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Lecture

# Appendixes A-C - Winter 2013

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University of Waterloo

Mathematics

MATH 127

Andrew Beltaos

Winter

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Math 127 - Winter 2013
Appendix A
Notation:
ℤ is the set of all integers
- To say x is an integer, we write x ⋲ ℤ, ℤ = { 0, ± 1, ± 2, ... )
- The natural numbers are the set of all the integers: N, N = { 1, 2, 3, ... }
- The set of all rational numbers is denoted ℚ = { p/q | p ⋲ ℤ, q ⋲ ℤ, and q ≠ 0 }
- the set of all real numbers is ℝ
If we have 2 sets, A and B, the intersection of A and B is denoted as A∩B - the set of
everything which is in A and B. A∩B = { x | x ⋲ A and x ⋲ B }
The union of A and B, denoted as A∪B, is the set of everything in A and everything in B.
A∪B = { x | x ⋲ A or x ⋲ B }
Example:
remove duplicates
Let A = { 1, 2, 3 } and B = { 2, 4, 6 }, then A∩B = { 2 } and A∪B = {1, 2, 3, 4, 6}
Linear Inequalities
Example 1:
Solve the inequality 2 + x < 5x - 3 and sketch the solution on the real number line.
1. Subtract x from both sides 2 < 4x - 3
2. Add 3 to both sides 2 + 3 < 4x → 5 < 4x
3. Divide both sides by 4 5/4 5/4 }
5/4
Example 2:
Solve: x² - 2x - 15 ≤ 0
Note: [ -3, 5 ] = [ -3, ∞) ∩ (-∞, 5 ]
(x - 5)(x +3) ≤ 0
Note: the LHS will be zero when: We make a table, breaking up the
▯ x - 5 = 0 or x + 3 = 0 (i.e x = 5 or x = -3) real number line at these points
1 x is in x - 5 x + 3 product
(- ∞, - 3) - - +
(-3, 5) - + -
(5, ∞) + + +
We know the LHS is negative when x ⋲ (-3, 5) and is zero when x = -3 or 5, therefore,
LHS is ≤ 0 when x ⋲ [-3, 5].
Alternate notation: { x ⋲ ℝ | x ≥ -3 and x ≤ 5 } or { x ⋲ ℝ | -3 ≤ x ≤ 5 }
-3 5
Example 3: 2 inequalities that need to be
Solve the inequality 2 ≤ -x + 1 < 3 satisﬁed at the same time!
2 ≤ -x + 1 -x + 1 < 3
1 ≤ - x x < 2
- 1 ≥ x x > - 2
Absolute Value ( non-negative)
Properties of Absolute Values
The absolute value of a real number x,
denoted |x|, is deﬁned as the distance 1. |ab| = |a| |b|
between x 2. |a/b| =|a|/|b| (b≠0)
3. |a | = |a|
|a| = a if a ≥ 0 2
|a| = -a if a < 0 √a = |a|
Example 1: (pg A7)
Express |3x - 2| without using the absolute value symbol
▯ ▯ 3x - 2 ▯▯ if 3x-2 ≥ 0▯ ▯ 3x - 2▯ if x ≥ 2/3
|3x - 2| = { ▯ ▯ ▯ ▯ = {
- (3x - 2)▯ if 3x - 2 < 0▯ ▯ 2 - 3x▯ if x < 2/3
2 Math 127 - Winter 2013
Example 2:
Solve |2x - 5|
|2x - 5| is equivalent to: ▯ ▯ 2x - 5 = 3▯ or▯ 2x - 5 = -3
▯ ▯ ▯ ▯ ▯ 2x = 8▯ ▯ 2x = 2
▯ ▯ ▯ ▯ ▯ x = 4▯ ▯ x = 1
Therefore, x = 4 or x = 1.
Example 3: (pg A8)
Solve |x - 5| < 2
|x - 5| < 2 is equivalent to: ▯▯ -2 < x - 5 < 2
▯ ▯ ▯ ▯ ▯ 3 < x < 7▯ ▯ ▯ add 5 to both sides
Therefore the solution is on the open interval (3,7).

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