### stigler’s law of eponymy

6.1.5

Let a, b, c, & d

be integers satisfying

b > 0, d > 0, and

ad – bc = 1.

Let n = max{b,d}.

Then a/b and c/d are consecutive

fractions in the n^th Farey Sequence.

Proof:

Since ad-bc=1, we have

ad/(bd) = (1+bc)/(bd), hence

a/b = 1/(bd) + c/d and so

a/b > c/d.

Note that (cf. Corollary 6.2)

ad-bc = 1 also implies

(a,b) = 1 and (c,d) = 1;

our fractions are in

“lowest terms” (and hence

appear in F_n).

Now suppose c/d & a/b are *not* adjacent.

Then there is at least one fraction,

which we may assume is in “lowest terms”,

“p/q”, say, with (p,q) = 1, such that

p/q is *between* c/d and a/b

in the Farey sequence F_n:

F_n = { …, c/d, …, p/q, …, a/b, …}.

Then

q = (ad-bc)q

= adq ___________ – bqc

= adq – bdp + bdp – bcq

= d(aq-bp) + b(dp-cq)

(*).

But a/b > p/q implies

aq – pb > 0, and so

(since the variables are integers)

we have that

aq – bp is greater-or-equal to 1

(aq – bp \ge 1).

The same computation with different

letters shows that

[p/q > c/d] \implies [dp – cq \ge 1].

These inequalities together with (*) give

q = d(aq-bp) + b(dp-cq)

q \ge d + b.

Recall that n = max{b,d}.

It follows that d + b > n.

This gives us q > n, a contradiction

(“p/q” cannot be in F_n with q > n).

Hence a/b and c/d *are* adjacent in F_n; done.

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