Prove $f(x)=x/\sqrt{x^2+1}$ for $f:\mathbb{R}\to(-1,1)$ is a bijection.

Question

I am a beginner. As per Professor's instructions, every step must be written explicitly. If you spot a mistake be sure to note it.

Question: Prove $f(x)=x/\sqrt{x^2+1}$, is a bijection onto $(-1,1)$.

Attempt: Suppose $f(x_1)=f(x_2)$, then $x_1/\sqrt{x_1^2+1}=x_2/\sqrt{x_2^2+1}$. Therefore, $x_1\sqrt{x_2^2+1}=x_2\sqrt{x_1^2+1}$. Thus, $x_1=x_2$ or $\sqrt{x_2^2+1}=\sqrt{x_1^2+1}$ which means $x_1=x_2$ and $x_2^2+1=x_1^2+1$, hence $x_1=x_2$ or $x_2^2=x_1^2$. Therefore, $x_1=x_2$ or $(x_1\neq x_2 \; \text{and} \; x_1=x_2)$. Thus ($x_1=x_2 \; \text{or} \; x_1\neq x_2$) and ($x_1= x_2$ or $x_1=x_2$). Therefore $x_1=x_2$ which means $f$ is one-to-one.

Now suppose $y\in(-1,1)$, then $\left[y=x/\sqrt{x^2+1}\right] \Rightarrow \left[y\sqrt{x^2+1}=x\right]\Rightarrow \left[y^2(x^2+1)=x^2 \right]\Rightarrow \left[x^2 y^2+y^2=x^2 \right]\Rightarrow \left[y^2=x^2-x^2y^2 \right]\Rightarrow \left[y^2=x^2(1-y^2)\right]\Rightarrow \left[y^2/(1-y^2)=x^2\right] \Rightarrow \left[\pm\sqrt{y^2/(1-y^2)}=x\right]$

This implies $\left[y^2,1-y^2>0\right] \lor \left[y^2,1-y^2<0 \right]$. Hence [$y^2>0$ and $y^2<1$] or [$y^2<0$ and $y^2>1$]. Hence [$y=0$ and $-1<y<1$] or [$y=0$ and $y>1$ and $y<-1$]. Hence $-1<y<1$. Therefore the $\text{Rng}(f)=(-1,1)$. Hence $f$ is onto $(-1,1)$.

Hence since $f$ is one-to-one and onto $(-1,1)$, f is a bijection onto $(-1,1)$.

Answer 1

I should point out that my first pass at this problem had an algebraic flaw that caused me to believe that the domain of $f$ had to be $(-1,1).$ It was only after seeing zwim's answer, and re-checking my math that I saw and corrected my error.

---

First see comments following the question.

This is how I would approach it:

Let $\displaystyle f(x) = x/\sqrt{x^2 + 1}.$

Suppose $\displaystyle f(x_1) = f(x_2)$, with $x_1 \neq 0 \neq x_2$.

Then $\displaystyle x_1/\sqrt{x_1^2 + 1} = x_2/\sqrt{x_2^2 + 1}.$

This implies that $\displaystyle x_1\sqrt{x_2^2 + 1} = x_2\sqrt{x_1^2 + 1}.$

Squaring both sides implies that $\displaystyle (x_1)^2(x_2^2 + 1) = (x_2)^2(x_1^2 + 1).$

This implies that $\displaystyle (x_1)^2 = (x_2)^2.$

Therefore, either $x_1 = x_2$ or $x_1 = -x_2.$

Suppose $x_1 = -x_2$.
This would imply that $f(x_1) = - f(x_2)$ which violates the assumption that $f(x_1) = f(x_2).$

Therefore, $f(x_1) = f(x_2)$ does imply that $x_1 = x_2,$
when $x_1 \neq 0 \neq x_2.$

Suppose that $x_1 = 0$ and $f(x_1) = f(x_2)$.

Then $f(x_2) = 0 \implies x_2 = 0.$

Therefore, $f(x)$ is 1-1.

---

To prove onto (i.e. a surjection), given -1 < y < 1, you want to find $x$ such that $f(x) = y$.

This means that you are looking for $x$ such that

$\displaystyle x/\sqrt{x^2 + 1} = y \implies$

$x^2/(x^2 + 1) = y^2 \implies$

$x^2 = y^2(x^2 + 1) \implies$

$(x^2)(1 - y^2) = y^2 \implies$

$x^2 = \frac{y^2}{1 - y^2}.$

So, a reasonable candidate value for $x$, in order to satisfy $f(x) = y$ would be

$\displaystyle x = \frac{y}{\sqrt{1 - y^2}}$.

In order to verify that the above specification works, you have to consider

$\displaystyle f(x) = \frac{\frac{y}{\sqrt{1 - y^2}}}{\sqrt{1 + \frac{y^2}{1 - y^2}}} = \frac{\frac{y}{\sqrt{1 - y^2}}}{\sqrt{\frac{1}{1-y^2}}} = y.$

Therefore, the candidate specification for $x$ above, causes $f(x) = y.$

Therefore $f(x)$ is onto.

---

Finally, after examining zwim's answer, I verified that for any $x \in \Bbb{R}$, $f(x) \in (-1,1).$

---

Therefore, $f(x)$ is both onto and 1-1, so $f(x)$ is a bijection.

Answer 2

Let $f(x)=\dfrac x{\sqrt{x^2+1}}$ then $f'(x)=\dfrac 1{\sqrt{x^2+1}}-\dfrac{x^2}{(x^2+1)^\frac 32}=\dfrac{1}{(x^2+1)^\frac 32}>0$

Since $f$ is strictly monotonic and continuous (denominator always defined) it is a bijection from its domain of definition, namely $\mathbb R$ to its image. Since it is $\nearrow$ its image is given by its limits in $\pm\infty$ which are $\pm 1$.

Therefore it is a bijection $\mathbb R\mapsto(-1,1)$

As a side note, notice that $f(\sinh(x))=\tanh(x)$ and since both $\sinh$ and $\tanh$ are bijections, $f$ must be too.

Answer 3

Solve the equation

$$y=\frac x{\sqrt{x^2+1}}$$ for $x$. We notice that $x$ and $y$ have the same sign. Then by squaring

$$y^2(x^2+1)=x^2,$$

$$x=\pm\frac{y}{\sqrt{1-y^2}},$$ and we always choose the $+$ sign. Hence $y(x)$ is a function from $\mathbb R$ to $(-1,1)$ and conversely, $x(y)$ is a function from $(-1,1)$ to $\mathbb R$.