either hspace.2cmx_1=x_2 hspace.2cm or hspace.2cm x_1=-frac1x_2.$$
The last equality is not possible due to the domain $(-1,1)$.

$f(x) = fracxx^2-1 = frac12 (frac11+x - frac11-x).$

$-1 < x < y < 1 implies frac11+x - frac11+y = fracy - x(1+x)(1+y) > 0.$
$-1 < x < y < 1 implies -frac11-x - (-frac11-y) = frac-(1-y)+(1-x)(1-x)(1-y) = fracy-x(1-x)(1-y) > 0.$

$therefore$
$-1 < x < y < 1 implies fracxx^2-1 > fracyy^2-1.$

$therefore f$ is injective.

Let $y in mathbbR$.

$$lim_x rightarrow -1+0 f(x) = +infty.$$
$$lim_x rightarrow 1-0 f(x) = -infty.$$

There is $delta_1 > 0$ such that $-1 < x < -1 + delta_1 implies f(x) < y$.

There is $delta_2 > 0$ such that $1 - delta_2 < x < 1 implies y < f(x)$.

Let $-1 < a < -1+delta_1$ and $1-delta_2 < b < 1$.

Then $f(a) < y < f(b)$.

$f$ is continuous on $[a, b]$.

So by the intermediate-value theorem, there exists $c in (a, b)$ such that $f(c) = y$.

$therefore f$ is surjective.