● 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”
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 .
