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REAL NUMBERS

by

Nittaya Noinan

Kanchanapisekwittayalai Phetchabun School

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.

Natural Numbers

• Natural numbers are the set of counting numbers.

{1, 2, 3,…}

Whole Numbers

• Whole numbers are the set of numbers that include 0 plus the set of natural numbers.

{0, 1, 2, 3, 4, 5,…}

Integers

• Integers are the set of whole numbers and their opposites.

{…,-3, -2, -1, 0, 1, 2, 3,…}

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

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

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

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

Terminating decimalsTo 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.

.52 1.0

Terminating Decimals

• Terminating decimals are decimals that contain a finite number of digits.

• Examples:36.80.1254.5

Repeating decimalsTo 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.

.3333 1.000

.3

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 numberrepeats.

Repeating decimals

.3

The bar tells us that it is a repeating decimal.

The bar extends over the entire pattern that repeats.

.09

Rational numbers as decimals

Rational numbers can be converted from fractions to either

• Terminating decimals or

• Repeating decimals

Rational numbers

The subsets of real numbers that we’ve discussed are “nested” like Russian dolls.

Examples of Rational Numbers

•16•1/2•3.56

•-8•1.3333…

•- 3/4

To show how these number are classified, use the Venn diagram. Place the number where it belongs

on the Venn diagram.

9

4

2

1

Rational Numbers

Integers

Whole Numbers

NaturalNumber

s

Irrational Numbers

-12.64039…

117

0

6.369

4

-3

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

2

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.

Example• Classify all the following numbers as natural, whole,

integer, rational, or irrational. List all that apply.a. 117b. 0c. -12.64039…d. -½e. 6.36f. g. -3

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.

9

4

2

1

To show how these number are classified, use the Venn diagram. Place the number where it belongs

on the Venn diagram.

9

4

2

1

Rational Numbers

Integers

Whole Numbers

NaturalNumber

s

Irrational Numbers

-12.64039…

117

0

6.369

4

-3

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.

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.

Examples of Irrational Numbers

• Square roots of non-perfect “squares”

• Pi

17

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

Irrational NumbersThere 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!

25Num bers like and 81

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.

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...

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

3.141593.1416

3.1415926...

CLASS WORK1. Given the set,

list the elements of the set that are:

a) Natural numbersb) Integersc) Rational numbersd) Irrational numbers

13 151.001,0.333..., , 11,11, , 16,3.14,

15 3

Properties of Real Numbers

For any real number a, b and c

Addition MultiplicationClose 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


• Distributive property

– For all real numbers a, b, and c

a(b+c) = ab + ac and (b+c)a = ba + ca

Solving Equations; 5 Properties of Equality

Reflexive For any real number a, a=a

SymmetricProperty

For all real numbers a and b, if a=b, then b=a

TransitiveProperty

For all reals, a, b, and c, if a=b and b=c, then a=c

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

Applications of Equations

Problem Solving Plan1. Explore the Problem2. Plan the solution3. Solve the problem4. Examine the solution

Absolute Value Equations

Absolute value: Distance from zero

For any real number a:If , then If , then

0a a a

a a0a


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

Distributive Property: a(b+c) = ab + ac

you can add then multiply or multiply then add.

Theory about Real Numbers

Theorem 1. Eliminated Rule for Additionwhen 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


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) = -ab2. (-a)b = -ab3. (-a)(-b) = ab

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)


Theorem 7. When a , b and c are real numbers.

1. a(b – c) = ab – ac 2. (a – b)c = ac – bc3. (-a)(b – c) = -ab + ac

Theorem 8. When a 0 then a-1 0



1. when b , c 0

2. when b , c 0

3. when b , d 0

4. when b , d 0



5. when b , c 0

6. when b , c 0

7. when b , d 0

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

TRUE OR FALSETRUE OR FALSE

1. The set of WHOLE numbers is closed with respect to multiplication.


2. The set of NATURAL numbers is closed with respect to multiplication.


3. The product of any two REAL numbers is a REAL number.


4. The quotient of any two REAL numbers is a REAL number.


5. Except for 0, the set of RATIONAL numbers is closed under division.


6. Except for 0, the set of RATIONAL numbers contains

the multiplicative inverse for each of its members.


7. The set of RATIONAL numbers is associative under multiplication.


8. The set of RATIONAL numbers contains the additive inverse for each of its members.


9. The set of INTEGERS is commutative under subtraction.


10. The set of INTEGERS is closed with respect to division.

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

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}

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)


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 = dHence remainder equal is p(c)


Example 1. Find the remainder when 9x3 + 4x - 1

is divided by x - 2Solve.

9x2 + 18x + 40

9x3 - 18x2

18x2 + 4x – 1 18x2 - 36x 40x – 1 40x – 8079

Thus remainder is 79


Example 1. Find the remainder when 9x3 + 4x - 1

is divided by x - 2Solve.

Give p(x) = 9x3 + 4x - 1 thus p(2) = 9(2)3 + 4(2) - 1 = 72 + 8 – 1 = 79

Thus the remainder is 79.


Example 2. Find the remainder when

2x4 - 7x3 + x2 + 7x – 3 is divided by x + 1Solve.

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.

Factor theorem.


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

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

Factor theorem.

Example. 1) Write x4 – x3 -2x2 – 4x – 24 be factor.

Solve.

Thus x + 2 is the factor of x4 – x3 -2x2 – 4x – 24

Take x + 2 divide x4 – x3 -2x2 – 4x – 24 then get x3 -3x2 + 4x – 12Thuse x4 – x3 -2x2 – 4x – 24 = (x + 2)(x3 -3x2 + 4x – 12)

= (x + 2){(x3 -3x2 )+ 4(x – 3)} = (x + 2){x2 (x -3) + 4(x – 3)} = (x + 2)(x – 3)(x2 + 4)

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

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)

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

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)


Rational factorization theorem.


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 .

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.

= 0


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)

Example. 2) Solving Equation 6x3 - 11x2 + 6x = 1

Solve.Since 6x3 - 11x2 + 6x = 1

Then we get 6x3 - 11x2 + 6x -1 = 0Give 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 = 0Thus x = 1 or x = or x = Answer {1 , , }


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)

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 .

Definition.

a b that mean a less than or equal to ba b that mean a more than or equal to ba < b < c that mean a < b and b < c a b c that mean a b and b ca < b c that mean a < b and b ca b < c that mean a b and b < c


Give a , b , c be real numbers1. 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


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 > bcCase 2 if a > b and c < o then ac < bc


Give a , b , c be real numbers

5. Properties excision for Addition. if a + c > b + c then a > bSuch as 5 + 2 > 3 + 2 then 5 > 3

6. Properties excision for Multiplication.Case 1 if ac > bc and c > 0 then a > bSuch as 6 3 > 4 3 and 3 > 0 then 6 > 4Case 2 if ac > bc and c < 0 then a < bSuch as 3 (-3) > 4 (-3) and -3 < 0 then 3 < 4

IntervalsNotation Graph Set-builder

Notation

(a, b)

[a, b]

[a, b)

(a, b]

(a, )

[a, )

(-, b)

(-, b]

(-, )

b

b

Solving inequalities.

Exmaple. Solving inequalities following problems.

1. 3x + 5 < x – 7Solve.

Since 3x + 5 < x – 73x – x < -7 – 5 2x < -12 x < -6

Answer {x/x < -6} or (- , -6)



2. 4y + 7 > 2(y + 1)Solve.

Since 4y + 7 > 2(y + 1) 4y + 7 > 2y + 24y – 2y > 2 - 7 2y > -5

y >

Answer {y/y > } or ( , )



3. x2 – x – 6 0 Solve.

Since x2 – x – 6 0 (x – 3)(x + 2) 0

Then x – 3 = 0 or x + 2 = 0We get x = 3 or x = -2 Thus + - + -2 3 Answer { x / -2 x 3 } or [-2 , 3]



4. 2x2 + 7x + 3 0 Solve.

Since 2x2 + 7x + 3 0 (2x + 1)(x + 3) 0

Then 2x + 1 = 0 or x + 3 = 0We get x = or x = -3 Thus + - + -3

Answer (- , -3) ( , )



5. Solve.

Since



5. Solve.

Take multiplicate all sides Then

(x – 4)2

then we take (-1) multiplicate all sidesWe get

Then x – 5 = 0 or x + 2 = 0 or x – 4 = 0 x = 5 or x = -2 or x = 4 - + - + -2 4 5

Answer [-2 , 4] [5 , )

(x2 – 3x – 10)(x – 4) 0

(x – 5)(x + 2)(x – 4) 0

And if the inequalities to a degree greater than two.

We can use the Remainder Theorem Method .

Absolute value inequalities.

Definition.give a be real number

a if a > 0 a = 0 if a = 0 -a if a < 0

Absolute value inequalities.

Theorem.when x and y be real number

1. x = -x 2. xy = x y 3. = , y 04. x - y = y - x 5. x = x 6. x + y x + y

2 2

Solving equations and inequalities in Absolute

value.

Theory 11. when a be positive number

set of the answers of a = a is {a , -a}

Example. Find the answer of the following

equations.

1. 2x - 3 = 9

Solve.since 2x - 3 = 9then 2x -3 = 9 or 2x – 3 = -9

x = 6 or x = -3

Answer {-3 , 6}


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}

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 }

Solving equations and inequalities in Absolute

value.

Theory 12. when a be positive number

1. a < a that mean -a < x < a2. a a that mean -a x a3. a > a that mean x <-a or x > a4. a a that mean x -a or x a


equations.

1. 4x - 3 < 1

Solve.since 4x - 3 < 1then -1 < 4x – 3 < 1

2 < 4x < 4 < x < 1

Answer ( , 1)


equations.

2. x - 3 2

Solve.since x - 3 2then x -3 -2 or x - 3 2

x 1 or x 5

Answer (- , 1] [ 5 , )


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 , )


4. 2x + 4 > x + 1

Solve.since 2x + 4 > x + 1

Then 2x + 4 <-(x + 1) or 2x + 4 > x + 13x < -5 or x > -3

x < - or x > -3 Answer (- , - ) (-3 , )


5. x < 2x - 1

Solve.since x < 2x - 1

Then x < (2x – 1)

22

x2 < 4x2 – 4x + 1

0 < 3x2 – 4x + 1

0 < (3x -1)(x – 1)


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)

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