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

Solutions for the Exercises at the End of the Sections and the Review Exercises 127
is nonnegative. (Because of the algebraic properties that determine
the sign of a fraction, we need to prove that both the numerator
and the denominator have the same sign, and consider the possibiUty
that the numerator equals 0.) Because c — x:^0, there are two possible
cases:
i. c — X > 0
ii. c - X < 0.
In the first case (namely, c - x > 0), ox. Because the function /
is nondecreasing, /(c) >/(x). Thus, f{c) —f(x) > 0 and c —x > 0.
This implies that the fraction ((/(c) -/(x))/(c - x)) is nonnegative.
In the second case (namely, c - x < 0), c < x. Because the function
/ is nondecreasing, /(c) </(x). Thus, /(c) -/(x) < 0 and c — x < 0.
This impUes that the fraction ((f(c) ~/(x))/(c - x)) is nonnegative.
23. To prove that/is one-to-one, we have to prove that, if xi ^ X2, then
/(Xl) #/(X2).
i. Direct method. Because xi ^ X2 and m^O, it follows that mxi ^
mx2. Thus, mxi -f- ft 7^ mx2 4- b. So/(xi) //(X2).
ii. Contrapositive. We will assume that/(xi) =/(x2). This assumption
allows us to set up an equaUty and use it as a starting point. Because
/(xi) =/(x2), it follows that mxi -j-b = mxi + b. This equality implies
that mx\ =mx2. Because m^^O, we can divide both sides of the
equahty by m, and we obtain xi = X2. Thus, we proved that if
/(xi) =f(x2\ then xi = X2. This is equivalent to proving that, if
xi ^ X2, then/(xi) 7^/(x2). So the function/is one-to-one.
24. We have to prove that for every real number y there is at least one
real number x such that/(^(x)) = y. By hypothesis/is onto. Therefore,
there exists at least one real number z such thsit f{z) = y. The function
g is onto. Therefore, there exists at least one real number t such that
g{t) — z. Therefore,/(^(0) =f(z) = y, and this proves that the function
f o g is onto.
"IF AND ONLY IF" OR EQUIVALENCE THEOREMS
1. Part 1. Assume that/is a nonincreasing function. We want to show
that ((/(x) -/(c))/(x - c) < 0) for all c and x in the domain of/with
c 7^ X. Because c / x, there are two possibiUties: either x < c or x > c.
If X < c, because/is nonincreasing, it follows that/(x) >/(c). These
two inequaUties can be rewritten as:
If X - c < 0, then/(x) -/(c) > 0; therefore, ^^^^ ~^^^^ < 0.
X — c
If X > c, because/ is nonincreasing, it follows that/(x) <f(c).