Concrete mathematics : a foundation for computer science

Skepticism is
healthy only to
a limited extent.
Being skeptical
about proofs and
programs (particularly
your own) will
probably keep your
grades healthy and
your job fairly secure.
But applying
that much skepticism
will probably
also keep you shut
away working all
the time, instead
of letting you get
out for exercise and
relaxation.
Too much skepticism
is an open invitation
to the state
of rigor mortis,
where you become
so worried about
being correct and
rigorous that you
never get anything
finished. -A skeptic
(This observation
was made by R. J.
McEliece when he
was an undergrad.)
3.2 FLOOR/CEILING APPLICATIONS 71
Incidentally, when we’re faced with a “prove or disprove,” we’re usually
better off trying first to disprove with a counterexample, for two reasons:
A disproof is potentially easier (we need just one counterexample); and nitpicking
arouses our creative juices. Even if the given assertion is true, our
search for a counterexample often leads us to a proof, as soon as we see why
a counterexample is impossible. Besides, it’s healthy to be skeptical.
If we try to prove that [m] = L&J with the help of calculus, we might
start by decomposing x into its integer and fractional parts [xJ + {x} = n + 0
and then expanding the square root using the binomial theorem: (n+(3)‘/’ =
n’/2 + n-‘/2(j/2 _ &/2@/g + . . . . But this approach gets pretty messy.
It’s much easier to use the tools we’ve developed. Here’s a possible strategy:
Somehow strip off the outer floor and square root of [ml, then remove
the inner floor, then add back the outer stuff to get Lfi]. OK. We let
m=llmj and invoke (3.5(a)), giving m 6 m < m + 1. That removes
the outer floor bracket without losing any information. Squaring, since all
three expressions are nonnegative, we have m2 6 Lx] < (m + 1)‘. That gets
rid of the square root. Next we remove the floor, using (3.7(d)) for the left
inequality and (3.7(a)) for the right: m2 6 x < (m + 1)2. It’s now a simple
matter to retrace our steps, taking square roots to get m 6 fi < m + 1 and
invoking (3.5(a)) to get m = [J;;]. Thus \m] = m = l&J; the assertion
is true. Similarly, we can prove that
[ml=
[J;;] ,
real x 3 0.
The proof we just found doesn’t rely heavily on the properties of square
roots. A closer look shows that we can generalize the ideas and prove much
more: Let f(x) be any continuous, monotonically increasing function with the
property that
f(x) = integer ===3 x = integer.
(The symbol ‘==+I means “implies!‘) Then we have
lf(x)J = lf(lxJ 11 and If(x)1 = Tf(Txl)l, (3.10)
whenever f(x), f(lxJ), and f( [xl) are defined. Let’s prove this general property
for ceilings, since we did floors earlier and since the proof for floors is
almost the same. If x = [xl, there’s nothing to prove. Otherwise x < [xl,
and f(x) < f ( [xl ) since f is increasing. Hence [f (x)1 6 [f ( [xl )I, since 11 is
nondecreasing. If [f(x)] < [f( [xl)], there must be a number y such that
x 6~ < [xl and f(y) = Tf(x)l, since f is continuous. This y is an integer, because
of f's special property. But there cannot be an integer strictly between
x and [xl. This contradiction implies that we must have [f (x)1 = If ( [xl )I.