[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 $T(x) \in (0,1)\setminus\mathbb{Q}$.
[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]
[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]
