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56 The Nuts and Bolts of Proof, Third Edition
Another Correct Proof
1. Case 1. The statement is true for k=l, because 7^ - 4^ = 3.
2. Case 2. Let fc > 1. Then it is possible to write k = n-\-1, with n> 1.
Using the factorization technique for differences of two powers, we can
use the following equality:
jn+l _ 4^+1 ^ (7 _ ^^^jn ^ jn-1 ^ 4 + . . . + 7 x 4"'^ + 4")
= 3(7" + 7"-^ X 4 + ... + 7 X 4"-^ + 4").
Because the number in parentheses is an integer, we proved that
7^ — 4^ = 3s for an integer number s. Thus, the number 7^ — 4^ is
divisible by 3. •
From a technical point of view, one could argue that the factorization
formula used to factor 7" - 4" must be proved using mathematical induction,
so we have a completely self-contained proof that does not invoke
factorization techniques estabHshed in some other context. We trust the
reader to be familiar with them.
Statements involving inequalities are, in general, more difficult to prove
than those involving equalities. Indeed, when working with inequalities,
there is a certain degree of freedom because the relation between the
mathematical expressions involved is not quite as unique as the one
determined by equality. A mathematical expression can only be equal to
some "variation" of itself, so all we can do is try to rewrite it in seemingly
different ways using the rules of algebra. But, a mathematical expression can
be larger (or smaller) than several other expressions, and we need to choose
the one that is useful for completing our task.
EXAMPLE 6.
For all integer numbers a>4:
a^ > 3a^ + 3(3 + 1
Proof: We will use the principle of mathematical induction.
1. Base case. The statement is true for the smallest number we can
consider, namely 4, because 4^>3x4^ + 3x4+l, since 64 > 61.
2. Inductive hypothesis. Assume the statement is true for an arbitrary
number n; that is,
n^ > 3n^ + 3n+l.
3. Deductive proof. Is the statement true for n + 1? Is
(n+l)^>3(n+l)^ + 3(n+l) + l?