1873_solutions

# That 0 xy 2 prove that if x and y are negative real

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Unformatted text preview: ||. Solution: The inequality |2|x | 5| |4 |x 1 || can be written as 4 |x 1| 2|x | 5 4 |x 1  |x 1| 2|x | 9 |x 1| 1| which says that 1 0 In order to express this inequality without any absolute value signes we shall look separately at the cases x  0 and x 0 and also the cases x  1 and x 1. When x  0, the inequality 1  |x 1| 2|x | 9 |x 1| becomes which says that x When 0 x 2 and 8/3 11 x 2x 9 1 x x. In other words, when x  0, we must have 8 x 2. 3 1, the inequality 1  |x 1| 2|x | 9 |x 1| becomes which says that x When x 2/3 and x 1  1 x 2x 9 1 x 8. Therefore the required inequality holds when 2 x 1. 3 1, the inequality 1  |x 1| 2|x | 9 |x 1| becomes which says that x 0 and x 1  x 1 2x 10/3. In other words 1 x 9 x 1 10 . 3 Thus the solution of the required inequality is the set 8 , 2 Þ 2 , 1 Þ 1, 10  3 3 3 8, 2 Þ 3 6. Prove that if a, b, c, x, y and z are any real numbers then ax  by  cz 2 a2  b2  c2 x2  y2  z2 . One way to produce this inequality is to oberve that 76 2 , 10 33 a2  b2  c2 x2  y2  z2 ax  by  cz  a2y2  a2z2  b2x2  b2z2  c2x2  c2y2  ay bx  az cx  bz There are several other possible approaches. 2 2 cy 2 2 2axby 2axcz 2bycz 0 7. Given that a, b and c are positive numbers and that c  a  b, prove that c  ab. 1a 1c 1b We observe that c  a 1b 1c b 1a 1c c 1a 1b ab 1c 1a 1b 1a 1b 1c 2ab  acb  0  2ab  acb  a  b c   0. 1a 1b 1c 1a 1b 1c Exercises on Integers and Rational Numbers 1. 2 a. Explain why the numbers 3  2 and 3 are both rational. 7 7 2 2 These facts become obvious when we simplify: 3  2  25 3 2  3. and 7 7 7 2 14 2 b. Explain why the number 2. 345 is rational. We need only observe that 2. 345  2345 . 1000 c. Explain why the sum, difference, product and quotient of rational numbers must always be rational. 2. For each of the following statements, say whether the statement is true or false and justify your assertion: a. If x is rational and y is irrational then x  y is irrational. This statement is true. To see why we should observe that an equivalent way of making this assertion is to say that if x is rational and x  y is rational then y must be rational; a statement that follows at once from the equation y  x  y x. b. If x is rational and y is irrational then xy is irrational. As long as x 0 we can use the approach used in part a to prove this assertion from the equation y y  x x. However, if x  0 then xy  0 for every y and so it would be false to say that y must be rational. c. If x is irrational and y is irrational then x  y is irrational. This statement is false. Observe that 3 2 2 2  5. Exercises on Upper And Lower Bounds 1. Suppose that A is a nonempty bounded set of real numbers that has no largest member and that a Explain why the sets A and A a have exactly the same upper bounds. 77 A. Solution: Since the set A a is a subset of A, it is clear that any upper bound of A must be an upper bound of A a . Now, to show that any upper bound of A a...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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