Exercise 3.10

Proof. Suppose that $x,y\in (-1,1)$ and that $x<y$. Now, since $-1<x<1$, we have $0\le \left|x\right|<1$ so that

$$\begin{array}{cc}{x}^2={\left|x\right|}^2=\left|x\right|\left|x\right|<\left|x\right|\cdot 1=\left|x\right|<1.& \end{array}$$

Hence $0<1-x^2$ so that also $0<1∕(1-x^2)$ as well. By the same argument $0<1-y^2$ so that $0<1∕(1-y^2)$. We then have the following cases:

Case: $x\ge 0$. Then also $y>0$ since $y>x\ge 0$. Thus

$$\begin{array}{lllllll}\hfill x^2=x\cdot x\le x& \cdot y<y\cdot y=y^2& \hfill \text{(since }x<y\text{, }x\ge 0\text{, and }y>0\text{)}& & \hfill & & \hfill \\ \hfill -x^2& >-y^2& \hfill & & \hfill & \\ \hfill 1-x^2& >1-y^2& \hfill & & \hfill & \\ \hfill \frac{1}{1-y^2}& >\frac{1}{1-x^2}.& \hfill & & \hfill & \end{array}$$

From this, it follows that

$$\begin{array}{cc}f(x)=\frac{x}{1-x^2}<\frac{y}{1-x^2}<\frac{y}{1-y^2}=f(y)& \end{array}$$

since $x<y$ and $1∕(1-x^2)$ and $1∕(1-y^2)$ are both positive.

Case: $x<0$. Then $f(x)=x∕(1-x^2)<0$ since $x<0$ and $1∕(1-x^2)>0$.

• Case: $y\le 0$. Then we have that $-x>-y\ge 0$ so that $f(-y)<f(-x)$ by the previous case. But we have $f(-x)=-x∕(1-{(-x)}^2)=-x∕(1-x^2)=-f(x)$, and similarly $f(-y)=-f(y)$. Hence $-f(y)=f(-y)<f(-x)=-f(x)$ so that $f(y)>f(x)$.
• Case: $y>0$. Then $f(y)=y∕(1-y^2)>0$ since $y>0$ and $1∕(1-y^2)>0$. Hence $f(x)<0<f(y)$.

Therefore in all cases we have $f(x)<f(y)$, which shows that $f$ preserves order. □

Proof. First we need to show that $g$ is even a function from $ℝ$ to $(-1,1)$. We note that

$$\begin{array}{cc}{y}^2\ge 0& \\ 4y^2\ge 0& \\ 1+4y^2\ge 1>0& \\ \sqrt{1+4y^2}>0& \\ 1+\sqrt{1+4y^2}>1>0& \end{array}$$

so that $g(y)$ is well-defined for all $y\in ℝ$ since the denominator of $g(y)$ is always nonzero. Now, if $y=0$ then clearly $\left|g(y)\right|=\left|g(0)\right|=\left|0\right|=0<1$. So in what follows assume that $y\ne 0$ so that $\left|y\right|>0$ and $y^2>0$. Then we have

$$\begin{array}{cc}1>0& \\ 1+4y^2>4y^2>0& \\ \sqrt{1+4y^2}>\sqrt{4y^2}=2\left|y\right|& \\ 1+\sqrt{1+4y^2}>1+2\left|y\right|>2\left|y\right|>0& \\ \frac{1}{2\left|y\right|}>\frac{1}{1+\sqrt{1+4y^2}}& \\ 1=\frac{2\left|y\right|}{2\left|y\right|}>\frac{2\left|y\right|}{1+\sqrt{1+4y^2}}=\left|\frac{2y}{1+\sqrt{1+4y^2}}\right|=\left|g(y)\right|& \end{array}$$

so that $-1<g(y)<1$ and hence $g(y)\in (-1,1)$. Thus $g$ is in fact a well-defined function from $ℝ$ into $(-1,1)$.

To show that $g$ is a left inverse of $f$ consider any $x\in (-1,1)$. First, we have

$$\begin{array}{llll}\hfill \sqrt{1+4f{(x)}^2}& =\sqrt{1+\frac{4x^2}{{(1-x^2)}^2}}=\sqrt{\frac{{(1-x^2)}^2+4x^2}{{(1-x^2)}^2}}& \hfill & \\ \hfill & =\sqrt{\frac{1-2x^2+x^4+4x^2}{{(1-x^2)}^2}}=\sqrt{\frac{1+2x^2+x^4}{{(1-x^2)}^2}}& \hfill & \\ \hfill & =\sqrt{\frac{{(1+x^2)}^2}{{(1-x^2)}^2}}=\frac{1+x^2}{1-x^2}& \hfill & \end{array}$$

so that

$$\begin{array}{llll}\hfill (g\circ f)(x)& =g(f(x))=\frac{2f(x)}{1+\sqrt{1+4f{(x)}^2}}=\frac{2x∕(1-x^2)}{1+\frac{1+x^2}{1-x^2}}& \hfill & \\ \hfill & =\frac{2x}{(1-x^2)\left(1+\frac{1+x^2}{1-x^2}\right)}=\frac{2x}{1-x^2+1+x^2}& \hfill & \\ \hfill & =\frac{2x}{2}=x,& \hfill & \end{array}$$

which shows that $g\circ f=i_{(-1,1)}$ since $x$ was arbitrary so that $g$ is a left inverse of $f$.

Now consider any $y\in ℝ$. Then we first have

$$\begin{array}{llll}\hfill g{(y)}^2& ={\left(\frac{2y}{1+\sqrt{1+4y^2}}\right)}^2=\frac{4y^2}{{(1+\sqrt{1+4y^2})}^2}& \hfill & \\ \hfill & =\frac{4y^2}{1+2\sqrt{1+4y^2}+1+4y^2}=\frac{4y^2}{2+4y^2+2\sqrt{1+4y^2}}& \hfill & \\ \hfill & =\frac{2y^2}{1+2y^2+\sqrt{1+4y^2}}& \hfill & \end{array}$$

so that

$$\begin{array}{llll}\hfill (f\circ g)(y)& =f(g(y))=\frac{g(y)}{1-g{(y)}^2}=\frac{2y∕(1+\sqrt{1+4y^2})}{1-\frac{2y^2}{1+2y^2+\sqrt{1+4y^2}}}& \hfill & \\ \hfill & =\frac{2y}{(1+\sqrt{1+4y^2})\left(\frac{1+2y^2+\sqrt{1+4y^2}-2y^2}{1+2y^2+\sqrt{1+4y^2}}\right)}& \hfill & \\ \hfill & =\frac{2y}{(1+\sqrt{1+4y^2})\left(\frac{1+\sqrt{1+4y^2}}{1+2y^2+\sqrt{1+4y^2}}\right)}& \hfill & \\ \hfill & =\frac{2y}{\frac{1+2\sqrt{1+4y^2}+1+4y^2}{1+2y^2+\sqrt{1+4y^2}}}=\frac{2y}{\frac{2+4y^2+2\sqrt{1+4y^2}}{1+2y^2+\sqrt{1+4y^2}}}& \hfill & \\ \hfill & =\frac{2y}{2\left(\frac{1+2y^2+\sqrt{1+4y^2}}{1+2y^2+\sqrt{1+4y^2}}\right)}=\frac{2y}{2}=y,& \hfill & \end{array}$$

which shows that $f\circ g=i_{ℝ}$ since $y$ was arbitrary so that $g$ is also a right inverse of $f$. □