epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

166 The Nuts and Bolts of Proof, Third Edition
In order for 1/2" to be smaller than s, we must have l/6:<2"
or n>(lnl/s)/ln2. To be sure that Ar>0, choose iV = max
{l,(lnlA)/ln2}.
28. Because a, b, and c are three consecutive integers, without loss of
generahty we can assume that a is the smallest of them and write
b = a-\-l Siud c = a-\-2. Then a-\-b-\-c = a + {a+l)-\-{a-{-2) = 3a-\-3
= 3{a + 1) == 3b. Because b is an integer number, the equahty proves
that a-\-b-\-c is divisible by 3. Note: We cannot use proof by
induction for this statement because the three numbers could be
negative. Therefore, there is no smallest number for which to check
that the statement is true. The statement "Let a, b, and c be three
consecutive positive integer numbers; then 3 divides the sum
a + b + c'' could be proved by induction. Try this method, and see
what happens. (Does this result relate in any way to finding the
average of three consecutive integer numbers?)
29. The proof is constructed by induction.
a. We need to check whether the statement is true for k — ^. Because
/c^ — /c = 0 — 0 = 0, and 0 is divisible by 3, the statement is indeed
true.
b. Let us assume that the statement is true for a generic number n > 1;
that is, n^ — n = 3p for some integer number p.
c. We now need to prove that (n + 1)^ — (n-\-1) = 3t for some integer
number t. Performing some algebraic steps we obtain:
(n + 1)^ -(n+l) = n^ -\-3n^-\-3n-n
= (n^ - n) + 3(n^ + n).
The number M^ + n is an integer. Call it q. Then, using the inductive
hypothesis yields:
(n + 1)^ -{n-\-l) = 3p-\-3q = 3(p + q).
Because the number p-\-q is an integer, we have proved that the
statement is true. Therefore, by the principle of mathematical
induction the statement is true for all whole numbers. {Note: There
is another way of proving this statement without using mathematical
induction. Indeed, n^ — n = n(n^ - 1) — n{n — l)(n + 1). The
three numbers n, n-\-l, and n-1 are consecutive. So, to complete
the proof, we could prove that one of them is divisible by 3. In
Exercise 28 we proved that the sum of three consecutive integers is