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 155
formula to calculate the cubic root of a sum. Therefore, the fact that n^
is an odd number does not seem to give us an effective starting point
for the proof, and we could try to use the contrapositive of
the statement to be proved.) Prove the statement: "If n is not an odd
number, then n^ is not an odd number." Because n is not an
odd number, it must be even. Therefore, we can write n = 2p for
some integer number p. Then n^ = (2p)^ = 8p^ = 2(4p^). Because 4p^ is
an integer number, the equahty above proves that n^ is an even
number; that is, n^ is not an odd number. Because the contrapositive
of the original statement is true, the original statement is true.
Part 2. If n is an odd number, then n^ is an odd number. Because n is
an odd number, we can write n = 2q + lfor some integer q. Therefore,
n^ =: 8^3 + 12q^ -\-6q+l= 2{4q^ + 6q^ + 3q) + 1. The number t =
4q^ + 6^^ + 3q is an integer. Thus, we have n^ = 2t+ i with t integer
number. This means that n^ is an odd number. So the statement is true.
7. Part 1. Statement (a) implies statement (b). Suppose that the two
inequahties in statement (a) hold true. We can combine them and we
obtain:
a < b < a (*)
Because the number a cannot be strictly smaller than itself, the chain
of inequalities (*) can be true only if the two relations are equahties.
Therefore, we have a — h = a. Because a = fe, we conclude that
a-b = 0.
Part 2. Statement (b) implies statement (a). Because a —b = 0, we
know that a and b are indeed equal. Thus, the inequalities a<b and
b < a are trivially true. The proof is now complete.
8. This is an existence and uniqueness theorem. Indeed the statement
can be read as:
a. Any nonzero number has a reciprocal. (This is an axiom.)
b. Such reciprocal is unique. (This has to be proved.)
We assume that there are at least two numbers, a~^ and s, with the
properties as = sa=\ and aa~^ = a~^a= 1. We want to prove that
a~^ = s. Therefore, we need to use the properties of these two
numbers to compare them. We can obtain the following chain of
equalities: s = si = s{aa~^) = {sa)a~^ = \a~^ = a~^. Thus, a~^ = s.
Therefore, a has a unique reciprocal.
9. Because the statement mentions "factors" and "division,"
we might consider p and q as products of their prime factors.
Thus, p = pip2 ... Pr-iPr and q = qxqi... qs-iqs- The numbers pi need