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The Algebra Buster


October 24th









October 24th

Distance on a Number Line

● Syllabus
● Distance on a number line , ordering
● Absolute value
● Distance in the plane & midpoint formula
● Homework 1

Distance on a number line

Real numbers
● Real number line
 – Distance on a number line
● Distance from x to the origin
● Distance between two points on the line
solve -second-order.html">Order properties : real numbers a,b,c,d
 – Trichotomy law: exactly one of the following is true
a < b, a > b, or a = b
 – Transitive law of inequality : if a < b and b < c, then a < c
 – Additive law of inequality: if a < c and b < d, then a+b < c+d
 – Multiplicative law of inequality:
if a < b, then ac < bc if c > 0,
and ac > bc if c < 0

Absolute value

● The number x is located |x| units from 0
– To the right if x > 0; to the left if x < 0

● Distance between numbers x1 and x2 is |x1 – x2|
– Note that |x1 – x2| = |x2 - x1|

Terminology : “p if and only if q” means that both the statement and its
converse are true, i.e.,
“if p, then q” and “if q, then p”

Properties of the absolute value

For a,b real numbers:

● |a| ≥ 0 absolute value is nonnegative
● |-a| = |a|  
● |a|2 = a2  
● |ab| = |a| |b|  
● |a/b| = |a| / |b|, b≠0  
● -|a| ≤ a ≤ |a| true because either |a| = a or |a| = -a
● Let b ≥ 0; |a| = b if and only if a = ±b  
● Let b > 0; |a| < b if and only if -b < a < b  
● Let b > 0; |a| > b if and only if a > b or a < -b (think of number line)
● |a + b| ≤ |a| + |b| Triangle Inequality (very useful in theory and
computation)

Interval notation

Closed interval

Open interval

Half-open

Real number line

Absolute value equations

Example: solve |2x – 6| = x

Solution:
if 2x – 6 ≥ 0, then |2x – 6| = 2x - 6
solve 2x – 6 = x
x = 6
if 2x – 6 < 0, then |2x – 6| = -(2x - 6)
solve -(2x – 6) = x
-3x = -6
x = 2
Two solutions : x = 6 and x = 2
(It is always a good idea to plug them back in and doubled check!)

Recall that |a – b| is the distance between a and b on the number line. So
|x – a| = b is satisfied by the two points x that are of distance b from a.
The equation above is satisfied by the two points x such that the
distance between 2x and 6 on the number line is x .

Absolute value inequality

Example: Solve |2x – 3| ≤ 4
Solution:
-4 ≤ 2x – 3 ≤ 4 (using one of our abs value properties)
-4 + 3 ≤ 2x ≤ 4 + 3
-1 ≤ 2x ≤ 7
-1/2 ≤ x ≤ 7/2
[-1/2, 7/2] in interval notation
See textbook for geometric solution

Absolute value as tolerance

● Let w be a measurement (e.g., weight)
● |w – a| ≤ b can be interpreted as “w being compared to a with absolute
error of measurement of b units.”
● Example: a bag of cement weighs 90 lbs plus or minus 2 lbs
So a given bag can weigh as much as 92 lbs or as little as 88 lbs.
State as an absolute value inequality.
● Solution: let w = weight of the bag of cement in pounds
90 – 2 ≤ w ≤ 90 + 2
-2 ≤ w – 90 ≤ 2
|w – 90| ≤ 2

Distance between points in the plane

● Theorem: distance between two points P1(x1,y1) and P2(x2,y2) is given
by

Δx = horizontal change x 2 – x1 (a.k.a. “run”)
Δy = vertical change y2 – y1 (a.k.a. “rise”)

Midpoint formula

● Midpoint of a line segment with endpoints P1(x1,y1) and P2(x2,y2) has
coordinates

Average 1st and 2nd components of coordinates of endpoints .

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