2. Real Numbers
• Real numbers consist of all the
rational and irrational numbers.
• The real number system has many
subsets:
– Natural Numbers
– Whole Numbers
– Integers
– Ect.
4. Whole Numbers
• Whole numbers are the set of numbers
that include 0 plus the set of natural
numbers.
{0, 1, 2, 3, 4, 5,…}
5. Integers
• Integers are the set of whole numbers and
their opposites.
{…,-3, -2, -1, 0, 1, 2, 3,…}
6. Rational Numbers
• Rational numbers are any numbers that can
be expressed in the form of , where a and
b are integers, and b ≠ 0.
• They can always be expressed by using
terminating decimals or repeating decimals.
b
a
7. The Rational Numbers
The word ratio means fraction.
Therefore rational numbers are any numbers which
can be written as fractions.
2
3
3
4
5
1
1
5
8. Integers are Rational
Numbers
Like the 5 in our example, any integer can be made
into a fraction by putting it over 1. Since it can
be a fraction, it is a rational number.
2
3
3
4
5
1
1
5
9. Changing fractions to
decimals
It’s easy to change a fraction to a decimal, so
rational numbers can also be written as decimals.
Rational numbers convert to two different types
of decimals:
Terminating decimals – which end
Repeating decimals – which repeat
10. Terminating decimals
To convert a fraction to a decimal, divide the top by the
bottom.
To convert ½ to a decimal you would do:
There is no remainder. The answer just ends – or terminates.
.5
2 1.0
12. Repeating decimals
To convert a fraction to a decimal, divide the top by the
bottom.
To convert 1
/3 to a decimal you would do:
=0.3333…
There is a remainder. The answer just keeps repeating.
.333
3 1.000
.3
−
13. Repeating Decimals
• Repeating decimals are decimals that contain a
infinite number of digits.
• Examples:
0.333…
0.19191919…
7.689689…
FYI…The line above the decimals indicate that number
repeats.
14. Repeating decimals
.3
−
The bar tells us that it is a repeating decimal.
The bar extends over the entire pattern that
repeats.
.09
18. To show how these number are classified, use the Venn
diagram. Place the number where it belongs on the Venn
diagram.
π
9
4
2
1
−
Rational Numbers
Integers
Whole Numbers
Natural
Numbers
Irrational Numbers
-12.64039…
117
0
6.36
9
4
-3
19. Irrational Numbers
• Irrational numbers are any numbers that cannot be
expressed as .
• They are expressed as non-terminating, non-
repeating decimals; decimals that go on forever
without repeating a pattern.
• Examples of irrational numbers:
– 0.34334333433334…
– 45.86745893…
– (pi)
–
b
a
π
2
20. Other Vocabulary Associated
with the Real Number System
• …(ellipsis)—continues without end
• { } (set)—a collection of objects or
numbers. Sets are notated by using
braces { }.
• Finite—having bounds; limited
• Infinite—having no boundaries or
limits
• Venn diagram—a diagram consisting
of circles or squares to show
relationships of a set of data.
21. Example
• Classify all the following numbers as natural, whole, integer,
rational, or irrational. List all that apply.
a. 117
b. 0
c. -12.64039…
d. -½
e. 6.36
f.
g. -3
π
22. Solution
• Now that all the numbers are placed where they belong
in the Venn diagram, you can classify each number:
– 117 is a natural number, a whole number, an integer,
and a rational number.
– is a rational number.
– 0 is a whole number, an integer, and a rational
number.
– -12.64039… is an irrational number.
– -3 is an integer and a rational number.
– 6.36 is a rational number.
– is an irrational number.
– is a rational number.
π
9
4
2
1
−
23. To show how these number are classified, use the Venn
diagram. Place the number where it belongs on the Venn
diagram.
π
9
4
2
1
−
Rational Numbers
Integers
Whole Numbers
Natural
Numbers
Irrational Numbers
-12.64039…
117
0
6.36
9
4
-3
24. FYI…For Your Information
• When taking the square root of any
number that is not a perfect square,
the resulting decimal will be non-
terminating and non-repeating.
Therefore, those numbers are
always irrational.
25. Irrational Numbers
• An irrational number is a
number that cannot be
written as a ratio of two
integers.
• Irrational numbers written as
decimals are non-terminating
and non-repeating.
27. Irrational Numbers
In English, the word “irrational” means not rational -
illogical, crazy, wacky.
In math, irrational numbers are not rational.
They usually look wacky!
…and their decimals never end or repeat!
3
175 π
28. Irrational Numbers
There is one trick you need to watch out for!
They look wacky but because the number in the house is
a perfect square, they are really the integers 5 and 9
in disguise!
Sort of like the wolf at Grandma’s house!
25Numbers like and 81
29. Rounding or truncating
Some decimals are much longer than we need. There
are two ways we can make them shorter.
Truncating – just lop the extra digits off.
Rounding – use the digit to the right of the one we
want to end with to determine whether to round
up or not. If that digit is 5 or higher, round up.
30. Truncating
Truncating – just lop the extra digits off.
If we want to use with just 4 decimal places.
We’d just chop off the rest!
3.1415/926…
3.1415
Truncate ~ tree trunk ~ chop!
3.1415926...π =
π
31. Rounding
If we want to round to 4 decimal places.
We’d look at the digit in the 5th
place
9 is “5 or bigger” so the digit in the 4th
spot goes up
3.14159
3.1416
3.1415926...π =
π
32. CLASS WORK
1. Given the set,
list the elements of the set that are:
a) Natural numbers
b) Integers
c) Rational numbers
d) Irrational numbers
13 15
1.001,0.333..., , 11,11, , 16,3.14,
15 3
π
−
33. Properties of Real Numbers
For any real number a, b and c
Addition Multiplication
Close a + b ∈ R ab ∈ R
Commutative a + b = b + a ab = ba
Associative (a + b) + c = a + (b + c) (ab)c = a(bc)
Identity A + 0 = a = 0 + a A1 = a = 1a
Inverse A + (-a) = 0 = (-a) + a If a in note zero then
a-1
.a = 1 =a. a-1
34. Properties of Real Numbers
• Distributive property
– For all real numbers a, b, and c
a(b+c) = ab + ac and (b+c)a = ba + ca
35. Solving Equations;
5 Properties of Equality
Reflexive For any real number a, a=a
Symmetric
Property
For all real numbers a and b,
if a=b, then b=a
Transitive
Property
For all reals, a, b, and c,
if a=b and b=c, then a=c
36. Solving Equations;
5 Properties of Equality
Addition and
Subtraction
For any reals a, b, and c, if
a=b then a+c=b+c and a-
c=b-c
Multiplication
and Division
For any reals a, b, and c, if
a=b then a*c=b*c, and, if c is
not zero, a/c=b/c
37. Applications of Equations
Problem Solving Plan
1. Explore the Problem
2. Plan the solution
3. Solve the problem
4. Examine the solution
39. Properties of Real Numbers
Commutative Property:
a + b = b + a
ab = ba order doesn’t matter
Associative Property:
(a+b)+c = a+(b+c)
(ab)c = a(bc)
order doesn’t change
41. Theory about Real Numbers
Theorem 1. Eliminated Rule for Addition
when a , b and c are real numbers.
(i) if a + c = b + c then a = b
(ii) if a + b = a + c then b = c
Theorem 2. Eliminated Rule for Multiplication
when a , b and c are real numbers.
(i) if ac = bc and c ≠ 0 then a = b
(ii) if ab = ac and a ≠ 0 then b = c
42. Theory about Real Numbers
Theorem 3. When a is real numbers. a.0 = 0
Theorem 4. When a is real numbers. (-1)a = -a
Theorem 5. When a and b are real numbers.
if ab = 0 then a = 0 or b = 0
Theorem 6. When a and b are real numbers.
1. a(-b) = -ab
2. (-a)b = -ab
3. (-a)(-b) = ab
43. Subtraction and Divisor of Real Numbers
Definition. When a and b are real numbers.
a – b = a + (-b)
Definition. When a and b are real numbers.
= a(b-1
)
44. Theory about Real Numbers
Theorem 7. When a , b and c are real numbers.
1. a(b – c) = ab – ac
2. (a – b)c = ac – bc
3. (-a)(b – c) = -ab + ac
Theorem 8. When a ≠ 0 then a-1
≠ 0
45. Theory about Real Numbers
Theorem 9. When a , b and c are real numbers.
1. when b , c ≠ 0
2. when b , c ≠ 0
3. when b , d ≠ 0
4. when b , d ≠ 0
46. Theory about Real Numbers
Theorem 9. When a , b and c are real numbers.
5. when b , c ≠ 0
6. when b , c ≠ 0
7. when b , d ≠ 0
47. CLASS WORK
State the property of real numbers
being used.
2.
3.
4.
( ) ( )2 3 5 3 5 2+ = +
( )2 2 2A B A B+ = +
( ) ( )2 3 2 3p q r p q r+ + = + +
48. TRUE OR FALSE
1. The set of WHOLE
numbers is closed
with respect to
multiplication.
49. TRUE OR FALSE
2. The set of NATURAL
numbers is closed with
respect to
multiplication.
50. TRUE OR FALSE
3. The product of any two
REAL numbers is a
REAL number.
51. TRUE OR FALSE
4. The quotient of any
two REAL numbers is a
REAL number.
52. TRUE OR FALSE
5. Except for 0, the set of
RATIONAL numbers is
closed under division.
53. TRUE OR FALSE
6. Except for 0, the set of
RATIONAL numbers
contains the
multiplicative inverse for
each of its members.
54. TRUE OR FALSE
7. The set of RATIONAL
numbers is associative
under multiplication.
55. TRUE OR FALSE
8. The set of RATIONAL
numbers contains the
additive inverse for
each of its members.
56. TRUE OR FALSE
9. The set of INTEGERS
is
commutative under
subtraction.
57. TRUE OR FALSE
10. The set of INTEGERS
is closed with respect
to division.
58. Solving polynomial Equations one variable.
We can write polynomial Equations of x variable that is
anxn
+ an-1xn-1
+ an-2xn-2
+ … + a1x + a0
when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be
coefficiants of the polynomail are real numbers by a ≠ 0
Then we can called anxn
+ an-1xn-1
+ an-2xn-2
+ … + a1x + a0
is polynomail of degree n.
The symbol is p(x) , q(x) , r(x) and if p(a) that mean we instead x in
p(x) by a .
Example. P(x) = x3
– 4x2
+ 3x + 2
p(1) = 13
– 4(1)2
+ 3(1) + 2 = 2
59. Solving polynomial Equations one variable.
Example. Find the answer of 3x3
+ 2x2
- 12x – 8 = 0
Solve.
by use Addition and Multiplication of real numbers. Then we
can multiplied by the following factor.
3x3
+ 2x2
- 12x – 8 = (3x3
+ 2x2
) – (12x + 8)
= x2
(3x + 2) – 4(3x + 2)
= (3x + 2)(x2
- 4)
= (3x + 2)(x - 2)(x + 2)
By Theory 5 then x = , or x = 2 or x = -2
Answer {-2 , , 2}
60. Solving polynomial Equations by remainder
Theorem Method.
Remainder theorem.
When p(x) is anxn
+ an-1xn-1
+ an-2xn-2
+ … + a1x + a0
when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be
real numbers by a ≠ 0 .
if p(x) is divied by x – c when c is real number then
the remainder is equal p(c)
61. Solving polynomial Equations by remainder
Theorem Method.
Proof.
Give p(x) is divied by x – c then we get quotient q(x)
And remainder be q(x)
Thus p(x) = (x – c)q(x) + r(x) ……(1)
Which r(x) is zero or polynomail of degree is less than
x – c that mean degree 0. hence r(x) is constant.
Give r(x) = d when d is constant.
Thus p(x) = (x – c)q(x) + d ……(2)
when instead x in (2) by c
We get p(c) = (c – c)q(x) + d = d
Hence remainder equal is p(c)
62. Solving polynomial Equations by remainder
Theorem Method.
Example 1. Find the remainder when 9x3
+ 4x - 1
is divided by x - 2
Solve.
9x2
+ 18x + 40
9x3
- 18x2
18x2
+ 4x – 1
18x2
- 36x
40x – 1
40x – 80
79
Thus remainder is 79
63. Solving polynomial Equations by remainder
Theorem Method.
Example 1. Find the remainder when 9x3
+ 4x - 1
is divided by x - 2
Solve.
Give p(x) = 9x3
+ 4x - 1
thus p(2) = 9(2)3
+ 4(2) - 1
= 72 + 8 – 1
= 79
Thus the remainder is 79.
64. Solving polynomial Equations by remainder
Theorem Method.
Example 2. Find the remainder when
2x4
- 7x3
+ x2
+ 7x – 3 is divided by x + 1
Solve.
Give p(x) = 2x4
- 7x3
+ x2
+ 7x – 3
Since x +1 = x – (-1) thus c = -1
thus p(-1) = 2(-1)4
– 7(-1)3
+ (-1)2
+ 7(-1) – 3
= 2 + 7 + 1 – 7 – 3
= 0
Thus the remainder is 0.
65. Factor theorem.
When p(x) is anxn
+ an-1xn-1
+ an-2xn-2
+ … + a1x + a0
when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be
real numbers by a ≠ 0 .
p(x) there is x – c that is factor iff p(c) = 0
66. Factor theorem.
Example. 1) Write x4
– x3
-2x2
– 4x – 24 be factor.
Solve.
Give p(x) = x4
– x3
-2x2
– 4x – 24
since integers that can divide -24 are
±1,±2,±3,±4,±6,±8,±12,
±24
Then consider p(1) ,p(-1) ,p(2) that is not equal zero.
But p(-2) = (-2)4
– (-2)3
-2(-2)2
– 4(-2) – 24
= 16 + 8 – 8 + 8 – 24
= 0
Thus x + 2 is the factor of x4
– x3
-2x2
– 4x – 24
68. Factor theorem.
Example. 2) Write x3
-5x2
+ 2x + 8 be factor.
Solve.
Give p(x) = x3
-5x2
+ 2x + 8
since integers that can divide 28 are
±1,±2,±4,±8
Then consider p(1) ,p(-1) ,p(-2) that is not equal zero.
But p(2) = (2)3
-5(2)2
+ 2(2) + 8
= 8 - 20 + 4 + 8
= 0
Thus x - 2 is the factor of x3
-5x2
+ 2x + 28
69. Factor theorem.
Example. 2) Write x3
-5x2
+ 2x + 8 be factor.
Solve.
Thus x - 2 is the factor of x3
-5x2
+ 2x + 8
Take x - 2 divide x3
-5x2
+ 2x + 8 then get x2
- 3x – 4
Thuse x3
-5x2
+ 2x + 8 = (x - 2)(x2
- 3x – 4)
= (x - 2)(x – 4)(x + 1)
70. Factor theorem.
Example. 3) Write x3
+ 2x2
- 5x - 6 be factor.
Solve.
Give p(x) = x3
+ 2x2
- 5x - 6
since integers that can divide 6 are
±1,±2,±3,±6
Then consider p(1) that is not equal zero.
But p(-1) = (-1)3
+ 2(-1)2
– 5(-1) - 6
= -1 + 2 + 5 - 6
= 0
Thus x + 1 is the factor of x3
+ 2x2
- 5x - 6
71. Factor theorem.
Example. 3) Write x3
+ 2x2
- 5x - 6 be factor.
Solve.
Thus x + 1 is the factor of x3
+ 2x2
- 5x - 6
Take x + 1 divide x3
+ 2x2
- 5x - 6 then get x2
+ x – 6
Thuse x3
+ 2x2
- 5x - 6 = (x + 1)(x2
+ x – 6)
= (x + 1)(x – 2)(x + 3)
72. Solving polynomial Equations by remainder
Theorem Method.
Rational factorization theorem.
When p(x) is anxn
+ an-1xn-1
+ an-2xn-2
+ … + a1x + a0
when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be
real numbers by a ≠ 0 .
if x - is factor of polynomial p(x) by m and k are integer
which m ≠ 0 and greatest common factor of m and k is equal 1
Then m can divied an and k can divided a0 .
73. Example. 1) Write 12x3
+ 16x2
- 5x – 3 be factor.
Solve.
Give p(x) = 12x3
+ 16x2
- 5x – 3
since integers that can divide -3 are ±1,±2,±3
And integers that can divide 12 are ±1,±2,±3,
±4,±6,±12
Then the rational number that p( ) = 0 in among
±1,±2,±3,
Then consider p( ) that is equal zero.
74. Example. 1) Write 12x3
+ 16x2
- 5x – 3 be factor.
Solve.
Thus x - is the factor of 12x3
+ 16x2
- 5x – 3
Take x - divide 12x3
+ 16x2
- 5x – 3 then get 12x2
+ 22x + 6
Thuse 12x3
+ 16x2
- 5x – 3 = (x - )(12x2
+ 22x + 6)
= (x - )(2) (6x2
+ 11x + 3)
= (2x – 1)(6x2
+ 11x + 3)
= (2x – 1)(3x + 1)(2x + 3)
75. Example. 2) Solving Equation 6x3
- 11x2
+ 6x = 1
Solve.
Since 6x3
- 11x2
+ 6x = 1
Then we get 6x3
- 11x2
+ 6x -1 = 0
Give p(x) = 6x3
- 11x2
+ 6x -1
p(1) = 6 – 11 + 6 – 1 = 0
Thus p(x) = (x -1)(6x2
- 5x + 1)
= (x – 1)(2x – 1)(3x – 1)
Since 6x3
- 11x2
+ 6x -1 = 0
(x – 1)(2x – 1)(3x – 1) = 0
Then x – 1 = 0 or 2x -1 = 0 or 3x – 1 = 0
Thus x = 1 or x = or x =
Answer {1 , , }
76. Example. 1) Write 12x3
+ 16x2
- 5x – 3 be factor.
Solve.
Thus x - is the factor of 12x3
+ 16x2
- 5x – 3
Take x - divide 12x3
+ 16x2
- 5x – 3 then get 12x2
+ 22x + 6
Thuse 12x3
+ 16x2
- 5x – 3 = (x - )(12x2
+ 22x + 6)
= (x - )(2) (6x2
+ 11x + 3)
= (2x – 1)(6x2
+ 11x + 3)
= (2x – 1)(3x + 1)(2x + 3)
77. Properties of inequalities.
IN real number system we use symbol < , > , ≤ ,≥ , ≠
be less than , more than , less than or equal to ,
more than or equal to ,not equal sort by order
if a and be be real number the symbol a < b that
mean a less than b and a > b that mean a more than b
Trichotomy property.
if a and b be real number then a = b , a < b and a > b
That was actually only one .
78. Definition.
a ≤ b that mean a less than or equal to b
a ≥ b that mean a more than or equal to b
a < b < c that mean a < b and b < c
a ≤ b ≤ c that mean a ≤ b and b ≤ c
a < b ≤ c that mean a < b and b ≤ c
a ≤ b < c that mean a ≤ b and b < c
79. Properties of inequalities.
Give a , b , c be real numbers
1. Transitive Property.
if a > b and b > c then a > c
Such as 5 > 3 and 3 > 1 then 5 > 1
2. Properties of Addition and Subtraction.
So adding (or subtracting) the same value to both a and
b will not change the inequality
if a > b then a + c > b + c
Such as 4 > 2 then 4 + 1 > 2 + 1
80. Properties of inequalities.
Give a , b , c be real numbers
3.Positive and Negative number when compare with zero
a is positive number iff a > 0
a is negative number iff a < 0
4. Property of Multiplication but not with zero.
Case 1 if a > b and c > 0 then ac > bc
Case 2 if a > b and c < o then ac < bc
81. Properties of inequalities.
Give a , b , c be real numbers
5. Properties excision for Addition.
if a + c > b + c then a > b
Such as 5 + 2 > 3 + 2 then 5 > 3
6. Properties excision for Multiplication.
Case 1 if ac > bc and c > 0 then a > b
Such as 6 × 3 > 4 × 3 and 3 > 0 then 6 > 4
Case 2 if ac > bc and c < 0 then a < b
Such as 3 × (-3) > 4 × (-3) and -3 < 0 then 3 < 4
85. Solving inequalities.
Exmaple. Solving inequalities following problems.
3. x2
– x – 6 ≤ 0
Solve.
Since x2
– x – 6 ≤ 0
(x – 3)(x + 2) ≤ 0
Then x – 3 = 0 or x + 2 = 0
We get x = 3 or x = -2
Thus + - +
-2 3
Answer { x / -2 ≤ x ≤3 } or [-2 , 3]
86. Solving inequalities.
Exmaple. Solving inequalities following problems.
4. 2x2
+ 7x + 3 ≥ 0
Solve.
Since 2x2
+ 7x + 3 ≥ 0
(2x + 1)(x + 3) ≤ 0
Then 2x + 1 = 0 or x + 3 = 0
We get x = or x = -3
Thus + - +
-3
Answer (-∝ , -3) ∪ ( , ∝)
89. then we take (-1) multiplicate all sides
We get
Then x – 5 = 0 or x + 2 = 0 or x – 4 = 0
x = 5 or x = -2 or x = 4
- + - +
-2 4 5
(x2
– 3x – 10)(x – 4) ≥ 0
(x – 5)(x + 2)(x – 4) ≥ 0
90. And if the inequalities to a degree
greater than two.
We can use the Remainder Theorem
Method .
92. Absolute value inequalities.
Theorem.
when x and y be real number
1.| x | = | -x |
2.| xy | = | x || y |
3. = , y ≠ 0
4.| x - y | = | y - x |
5.| x | = x
6.| x + y | ≤ | x | + | y |
2 2
94. Example.
Find the answer of the following equations.
1. |2x - 3| = 9
Solve.
since |2x - 3| = 9
then 2x -3 = 9 or 2x – 3 = -9
x = 6 or x = -3
Answer {-3 , 6}
95. Example.
Find the answer of the following equations.
2. |3x - 1| = |x + 5|
Solve.
since |3x - 1| = |x + 5|
then 3x -1 = x + 5 or 3x -1 = -(x + 5)
2x = 6 or 4x = -4
x = 3 or x = -1
Answer {-1 , 3}
96. Example.
Find the answer of the following equations.
3. |2x + 1| = 3x - 5
Solve.
since |2x + 1| = 3x - 5
then 3x – 5 ≥ 0 and [2x +1 = 3x - 5 or 2x +1 = -(3x - 5)
x ≥ and x = 6 0r x =
Answer { 6 }
97. Solving equations and
inequalities in Absolute value.
Theory 12. when a be positive number
1. |a| < a that mean -a < x < a
2. |a| ≤ a that mean -a ≤ x ≤ a
3. |a| > a that mean x <-a or x > a
4. |a| ≥ a that mean x ≤-a or x ≥ a
98. Example.
Find the answer of the following equations.
1. |4x - 3| < 1
Solve.
since |4x - 3| < 1
then -1 < 4x – 3 < 1
2 < 4x < 4
< x < 1
Answer ( , 1)
99. Example.
Find the answer of the following equations.
2. |x - 3| ≥ 2
Solve.
since |x - 3| ≥ 2
then x -3 ≤ -2 or x - 3 ≥ 2
x ≤ 1 or x ≥ 5
Answer (-∝ , 1] ∪ [ 5 , ∝)
100. Example.
Find the answer of the following equations.
3. |x + 1| ≤ 2x - 3
Solve.
since |x + 1| ≤ 2x – 3
Then -(2x – 3) ≤ x + 1 ≤ 2x – 3
-(2x – 3) ≤ x + 1 and x + 1 ≤ 2x – 3
-3x ≤ -2 and -x ≤ -4
x ≥ and x ≥ 4
Answer [4 , ∝)
101. Example.
Find the answer of the following equations.
4. |2x + 4| > x + 1
Solve.
since |2x + 4| > x + 1
Then 2x + 4 <-(x + 1) or 2x + 4 > x + 1
3x < -5 or x > -3
x < - or x > -3
Answer (-∝ , - ) ∪ (-3 , ∝)
102. Example.
Find the answer of the following equations.
5. |x| < |2x - 1|
Solve.
since |x| < |2x - 1|
Then x < (2x – 1)2
2
x2
< 4x2
– 4x + 1
0 < 3x2
– 4x + 1
0 < (3x -1)(x – 1)
103. Example.
Find the answer of the following equations.
5. |x| < |2x - 1|
Solve.
Then 3x – 1 = 0 or x – 1 = 0
x = or x = 1
+ - +
1
Answer (-∝ , ) ∪ (1 , ∝)
0 < (3x -1)(x – 1)
