● 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 secondorder.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 x_{1} and x_{2} is x_{1} – x_{2}
– Note that x_{1} – x_{2} = x_{2}  x_{1}
● 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} = a^{2} 

● 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
Halfopen
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 P_{1}(x_{1},y_{1}) and P_{2}(x_{2},y_{2}) is given
by
Δx = horizontal change x _{2} – x_{1} (a.k.a. “run”)
Δy = vertical change y_{2} – y_{1} (a.k.a. “rise”)
Midpoint formula
● Midpoint of a line segment with endpoints P_{1}(x_{1},y_{1}) and P_{2}(x_{2},y_{2}) has
coordinates
Average 1^{st} and 2^{nd} components of coordinates of endpoints .