Exercise 6.1.3 (Second property of Farey fractions)

Question

Let $a/b,\ a'/b',\ a''/b''$ be any three consecutive fractions in the Farey sequence of order $n$. Prove that $a'/b' = (a+a'')/(b+b'')$.

Richard Ganaye · Accepted Answer

\begin{proof} 
By Theorem 6.1,
\begin{align*}
\left\{
\begin{array}{rrl}
b\,  a' &-\,a \, b' &= 1,\
-b''a '&+\, a'' b'&= 1.
\end{array}
ight.
\end{align*}
If we solve this system for $a'$ and $b'$, we obtain
\begin{align*}
\left\{
\begin{array}{cc}
(a'' b - a b'') a' &= a + a'',\
(a'' b - a b'') b' &= b + b'.
\end{array}
ight.
\end{align*}
Since $b+b'>0$, $a'' b - a b'' 
e 0$, so
$$\frac{a'}{b'} = \frac{a+a''}{b+b''}.$$
\end{proof}
(See Hardy \& Wright,  ``The Theory of Numbers'', Ch. 3.)

Exercise 6.1.3 (Second property of Farey fractions)

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Exercise 6.1.3 (Second property of Farey fractions)

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