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16 The Nuts and Bolts of Proof, Third Edition
Because the statement to be proved is already written in the form "If A,
then B" with the hypothesis and conclusion already separated, we will
proceed with the proof.
Proof: There are two possibihties: Either a is a multiple of b or a is not a
multiple of b. We will consider them separately.
Case 1.
r = 0.
If a is a multiple of b, then the statement is proved as a = qh, and
Case 2. We will assume that a is not a multiple of b. This means that if
we consider all the multiples of b, none of them will be equal to a. The
multiples of b are numbers of the form:
b, lb, 3b, 4b, 5b, , nb, (n + l)b,
This collection is infinitely large, and the values of the numbers get larger
and larger. They divide the number fine in separate consecutive intervals of
size b. As a is a finite number, and these intervals cover the whole positive
number line, then a will be in one of the intervals determined by these
multiples of b.
I — \ — \ — \ — \ — I — I — ^
0 b 2b 3b qb {c|+^)b
Thus,
qb < a <{q+
for some positive integer q. To show that this number satisfies the conclusion
to be reached, we need to show that we can find the other number r. If we
subtract qb from these inequahties, we obtain:
\)b
0 < a- qb < b. (*)
If we now set r = a-qb, we can show that this number satisfies the
conditions hsted in the conclusion.
By the previous inequalities (*), 0 <r <b. By its definition, a = qb-{-r.
Because the two cases presented cover all the possibilities, we proved that
the statement is true. •
The statement in Example 4 is part of the theorem known as the Division
Algorithm. Later on we will prove that the numbers q and r we just found
are the only ones satisfying the required properties. (See the exercises at the
end of the section on Uniqueness Theorems.) To be accurate, Example 4
could have been included in the section on Existence Theorems, because it
states that there exist two numbers having certain properties. Moreover, its