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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/