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ORDERING REAL NUMBERS
THANOMSAK LAOKUL
Mathematics Teacher
Mahidol Wittayanusorn School
MATH 30102
T.Laokul – p. 1/
Ordering Real Numbers
Definition of order on the real number line
If a and b are real numbers, a is less than b if b − a is
positive. The order of a and b is denoted by the inequality
a < b.
T.Laokul – p. 2/
Ordering Real Numbers
Definition of order on the real number line
If a and b are real numbers, a is less than b if b − a is
positive. The order of a and b is denoted by the inequality
a < b.
This relationship can also be described by saying that b is
greater than a and writing b > a. The inequality a ≤ b
means that a is less than or equal to b, and the inequality
b ≥ a means that b is greater than or equal to a.
T.Laokul – p. 2/
Ordering Real Numbers
Definition of order on the real number line
If a and b are real numbers, a is less than b if b − a is
positive. The order of a and b is denoted by the inequality
a < b.
This relationship can also be described by saying that b is
greater than a and writing b > a. The inequality a ≤ b
means that a is less than or equal to b, and the inequality
b ≥ a means that b is greater than or equal to a.
The symbols <, >, ≤, and ≥ are inequality symbols.
T.Laokul – p. 2/
Intervals
Bounded Intervals
T.Laokul – p. 3/
Intervals
Bounded Intervals
[a, b] :
a≤x≤b
T.Laokul – p. 3/
Intervals
Bounded Intervals
[a, b] :
a≤x≤b
(a, b) :
a<x<b
T.Laokul – p. 3/
Intervals
Bounded Intervals
[a, b] :
a≤x≤b
(a, b) :
a<x<b
[a, b) :
a≤x<b
T.Laokul – p. 3/
Intervals
Bounded Intervals
[a, b] :
a≤x≤b
(a, b) :
a<x<b
[a, b) :
a≤x<b
(a, b] :
a<x≤b
T.Laokul – p. 3/
Intervals
Unbounded Intervals
T.Laokul – p. 4/
Intervals
Unbounded Intervals
[a, ∞) :
x≥a
T.Laokul – p. 4/
Intervals
Unbounded Intervals
[a, ∞) :
x≥a
(a, ∞) :
x>a
T.Laokul – p. 4/
Intervals
Unbounded Intervals
[a, ∞) :
x≥a
(a, ∞) :
x>a
(−∞, b] :
x≤b
T.Laokul – p. 4/
Intervals
Unbounded Intervals
[a, ∞) :
x≥a
(a, ∞) :
x>a
(−∞, b] :
x≤b
(−∞, b) :
x<b
T.Laokul – p. 4/
Intervals
Unbounded Intervals
[a, ∞) :
x≥a
(a, ∞) :
x>a
(−∞, b] :
x≤b
(−∞, b) :
x<b
(−∞, ∞) :
−∞ < x < ∞
T.Laokul – p. 4/
Properties of Inequality
Theorem. For any real number a, b and c.
T.Laokul – p. 5/
Properties of Inequality
Theorem. For any real number a, b and c.
1. If a < b and
b < c then a < c (transivity)
T.Laokul – p. 5/
Properties of Inequality
Theorem. For any real number a, b and c.
1. If a < b and
b < c then a < c (transivity)
2. If a < b then
a+c<b+c
T.Laokul – p. 5/
Properties of Inequality
Theorem. For any real number a, b and c.
1. If a < b and
b < c then a < c (transivity)
2. If a < b then
a+c<b+c
3. If a < b and
c > 0 then ac < bc
T.Laokul – p. 5/
Properties of Inequality
Theorem. For any real number a, b and c.
1. If a < b and
b < c then a < c (transivity)
2. If a < b then
a+c<b+c
3. If a < b and
c > 0 then ac < bc
4. If a < b and
c < 0 then ac > bc
T.Laokul – p. 5/
Properties of Inequality
Theorem. For any real number a, b and c.
1. If a < b and
b < c then a < c (transivity)
2. If a < b then
a+c<b+c
3. If a < b and
c > 0 then ac < bc
4. If a < b and
c < 0 then ac > bc
5. If ab > 0 then a > 0 and b > 0
or a < 0 and b < 0
T.Laokul – p. 5/
Theorem
Theorem.
If a be a real number and a 6= 0 then a2 > 0
T.Laokul – p. 6/
Theorem
Theorem.
If a be a real number and a 6= 0 then a2 > 0
Law of Trichotomy
For any two real numbers a and b, precisely one of three
relationships is possible:
a = b, a < b, or a > b
T.Laokul – p. 6/
Theorem
Theorem.
If a be a real number and a 6= 0 then a2 > 0
Law of Trichotomy
For any two real numbers a and b, precisely one of three
relationships is possible:
a = b, a < b, or a > b
Theorem.
1
If a < b then a < (a + b) < b
2
T.Laokul – p. 6/
Theorem
Theorem.
If a be a real number and a 6= 0 then a2 > 0
Law of Trichotomy
For any two real numbers a and b, precisely one of three
relationships is possible:
a = b, a < b, or a > b
Theorem.
1
If a < b then a < (a + b) < b
2
proof a < b =⇒ a + a < a + b < b + b
T.Laokul – p. 6/
Theorem
Theorem.
If a be a real number and a 6= 0 then a2 > 0
Law of Trichotomy
For any two real numbers a and b, precisely one of three
relationships is possible:
a = b, a < b, or a > b
Theorem.
1
If a < b then a < (a + b) < b
2
proof a < b =⇒ a + a < a + b < b + b
a+b
b+b
=⇒ a = a+a
<
<
=b
2
2
2
T.Laokul – p. 6/
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