Concrete mathematics : a foundation for computer science

. . . without MS
of generality. .
“If x be an incommensurable
number less than
unity, one of the
series of quantities
m/x, m/(1 -x),
where m is a whole
number, can be
found which shall
he between any
given consecutive
integers, and but
one such quantity
can be found.”
- Rayleigh [245]
Right, because
exact/y one of
the counts must
increase when n
increases by 1 .
3.2 FLOOR/CEILING APPLICATIONS 77
Our last application in this section looks at so-called spectra. We define
the spectrum of a real number a to be an infinite multiset of integers,
Sped4 = 114, 12a1, 13a1, . . .I.
(A multiset is like a set but it can have repeated elements.) For example, the
spectrum of l/2 starts out (0, 1, 1,2,2,3,3,. . .}.
It’s easy to prove that no two spectra are equal-that a # (3 implies
Spec(a) # Spec((3). For, assuming without loss of generality that a < (3,
there’s a positive integer m such that m( l3 - a) 3 1. (In fact, any m 3
[l/( (3 - a)] will do; but we needn’t show off our knowledge of floors and
ceilings all the time.) Hence ml3 - ma 3 1, and LrnSl > [ma]. Thus
Spec((3) has fewer than m elements < lrnaj, while Spec(a) has at least m.
Spectra have many beautiful properties. For example, consider the two
multisets
Spec(&) = {1,2,4,5,7,8,9,11,12,14,15,16,18,19,21,22,24 ,... },
Spec(2+fi) = {3,6,10,13,17,20,23,27,30,34,37,40,44,47,51,... }.
It’s easy to calculate Spec( fi ) with a pocket calculator, and the nth element
of Spec(2+ fi) is just 2n more than the nth element of Spec(fi), by (3.6).
A closer look shows that these two spectra are also related in a much more
surprising way: It seems that any number missing from one is in the other,
but that no number is in both! And it’s true: The positive integers are the
disjoint union of Spec( fi ) and Spec(2+ fi ). We say that these spectra form
a partition of the positive integers.
To prove this assertion, we will count how many of the elements of
Spec(&!) are 6 n, and how many of the elements of Spec(2+fi) are 6 n. If
the total is n, for each n, these two spectra do indeed partition the integers.
Let a be positive. The number of elements in Spec(a) that are < n is
N(a,n) = x[lkaJ O
= x[[kaj O
= tr kaO
= x[O