Find every bijection of non-zero real numbers whose inverse is its reciprocal

Question

Find all functions $f $: $\mathbb R^{×}$ $\to$ $\mathbb R^{×}$ that are one-to-one and onto and such that $f^{−1}(x)= 1/ f(x)$, Where $\mathbb R^{×}=\mathbb R-${0}

My approach:=

At first I have consider $f(x)=x$ and $f(x)=1/x$ then this does not satisfy $f^{−1}(x)= 1/ f(x)$

Now I am unable to construct such functions.

Answer 1

Partial answer that gives a very large family of solutions:

For any subset $A \subset ]0, +\infty[$, let's denote $A^{-1} = \left\lbrace \dfrac{1}{x} \mid x \in A \right\rbrace$.

Let $A$ be any subset of $]0, +\infty[$ satisfying the two following conditions : $$A \cap A^{-1} = \emptyset \quad \text{and} \quad A \cup A^{-1} = ]0, +\infty[ \setminus \lbrace 1 \rbrace$$

Let $\sigma : A \rightarrow A^{-1}$ be any bijection between $A$ and $A^{-1}$, and let $g : ]0, +\infty[ \rightarrow ]0, +\infty[$ be the bijection defined by

$$\forall x > 0, \quad g(x)= \left\lbrace \begin{array}{cl} \sigma(x) & \text{if } x \in A \\ \dfrac{1}{\sigma(1/x)} &\text{if } x \in A^{-1} \\ 1 & \text{if } x=1\end{array} \right.$$

Then the function $$f : x \longmapsto \left\lbrace \begin{array}{cc} -g(x) & \text{if } x>0 \\ \dfrac{1}{g^{-1}(-x)}& \text{if } x<0\end{array}\right.$$

satisfies the condition.

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For example, if $A = ]1, +\infty[$, and $\sigma : x \mapsto \dfrac{1}{x}$, one obtains the function $$f : x \longmapsto \left\lbrace \begin{array}{cc} -\dfrac{1}{x} & \text{if } x>0 \\ -x& \text{if } x<0\end{array}\right.$$

which is a solution.

Answer 2

First, if $f^{-1}(x) = 1/f(x)$, then we have $$x = f^{-1}(f(x)) = 1/f(f(x)),$$ which is equivalent to requiring $f(f(x)) = 1/x$. This means that every value of $x$ is periodic, belonging to a cycle of the form $$a \mapsto b \mapsto 1/a \mapsto 1/b \mapsto a.$$

Now if $a = \pm 1$, then $a = 1/a$ so we must have $1/b = f(1/a) = f(a) = b$. So such an $f$ will satisfy $f(1) = \pm 1$ and $f(-1) = \mp 1$. It then remains to describe $f(x)$ for $x \in \mathbb{R} \setminus \{0, 1, -1\}$.

Since $x \neq 1/x$ for $x \neq \pm 1$, every element of $\mathbb{R} \setminus \{0, 1, -1\}$ occurs in a 4-cycle $\{a, b, 1/a, 1/b\}$. So let $A$ and $B$ be any uncountable subsets of $\mathbb{R} \setminus \{0, 1, -1\}$ such that the four sets $A$, $B$, $A^{-1}$, and $B^{-1}$ partition $\mathbb{R}\setminus\{0, 1, -1\}$, where $$A^{-1} = \{1/x\ |\ x \in A\},$$ and similarly for $B^{-1}$. Let $g\colon A \to B$ be any bijection. Having fixed $f(1) = c = \pm 1$, define $f \colon \mathbb{R}^\times \to \mathbb{R}^\times$ by $$ f(x) = \begin{cases} c & \text{if $x = 1$} \\ -c & \text{if $x = -1$} \\ g(x) & \text{if $x \in A$} \\ 1/g^{-1}(x) & \text{if $x \in B$} \\ 1/g(1/x) & \text{if $x \in A^{-1}$} \\ g^{-1}(1/x) & \text{if $x \in B^{-1}$} \\ \end{cases} $$

An an example, you could take $A = (1, \infty)$, $B = (-\infty, -1)$, and $g(x) = -x$.

Note that each such tuple $(c, A, B, g)$ will give a suitable function $f$, and each such function $f$ arises from a tuple of this form (by choosing for each 4-cycle a "first" element $a$, and declaring $A$ to be the set of these elements), but multiple tuples may correspond to the same function.

Answer 3

If $$f(x) = y$$ then, $$f^{-1}(y) = x = 1/f(y) \implies f(f(x)) = 1/x$$

$$f(f(x)) = 1/x \implies f(f(f(x))) = 1/f(x) = f(1/x) $$

Hence $$f(x) f(1/x) = 1 = f(x) f^{-1}(x)$$ Hence $$f^{-1}(x) = f(1/x)$$

$$f(f^{-1}(x)) = x \implies f(f(1/x)) = x$$

So functions such that $f(f(1/x)) = x$ and $f(x)f(1/x) = 1$ is possible.

For $x>0$, let $$x = e^y \implies f(e^y) f(e^{-y}) = 1 \implies g(y)g(-y) = 1$$ where $g(y) = f(e^y)$.

So any solution is of the form: $$g(y)g(-y) = 1$$.

An example: $$g(y) = e^{-h(y)}$$ where $h(y)$ is any odd function. So now, $$f(e^y) = e^{-h(y)} \implies f(x) = e^{-h(\ln(x))}.$$

Define: $$For \ x>0, f(x) = -e^{h(\ln(|x|))}$$. $$For \ x<0, f(x) = e^{-h(\ln(|x|))}$$.

Now check $$f(f(x)) = 1/x \implies f(f(x)) = e^{-h(\ln(|e^{h(\ln(x))}|))} = e^{-h(h(\ln(x)))} = 1/x$$ Hence $$h(h(\ln(x))) = \ln(x)$$

So any odd function of the form $h(h(y)) = y$ will work in:

$$For \ x>0, f(x) = -e^{h(\ln(|x|))}$$. $$For \ x<0, f(x) = e^{-h(\ln(|x|))}$$.

The following are atleast possible: $h(y) = y$, $h(y) = -y$ ,$h(y) = 1/y$,$h(y) = -1/y$.