Notesale: Turn your study into money

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Numbers,
Inequalities
and Absolute
Values

Numbers and Sets
A number is a mathematical object used to count, measure and label
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 We start with the natural numbers:
N = {0, 1, 2,
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, −3, −2, −1, 0, 1, 2, 3,
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Thus, any
𝑚
rational number r can be expressed as r =
 where m and n are integers and n ≠ 0
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Some

numbers, such as √ 2 and π, can’t be expressed as a ratio of integers and are
therefore called irrational numbers
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Equivalently, we say b
is greater than
a and write b > a
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We use this order property of R to represent real numbers as points on a line,
which is called a real number line, or simply a real line
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 A set is a collection of objects, and these objects are called the elements
of the set
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 Some sets can be described by listing their elements between braces
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 We can also write A in set-builder notation as A = {x ∈ Z | 0 < x < 7}
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If A and B are sets, then their union A ∪ B is the set consisting of all
elements that are in A or B (or in both A and B)
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In other words, A ∩ B is the common part of A
and B
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Here every element of A is also an element of B
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For example, N ⊆ Q and Q ⊆ R
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The
empty set is a subset of every other set

ATTEMPT THIS ON YOUR
OWN BEFORE LOOKING AT
THE SOLUTION

 An interval is a subset of R that contains all real numbers between
two endpoints
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 Notice that the endpoints of the interval a and b are excluded
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The closed interval from a to b is the set [a, b] = {x ∈ R | a ≤ x ≤ b}
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Distances are always positive or 0, so we have |a| ≥ 0 for every number a
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Example: Solve |3x + 5| = 1
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Thus, x =

−4
3

or x = −2

2
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Solution: By Property 5, |x − 4| < 1 is equivalent to −1 < x − 4 < 1
Therefore, adding 4 to each side, we have 3 < x < 5
Hence, the solution set is the open interval (3, 5)

3
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Solution: By Property 4 and 6, |2x + 3| ≥ 5 is equivalent to:

2x + 3 ≤ −5

or

2x + 3 ≥ 5

In the first case 2x ≤ −8, which gives x ≤ −4
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Hence, the solution set is S = {x ∈ R | x ≤ −4 or x ≥ 1}
= (−∞, −4] ∪ [1, ∞)

Example: If |x − 4| < 0
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2, use the Triangle Inequality to
estimate |(x + y) − 11|
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1 + 0
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3
Thus, |(x + y) − 11| < 0
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