Omancollege of Managment and Techonolgy

OMANCOLLEGE OF MANAGMENT AND TECHONOLGY

MATHEMATICS

Handouts

Real Numbers

Rational Numbers Irrational Numbers

Set of Real Numbers is defined as the union of Rational and Irrational Numbers

Rational Numbers: Numbers which can be written in the form p ∕ q

For example ,, 2.5 and 3.777…

Irrational Numbers: Numbers which can not be written in the form p ∕ q

For example , and 1.709975947…

Examples

Differentiate between Rational and Irrational Numbers

2.5 

√2 
3.66666…
5 ∕ 3 
3 ∕ 5 
7 ∕ 5 
6.8 ∕ 3.4 
√93 
56 
7 ∕ 3 
3 ∕ 7 

Note:

Set of Irrational Numbers has no subset
Set of Rational Numbers has so many sub sets

for example

Set of Natural Numbers:N= {1, 2, 3, …}

Set of Whole Numbers:W= {0, 1, 2, 3, …}

Set of Odd Numbers:O= {1, 3, 5, …}

Set of Even Numbers:E= {2, 4, 6, …}

Set of Prime Numbers:P= {2, 3, 5, 7, 11, 13, 17, …}

Set of Integers:Z= {0, 1, 2, 3, …}

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Operations on Two Sets

Two sets can be combined in many different ways like

Union
Intersection
Difference
Complement

Example1
If A={1, 2, 3, 4, 5, 6}
B={1, 2, 3, 4, 5, 6, 7, 8}
Then find
A U B = {1, 2, 3, 4, 5, 6, 7, 8}
A ∩ B = {1, 2, 3, 4, 5, 6}
A - B = ¢ or {}
B - A = {7, 8} / Example2
If A={2, 4, 6, 8, 10}
B={1, 3, 5, 7, 9}
Then find
A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A ∩ B = ¢ or {}
A - B = {2, 4, 6, 8, 10}
B - A = {1, 3, 5, 7, 9}
Example3
If A={5, 10, 15, 20, 25, 30}
B={10, 20, 30, 40, 50}
Then find
A U B = {5, 10, 15, 20, 25, 30, 40, 50}
A ∩ B = {10, 20, 30}
A - B = {5, 15, 25}
B - A = {40, 50} / Example4
If U={1, 2, 3, 4, 5, 6, 7}
A={1, 2, 3}
B={4, 5, 6, 7}
Then find
Ac = U – A = {4, 5, 6, 7}
Bc = U – B = {1, 2, 3}

Example51

If U={11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

A={11, 12, 13, 14, 15, 16}

B={15, 16, 17, 18, 19, 20}

Then find

A U B = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

A ∩ B = {15, 16}

A - B = {11, 12, 13, 14}

B - A = {17, 18, 19, 20}

Ac = U – A = {17, 18, 19, 20}

Bc= U – B = {11, 12, 13, 14}

Ac ∩ Bc = ¢ or {}

(A U B) c = U –(A U B) = ¢ or {}

Example6

If U={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A={5, 6, 7, 8, 9}

B={1, 2, 3, 4, 5, 8, 9, 10}

Then find

A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A ∩ B = {5, 8, 9}

A - B = {6, 7}

B - A = {1, 2, 3, 4, 10}

Ac = {1, 2, 3, 4, 10}

Bc= {6, 7}

(A ∩ B) c = {1, 2, 3, 4, 6, 7, 10}

Ac U Bc = {1, 2, 3, 4, 6, 7, 10}

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De Morgan's Laws

(A ∩ B) c = Ac U Bc
(A U B) c = Ac ∩Bc

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Rationalization

It is a process in which we remove the radical sign (from Numerator or Denominator)

Example1
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
= =
D = = / Example2
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
= =
D = =
Example3
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
= =
D =
= / Example4
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
= =
D = = =
Example5
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
=
=
= / Example6
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
=
=
=
Example7
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
=
=
= / Example8
Rationalize the denominator of
Solution:
=
Multiplying and dividing with
= ×
=
=
=
=

Similarly we can remove radical sign from numerator
Rationalization is an important process in various calculations
Rationalization is widely used in higher studies

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Exponents (Power)

There are some rules to solve problems including exponents. We have to learn and follow these rules in different calculations

Example1
Simplify 2
Solution:
= 2 × 2 × 2
= 8 / Example2
Simplify 3
Solution:
= 3 × 3 × 3 × 3 × 3
= 243
Example3
Simplify 27
Solution:
= (3 × 3 × 3)
= (3)
= 3 / Example4
Simplify 32
Solution:
= (2× 2 ×2 × 2 × 2)
= (2)
= 2
Example5
Simplify 64
Solution:
= (2× 2 ×2 × 2 × 2 × 2 )
= (2)
= 2
= 8 / Example6
Simplify 125
Solution:
= (5 × 5 × 5)
= (5)
= 5= 5fde
Example7
Simplify 9
Solution:
= (3 × 3)
= (3)
= 3
= 27 = / Example8
Simplify 32
Solution:
= (2 × 2 ×2 × 2 × 2)
= (2)
= 2
= 4
Example9
Simplify 16
Solution:
= (2 × 2 ×2 × 2)
= (24)
= 2
= 8 / Example10
Simplify 125
Solution:
= (5 × 5 × 5)
= (5)
= 5

Rules for Multiplication and Division of Exponents

(i)If same numbers are multiplying with each other then add their powers

(ii)If same numbers are dividing with each other then subtract their powers

(power in the numerator minus power in the denominator)

Example1
Simplify 2 × 2
Solution:
= 2
= 2 / Example2
Simplify 3 × 3
Solution:
= 3
= 3
Example3
Simplify 2 × 2
Solution:
= 2
= 2
= 2 / Example4
Simplify 3 × 3
Solution:
= 3
= 3
= 3
Example5
Simplify
Solution:
= 5
= 5 / Example6
Simplify
Solution:
= 7
= 7= 5fde
Example7
Simplify 2 × 2
Solution:
= 2
= (2)
= 2
= 4 = / Example8
Simplify 2 × 2
Solution:
= 2
= (2)
= (2)
= (2)
Example9
Simplify
Solution:
= 7
= 7
= 7
= 7 = / Example10
Simplify
Solution:
= 11
= 11
= 11

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Algebraic Expressions

Algebraic Term: 4 x

Degree of Polynomial:

Highest power of the variable in a polynomial is called the degree of polynomial

Example1
What is the degree of Polynomial
x - x+ 3
Solution:
Degree = 5 / Example2
What is the degree of Polynomial
2 - y - y+ 2 y
Solution:
Degree = 8
Example3
What is the degree of Polynomial
5t -
Solution:
Degree = 1 / Example4
What is the degree of Polynomial
3
Solution:
Degree = 0

Types of Polynomial:

There are so many types of polynomial but these three types are important

(i)Linear

(ii)Quadratic

(iii)Cubic

Linear:If degree of the polynomial is 1 then the polynomial is called Linear

Quadratic:If degree of the polynomial is 2 then the polynomial is called Quadratic

Cubic:If degree of the polynomial is 3 then the polynomial is called Cubic

Examples

Write types of following polynomials:

x + x 
x - x
y + y + 4
1 + x
3t 
r 
7x 

Value of Polynomial:

Example1
Find value of
x + 5 at x = 4
Solution:
= x + 5
= (4)+ 5
= 16 + 5
= 21 / Example2
Find value of
x – 5x at x = -1
Solution:
= x – 5x
= -1 – 5(-1)
= -1 – 5(-1)
= -1 + 5
= 4
Example3
Find value of
x - x+ 3 at x = 2
Solution:
= x - x+ 3
= (2) - (2) + 3
= 32 – 16 + 3
= 19 / Example4
Find value of
5x - 3x+ 3 at x = 2
Solution:
= 5x - 3x+ 3
= 5(2) - 3(2) + 3
= 160 – 48 + 3
= 115
Example5
Find value of
x+ 2xy - y at x = 3
and y =-2
Solution:
= x+ 2xy - y
= (3)+ 2(3)(-2) –(-2)
= 9 – 12 – (4)
= 9 – 12 – 4
= -7 / Example6
Find value of
C = 2 r at r = 14
Solution:
C = 2 r
C = 2(3.14)(14)
C = 87.9

Solving Linear Equations

There are some steps to solve linear equations in one variable:

Remove small brackets
Remove fractions
Collect the like terms
Solve like terms and find value of given variable

Example1
Solve x + 7 = 4
Solution:
x + 7 = 4
x = 4 – 7
x = -3 / Example2
Solve 4x - 3 = 9
Solution:
4x - 3 = 9
4x = 9 + 3
4x = 12

x = 3
Example3
Solve 6(x + 3) = 9
Solution:
6(x + 3) = 9
6x + 18 = 9
6x = 9 - 18
6x = -9
or x = / Example4
Solve 13x + 15 = 5x - 9
Solution:
13x + 15 = 5x - 9
13x – 5x = -9 -15
8x = -24

x = -3
Example4
Solve 4x - 3 = 4 + 3x
Solution:
4x - 3 = 4 + 3x
4x – 3x = 4 + 3
x = 7 / Example5
Solve + 4 = 15
Solution:
+ 4 = 15
= 15 - 4
= 11
Taking square on both sides
() = (11)
x = 121
Example6
Solve = 6
Solution:
= 6
Taking square on both sides
() = (6)
x + 3 = 36
x = 36 – 3
x = 33 / Example7
Solve + 3 = 1
Solution:
+ 3 = 1
= 1 – 3
= -2
Taking square on both sides
() = (-2)
5x + 4 = 4
5x = 4 – 4
5x = 0

x = 0
Example8
Solve = 6
Solution:
= 6
Taking square on both sides
() = (6)
= 36
= 36 – 3
= 33
2x = 33 × 5
2x = 165

x = 82.5 / Example9
Solve = 4
Solution:
= 4
Taking square on both sides
() = (4)
= 16
= 16 + 4
= 20
3x = 20 × 2
3x = 40

x = 13.3
Example10
Solve
Solution:

Multiplying both sides with 10

2(2x - 3) = 5(x-2)
4x – 6 = 5x – 10
4x – 5x = -10 + 6
- x = -4
x = 4 / Example11
Solve
Solution:

Multiplying both sides with 24

4(2x - 5) = 6(x-3)
8x – 20 = 6x – 18
8x – 6x = -18 + 20
2x = 2

x = 1
Example12
Solve
Solution:

Multiplying both sides with 20

4(2x) +5(x) = 4(1)
8x + 5x = 4
13x = 4

x = 0.3 / Example13
Solve
Solution:

Multiplying both sides with 20

2(2x - 1) + 3(x + 2) = 18
4x – 2 + 3x + 6 = 18
7x + 4 = 18
7x = 18 – 4
7x = 14

x = 2
Example14
Solve
Solution:

Multiplying both sides with 6

2(2x) – 3(4x - 2) = 30
4x – 12x + 6 = 30
4x – 12x = 30 – 6
-8x = 24

x = -3 / Example15
Solve
Solution:

Multiplying both sides with 24

6(x-1) – 4(2x - 1) = 72
6x - 6 – 8x + 4 = 72
- 2x – 2 = 72
-2x = 72 + 2
-2x = 74

x = -37

Modeling with Linear Equations

This topic will enable the students to translate worded problems into mathematical expressions.

Example1
When 8 is added to a number, the result is equal to 21. Find the number
Solution:
Let x is the required number:
According to the given condition
x + 8 = 21
x = 21 – 8
x = 13
so the required number is 13 / Example2
When 67 is subtracted from 2 times of a number, the result is equal to 33. Find the number
Solution:
Let x is the required number:
According to the given condition
2x - 67 = 33
2x = 33 + 67
2x = 100

x = 50
so the required number is 50
Example3
5 is subtracted from 4 times a number and the result is doubled. If the answer is equal to 6, what is the number?
Solution:
Let x is the required number:
According to the given condition
2(4x – 5) = 6
8x – 10 = 6
8x = 6 + 10
8x = 16

x = 2
so the required number is 2 / Example4
What is the number which when multiplied by 2 and added to 8 gives same result as when it is divided by 2 and added to 32
Solution:
Let x is the required number:
According to the given condition
2x + 8 =

4x + 16 = x + 64
4x – x = 64 – 16
3x = 48

x = 16
so the required number is 16
Example5
The sum of three consecutive natural numbers is 192. Find the numbers.
(Hint:three consecutive natural numbers are x - 1, x, x + 1)
Solution:
According to the given condition
(x - 1) + (x) + (x + 1) = 192
x – 1 + x + x + 1 = 192
3x = 192

x = 64
so the required numbers are 63, 64 and 65 / Example6
The sum of three consecutive odd numbers is 279. Find the numbers.
(Hint:three consecutive natural numbers are x - 2, x, x + 2)
Solution:
According to the given condition
(x - 2) + (x) + (x + 2) = 279
x – 2 + x + x + 2 = 279
3x = 279

x = 93
so the required numbers are 91, 93 and 95

Solving Inequalities

An open sentence containing the symbol oris called inequality or inequation.

Example1
Find solution set of ()
Solution:

so the solution set is { and } / Example2
Find solution set of ()
Solution:

so the solution set is { and }
Example3
Find solution set of ()
Solution:

so the solution set is { and } / Example4
Find solution set of ()
Solution:

so the solution set is { and }