Download equality are frequently used to derive equations. Can these

3.3 Statements of the Form P → Q
69
equality are frequently used to derive equations. Can these substitution properties
be extended to the inequality relation? That is, suppose that you working with the
expressions a + x and px and you know that x < y. Then, by substitution, can you
conclude that a + x < a + y and px < py?
We now identify four substitution properties of inequality that we will use to
derive inequalities. These substitution properties offer a slightly different way of
viewing the laws of inequality given in 3.1.5. Note that a summand is a value that is
to be added; for example, in 1 + 2 the summands are 1 and 2.
Substitution Properties of Inequality 3.3.3. Let a, p, x, y real numbers with p > 0.
Then the following hold:
(1) Given the sum a + x, if x < y, then you can conclude that a + x < a + y.
(Replacing a summand with a larger value yields a larger sum.)
(2) Given the sum a + x, if x > y, then you can conclude that a + x > a + y.
(Replacing a summand with a smaller value yields a smaller sum.)
(3) Given the product px, if x < y, then you can conclude that px < py.
(Replacing a factor with a larger value yields a larger product.)
(4) Given the product px, if x > y, then you can conclude that px > py.
(Replacing a factor with a smaller value yields a smaller product.)
Before applying (3) or (4) in the above substitution properties of inequality, one
must ensure that p > 0. Similar properties hold if we replace each occurrence of <
with ≤ in (1) and (3), and replace each occurrence of > in (2) and (4) with ≥. For
example, given the sum a + x, if x ≤ y, then you can conclude that a + x ≤ a + y.
Similarly, given the product px, if x ≥ y, then you can conclude that px ≥ py when
p > 0. In our next example, we apply properties (2) and (4) of 3.3.3 to derive an
inequality.
Example 1. Prove that if 4n > 4n + 1, then 4n+1 > 4(n + 1) + 1, when n ≥ 2 is a
natural number.
Solution. Let n ≥ 2 be a natural number and assume 4n > 4n + 1. We prove that
4n+1 > 4(n + 1) + 1 as follows:
4n+1 = 4 · 4n
by property of exponents
> 4(4n + 1)
by (4) of 3.3.3 because 4n > 4n + 1
= 16n + 4
by distributivity
= 4n + 4 + 12n
because 16n = 4n + 12n
= 4(n + 1) + 12n by distributivity
> 4(n + 1) + 1
by (2) of 3.3.3 because 12n ≥ 12 · 2 > 1.
Therefore, 4n+1 > 4(n + 1) + 1 and this completes the proof.
!
9.4 Convergence of Sequences
313
To apply Proof Strategy 9.4.3 on a specific sequence, first let ε > 0. We must find
a natural number N such that when n > N, we can prove that |sn − ℓ| < ε . To find
the desired N, we will first attempt the following:
Using algebra and properties of inequality on the expression |sn − ℓ|, ‘extract
out’ a larger value that resembles 1n .
We shall then use this larger value to find N so that when n > N we will
have that |sn − ℓ| < ε . We shall illustrate this idea in our proof analysis of
Theorems 9.4.5–9.4.8. Before we discuss these theorems, we identify three substitution properties of inequality that follow from Exercise 9 on page 297. These
properties are very useful for proving theorems about convergence and extend the
substitution properties 3.3.3 given on page 69.
Substitution Properties of Inequality 9.4.4. Let a, b, c, d be positive real numbers. Then the following hold:
(1) Given the ratio ab , if a < c, then you can conclude that
a
b
< bc .
(2) Given the ratio ab , if d < b, then you can conclude that
a
b
< da .
(Replacing a numerator with a larger value yields a larger ratio.)
(Replacing a denominator with a smaller value yields a larger ratio.)
(3) Given the ratio ab , if a ≤ c and d ≤ b, then you can conclude that
a
b
≤ dc .
(Replacing a numerator with a larger value and denominator with a smaller value yields a
larger ratio.)
The Substitution Properties 9.4.4 and 3.3.3 will be implicitly used in nearly all
of the remaining proofs of this chapter. To apply the above 1–3, you must first
know that a, b, c, d are positive. To illustrate how we will be using these properties,
suppose x > 0 and that we working with the ratio 4x . It follows that 4x < x+1
4 by item
(1) of 9.4.4 because x < x + 1. We can also conclude from item (2) of 9.4.4 that
x
x
4 < 2 since 2 < 4.
Example 1. Property (2) in 9.4.4 implies the following assertions:
1.
2.
3.
1
1
n < N when n > N > 0.
√1 < √1 when n > N > 0,
n
N
1
1
n
2n < n when 2 > n > 0.
by Theorem 3.8.6 on page 95.
We shall now apply Proof Strategy 9.4.3 in the proofs of our next four theorems.
In each such proof we will first let ε > 0 and then we will tell the reader the value
for N that we will use to complete the proof. Prior to each of these proofs, we
shall also present a proof analysis in which we discuss how we actually found
N. In these proofs we will also be using some of the inequalities identified in
Example 1.