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Absolute Value and Distance Consider a coordinate system on a line L and let A1 and A2 be points on L with coordinates a1 and a2 . Then |a1 a2 | = A1 A2 = distance between A1 and A2 EXAMPLES
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(1.6)
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|a1 a2 | = |2 5| = | 3| = 3 = A1 A2 (b)
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|a1 a2 | = |4 ( 3)| = |4 + 3| = |7| = 7 = A1 A2 A special case of (1.6) is very important. If a is the coordinate of A, then |a| = distance between A and the origin Notice that, for any positive number c, |u| c is equivalent to c u c (1.8) (1.7)
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EXAMPLE |u| 3 if and only if 3 u 3.
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Similarly, |u| < c is equivalent to c <u<c (1.9)
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EXAMPLE To nd a simpler form for the condition |x 3| < 5, substitute x 3 for u in (1.9), obtaining 5 < x 3 < 5. Adding 3, we have 2 < x < 8. From a geometric standpoint, note that |x 3| < 5 is equivalent to saying that the distance between the point A having coordinate x and the point having coordinate 3 is less than 5.
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It follows immediately from the de nition of the absolute value that, for any two numbers a and b, |a| a |a| and |b| b |b|
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(In fact, either a = |a| or a = |a|.) Adding the inequalities, we obtain ( |a|) + ( |b|) a + b |a| + |b| (|a| + |b|) a + b |a| + |b| and so, by (1.8), with u = a + b and c = |a| + |b|, |a + b| |a| + |b| (1.10)
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The inequality (1.10) is known as the triangle inequality. In (1.10) the sign < applies if and only if a and b are of opposite signs. EXAMPLE |3 + ( 2)| = |1| = 1, but |3| + | 2| = 3 + 2 = 5.
COORDINATE SYSTEMS ON A LINE
[CHAP. 1
Solved Problems
1.1 Recalling that u always denotes the nonnegative square root of u, (a) evaluate 32 ; (b) evaluate (c) show that x 2 = |x|. (d) Why isn t the formula x 2 = x always true
( 3)2 ;
32 = 9 = 3. (b) ( 3)2 = 9 = 3. (c) By (1.4), x 2 = |x|2 ; hence, since |x| 0, x 2 = |x|. (d) By part (c), x 2 = |x|, but |x| = x is false when x < 0. For example, ( 3)2 = 9 = 3 = 3.
1.2 Solve |x + 3| 5; that is, nd all values of x for which the given relation holds.
By (1.8), |x + 3| 5 if and only if 5 x + 3 5. Subtracting 3, 8 x 2.
1.3 Solve |3x + 2| < 1.
By (1.9), |3x + 2| < 1 is equivalent to 1 < 3x +2 < 1. Subtracting 2, we obtain the equivalent relation 3 < 3x < 1. This is equivalent, upon division by 3, to 1 < x < 1 . 3
1.4 Solve |5 3x| < 2.
By (1.9), 2 < 5 3x < 2. Subtracting 5, 7 < 3x < 3. Dividing by 3, 7 > x > 1. 3 algebra review Multiplying or dividing both sides of an inequality by a negative number reverses the inequality: if a < b and c < 0, then ac > bc. To see this, notice that a < b implies b a > 0. Hence, (b a)c < 0, since the product of a positive number and a negative number is negative. So bc ac < 0, or bc < ac.
1.5 Solve
x+4 <2 x 3
We cannot simply multiply both sides by x 3, because we do not know whether x 3 is positive or negative. Case 1: x 3 > 0. Multiplying (1) by this positive quantity preserves the inequality: x + 4 < 2x 6 4<x 6 10 < x Thus, if x > 3, (1) holds if and only if x > 10. Case 2: x 3 < 0. Multiplying (1) by this negative quantity reverses the inequality: x + 4 > 2x 6 4>x 6 10 > x [subtract x] [add 6] [subtract x] [add 6]
Thus, if x < 3, (1) holds if and only if x < 10. But x < 3 implies that x < 10. Hence, when x < 3, (1) is true.
CHAP. 1]
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