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154 The Nuts and Bolts of Proof, Third Edition
d. We will consider the function g{x) = Inx - x for x > 1. We know that
g{l) = -l. Let's study the behavior of this function using the first
derivative test. Because gXx) = -l + 1/x, it follows that g' is always
negative for x>l and that g\l) = 0. Thus, g{l) = -l is the
maximum value of g. This implies that g{x)<g{l) < 0 for all x>l.
This proves that Inx < x for all x> 1. In particular, Inn<n for all
positive integers, as they are just real numbers larger than or equal
to 1.
4. a. To prove that the two sets are equal, we need to prove that they have
the same elements.
Part 1. A^B, Let x be an element of ^4. Then x is a multiple of 15;
that is, X = 15n for some integer number n. Therefore, x is a multiple
of 5, because x = 5(3n), and x is a multiple of 3, because x = 3(5n).
This means that x is an element of B.
Part 2. Be.A. Let y be an element of B. Then y is a multiple of 5
and a multiple of 3. Thus, we can write y = 3p for some integer
number p and y = 5s for some integer number s. Therefore,
y = 3p = 55. Because 3 is not divisible by 5, p must be divisible by
5. So, p = 5q for some integer number q. Thus, y = 3p = 3{5q)= I5q
for some integer number q. This proves that y is a multiple of 15 and
it belongs to ^.The two parts of the proof imply that A = B.
b. Part L A<ZB. See part 1 above.
Part 2. Be.A. This inclusion is not true. Consider the number 6. It
is a multiple of 3; therefore, it belongs to B. But it is not an element
of ^, as it is not a multiple of 15. Therefore, the set A cannot contain
the set B. Moreover, because of part 1, we know that A is contained
in B. This means that ^ is a proper subset of B.
5. i. A solution exists. Because the number a is not equal to 0, it has a
reciprocal, a~^. Then we can multiply both sides of the equation
ax = bhy a~^to obtain a~^(ax) = a~^b. So, the solution is x == a~^b.
ii. We can prove that the solution t = a~^b is unique in two ways: (1)
The solution is unique because of the algebraic procedure used to
find it and the fact that the reciprocal of a number is unique. (2) Let
5 be another solution of the equation. Then as = b and at = b. This
implies that as = at. If we multiply both sides of the equation by
fl~^ we obtain s = t. This means that there is only one solution.
6. Part L If n^ is an odd number, then n is an odd number. (Because n^ is
an odd number, we can write n^ = 2q-\-l for some integer number q.
To find any information about n we need to calculate the cubic root
of 2q-\-l; however this is no easy task because there is no easy