(1)

\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\cdots,a_{n}\right] & =a_{0}+\frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}+\frac{b_{2}}{a_{3}+\cdots}}}\\ & =a_{0}+\frac{1}{\frac{a_{1}}{b_{0}}+\frac{\frac{b_{1}}{b_{0}}}{a_{2}+\frac{b_{2}}{a_{3}+\cdots}}}\\ & =\left[a_{0};\left(\frac{1}{b_{0}}a_{1},\frac{b_{1}}{b_{0}}\right),\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]\right]\\ & =\left[a_{0};\left(\frac{1}{b_{0}}a_{1},\frac{b_{1}}{b_{0}}\right),\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right] \end{align*} となるのでこれを繰り返すと
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left(a_{2},b_{2}\right),\left(a_{3},b_{3}\right),\cdots,a_{n}\right] & =\left[a_{0};\left(\frac{1}{b_{0}}a_{1},\frac{b_{1}}{b_{0}}\right),\left(a_{2},b_{2}\right),\left(a_{3},b_{3}\right),\cdots,a_{n}\right]\\ & =\left[a_{0};\frac{1}{b_{0}}a_{1},\left(\frac{b_{0}}{b_{1}}a_{2},\frac{b_{2}b_{0}}{b_{1}}\right),\left(a_{3},b_{3}\right),\cdots,a_{n}\right]\\ & =\left[a_{0};\frac{1}{b_{0}}a_{1},\frac{b_{0}}{b_{1}}a_{2},\left(\frac{b_{1}}{b_{2}b_{0}}a_{3},\frac{b_{1}}{b_{2}b_{0}}b_{3}\right),\cdots,a_{n}\right]\\ & =\left[a_{0};\frac{1}{b_{0}}a_{1},\frac{b_{0}}{b_{1}}a_{2},\frac{b_{1}}{b_{2}b_{0}}a_{3},\cdots,\frac{b_{n-2}b_{n-4}\cdots b_{\mod\left(n,2\right)}}{b_{n-1}b_{n-3}\cdots b_{\mod\left(n+1,2\right)}}a_{n}\right] \end{align*} となるので与式は成り立つ。

-

\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のとき、
左辺は
\[ \left[\left(a_{0},r\right);\left(a_{1},r\right),\left(a_{2},r\right),\left(a_{3},r\right),\cdots,a_{n}\right]=\left[a_{0};a_{1},a_{2},a_{3},\cdots,a_{n}\right]_{r} \] 右辺は
\begin{align*} \left[a_{0};\frac{1}{r}a_{1},\frac{r}{r}a_{2},\frac{r}{r^{2}}a_{3},\cdots,\frac{r^{\frac{n-2-\mod\left(n,2\right)}{2}+1}}{r^{\frac{n-1-\mod\left(n+1,2\right)}{2}+1}}a_{n}\right] & =\left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},\cdots,\frac{r^{\frac{n-2-\mod\left(n,2\right)}{2}+1}}{r^{\frac{n-1-\mod\left(n+1,2\right)}{2}+1}}a_{n}\right]\\ & =\begin{cases} \left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},\cdots,\frac{r^{\frac{n-2}{2}+1}}{r^{\frac{n-1-1}{2}+1}}a_{n}\right] & n\in2\mathbb{N}_{0}\\ \left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},\cdots,\frac{r^{\frac{n-2-1}{2}+1}}{r^{\frac{n-1}{2}+1}}a_{n}\right] & n\in2\mathbb{N}_{0}+1 \end{cases}\\ & =\begin{cases} \left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},\cdots,a_{n}\right] & n\in2\mathbb{N}_{0}\\ \left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},\cdots,\frac{1}{r}a_{n}\right] & n\in2\mathbb{N}_{0}+1 \end{cases}\\ & =\left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},a_{4},\frac{1}{r}a_{5},a_{6},\cdots,r^{-\mod\left(n,2\right)}a_{n}\right] \end{align*} となるので、
\[ \left[a_{0};a_{1},a_{2},a_{3},\cdots,a_{n}\right]_{r}=\left[a_{0};\frac{1}{r}a_{1},a_{2},\frac{1}{r}a_{3},a_{4},\frac{1}{r}a_{5},a_{6},\cdots,r^{-\mod\left(n,2\right)}a_{n}\right] \] となり与式は成り立つ。

(2)

\(m=0\)のとき、明らかに成り立つ。
\(m=1\)のとき、
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,a_{n}\right] & =\left[a_{0}+\frac{b_{0}}{a_{1}+\frac{b_{1}}{a_{2}+\frac{b_{2}}{a_{3}+\cdots}}}\right]\\ & =\left[a_{0}+\frac{b_{0}}{\left[\left(a_{1},b_{1}\right);\left(a_{2},b_{2}\right),\cdots,a_{n}\right]}\right]\\ & =\left[\left(a_{0},b_{0}\right);\left[\left(a_{1},b_{1}\right);\left(a_{2},b_{2}\right),\cdots,a_{n}\right]\right] \end{align*} となるので成り立つ。
\(m=k\) のとき成り立つとすると
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,a_{n}\right] & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[\left(a_{k},b_{k}\right),\cdots,a_{n}\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[\left(a_{k},b_{k}\right);\left[\left(a_{k+1},b_{k+1}\right);\left(a_{k+2},b_{k+2}\right),\cdots,a_{n}\right]\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(a_{k},b_{k}\right),\left[\left(a_{k+1},b_{k+1}\right);\left(a_{k+2},b_{k+2}\right),\cdots,a_{n}\right]\right] \end{align*} となるので\(m=k+1\)でも成り立つ。
従って数学的帰納法より任意の\(m\in\mathbb{N}_{0}\)に対して題意が成り立つ。

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\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のとき、
左辺は、
\[ \left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,a_{n}\right]=\left[a_{0};a_{1},\cdots,a_{n}\right]_{r} \] となる。
右辺は、
\begin{align*} \left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,\left[\left(a_{m},r\right);\left(a_{m+1},r\right),\cdots,a_{n}\right]\right] & =\left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,\left[a_{m};a_{m+1},\cdots,a_{n}\right]_{r}\right]\\ & =\left[a_{0};a_{1},\cdots,\left[a_{m};a_{m+1},\cdots,a_{n}\right]_{r}\right]_{r} \end{align*} となる。
これより、
\[ \left[a_{0};a_{1},\cdots,a_{n}\right]_{r}=\left[a_{0};a_{1},\cdots,\left[a_{m};a_{m+1},\cdots,a_{n}\right]_{r}\right]_{r} \] となるので与式は成り立つ。

(3)

(2)より、
\begin{align*} c\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,a_{n}\right] & =c\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]\right]\\ & =c\left(a_{0}+\frac{b_{0}}{a_{1}+\frac{b_{1}}{\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]}}\right)\\ & =ca_{0}+\frac{cb_{0}}{a_{1}+\frac{b_{1}}{\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]}}\\ & =ca_{0}+\frac{b_{0}}{\frac{1}{c}a_{1}+\frac{b_{1}}{c\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]}}\\ & =\left[\left(ca_{0},b_{0}\right);\left(\frac{1}{c}a_{1},b_{1}\right),c\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]\right] \end{align*} となるのでこれを繰り返すと、
\begin{align*} c\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,a_{n}\right] & =\left[\left(ca_{0},b_{0}\right);\left(\frac{1}{c}a_{1},b_{1}\right),c\left[\left(a_{2},b_{2}\right);\left(a_{3},b_{3}\right),\cdots,a_{n}\right]\right]\\ & =\begin{cases} \left[\left(ca_{0},b_{0}\right);\left(\frac{1}{c}a_{1},b_{1}\right),\cdots,\left(ca_{2m},b_{2m}\right),\left(\frac{1}{c}a_{2m+1},b_{2m+1}\right),\cdots,ca_{n}\right] & n\in2\mathbb{N}_{0}\\ \left[\left(ca_{0},b_{0}\right);\left(\frac{1}{c}a_{1},b_{1}\right),\cdots,\left(ca_{2m},b_{2m}\right),\left(\frac{1}{c}a_{2m+1},b_{2m+1}\right),\cdots,\frac{1}{c}a_{n}\right] & n\in2\mathbb{N}_{0}+1 \end{cases}\\ & =\left[\left(ca_{0},b_{0}\right);\left(\frac{1}{c}a_{1},b_{1}\right),\cdots,\left(ca_{2m},b_{2m}\right),\left(\frac{1}{c}a_{2m+1},b_{2m+1}\right),\cdots,c^{1-2\mod\left(2,n\right)}a_{n}\right] \end{align*} となるので与式は成り立つ。

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\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のとき、
左辺は、
\[ c\left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,a_{n}\right]=c\left[a_{0};a_{1},\cdots,a_{n}\right]_{r} \] となり、
右辺は、
\[ \left[\left(ca_{0},r\right);\left(\frac{1}{c}a_{1},r\right),\cdots,\left(ca_{2m},r\right),\left(\frac{1}{c}a_{2m+1},r\right),\cdots,c^{1-2\mod\left(2,n\right)}a_{n}\right]=\left[ca_{0};\frac{1}{c}a_{1},\cdots,a_{2m},\frac{1}{c}a_{2m+1},\cdots,c^{1-2\mod\left(2,n\right)}a_{n}\right]_{r} \] となる。
これより、
\[ c\left[a_{0};a_{1},\cdots,a_{n}\right]_{r}=\left[ca_{0};\frac{1}{c}a_{1},\cdots,a_{2m},\frac{1}{c}a_{2m+1},\cdots,c^{1-2\mod\left(2,n\right)}a_{n}\right]_{r} \] となるので与式は成り立つ。

(4)

(2)より、
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,a_{n}+c\right] & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[a_{n}+c\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[a_{n}+\frac{b_{n}}{\frac{b_{n}}{c}}\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[\left(a_{n},b_{n}\right);\frac{b_{n}}{c}\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(a_{n},b_{n}\right),\frac{b_{n}}{c}\right] \end{align*} となるので与式は成り立つ。

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\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のとき、
左辺は、
\[ \left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,a_{n}+c\right]=\left[a_{0};a_{1},\cdots,a_{n}+c\right]_{r} \] 右辺は、
\[ \left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,\left(a_{n},r\right),\frac{r}{c}\right]=\left[a_{0};a_{1},\cdots,a_{n},\frac{r}{c}\right]_{r} \] となる。
これより、
\[ \left[a_{0};a_{1},\cdots,a_{n}+c\right]_{r}=\left[a_{0};a_{1},\cdots,a_{n},\frac{r}{c}\right]_{r} \] となるので与式は成り立つ。

(5)

(2)より、
\begin{align*} \left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\frac{n}{m}\right] & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left\lfloor \frac{n}{m}\right\rfloor +\frac{\mod\left(n,m\right)}{m}\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[\left\lfloor \frac{n}{m}\right\rfloor +\frac{\mod\left(n,m\right)}{m}\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left[\left(\left\lfloor \frac{n}{m}\right\rfloor ,b\right);\frac{mb}{^{\mod\left(n,m\right)}}\right]\right]\\ & =\left[\left(a_{0},b_{0}\right);\left(a_{1},b_{1}\right),\cdots,\left(\left\lfloor \frac{n}{m}\right\rfloor ,b\right),\frac{mb}{^{\mod\left(n,m\right)}}\right] \end{align*} となるので与式は成り立つ。

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\(\forall n\in\mathbb{N}_{0},b_{n}=r\)のとき、
左辺は、
\[ \left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,\frac{n}{m}\right]=\left[a_{0};a_{1},\cdots,\frac{n}{m}\right]_{r} \] 右辺は、
\[ \left[\left(a_{0},r\right);\left(a_{1},r\right),\cdots,\left(\left\lfloor \frac{n}{m}\right\rfloor ,r\right),\frac{mr}{^{\mod\left(n,m\right)}}\right]=\left[a_{0};a_{1},\cdots,\left\lfloor \frac{n}{m}\right\rfloor ,\frac{mr}{^{\mod\left(n,m\right)}}\right]_{r} \] となる。
これより、
\[ \left[a_{0};a_{1},\cdots,\frac{n}{m}\right]_{r}=\left[a_{0};a_{1},\cdots,\left\lfloor \frac{n}{m}\right\rfloor ,\frac{mr}{^{\mod\left(n,m\right)}}\right]_{r} \] となるので与式は成り立つ。