We shall use the notation $\equiv$ for equivalence, as in $\S 1.6$.
(i) Reflexivity. For all $a, b \in \mathbf{Z}, a b=b a$ and so $\frac{a}{b} \equiv \frac{a}{b}$.
(ii) Symmetry. Suppose $\frac{a}{b} \equiv \frac{c}{d}$. Then $a d=b c$, and so, by commutativity for the integers, $c b=d a$, showing that $\frac{c}{d} \equiv \frac{a}{b}$.
(iii) Transitivity. Suppose $\frac{a}{b} \equiv \frac{c}{d}$ and $\frac{c}{d} \equiv \frac{e}{f}$. Then $a d=b c$ and $c f=$ de. Multiplying both left-hand sides and both right-hand sides together gives adcf $=b c d e$. This factorizes to give $(a f-b e) c d=0$. Now $b, d, f$ are non-zero, being denominators. Since $a d=b c$ and $c f=$ de we conclude that if $c=0$ then $a=0$ and $e=0$, giving $a f=b e$. If $c \neq 0$ then since $d \neq 0,(a f-b e) c d=0$ implies that $a f-b e=0$, i.e. $a f=b e$. This shows that $\frac{a}{b} \equiv \frac{e}{f^{\prime}}$, establishing transitivity.
4CCM115A COURSE NOTES ：
(i) From the calculation of the first four convergents earlier in this section the result can be verified for small values of $n$. (The algebra is fairly involved for $n=3$ and $n=4$.) For $n=1$ the conventions introduced in Proposition 3 tell us that
$$
p_{1} q_{0}-q_{1} p_{0}=a_{1} \times 0-1 \times 1=-1=(-1)^{1},
$$
so the equation is satisfied for $n=1$.
Suppose that the equation is valid for $n-1$. Using the recurrence relations of proposition 3 gives
$$
\begin{aligned}
p_{n} q_{n-1}-q_{n} p_{n-1} &=\left(a_{n} p_{n-1}+p_{n-2}\right) q_{n-1}-\left(a_{n} q_{n-1}+q_{n-2}\right) p_{n-1} \
&=-\left(p_{n-1} q_{n-2}-q_{n-1} p_{n-2}\right)=(-1)^{n}
\end{aligned}
$$