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 129
Because r > s, we can write a^ — d — a^(\ — d~^^ — a^(\ — a% where
t = r — s is a positive number. Let us consider the two factors of this
product: a^ and 1 - aK The first factor, a\ is positive because a> I, The
second factor, 1 — a\ is negative for the same reason. Therefore,
a%l — a^)<0, which is the conclusion we wanted to obtain.
(ii) implies (Hi): Let a be a real number such that a < 1. The
hypothesis we are using is about numbers larger than L So, we need to
relate a to a number larger than L The inverse of any number smaller
than 1 is a number larger than L If we consider b = a~^, then b > 1.
Moreover a = {a~^)~ = b~^ when b> 1. Then, by hypothesis, b^<b\
Therefore, (a~^/<(a~^/; that is, -^ <-T' This implies that a^>a\
(Hi) implies (ii): Any number larger than 1 is the inverse of a number
smaller than L Thus, a= («~^)~ =b~^ when b < I. Then, by
hypothesis, b' < b\ Therefore, {a'^Y < (a"^)'; that is, -^ < -^. This
implies that a^<a\
(Hi) implies (i): By hypothesis d < a^ for all real numbers a < 1; or
fl'" — a^<0 for all real numbers a < 1. We can rewrite this difference
as a^ — a^ = a\l — a^~^). This product is negative and its first factor
is positive. Therefore, the other factor, (1 — a^'^), must be negative.
So l<a^"^ Because a < 1, this will happen only if s —r<0.
Thus, s <r.
5. (i) implies (ii): Already proved; see Example 3 in this section.
(ii) implies (Hi): From the inequahty (a + b)/2 > a, we obtain
a 4- fo > 2a. This implies b > a. Because b > a, it follows that
2b> a + b. Thus, fc > (a -h b)/2.
(Hi) implies (i): Already proved; see Example 3 in this section.
6. (i) implies (ii): The numbers x and y are both negative. Thus, by
definition of absolute value, |x| = -x and \y\ = —y. The inequality
X < y implies —x > -y. So |x| > \y\.
(ii) implies (i): Because x and y are both negative numbers, \x\ = —x
and \y\ = —y. The inequality |x| > \y\ implies —x > —y. Therefore,
X < y.
(iH) implies (i): By hypothesis x^>y^. So x^ — y^>0, or
(x - y)(x + y) > 0. The number x-i-y is negative because x and y are
both negative numbers. The product {x — y){x + y) can be positive
only if the number x - y is negative as well. Then, x-y <0,OT x < y,
(i) implies (iH): Because x < y,it follows that x — y<0. Because we
want to obtain information about x^ — y^, we can use factorization
techniques to write x^ - y^ = (x- y)(x + y). The first factor is negative
by hypothesis. The second factor is negative because it is the sum of
two negative numbers. Therefore, x^ — y^>0, or x-^>y^.