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# 6 - ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W L...

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ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W L Chen, X T Duong and Macquarie University, 1999. This work is available free, in the hope that it will be useful. Any part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, with or without permission from the authors. Chapter 6 INEQUALITIES AND ABSOLUTE VALUES 6.1. Some Simple Inequalities Basic inequalities concerning the real numbers are simple, provided that we exercise due care. We begin by studying the effect of addition and multiplication on inequalities. ADDITION AND MULTIPLICATION RULES. Suppose that a, b R and a < b . Then (a) for every c R , we have a + c < b + c ; (b) for every c R satisfying c > 0 , we have ac < bc ; and (c) for every c R satisfying c < 0 , we have ac > bc . In other words, addition by a real number c preserves the inequality. On the other hand, multipli- cation by a real number c preserves the inequality if c > 0 and reverses the inequality if c < 0. Remark. We can deduce some special rules for positive real numbers. Suppose that a, b, c, d R are all positive. If a < b and c < d , then ac < bd . To see this, note simply that by part (b) above, we have ac < bc and bc < bd . SQUARE AND RECIPROCAL RULES. Suppose that a, b R and 0 < a < b . Then (a) a 2 < b 2 ; and (b) a 1 > b 1 . Proof. Part (a) is a special case of our Remark if we take c = a and d = b . To show part (b), note that a 1 b 1 = 1 a 1 b = b a ab > 0 . This chapter was written at Macquarie University in 1999.

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7 6 5 4 3 2 1 -1 -1 -2 1 x y y = 3 y = 4 x + 7 6–2 W W L Chen and X T Duong : Elementary Mathematics CAUCHY’S INEQUALITY. For every a, b R , we have a 2 + b 2 2 ab . Furthermore, equality holds precisely when a = b . Proof. Simply note that a 2 + b 2 2 ab = a 2 2 ab + b 2 = ( a b ) 2 0 , and that equality holds precisely when a b = 0. We now use some of the above rules to solve inequalities. We shall illustrate the ideas by considering a few examples in some detail. Example 6.1.1. Consider the inequality 4 x + 7 < 3. Using the Addition rule and adding 7 to both sides, we obtain 4 x < 4. Using one of the Multiplication rules and multiplying both sides by the positive real number 1 / 4, we obtain x < 1. We have shown that 4 x + 7 < 3 = x < 1 . Suppose now that x < 1. Using one of the Multiplication rules and multiplying both sides by the positive real number 4, we obtain 4 x < 4. Using the Addition rule and adding 7 to both sides, we obtain 4 x + 7 < 3. Combining this with our earlier observation, we have now shown that 4 x + 7 < 3 ⇐⇒ x < 1 . We can confirm our conclusion by drawing a graph of the line y = 4 x +7 and observing that the part of the line below the horizontal line y = 3 corresponds to x < 1 on the x -axis.
3 2 1 -1 -1 1 x y y = -1 y = -5 x + 4 2 -2 -3 4 6 5 Chapter 6 : Inequalities and Absolute Values 6–3 Example 6.1.2. Consider the inequality 5 x + 4 > 1. Using one of the Multiplication rules and multiplying both sides by the negative real number 1, we obtain 5 x 4 < 1. Using the Addition rule

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6 - ELEMENTARY MATHEMATICS W W L CHEN and X T DUONG c W W L...

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