Farey Series

If \( h/k \) and \( h'/k' \) are two successive terms of \( \mathfrak{F}_n \), then

The ’mediant’

\[ \frac{h+h'}{k+k'} \]

of \( h/k \) and \( h'/k' \) falls in the interval

\[ \left( \frac{h}{k},\; \frac{h'}{k'} \right). \]

Hence, unless (3.1.4) is true, there is another term of \( \mathfrak{F}_n \) between \( h/k \) and \( h'/k' \).

If \( n > 1 \), then no two successive terms of \( \mathfrak{F}_n \) have the same denominator.

If \( k > 1 \) and \( h'/k \) succeeds \( h/k \) in \( \mathfrak{F}_n \), then \( h+1 \le h' < k \). But then

\[ \frac{h}{k} < \frac{h}{k-1} < \frac{h+1}{k} \le \frac{h'}{k}; \]

and \( h/(k-1) \) comes between \( h/k \) and \( h'/k \) in \( \mathfrak{F}_n \), a contradiction.

The aim is now to prove that the middle fraction is exactly the mediant:

\[ \frac{h''}{k''} = \frac{h+h'}{k+k'}. \]

Since we already have

\[ h'' = sh + rh', \qquad k'' = sk + rk', \]

it is enough to show that

\[ r=s=1. \]

Then (3.3.2) immediately becomes

\[ h'' = h+h', \qquad k'' = k+k'. \]

Consider the set \(S\) of all fractions of the form

This is a natural family to consider because \( h''/k'' \) already has exactly this form: by (3.3.2),

\[ \frac{h''}{k''} = \frac{sh+rh'}{sk+rk'}, \]

so one can take

\[ \mu=s,\qquad \lambda=r. \]

Thus \( h''/k'' \in S \).

Let

\[ \frac{H}{K} = \frac{\mu h + \lambda h'}{\mu k + \lambda k'} \in S. \]

To compare it with \( h/k \), compute

\[ kH - hK = k(\mu h + \lambda h') - h(\mu k + \lambda k') = \lambda(kh' - hk'). \]

Since \( h/k \) and \( h'/k' \) are consecutive in \( \mathfrak{F}_{n-1} \), the induction hypothesis gives

\[ kh' - hk' = 1. \]

Hence

\[ kH - hK = \lambda > 0, \]

so

\[ \frac{h}{k} < \frac{H}{K}. \]

Similarly,

\[ h'K - k'H = h'(\mu k + \lambda k') - k'(\mu h + \lambda h') = \mu(kh' - hk') = \mu > 0, \]

and therefore

\[ \frac{H}{K} < \frac{h'}{k'}. \]

So every fraction in \( S \) satisfies

\[ \frac{h}{k} < \frac{H}{K} < \frac{h'}{k'}. \]

Suppose \( d \) divides both \( H \) and \( K \). Then \( d \) divides every integer linear combination of \( H \) and \( K \). In particular,

\[ kH - hK = \lambda, \qquad h'K - k'H = \mu. \]

So \( d \mid \lambda \) and \( d \mid \mu \). But \( (\lambda,\mu)=1 \), hence \( d=1 \). Therefore

\[ (H,K)=1. \]

So every fraction of \( S \) is irreducible.

A reduced fraction \( H/K \) belongs to \( \mathfrak{F}_q \) as soon as \( q \ge K \). Therefore every element of \( S \) eventually appears in some Farey series.

Its first appearance is in \( \mathfrak{F}_K \), so among all fractions in \( S \), the one that appears earliest is the one with the smallest denominator \( K \).

For fractions in \( S \),

\[ K = \mu k + \lambda k'. \]

Since \( \lambda,\mu \) are positive integers, the smallest possible value of \( K \) is obtained when both are as small as possible, namely

\[ \lambda=1,\qquad \mu=1. \]

Thus the earliest possible fraction from \( S \) is

\[ \frac{h+h'}{k+k'}. \]

The fraction \( h''/k'' \) lies in \( S \), and it is the fraction that first appears between the previously consecutive terms \( h/k \) and \( h'/k' \) when passing from \( \mathfrak{F}_{n-1} \) to \( \mathfrak{F}_n \).

But among all fractions in \( S \), the earliest one is uniquely determined by the choice \( \lambda=\mu=1 \). Hence the middle fraction must be exactly that one:

Therefore

\[ h'' = h+h', \qquad k'' = k+k'. \]

This proves Theorem 29 in this situation.

The proof uses induction on \( n \). In the induction step, it focuses on a special configuration: a fraction that makes its first appearance in \( \mathfrak{F}_n \), i.e. a fraction that belongs to \( \mathfrak{F}_n \) but not to \( \mathfrak{F}_{n-1} \).

Because of this special setup, the proof obtains a stronger conclusion than is needed for Theorem 29. Namely, in this first-appearance case one gets

\[ h'' = h+h', \qquad k'' = k+k'. \]

This is stronger than merely saying

\[ \frac{h''}{k''} = \frac{h+h'}{k+k'}. \]

Indeed, the equality of numerators and denominators term-by-term implies the equality of the two fractions, but the converse is not generally true, since the fraction \( (h+h')/(k+k') \) may need to be reduced.

So the logical structure is:

\[ \text{first-appearance case} \;\Longrightarrow\; h'' = h+h',\ k'' = k+k' \;\Longrightarrow\; \frac{h''}{k''} = \frac{h+h'}{k+k'}. \]

However, the stronger statement

\[ h'' = h+h', \qquad k'' = k+k' \]

should not be regarded as a general property of every three successive terms of \( \mathfrak{F}_n \). It is only valid in the special inductive situation where the central fraction is appearing for the first time in \( \mathfrak{F}_n \).

What is generally true, and what Theorem 29 actually asserts, is only the weaker statement

\[ \frac{h''}{k''} = \frac{h+h'}{k+k'}. \]

Thus the proof establishes the theorem by proving a stronger fact in a special case, but that stronger fact is not itself the full general theorem.