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Notes for Inequalities.pdf

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Department
Mathematics
Course
MATH 135
Professor
John Paul
Semester
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
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