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Title: Pre Calculus - Numbers and Inequalities with solved examples
Description: This chapter reviews real numbers, inequalities, intervals, and absolute values (9 pages).

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Chapter 1
Numbers and Inequalities
This chapter reviews real numbers, inequalities, intervals, and absolute values
...
1 Real Numbers
The real numbers can be represented geometrically as points on a number line called
the real line
...

We distinguish four special subsets of real numbers
...
};
2) The integers ℤ = {
...
};
3) The rational numbers ℚ, namely the numbers that can be expressed in the form
of a fraction , where
1
,
2

are integers and  0
...
For
examples
, √2, √5,

log 3
...
2 Inequalities
If

and

are real numbers and if

greater than or equal to


or

or



is a nonnegative number, we say that

is less than or equal to , and write, respectively,


...

1

>

Geometrically,

if the point on the real axis corresponding to

lies to the right

of the point corresponding to
...


Rules of inequalities
If , , and

are any given real numbers, then:

1) Either

< ,

2) < ⟹

=
±

<

> ;
± ;

3)

<

>0⟹

<

;

4)

<

<0⟹

>



5)

>0⟹



: <

⟹− >− ;

> 0;

6) If a and b are both positive or negative, then

< ⟹

>
...
3 Intervals
A subset of the real line is called an interval if it contains at least two numbers and
contains all the real numbers lying between any two of its elements
...

Types of intervals
Notation
( , )
Finite:

Infinite:

Set description
{ : < < }

Type
Open

[ , ]

{ :



≤ }

Closed

[ , )

{ :



< }

Half-open

( , ]

{ :

<

≤ }

Half-open

( , ∞)

{ :

> }

Open

[ , ∞)

{ :

≥ }

Closed

Mathematical Analysis

Picture

2

(−∞, )

{ :

< }

Open

(−∞, ]

{ :

≤ }

Closed

(−∞, ∞)


Real number

Open
Closed

1
...

Example 1
...

6
) 2 − 1 < + 3
) − < 2 + 1
)
≥5
3
−1
Solution
2 −1<

a)

+3

Add 1 to both sides, we have
2 <
Subtract

+ 4

from both sides, we get
<4

The solution set is the open interval (−∞, 4) (see Figure 1
...
1

b)



3

<2 +1

Multiply both sides by 3, we obtain
− < 6 +3
Add

to both sides, we have

Mathematical Analysis

3

0 <7 +3
Subtract 3 from both sides, we get
−3 < 7
Divide by 7, we have


3
<
7

The solution set is the open interval − , ∞ (see Figure 1
...
2

c)

The inequality

≥ 5 can hold only if

> 1, because otherwise

is

undefined or negative
...

5
The solution set is the half-open interval 1,

(see Figure 1
...
3
Mathematical Analysis

4

Example 1
...


=

or If 2 − 7 < 0 ⟹

>

and

<

and
3 +2
< 0
2 −7

3 +2
< 0
2 −7

Multiply both sides by 2 − 7

Multiply both sides by 2 − 7

3 +2< 0⟹

<−

2
3

3 +2> 0⟹

Mathematical Analysis

2
3

2 7
= − ,
3 2


The solution set is the interval

>−

=



= − ,


...
5 Absolute Value
The absolute value of a number , denoted by | |, is defined by the formula
| |=

,
− ,

≥ 0;
< 0
...

to 0 on the real number

line
...


2) | − | = the distance between x and y on the real line
...


The absolute value has the following properties
...

The following statements are all consequences of the definition of absolute value and
are often helpful when solving equations or inequalities involving absolute values
...


Example 1
...
4)
...
4
|2 − 3| ≥ 1

b)
2 −3≥1

2 − 3 ≤ −1

or

2 ≥4

2 ≤2

≥2

≤1

Add 3 to both sides
Divide by 2

The solution set is the interval (−∞, 1]⋃[2, ∞) (see Figure 1
...


Figure 1
...
5
...


b) We have
+
If

>2⟹

+ 2 > 0


+

− 2 > 0 ⟹ ( + 2)( − 1) > 0

−1> 0

> −2

or If

>1



= (1, ∞)
The solution set is the interval

+ 2 < 0
< −2

−1<0
<1

= (−∞, −2)
=



= (1, ∞) ∪ (−∞, −2)
...




− 6 = ( − 3)( + 2) ≥ 0,
8

it means that
( − 3) ≥ 0
≥ 3
So

( + 2) ≥ 0
≥ −2
= [3, ∞)

Then the solution set is

Mathematical Analysis



or

( − 3) ≤ 0
≤ 3
So

( + 2) ≤ 0
≤ −2

= (−∞, −2]

= [3, ∞)⋃(−∞, −2]
Title: Pre Calculus - Numbers and Inequalities with solved examples
Description: This chapter reviews real numbers, inequalities, intervals, and absolute values (9 pages).