Proof. This follows immediately from the construction, since $f(e_{\ell -1})\neq f(e_{\ell })$ .◻

Proof. By construction, the interior edges of ${\mathcal{H}}_{i}$ all have the same sign $\operatorname{sgn}(a_{i})$ , and ${\mathcal{H}}_{i}$ has $a_{i}-1$ tiles. This implies that ${\mathcal{H}}_{i}\cong {\mathcal{G}}[a_{i}]$ and that it is a zigzag snake graph.◻

Proof. (a) If $n=1$ , then ${\mathcal{G}}[a_{1}]$ is a zigzag snake graph with $a_{1}-1$ tiles. If $a_{1}=1$ , this graph is a single edge and it has exactly one perfect matching. If $a_{1}>1$ , then there is exactly one perfect matching of ${\mathcal{G}}[a_{1}]$ that contains the edge $e_{0}$ . On the other hand, if a perfect matching does not contain $e_{0}$ , then it must contain the other edge $b_{0}$ in $_{\text{SW}}{\mathcal{G}}[a_{1}]$ and it must be a perfect matching on the remaining part which is a zigzag snake graph with $a_{1}-2$ tiles. By induction this gives a total of $a_{1}$ perfect matchings.

Now suppose that $n>1$ , and let $P$ be a perfect matching of the snake graph ${\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}]$ . We will denote by ${\mathcal{N}}[a_{1},a_{2},\ldots ,a_{n}]$ the numerator of the continued fraction $[a_{1},a_{2},\ldots ,a_{n}]$ . By Lemma 3.2, the tiles $G_{\ell _{1}-1},G_{\ell _{1}}$ and $G_{\ell _{1}+1}$ form a straight subsnake graph of ${\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}]$ . If $P$ does not contain the two boundary edges of $G_{\ell _{1}}$ then the restrictions of $P$ to ${\mathcal{H}}_{1}\cong {\mathcal{G}}[a_{1}]$ and ${\mathcal{G}}\setminus {\mathcal{H}}_{1}\cong {\mathcal{G}}[a_{2},a_{3},\ldots ,a_{n}]$ are perfect matchings of these subgraphs. By induction this gives $a_{1}+{\mathcal{N}}[a_{2},a_{3},\ldots ,a_{n}]$ perfect matchings. On the other hand, if $P$ contains the boundary edges of $G_{\ell _{1}}$ then, since ${\mathcal{H}}_{1}$ and ${\mathcal{H}}_{2}$ are zigzag snake graphs, the restriction of $P$ to ${\mathcal{H}}_{1}$ and ${\mathcal{H}}_{2}$ consists of boundary edges only, and there is a unique such matching on ${\mathcal{H}}_{1}$ and on ${\mathcal{H}}_{2}$ . Moreover, the restriction of $P$ to ${\mathcal{G}}[a_{3},a_{4},\ldots ,a_{n}]$ is a perfect matching of that subgraph. By induction, this gives ${\mathcal{N}}[a_{3},a_{4},\ldots ,a_{n}]$ perfect matchings. Putting these two cases together, we get a total of

(3.3) $$\begin{eqnarray}a_{1}\,{\mathcal{N}}[a_{2},a_{3},\ldots ,a_{n}]+{\mathcal{N}}[a_{3},a_{4},\ldots ,a_{n}]\end{eqnarray}$$

perfect matchings.

For the remainder of this proof, denote by $N$ and $D$ the numerator and the denominator of the continued fraction $[a_{3},\ldots ,a_{n}]$ . Thus $[a_{3},\ldots ,a_{n}]=N/D$ and $N,D$ are relatively prime. Then

$$\begin{eqnarray}[a_{1},a_{2},\ldots ,a_{n}]=a_{1}+\displaystyle \frac{1}{a_{2}+\displaystyle \frac{D}{N}}=a_{1}+\displaystyle \frac{N}{a_{2}N+D}=\displaystyle \frac{a_{1}(a_{2}N+D)+N}{a_{2}N+D}.\end{eqnarray}$$

Note that this fraction is reduced, because $N$ and $D$ are relatively prime. Thus

(3.4) $$\begin{eqnarray}{\mathcal{N}}[a_{1},a_{2},\ldots ,a_{n}]=a_{1}(a_{2}N+D)+N.\end{eqnarray}$$

Furthermore, from the identity

$$\begin{eqnarray}[a_{2},\ldots ,a_{n}]=a_{2}+\frac{D}{N}=\frac{a_{2}N+D}{N}\end{eqnarray}$$

we see that ${\mathcal{N}}[a_{2},\ldots ,a_{n}]=a_{2}N+D$ , and hence (3.4) implies that

$$\begin{eqnarray}{\mathcal{N}}[a_{1},a_{2},\ldots ,a_{n}]=a_{1}{\mathcal{N}}[a_{2},\ldots ,a_{n}]+{\mathcal{N}}[a_{3},\ldots ,a_{n}].\end{eqnarray}$$

This completes the proof of (a).

Statement (b) follows directly from (a) because the denominator of $[a_{1},a_{2},\ldots ,a_{n}]$ is equal to the numerator of $[a_{2},\ldots ,a_{n}]$ , and (c) is obtained by combining (a) and (b).◻

Proof. Note first that all maps are well defined, since the continued fraction $[1]$ is excluded.

To show that the diagram commutes, suppose first that $f(e_{d})=f(e_{d-1})$ . Then $\mathsf{F}({\mathcal{G}},e_{d})=\mathsf{F}^{\prime }({\mathcal{G}})$ . Moreover, if $[a_{1},a_{2},\ldots ,a_{n}]=\mathsf{F}({\mathcal{G}},e_{d})$ then $a_{n}>1$ and thus $g(\mathsf{F}({\mathcal{G}},e_{d}))=\mathsf{F}^{\prime }({\mathcal{G}})$ . Now suppose that $f(e_{d})=-f(e_{d-1})$ , and let $[a_{1},a_{2},\ldots ,a_{n}]=\mathsf{F}({\mathcal{G}},e_{d})$ . Then $a_{n}=1$ and thus $g({\mathcal{F}}({\mathcal{G}},e_{d}))=[a_{1},a_{2},\ldots ,a_{n-1}+1]$ , and this is also equal to $\mathsf{F}^{\prime }({\mathcal{G}})$ . The lower triangle commutes by Theorem 3.4.

Next, we show that $\mathsf{F}$ and $\mathsf{G}$ are bijections. Let ${\mathcal{G}}$ be a snake graph with $d$ tiles, $e_{d}\in {\mathcal{G}}^{\text{NE}}$ , and $f$ a sign function on ${\mathcal{G}}$ . Then $\mathsf{F}({\mathcal{G}},e_{d})$ is the continued fraction $[a_{1},a_{2},\ldots ,a_{n}]$ defined by the sign sequence

$$\begin{eqnarray}(f(e_{0}),f(e_{1}),\ldots ,f(e_{d-1}),f(e_{d})),\end{eqnarray}$$

where $e_{1},\ldots ,e_{d-1}$ are the interior edges of ${\mathcal{G}}$ and $e_{0}$ is the south edge of ${\mathcal{G}}$ . The map $\mathsf{G}$ sends this continued fraction to the pair $({\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}],e_{d}^{\prime })$ , where ${\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}]={\mathcal{G}}$ by construction, and $e_{d}^{\prime }\in {\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}]^{\text{NE}}$ is the unique edge such that $f(e_{d}^{\prime })=f(e_{d})$ ; thus $e_{d}^{\prime }=e_{d}$ . This shows that $\mathsf{G}\circ \mathsf{F}$ is the identity. Conversely, let $[a_{1},a_{2},\ldots ,a_{n}]$ be a positive continued fraction. Then $\mathsf{G}([a_{1},a_{2},\ldots ,a_{n}])=({\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}],e_{d})$ , where $e_{d}$ is determined by the sign sequence (3.2). Applying $\mathsf{F}$ to this pair yields the continued fraction determined by the same sign sequence. Thus $\mathsf{F}\circ \mathsf{G}$ is the identity.

To show that $\mathsf{F}^{\prime }$ is a bijection, let $[a_{1},a_{2},\ldots ,a_{n}]$ be a positive continued fraction with $a_{n}>1$ . Applying the map $\mathsf{G}$ yields a pair $({\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}],e_{d})$ , where $e_{d}\in {\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}]^{\text{NE}}$ is the unique edge whose sign is equal to the sign of the last interior edge $e_{d-1}$ , because $a_{n}>1$ . Thus $f(e_{d})=f(e_{d-1})=\operatorname{sgn}(a_{n})$ . Forgetting the edge $e_{d}$ and applying the map $\mathsf{F}^{\prime }$ will produce the continued fraction given by the sign sequence

$$\begin{eqnarray}(f(e_{0}),f(e_{1}),\ldots ,f(e_{d-1}),f(e_{d-1})),\end{eqnarray}$$

and since $f(e_{d-1})=f(e_{d})$ , this continued fraction is $[a_{1},a_{2},\ldots ,a_{n}]$ . This shows that $\mathsf{F}^{\prime }$ is surjective. Now assume that there are two snake graphs ${\mathcal{G}},{\mathcal{G}}^{\prime }$ , with respective sign functions $f,f^{\prime }$ such that $\mathsf{F}^{\prime }({\mathcal{G}})=\mathsf{F}^{\prime }({\mathcal{G}}^{\prime })=[a_{1},a_{2},\ldots ,a_{n}]$ . Then the sign sequences of ${\mathcal{G}}$ and ${\mathcal{G}}^{\prime }$ have the same length $d+1=\sum a_{j}$ , and either $f(e_{i})=f^{\prime }(e_{i}^{\prime })$ for each $i=1,2,\ldots ,d$ , or $f(e_{i})=-f^{\prime }(e_{i}^{\prime })$ for each $i=1,2,\ldots ,d$ . Since $e_{0}$ (respectively, $e_{0}^{\prime }$ ) is the south edge of ${\mathcal{G}}$ (respectively, ${\mathcal{G}}^{\prime }$ ), this implies that ${\mathcal{G}}={\mathcal{G}}^{\prime }$ . Thus $\mathsf{F}^{\prime }$ is a bijection.

Figure 4.

An example of the action of the maps in Theorem 4.1.

The map $Ev$ is a bijection by Proposition 2.1, and therefore the map $\unicode[STIX]{x1D712}$ is a bijection by commutativity of the diagram.◻

Proof. The snake graph consisting of a single edge has $1=\unicode[STIX]{x1D719}(1)$ perfect matching. For $N\geqslant 2$ , the map $\unicode[STIX]{x1D712}$ of the theorem induces a bijection

$$\begin{eqnarray}\{{\mathcal{G}}\mid m({\mathcal{G}})=N\}\rightarrow \{N/q\mid q\in \{1,2,\ldots ,N-1\},\gcd (N,q)=1\},\end{eqnarray}$$

and the set on the right-hand side has cardinality $\unicode[STIX]{x1D719}(N)$ .◻

Proof. Part (a) follows from a construction called grafting with the single edge $b_{i}$ ; see [Reference Çanakçı and SchifflerCS15, §3.3, case 3]. Part (b) is the resolution of two snake graphs with a crossing overlap ${\mathcal{G}}[a_{i},\ldots ,a_{i+j}]$ ; see [Reference Çanakçı and SchifflerCS13, §2.4]. If $j$ is even, then the overlap is crossing in the pair on the left ${\mathcal{G}}[a_{1},\ldots ,a_{i+j}]{\mathcal{G}}[a_{i},\ldots ,a_{n}]$ , and if $j$ is odd, then the overlap is crossing in the pair on the right ${\mathcal{G}}[a_{1},\ldots ,a_{n}]{\mathcal{G}}[a_{i},\ldots ,a_{i+j}]$ . Part (c) is the resolution of two snake graphs with a crossing overlap ${\mathcal{G}}[a_{i-1},a_{i},\ldots ,a_{i+k},d]={\mathcal{G}}[a_{i-1},b_{j},\ldots ,b_{j+k},d]$ where $d=\min \{a_{i+k+1},b_{j+k+1}\}$ ; see [Reference Çanakçı and SchifflerCS15, §3.1].◻

Proof. (a) The equation in part (a) of Theorem 5.1 is an equation of Laurent polynomials in several variables $x_{i}$ . Setting each of these variables equal to 1, we get the identity

$$\begin{eqnarray}\begin{array}{@{}c@{}}m({\mathcal{G}}[a_{1},\ldots ,a_{n}])=m({\mathcal{G}}[a_{1},\ldots ,a_{i}])\,m({\mathcal{G}}[a_{i+1},\ldots ,a_{n}])+m({\mathcal{G}}[a_{1},\ldots ,a_{i-1}])\,m({\mathcal{G}}[a_{i+2},\ldots ,a_{n}]).\end{array}\end{eqnarray}$$

Now the result follows from Theorem 3.4. Part (b) and (c) are proved in the same way. ◻

Proof. We prove the formula for $i=1$ . For arbitrary $i$ , the proof is the same up to a shift in the subscripts. If $a_{1}=1$ then ${\mathcal{H}}_{1}$ is a single edge $e_{0}$ and thus $x({\mathcal{H}}_{1})=x_{e_{0}}$ . On the other hand, $\ell _{1}=\ell _{0}+1$ and thus the formula in the proposition becomes $x_{0}(x_{e_{0}}/x_{0}x_{1})x_{1}=x_{e_{0}}$ , proving the result for $a_{1}=1$ .

Now suppose $a_{1}>1$ . According to Lemma 3.3, ${\mathcal{H}}_{1}={\mathcal{G}}[a_{1}]$ is a zigzag snake graph with $a_{1}-1$ tiles $G_{1},G_{2},\ldots ,G_{a_{1}-1}$ and $a_{1}$ perfect matchings. An example with $a_{1}=7$ is given in Figure 6. Note that $\ell _{1}=a_{1}$ . Let ${\mathcal{H}}_{1}^{\prime }$ be the subgraph consisting of the tiles $G_{1},G_{2},\ldots ,G_{a_{1}-2}$ . Then ${\mathcal{H}}_{1}^{\prime }$ has $a_{1}-1$ perfect matchings each of which can be completed to a perfect matching of ${\mathcal{H}}_{1}$ by adding the unique edge $e\in {\mathcal{H}}_{1}^{\text{NE}}$ that is not equal to $e_{\ell _{1}-1}$ . Note that the edge $e$ has label $\ell _{1}$ . The only remaining perfect matching $P_{0}$ of ${\mathcal{H}}_{1}$ does not contain the edge $e$ and it does contain the edge $e_{\ell _{1}-1}$ and every other boundary edge of ${\mathcal{H}}_{1}$ . In the example in Figure 6, the matching $P_{0}$ is marked in thick black. Since ${\mathcal{H}}_{1}$ is a complete zigzag snake graph, the weight of $P_{0}$ is $x(P_{0})=x_{0}x_{1}x_{2}\cdots x_{\ell _{1}-2}x_{e_{\ell _{1}-1}}$ , and thus

$$\begin{eqnarray}\frac{x(P_{0})}{x_{1}x_{2}\cdots x_{\ell _{1}-1}}=\frac{x_{0}x_{e_{\ell _{1}-1}}}{x_{\ell _{1}-1}}.\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}x({\mathcal{H}}_{1})=x({\mathcal{H}}_{1}^{\prime })\frac{x_{\ell _{1}}}{x_{\ell _{1}-1}}+\frac{x_{0}x_{e_{\ell _{1}-1}}}{x_{\ell _{1}-1}},\end{eqnarray}$$

and by induction, this is equal to

$$\begin{eqnarray}x_{0}\biggl(\mathop{\sum }_{j=0}^{\ell _{1}-2}\frac{x_{e_{j}}}{x_{j}x_{j+1}}\biggr)x_{\ell _{1}-1}\frac{x_{\ell _{1}}}{x_{\ell _{1}-1}}+\frac{x_{0}x_{e_{\ell _{1}-1}}}{x_{\ell _{1}-1}}=x_{0}\biggl(\mathop{\sum }_{j=0}^{\ell _{1}-1}\frac{x_{e_{j}}}{x_{j}x_{j+1}}\biggr)x_{\ell _{1}}.\Box\end{eqnarray}$$

Proof. If $n=1$ then $\unicode[STIX]{x1D6FE}^{\prime }=b_{1}$ by construction. Therefore $x_{\unicode[STIX]{x1D6FE}}/x_{\unicode[STIX]{x1D6FE}^{\prime }}=x({\mathcal{G}}_{\unicode[STIX]{x1D6FE}})/b_{1}=x({\mathcal{H}}_{1})/b_{1}=L_{1}$ . If $n=2$ , then

$$\begin{eqnarray}[L_{1},L_{2}]=L_{1}+\frac{1}{L_{2}}=\frac{L_{1}L_{2}+1}{L_{2}}=\frac{x({\mathcal{H}}_{1})x({\mathcal{H}}_{2})+b_{0}b_{2}}{b_{1}\,x({\mathcal{H}}_{2})}=\frac{x({\mathcal{G}}[a_{1}])x({\mathcal{G}}[a_{2}])+b_{0}b_{2}}{b_{1}\,x({\mathcal{G}}[a_{2}])},\end{eqnarray}$$

and, using Theorem 5.1(a) with $i=1$ and $n=2$ , this is equal to $x({\mathcal{G}}[a_{1},a_{2}])/x({\mathcal{G}}[a_{2}])=x_{\unicode[STIX]{x1D6FE}}/x_{\unicode[STIX]{x1D6FE}^{\prime }}$ .

Now suppose that $n>2$ . Again using Theorem 5.1(a), we have

$$\begin{eqnarray}b_{1}\,{\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}]={\mathcal{H}}_{1}\,{\mathcal{G}}[a_{2},a_{3},\ldots ,a_{n}]+b_{0}\,{\mathcal{G}}[a_{3},a_{4},\ldots ,a_{n}].\end{eqnarray}$$

Passing to the Laurent polynomials and dividing by $x({\mathcal{G}}[a_{2},a_{3},\ldots ,a_{n}])$ and $b_{1}$ , we get

$$\begin{eqnarray}\frac{x_{\unicode[STIX]{x1D6FE}}}{x_{\unicode[STIX]{x1D6FE}^{\prime }}}=\frac{x({\mathcal{G}}[a_{1},a_{2},\ldots ,a_{n}])}{x({\mathcal{G}}[a_{2},a_{3},\ldots ,a_{n}])}=\frac{x({\mathcal{H}}_{1})}{b_{1}}+\frac{1}{(b_{1}/b_{0})(x({\mathcal{G}}[a_{2},a_{3},\ldots ,a_{n}])/x({\mathcal{G}}[a_{3},a_{4},\ldots ,a_{n}]))}.\end{eqnarray}$$

Using the definition of $L_{1}$ and induction, this is equal to

(6.1) $$\begin{eqnarray}L_{1}+\frac{1}{(b_{1}/b_{0})[L_{2}^{\prime },L_{3}^{\prime },\ldots ,L_{n}^{\prime }]},\end{eqnarray}$$

where $L_{i}^{\prime }=(b_{0}/b_{1})L_{i}$ if $i$ is even, and $L_{i}^{\prime }=(b_{1}/b_{0})L_{i}$ if $i$ is odd. Lemma 2.3 implies that (6.1) is equal to

$$\begin{eqnarray}L_{1}+\frac{1}{[L_{2},L_{3},\ldots ,L_{n}]}\end{eqnarray}$$

which in turn equals $[L_{1},L_{2},\ldots ,L_{n}]$ . Finally, if a Laurent polynomial $f$ is a common divisor of $x_{\unicode[STIX]{x1D6FE}}$ and $x_{\unicode[STIX]{x1D6FE}^{\prime }}$ then the number of terms in $f$ is a common divisor of the number of terms in $x_{\unicode[STIX]{x1D6FE}}$ and $x_{\unicode[STIX]{x1D6FE}^{\prime }}$ , and Theorem 3.4 implies that $f$ is a Laurent monomial. Thus $f$ is a unit in the ring of Laurent polynomials. This completes the proof.◻

Proof. Let $x_{1},x_{2},x_{3}\in \mathbb{R}_{{>}0}$ , and denote by $r_{i}$ the value of $L_{i}$ at the point $(x_{1},x_{2},x_{3})$ . From the formulas above, we see that $r_{i}\in \mathbb{R}_{{>}0}$ . By a classical result of Seidel and Stern (see [Reference PerronPer50, §50, Satz 8]), the continued fraction $[r_{1},r_{2},\ldots ]$ with $r_{i}\in \mathbb{R}_{{>}0}$ converges if and only if the series $\sum r_{i}$ diverges. Thus, using the divergence test, it suffices to show that $\lim _{i\rightarrow \infty }r_{i}\neq 0$ . Suppose to the contrary that the sequence $r_{i}$ converges to zero. Then also the subsequence $r_{2i}$ with even indices converges to zero. Thus equation (7.1) yields

$$\begin{eqnarray}0=\lim _{i\rightarrow \infty }r_{2i}=\lim _{i\rightarrow \infty }\frac{x_{3}}{x_{2}}\biggl(\frac{x_{1}}{x_{2}}\biggr)^{2i-1},\end{eqnarray}$$

which implies that $x_{1}<x_{2}$ . However, the same argument for the odd indices gives

$$\begin{eqnarray}0=\lim _{i\rightarrow \infty }r_{2i+1}=\lim _{i\rightarrow \infty }\frac{x_{3}}{x_{1}}\biggl(\frac{x_{2}}{x_{1}}\biggr)^{2i-1},\end{eqnarray}$$

which implies that $x_{1}>x_{2}$ , a contradiction.◻

Proof. Part (a) has already been proved above.

(b) To study the quotient of the cluster variables $u(i)/u(i-1)$ , we need to consider the snake graph in Figure 9, except that we interchange the labels $e_{0}$ and $b_{0}$ . The continued fraction of this snake graph is $[2,\overline{1}]=(3+\sqrt{5})/2$ . The subgraph ${\mathcal{H}}_{1}$ is equal to the first tile and $x({\mathcal{H}}_{1})=(x_{2}^{2}+x_{3}^{2})/x_{1}=x_{1}^{\prime }$ . Denoting the corresponding Laurent polynomials by $L_{i}^{\prime }$ , we have

$$\begin{eqnarray}L_{1}^{\prime }=\frac{x_{1}^{\prime }}{x_{2}},\quad L_{2}^{\prime }=\frac{e_{2}b_{1}}{b_{0}b_{2}}=\frac{x_{2}}{x_{1}}=L_{1}\frac{x_{2}}{x_{3}},\quad L_{3}^{\prime }=\frac{e_{3}b_{0}b_{2}}{b_{1}^{2}b_{3}}=\frac{x_{1}x_{3}^{2}}{x_{2}^{3}}=L_{2}\frac{x_{3}}{x_{2}},\end{eqnarray}$$

and, for $i\geqslant 2$ ,

$$\begin{eqnarray}L_{i}^{\prime }=\left\{\begin{array}{@{}ll@{}}\displaystyle L_{i-1}\frac{x_{2}}{x_{3}}\quad & \text{if }i\text{ is even,}\\ \displaystyle L_{i-1}\frac{x_{3}}{x_{2}}\quad & \text{if }i\text{ is odd.}\end{array}\right.\end{eqnarray}$$

Let $\unicode[STIX]{x1D6FD}=[L_{1}^{\prime },L_{2}^{\prime },\ldots ]$ and $\unicode[STIX]{x1D6FC}^{\prime }=[L_{2}^{\prime },L_{3}^{\prime },\ldots ]$ . Then, using Lemma 2.3, we have $\unicode[STIX]{x1D6FC}^{\prime }=(x_{2}/x_{3})\unicode[STIX]{x1D6FC}$ and

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}=L_{1}^{\prime }+\frac{1}{\unicode[STIX]{x1D6FC}^{\prime }}=\frac{x_{1}^{\prime }+x_{3}/\unicode[STIX]{x1D6FC}}{x_{2}}=\frac{x_{1}^{\prime }+x_{1}+\sqrt{(x_{1}^{\prime }-x_{1})^{2}+4x_{3}^{2}}}{2x_{2}},\end{eqnarray}$$

where the last identity holds because $1/\unicode[STIX]{x1D6FC}=(-x_{1}^{\prime }+x_{1}+\sqrt{(x_{1}^{\prime }-x_{1})^{2}+4x_{3}^{2}})/2x_{3}$ . This shows the first identity in (b) and the second is a direct computation.

(c) If $z>0$ , then $[\overline{z}]>0$ and

$$\begin{eqnarray}[\overline{z}]=z+\frac{1}{[\overline{z}]}.\end{eqnarray}$$

Therefore

$$\begin{eqnarray}[\overline{z}]^{2}-z[\overline{z}]-1=0,\end{eqnarray}$$

and from the quadratic formula we get

$$\begin{eqnarray}[\overline{z}]=\frac{z+\sqrt{z^{2}+4}}{2}=\unicode[STIX]{x1D6FC},\end{eqnarray}$$

where the sign in front of the radical is + because $[\overline{z}]>0$ . On the other hand, if $z<0$ , then the computation above yields

$$\begin{eqnarray}[\overline{-z}]=\frac{-z+\sqrt{z^{2}+4}}{2},\end{eqnarray}$$

and its multiplicative inverse is

$$\begin{eqnarray}[0,\overline{-z}]=\frac{z+\sqrt{z^{2}+4}}{2}=\unicode[STIX]{x1D6FC}.\end{eqnarray}$$

This shows (c), and (d) is obtained by replacing $x_{1}^{\prime }$ by $(x_{2}^{2}+x_{3}^{2})/x_{1}$ .◻

Proof. This follows directly from part (d) of Theorem 7.3. ◻

Proof. This follows from Theorem 7.3 by specializing $x_{3}=1$ .◻