[proofplan] The interval $I_k$ is exactly the set of points $x \in (0,1)$ for which $k < 1/x < k+1$, so the integer part of $1/x$ is $k$. This identifies the first partial quotient and gives the formula $T(x)=1/x-k$. The inverse branch is obtained by solving $y=1/x-k$ for $x$, and rationality is checked directly from the formula $x=1/(k+y)$. [/proofplan]

[step:Identify the integer part of $1/x$ on $I_k$]Fix an integer $k \geq 1$ and let $x \in I_k \setminus \mathbb{Q}$. Let $\lfloor \cdot \rfloor: \mathbb{R} \to \mathbb{Z}$ denote the floor function, so $\lfloor r \rfloor$ is the greatest integer less than or equal to $r$ for each $r \in \mathbb{R}$. Since $x>0$, the inequalities defining $I_k$ may be inverted with the order reversed. The defining inequalities are \begin{align*} \frac{1}{k+1} < x < \frac{1}{k}. \end{align*} After taking reciprocals, this is equivalent to \begin{align*} k < \frac{1}{x} < k+1. \end{align*} Because $k$ is an integer, this gives \begin{align*} \left\lfloor \frac{1}{x} \right\rfloor = k. \end{align*} Thus, by the definition of the [Gauss map](/page/Gauss%20Map), \begin{align*} T(x) = \frac{1}{x} - k. \end{align*} The same equality shows $0<T(x)<1$. If $T(x)$ were rational, then $1/x=T(x)+k$ would be rational, and hence $x$ would be rational, contradicting $x \notin \mathbb{Q}$. Therefore $T(x) \in (0,1)\setminus\mathbb{Q}$.[/step]

[guided]Fix an integer $k \geq 1$ and let $x \in I_k \setminus \mathbb{Q}$. Let $\lfloor \cdot \rfloor: \mathbb{R} \to \mathbb{Z}$ denote the floor function, so $\lfloor r \rfloor$ is the greatest integer less than or equal to $r$ for each $r \in \mathbb{R}$. The interval $I_k$ was chosen so that taking reciprocals isolates one integer. Since every number involved is positive, taking reciprocals reverses the inequalities. First we have \begin{align*} \frac{1}{k+1} < x < \frac{1}{k}. \end{align*} Taking reciprocals gives the equivalent inequality \begin{align*} k < \frac{1}{x} < k+1. \end{align*} The number $1/x$ therefore lies strictly between the consecutive integers $k$ and $k+1$. By the defining property of the floor function, the greatest integer not exceeding $1/x$ is exactly $k$: \begin{align*} \left\lfloor \frac{1}{x} \right\rfloor = k. \end{align*} Substituting this into the definition of the [Gauss map](/page/Gauss%20Map) gives \begin{align*} T(x) = \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor. \end{align*} Using $\lfloor 1/x \rfloor=k$, this becomes \begin{align*} T(x) = \frac{1}{x} - k. \end{align*} The inequality $k < 1/x < k+1$ also gives \begin{align*} 0 < \frac{1}{x} - k < 1, \end{align*} so $T(x) \in (0,1)$. It remains to check that the image point is irrational. If $T(x)$ were rational, then $1/x=T(x)+k$ would be rational because $k \in \mathbb{Z} \subset \mathbb{Q}$. Taking reciprocals of the nonzero rational number $1/x$ would imply $x \in \mathbb{Q}$, contradicting the hypothesis $x \notin \mathbb{Q}$. Hence $T(x) \in (0,1)\setminus\mathbb{Q}$.[/guided]

[step:Recover the first partial quotient] Let $x=[0;a_1,a_2,\dots]$ be the [continued fraction expansion](/page/Continued%20Fraction) of $x$. For an irrational $x \in (0,1)$, the first partial quotient is defined by $a_1=\lfloor 1/x\rfloor$. The previous step gives $\lfloor 1/x \rfloor = k$, hence \begin{align*} a_1 = k. \end{align*} [/step]

[step:Show that the inverse formula maps irrational points into the correct interval] Define the inverse-branch candidate $\nu_k: (0,1)\setminus \mathbb{Q} \to (0,1)$ by \begin{align*} \nu_k(y)=\frac{1}{k+y}. \end{align*} Let $y \in (0,1)\setminus \mathbb{Q}$ and define $x \in (0,1)$ by \begin{align*} x = \nu_k(y). \end{align*} Since $0<y<1$, we have \begin{align*} k < k+y < k+1. \end{align*} Taking reciprocals gives \begin{align*} \frac{1}{k+1} < \frac{1}{k+y} < \frac{1}{k}, \end{align*} so $x \in I_k$. It remains to check irrationality. If $x$ were rational, then $1/x$ would be rational, and therefore \begin{align*} y = \frac{1}{x} - k \end{align*} would be rational, contradicting $y \notin \mathbb{Q}$. Hence $x \notin \mathbb{Q}$, and so $\nu_k(y) \in I_k \setminus \mathbb{Q}$. [/step]

[step:Verify that $T$ and $\nu_k$ are inverse maps] Let $y \in (0,1)\setminus \mathbb{Q}$ and define $x \in (0,1)$ by \begin{align*} x = \nu_k(y). \end{align*} By the definition of $\nu_k$, this means \begin{align*} x = \frac{1}{k+y}. \end{align*} From the previous step, $x \in I_k\setminus\mathbb{Q}$, so $\lfloor 1/x \rfloor=k$. Therefore \begin{align*} T(\nu_k(y))=T\left(\frac{1}{k+y}\right). \end{align*} Since $1/\nu_k(y)=k+y$, the definition of $T$ on $I_k\setminus\mathbb{Q}$ gives \begin{align*} T(\nu_k(y))=(k+y)-k=y. \end{align*} Let $\operatorname{id}_{(0,1)\setminus\mathbb{Q}}$ denote the identity map on $(0,1)\setminus\mathbb{Q}$. Thus $T \circ \nu_k = \operatorname{id}_{(0,1)\setminus\mathbb{Q}}$. Conversely, let $x \in I_k\setminus\mathbb{Q}$. By the first step, $T(x)=1/x-k$. Hence \begin{align*} \nu_k(T(x))=\frac{1}{k+T(x)}. \end{align*} Substituting $T(x)=1/x-k$ gives \begin{align*} \nu_k(T(x))=\frac{1}{k+\frac{1}{x}-k}=x. \end{align*} Let $\operatorname{id}_{I_k\setminus\mathbb{Q}}$ denote the identity map on $I_k\setminus\mathbb{Q}$. Thus $\nu_k \circ T|_{I_k\setminus\mathbb{Q}}=\operatorname{id}_{I_k\setminus\mathbb{Q}}$. Therefore $T|_{I_k\setminus\mathbb{Q}}$ is a bijection from $I_k\setminus\mathbb{Q}$ onto $(0,1)\setminus\mathbb{Q}$, with inverse branch $\nu_k$. [/step]