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

118 The Nuts and Bolts of Proofs, Third Edition
Because 0 is an even number, the statement is true for the smallest number
included in the statement.
Step 2: Assume that k^-k is even for an integer k.
Step 3: Let's prove that the statement is true for /c+ 1.
(fe + 1)2 - (fc -h 1) = fe^ + 2/c + 1 - fc - 1 = (fc^ - fe) + 2k.
Because k^ — k is even and so is 2/c, we can conclude that the statement is
true for /c+ 1. •
Alleged Proof 2.
We will use proof by cases.
Case 1. Let n be an even number. Then n = 2k for some integer number k.
Thus,
n^-n = {2kf - 2/c
= 4fc2 - 2k
= 2(2fc2 - fe).
Because the number 2k^ — /c is an integer, n^ — n is even.
Case 2. Let n be an odd number. Then /t = 2/c + 1 for some integer number
k. Thus,
n2-n = (2/c+l)2-(2/c+l)
= 4/c2 + 4/c+l-2/c-l
= 2{2k^ + k).
Because the number 2k^ + /c is an integer, n^ ~ n is even.
Alleged Proof 3.
We can write:
•
n^ — n = n(n — 1).
The product of two consecutive numbers is always even, so n^ — n is
even. •
Alleged Proof 4. Assume there is a positive integer s such that s^ — s
is odd. Then we can write:
s^ -s = 2k-\-l