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 159
Part 1: Statement 1 implies statement 2. Because degree P{x) > degree
(x —a)=l, the polynomial P{x) can be divided by the polynomial
x — a. Therefore, P{x) = (x — a)q(x)-\~ r (see Exercise 15); however,
P{a) = 0. So 0 = P{a) = {a — a)q(a) -i-r = r. Thus, the remainder of
the division is 0 and P{x) = {x — a)q{x). This means that the
polynomial P{x) is divisible by the monomial x — a.
Part 2: Statement 2 implies statement 3. By hypothesis, the remainder
of the division of P{x) by the polynomial x - a is zero. So, P{x) =
(x - a)q(x). By definition, this means that x - a is a factor of P(x).
Part 3: Statement 3 implies statement 1. Because x — a is a
factor of P(x), we can write P(x) = {x - a)q{x). Therefore,
P{a) = {a- a)q{a) = 0. This proves that the number a is a root of
the polynomial P{x),
17. By hypothesis, the number:
l.mM-^=/Xa) (*)
x-^a X — a
exists and is finite. We want to show that limx_>a/(x) =f{a)
equivalently, by the properties of limits, that
or,
limr/(x)-/(a)l=0
x-^a L
J
To reconstruct the fraction in (*) and thus be able to use the
hypothesis, divide and multiply the expression in the brackets by
x — a. Observe that it is algebraically correct to do so because x^^a;
therefore, x — a^O. In this way, we obtain:
lim f/(x) -/(a)l = lim^S^^LJm^^ _ a)
x-^a L J x^a X — a
= lim -^^^—^^-^ lim(x - a).
x->a X — a x-^a
Therefore,
lim f/(x) -/(a)l =f\a) lim(x - a)
x-^a L J x-^a
=fXa) X 0 = 0.