Concrete mathematics : a foundation for computer science

Fdrey ‘nough.
4.5 RELATIVE PRIMALITY 119
pruning off unwanted branches.) It follows that m’n - mn’ = 1 whenever
m/n and m//n’ are consecutive elements of a Farey series.
This method of construction reveals that 3~ can be obtained in a simple
way from 3~~1: We simply insert the fraction (m + m’)/N between consecutive
fractions m/n, m//n’ of 3~~1 whose denominators sum to N. For
example, it’s easy to obtain 37 from the elements of 36, by inserting f , 5,
. . . , f according to the stated rule:
3, = 0 111 I 112 I 14 3 1s 3 4 5 6 1
1'7'6'5'4'7'3'5'7'2'7'5'3'7'4'5'6'7'1'
When N is prime, N - 1 new fractions will appear; but otherwise we’ll have
fewer than N - 1, because this process generates only numerators that are
relatively prime to N.
Long ago in (4.5) we proved-in different words-that whenever m I n
and 0 < m 6 n we can find integers a and b such that
ma-nb = 1. (4.32)
(Actually we said m’m + n’n = gcd( m, n), but we can write 1 for gcd( m, n),
a for m’, and b for -n’.) The Farey series gives us another proof of (4.32),
because we can let b/a be the fraction that precedes m/n in 3,,. Thus (4.5)
is just (4.31) again. For example, one solution to 3a - 7b = 1 is a = 5, b = 2,
since i precedes 3 in 37. This construction implies that we can always find a
solution to (4.32) with 0 6 b < a < n, if 0 < m < n. Similarly, if 0 6 n < m
and m I n, we can solve (4.32) with 0 < a 6 b 6 m by letting a/b be the
fraction that follows n/m in 3m.
Sequences of three consecutive terms in a Farey series have an amazing
property that is proved in exercise 61. But we had better not discuss the
Farey series any further, because the entire Stern-Brocot tree turns out to be
even more interesting.
We can, in fact, regard the Stern-Brocot tree as a number system for
representing rational numbers, because each positive, reduced fraction occurs
exactly once. Let’s use the letters L and R to stand for going down to the
left or right branch as we proceed from the root of the tree to a particular
fraction; then a string of L’s and R’s uniquely identifies a place in the tree.
For example, LRRL means that we go left from f down to i, then right to 5,
then right to i, then left to $. We can consider LRRL to be a representation
of $. Every positive fraction gets represented in this way as a unique string
of L’s and R’s.
Well, actually there’s a slight problem: The fraction f corresponds to
the empty string, and we need a notation for that. Let’s agree to call it I,
because that looks something like 1 and it stands for “identity!’