false

Class Notes
(836,580)

Canada
(509,856)

University of Waterloo
(18,584)

Mathematics
(1,919)

MATH 135
(334)

John Paul
(2)

Lecture

Unlock Document

Mathematics

MATH 135

John Paul

Fall

Description

MATH 135 Fall 2013
Optional Reading I: Equalities of Real Numbers
1 Review of inequalities
We agree that there is a standard ordering of the real numbers. Given two real numbers a and b, the sentence
a ▯ b
is true if a is less than or equal to b. We say that a is less than b, written a < b, if a ▯ b but a , b. Also, the sentence
a ▯ b as another way of expressing b ▯ a. The following properties are accepted as true without proof.
(a) ▯ is reﬂexive: For any real number a, a ▯ a.
(b) ▯ is transitive: For any real numbers a, b and c, if a ▯ b and b ▯ c, then a ▯ c.
(c) ▯ is antisymmetric: For any real numbers a and b, if a ▯ b and b ▯ a, then a = b.
(d) For any real numbers a, b and c, a ▯ b implies that a + c ▯ b + c.
(e) For any real numbers a, b and c > 0, a ▯ b implies that ac ▯ bc. For any c < 0, a ▯ b implies that ca ▯ cb.
We can immediately strengthen property (d) to:
(d’) For any real numbers a, b and c, a ▯ b if and only if a + c ▯ b + c.
Obviously, property (d) yields one direction (if a ▯ b, then a + c ▯ b + c). Conversely, if a + c ▯ b + c, then adding ▯c
to both sides and using (d), we can derive
(a + c) + (▯c) ▯ (b + c) + (▯c);
or simply a ▯ b.
1 As an example of proofs of inequalities, we can demonstrate:
2
Proposition 1. For any real number x, x ▯ 0.
Proof. If x = 0 is true, then x = 0 and so x ▯ 0. If x > 0 is true, then by property (e),
2
x = x ▯ x > x ▯ 0 = 0:
Finally, if x < 0 is true, then invoking property (e) again,
x = x ▯ x > x ▯ 0 = 0:
We have proved that x ▯ 0 for all x. ▯
Proposition 2. Let x be a non-zero real number. The following hold:
▯1
(i) x > 0 if and only if x > 0.
(ii) x < 0 if and only if x▯1 < 0.
Proof. (i) Let x > 0. If x1 < 0 were true, then by property (e),
▯1
1 = x ▯ x < x ▯ 0 = 0:
▯1
This is a contradiction. Hence, x > 0 must be true.
(ii) Let x < 0. Then ▯x > 0. Using (i), we have (▯x) ▯1 > 0. Since (▯x) ▯1 = ▯x , this means that ▯x ▯1 > 0, or
equivalently, x ▯1 < 0. ▯
As a corollary of Proposition 2, we can strengthen property (e) listed earlier to:
(e’) For any real numbers a, b and c > 0, a ▯ b if and only if ac ▯ bc. For any c < 0, a ▯ b if and only if ca ▯ cb.
The argument is analogous to the one given to prove (d’).
2 Practice
1. Suppose that x; y 2 R satisfy x ▯ 0 and y ▯ 0. Prove that if x + y = 0, then x = y = 0.
Example 1. Solve the inequality
2x ▯ 3
▯ 1
x + 1
for real numbers x. (This means to determine exactly which real numbers x satisfy the inequality stated above.)
Solution. Since the fraction is undeﬁned when x = ▯1, we may immediately assume that x , ▯1, so that x + 1 , 0. It
is natural to try multiplying both sides by x+1 and then using algebraic manipulation. To do this, we must be aware
that the possibility x+1 < 0 results with a reversal of the inequality. With this in mind, we may argue separately for
the cases x < ▯1 and x > ▯1, as follows.
Case 1: x > ▯1, or, x + 1 > 0. We obtain a chain of equivalent statements through algebraic manipulation:
2x ▯ 3
▯ 1 , 2x ▯ 3 ▯ x + 1 by (e’)
x + 1
, 2x ▯ x ▯ 1 + 3 by (d’)
, x ▯ 4:
What we have shown is that among those x with x > ▯1, the given inequality holds precisely for those satisfying
x ▯ 4 in addition to x > ▯1. Simply put, the part of the solution set within x > ▯1 is described by the inequality
▯1 < x ▯ 4.
Case 2: x < ▯1, or, x + 1 < 0. Multiplication by x + 1 reverses the inequality:
2x ▯ 3
▯ 1 , 2x ▯ 3 ▯ x + 1 by (e’)
x + 1
, x ▯ 4 by (d’):
Among those x with x < ▯1, the inequality holds only for those satisfying x ▯ 4 and x < ▯1; but no such x exists.
Summarizing, for x 2 R we have
2x ▯ 3
▯ 1 , ▯1 < x ▯ 4:
x + 1
In other words, the solution set is the interval (▯1;4]. ▯
Example 2. Prove that if a;b;c;d > 0 are such that a=b < c=d and b > d, then
a ▯ c c
b ▯ d < d :
a ▯ c c
Solution. Since x < y if and only if y ▯ x > 0, in order to establish < , we can just as well show that
b ▯ d d
c▯ a ▯ c> 0:
d b ▯ d
The left-hand side can be expressed as:
c a ▯ c c(b ▯ d) ▯ d(a ▯ c) cb ▯ cd ▯ da + dc cb ▯ da
▯ = = = : (1.1)
d b ▯ d d(b ▯ d) d(b ▯ d) d(b ▯ d)
3 We know that d > 0 and b ▯ d > 0, so that d(b ▯ d) > 0, but what about the sign of the numerator cb ▯ da? Here
is where the condition a=b < c=d comes into play. Since multiplying by positive numbers preserves the ordering,
multiplying both sides of a=b < c=d by bd, we get
ad < bc; or cb ▯ da > 0:
Returning to (1.1), we have all the needed inequalities to deduce that
c ▯ a ▯ c= cb ▯ da > 0:
d b ▯ d d(b ▯ d)
This proves the original inequality. ▯
2 Review of absolute values
Deﬁnition 2.1. For any x 2 R, the absolute value jxj of x is the real number deﬁned as follows:
8
:▯x if x ▯ 0:
More explicitly, jxj = x whenever x ▯ 0, and jxj = ▯x whenever x ▯ 0.
As may be familiar with the reader, this is an example of a piecewise deﬁnition. One may probably worry that the
two cases x ▯ 0 and x ▯ 0 are not mutually exclusive. But the two branches of the deﬁnition clearly agree when
x = 0 (since x = 0 = ▯x). So, our deﬁnition does not cause any real concern.
The following properties of the absolute values are easily proved for any real number x:
(a) jxj = j ▯ xj.
(b) jxj = x .
Example 3. Solve the inequality
jx ▯ 1j < j2x + 1j
for real numbers x.
Solution. To those who are familiar with graphs of absolute value functions, solving the inequality simply means to
ﬁnd those x at which the graph of jx ▯ 1j lies below the graph of j2x + 1j. We will not

More
Less
Related notes for MATH 135

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.