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 165
The statement is now proved. Therefore, by the principle of mathematical
induction, the inequality holds true for all integers k>2.
25. Because a is a multiple of b, we can write a = bn for some integer n.
Because h is a multiple of c, we can write b = cm for some integer m.
We will now combine this information to find a direct relation
between a and c. Thus, we obtain a = bn — {cm)n = c(mn). Since the
number mn is an integer (as it is the product of two integers), we can
conclude that a is a multiple of c.
26. The proof of this statement has two components—namely, the proofs
of the following statements:
a. If p is a nonzero rational number, then its reciprocal is a rational
number,
b. If the reciprocal of a nonzero number p is a rational number, then
the number itself is a rational number.
The reciprocal of a nonzero number p is the number q such that
pq=L
Proof of part a. Because the number p is rational and nonzero, we can
write p = a/b, where a and b are relatively prime numbers, both not
equal to zero. Therefore, {a/b)q=l. If we multiply both sides of the
equation by b and divide them by a, we obtain q = b/a. This means
that g is a rational number.
Proof of part b. Because the inverse of the nonzero number p, which
we will indicate with p~^, is a rational number, we can write it as
p~^ = c/d, where c and d are relatively prime numbers, both not equal
to zero. Thus,
-1 ^ 1
PP ^^2^
If we multiply the equality by d and divide it by c, we obtain p = d/c,
where c and d are relatively prime numbers, both not equal to zero.
Thus, p is a rational number. {Note: In this proof we assume that both
p and q are in reduced form, by stating that a and b are relatively
prime, and c and d are relatively prime. While it is correct to make
these assumptions, in this case there is no need for them. The proof is
still correct if the "relatively prime" requirement is removed.)
27. Let £>0 be given. Is there an N>0 such that |a„-0|<£ for all
n > AT? Observe that:
\an - 0| =
(4H=1(4)1=©"4-