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

138 The Nuts and Bolts of Proof, Third Edition
integer, (n + 2)^-1 is divisible by 4. Therefore, the statement is
true for all fc > 1 by the principle of mathematical induction.
There is another proof of this statement that does not use the principle
of mathematical induction. Let n be an odd counting number. Then
n = 2/c + 1, where fe is a counting number. Therefore,
n2-l = (2fc+1)^-1
=:4/c2 + 4/c+l-l=4(/c2 + /c).
Because fc^ + fc is a counting number, n^ — 1 is divisible by 4. One can
observe that n^ — 1 is divisible by 8 because fc-^4- fc = fc(fc 4-1). The
sum of two consecutive numbers is always an even number. Thus,
/c2 + fc = fc(fc + 1) = 25, and n^ - 1 = 8s.
8. a. The statement is true for fc = 1, as 10 — 1 = 9.
b. Assume it is true for a generic number n. Thus, 10" — 1 = 9s, where s
is an integer.
c. Is the statement true for the next number (namely, n +1)? Using
algebra and the inductive hypothesis, we obtain:
10"+^ - 1 z= (9 + 1)10" - 1
= 9 X 10" + (10" - 1)
= 9 X 10" + 9s = 9(10" + s).
Therefore, the conclusion is true for all fc > 1 by the principle of
mathematical induction. It is possible to prove this statement using
factorization techniques for differences of powers because
10^ - 1 = 10^ - 1^
9. a. Check the statement for n = 5. Because 3^ = 243, the next to the last
digit from the right is 4, an even number.
b. Assume the statement is true for a generic number n. Thus,
3" = akttk-i... ai^o, where ai = 0, 2, 4, 6, 8 and ao = 1, 3, 7, 9
(check the information regarding ao)-
c. Is the statement true for the next number (namely, n + 1)? We have:
3"+^ = 3(akak-i... aiao) = btbt-i... fci^o-
If ao = 1 or 3, then bi is the unit digit of 3ai. Because 3ai is an even
number (since ai is even), bi is even.
If ao — 9 or 7, then bi is the unit digit of 3ai + 2, which is an even
number, as ai is even.
Thus, the statement is true for all n>5 by the principle of
mathematical induction.