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152 The Nuts and Bolts of Proof, Third Edition
some algebraic expression that uses all the numbers we are considering,
namely a, a\ and b. It makes sense to keep using addition as this is the
only operation used to define the opposite of a number. These are some
of the reasons for trying to start from a-\-a! -\-b. Which conclusion
can we reach?) If we use the associative property of addition and
Equation (1), we obtain a-\-a'-\-b = {a-\-a!)-\-h = Q-\-b = h. If we use
the associative and commutative property of addition and Equation
(2), we obtain a-\-a'-\-b = a'-\-{a-\-b) = a'-\-0 = a\ Therefore, we have
l) = a-\-a!-\-b — a'. Thus, b = a', and the opposite of a is unique.
3. a. Proof by induction:
i. The smallest positive integer number to use is 1. Because Inl = 0,
it is true that Inl < 1.
ii. Let us assume that the inequality is true for n. Thus, Inn < n.
iii. We have to prove that ln(n + 1) < (n + 1).
{Remember. ln(n + 1) 7^ Inn + Inl.)
We have to try to use what we know about n and n + 1. One possible
relation is:
n+1
n + 1 = n.
n
Therefore, using the properties of the natural logarithm, we have:
ln(n+l) = ln(^?-^nj
= ln(^)+lnn
= ln(l + l/n) + lnn.
By the inductive hypothesis, Inn < n. Thus, we have ln(n + 1) =
ln(l + 1/n) + lnn< ln(l + l/n) + n. To show that the conclusion is
true, we need to prove that ln(l + 1/n) < 1. Because n> 1, 1/n < 1.
Therefore, 1 +1/n < 1-^ 1 =2 < e = 2.72.... So 1 + 1/n < e. The
function natural logarithm is an increasing function; thus, the larger
its input, the larger the corresponding output. Thus, In (1 + (1/n)) <
Ine = 1. If we use this information in the chain of inequalities, we
obtain ln(n + 1) < ln(l + 1/n) + n < 1 + n, or, equivalently,
ln(n + 1)< n + 1. By the principle of mathematical induction the
proof is now complete.
b. By graphing—The straight Une is the graph of g{x) = x, the other is
the graph off{x) = lnx. From the graph we can say that it seems
plausible that In n < n for all positive integers.