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158 The Nuts and Bolts of Proof, Third Edition
13. (Because it impossible to check directly all negative numbers, we have
to find a different way to prove that the statement is true. Try using
the contrapositive.) Assume that there exists a negative number z
whose reciprocal, z~\ is not negative. By definition of reciprocal of a
number, z x z~^ = 1. By the rules of algebra, z~^ ^0. Therefore, z~^
must be a positive number; however, the product of a negative and
a positive number is a negative number. This conclusion generates a
contradiction because 1 is positive.
14. Let £>0 be given. Is it possible to find an N>0 such that |a„ -3\<£
for all n>Nl Observe that:
\an-3\ = 3n + 2 -3
n
3n + 2--3n
To have ^ < ^ one must have f < n. Therefore, let N = ^.
15. The conclusion has two parts:
a. The remainder is a number.
b. The remainder is the number P{a).
Because we want to evaluate the remainder of the division between
P{x) and X — a, we need to start from the division algorithm. If we are
performing long division, we have the following diagram:
x-a
q(x)
\P(x).
7(x)
The polynomial q{x) represents the quotient, and the polynomial r(x)
the remainder of the division. Then we can write P(x) = (x — a)q(x)-\-
r{x). The degree of the remainder must be smaller than the degree
of the divisor, x — a; otherwise, the division is not complete. Because
the degree of x - a is 1, the degree of r{x) must be 0. Thus, r{x) is a
number, and we can write r{x) — r. Therefore, P(x) = (x- a)q(x) + r.
This equahty is true for all values of the variable x; in particular, we
can evaluate it for x = a, and we obtain P(a) = (a- a)q{a) -\-r = r.
The proof is now complete.
16. There are several ways of proving that these statements are
equivalent. We will show that statement 1 implies statement 2,
statement 2 imphes statement 3, and statement 3 implies statement 1.