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

128 The Nuts and Bolts of Proof, Third Edition
These two inequalities can be rewritten as:
If X - c> 0, then/(x) -/(c) < 0; therefore, ^^^^ 'J^^"^ < 0.
Part 2. The hypothesis is that ((/(x) -f(c))/(x - c) < 0 for all c
and X in the domain of/with c^/^x. Does this inequality imply that/
is a nonincreasing function? A quotient between two real number is
nonpositive when one of the two numbers is negative and the other
is positive (or the dividend is zero). Suppose that the denominator is
positive. Then the numerator must be either negative or zero. This means
that if X - c> 0, then f(x) -/(c) < 0. Therefore, if x > c, it follows
that/(x) <f{c). Suppose that the denominator is negative. Then the
numerator is either positive or zero. This means that if x - c < 0, then
/(x) -/(c) > 0. Therefore, if x < c, it follows that/(x) >/(c).
So, in either case we can conclude that/is a nonincreasing function.
2. Part 1. Let a and b be two odd numbers. Then a = 2/c+l and
b = 2t-\-U with k and t integers, and ah = {2k + l)(2t + 1) = 4kt + 2/c 4-
2t-hl = 2{2k + fc +1) + 1. Because the number 2kt + /c +1 is an integer,
then ah is an odd number.
Part 2. We want to prove that, if ab is an odd number, then a and b
are both odd. We will use the contrapositive of this statement. Assume
that either a or b is an even number. Will this imply that the product
ab is an even number? Without loss of generality, let us assume that a
is even. Then a = 2/c, where k is an integer. Thus, ab = 2{kb), where kb is
an integer. So ab is even. Because the contrapositive of the original
statement is true, the statement itself is true as well.
3. Part 1. Assume that n is divisible by 3. Thus, n = 2>k, with k integer.
Therefore, n^ = 9k^ = 3(3/c^). Because 3/c^ is an integer, we conclude
that n^ is a multiple of 3.
Part 2. We will prove this part using the contrapositive. Assume
that n is not divisible by 3. Then n = 3t + m when m=l or m = 2.
Therefore, n^ = (3t + mf = 9t^ + 6tm + m^ = 3(3t^ + 2tm) + ml The
number 3t^ + 2tm is an integer. The number m^ is equal to either 1 or
to 4, and it is not divisible by 3. Thus, n^ is not a multiple of 3. Because
the contrapositive of the original statement is true, the statement itself
is true as well.
4. We will prove that (i) implies (ii), (ii) is equivalent to (iii), and (iii)
implies (i). (This is only one of several possible ways of constructing
this proof.)
(i) implies (ii): We have to prove that a^<a^ for all real numbers
a > 1, or, equivalently, that a^ — a^<0 for all real numbers a > 1.