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OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
MATHEMATICS
Handouts
Real Numbers
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
5
For example , 25 , 2.5 and 3.777…
2
Irrational Numbers: Numbers which can not be written in the form p ∕ q
For example  , 2 and 1.709975947…
Examples
Differentiate between Rational and Irrational Numbers
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
xii.
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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OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Operations on Two Sets
Two sets can be combined in many different ways like
1. Union
2. Intersection
3. Difference
4. Complement
Example1
Example2
If
A={1, 2, 3, 4, 5, 6}
If
A={2, 4, 6, 8, 10}
B={1, 2, 3, 4, 5, 6, 7, 8}
B={1, 3, 5, 7, 9}
Then find
Then find
A U B = {1, 2, 3, 4, 5, 6, 7, 8}
A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A ∩ B = {1, 2, 3, 4, 5, 6}
A ∩ B = ¢ or {}
A - B = ¢ or {}
A - B = {2, 4, 6, 8, 10}
B - A = {7, 8}
B - A = {1, 3, 5, 7, 9}
Example3
Example4
If
A={5, 10, 15, 20, 25, 30}
If
U={1, 2, 3, 4, 5, 6, 7}
B={10, 20, 30, 40, 50}
A={1, 2, 3}
Then find
B={4, 5, 6, 7}
A U B = {5, 10, 15, 20, 25, 30, 40, 50}
Then find
A ∩ B = {10, 20, 30}
Ac = U – A = {4, 5, 6, 7}
A - B = {5, 15, 25}
Bc = U – B = {1, 2, 3}
B - A = {40, 50}
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}
c
B
= {6, 7}
(A ∩ B) c = {1, 2, 3, 4, 6, 7, 10}
Ac U Bc = {1, 2, 3, 4, 6, 7, 10}
-------------------------------------------------------------------------------------------------------------------------------De Morgan's Laws
1.
(A ∩ B) c = Ac U Bc
2. (A U B) c = Ac ∩ Bc
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OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Rationalization
It is a process in which we remove the radical sign (from Numerator or Denominator)
Example1
Example2
Rationalize the denominator of
1
7
Rationalize the denominator of
Solution:
=
=
1
7
7
49
7
=
7
D
11
Solution:
1
7
Multiplying and dividing with
=
3
=
7
7
×
7
=
D
=
3
11
Multiplying and dividing with 11
3
11
=
×
11
11
3 11
=
=
121
3 11
=
11
=
Example3
Example4
Rationalize the denominator of
3
Rationalize the denominator of
18
Solution:
6
Solution:
=
3
=
18
Multiplying and dividing with 18
=
=
D
4
3
×
18
3 18
324
3 18
=
18
18
=
6
4
6
Multiplying and dividing with
18
=
18
=
=
D
=
4
6
×
4 6
36
4 6
=
6
2 6
=
3
=
6
6
6
OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Example5
Example6
Rationalize the denominator of
1
5 2
Solution:
1
11  7
Solution:
=
1
5 2
Multiplying and dividing with
=
=
=
=
1
5 2
5 2
×
1
=
5 2
11  7
Multiplying and dividing with 11  7
5 2
=
5 2
=
25  4
5 2
52
5 2
3
=
=
Example7
1
11  7
11  7
×
11  7
11  7
121  49
11  7
11  7
11  7
4
Example8
Rationalize the denominator of
3
7 2
Solution:
Rationalize the denominator of
4
18  12
Solution:
=
3
7 2
Multiplying and dividing with
=
=
3
×
7 2
3( 7  2 )
49  4
3( 5  2 )
=
72
3( 7  2 )
=
5



Rationalize the denominator of
7 2
7 2
4
=
7 2
18  12
Multiplying and dividing with 18  12
=
=
4
×
18  12
4( 18  12 )
18  12
18  12
324  144
4( 18  12 )
=
18  12
4( 18  12 )
=
6
2( 18  12 )
=
3
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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OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Exponents (Power)
There are some rules to solve problems including exponents. We have to learn and follow
these rules in different calculations
Example1
Example2
Simplify 2
3
Simplify 3 5
Solution:
Solution:
=2×2×2
=8
=3×3×3×3×3
= 243
Example3
Example4
Simplify 27
1
3
1
Simplify 32 5
Solution:
Solution:
= (3 × 3 × 3)
1
3
1
= (2 × 2 ×2 × 2 × 2) 5
1
1
= (3 3 ) 3
=3
= (2 5 ) 5
=2
Example5
Example6
Simplify 64
1
2
Simplify 125
Solution:
1
3
Solution:
1
1
= (2 × 2 ×2 × 2 × 2 × 2 ) 2
= (5 × 5 × 5) 3
6
= (2 )
= 23
=8
1
2
3
1
3
= (5 )
= 5= 5fde
Example7
Example8
Simplify 9
2
3
2
Simplify 32 5
Solution:
Solution:
= (3 × 3)
2
= (3 )
= 33
= 27
3
2
= (2 × 2 ×2 × 2 × 2)
3
2
5
= (2 )
= 22
=4
=
Example9
2
5
Example10
Simplify 16
3
4
Simplify 125
Solution:
1
3
Solution:
= (2 × 2 ×2 × 2)
3
4
= (5 × 5 × 5)
3
= (24) 4
= 23
=8
= (5 3 )
= 5 1
1
3
1
3
2
5
OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
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
Example2
Simplify 2 × 2
3
Simplify 3 5 × 3 7
4
Solution:
Solution:
3 4
= 3 5 7
= 3 12
=2
= 27
Example3
Example4
Simplify 2 × 2
3
5
Solution:
Simplify 3 2 × 3 7
Solution:
3 ( 5)
= 3 2( 7 )
= 3 27
= 3 9
=2
= 2 35
= 2 2
Example5
Example6
7
Simplify
5
53
Simplify
Solution:
7 11
75
Solution:
73
= 7 115
= 7 6 = 5fde
=5
= 54
Example7
Example8
Simplify 2
1
2
Solution:
×2
3
2
Simplify 2
2
3
×2
1
5
Solution:
2 1

5
1 3

2
=23
=22
4
2
= (2)
= 22
=4 =
= (2)
= (2)
= (2)
Example9
103
15
103
15
13
15
Example10
Simplify
7
7
Solution:
1
3
1
Simplify
7
3
11 2
1
11 4
Solution:
=7
=7
1 7

3 3
1 7
3
6
1 1

4
= 11 2
= 11
21
4
1
=7 3
= 11 4
= 7 2
=
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OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Algebraic Expressions
4 x2
Algebraic Term:
Degree of Polynomial:
Highest power of the variable in a polynomial is called the degree of polynomial
Example1
Example2
What is the degree of Polynomial
x5 - x4 + 3
What is the degree of Polynomial
2 - y 2 - y3 + 2 y8
Solution:
Degree = 5
Solution:
Degree = 8
Example3
Example4
What is the degree of Polynomial
5t - 7
What is the degree of Polynomial
3
Solution:
Degree = 0
Solution:
Degree = 1
Types of Polynomial:
There are so many types of polynomial but these three types are important
(i)
(ii)
(iii)
Linear
Quadratic
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:
i.
ii.
iii.
iv.
v.
vi.
vii.
x2 + x
x - x3
y + y2 + 4
1+x
3t
r2
7x 3







OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Value of Polynomial:
Example1
Find value of
x2 + 5
at x = 4
Solution:
= x2 + 5
= (4) 2 + 5
= 16 + 5
= 21
Example2
Find value of
x – 5x 3
at x = -1
Solution:
= x – 5x 3
= -1 – 5(-1) 3
= -1 – 5(-1)
= -1 + 5
=4
Example3
Example4
Find value of
x5 - x4 + 3
Find value of
5x 5 - 3x 4 + 3
at x = 2
Solution:
at x = 2
Solution:
= x5 - x4 + 3
= (2) 5 - (2) 4 + 3
= 32 – 16 + 3
= 19
= 5x 5 - 3x 4 + 3
= 5(2) 5 - 3(2) 4 + 3
= 160 – 48 + 3
= 115
Example5
Example6
Find value of
x 2 + 2xy - y 2
at
and
Solution:
2
2
x= 3
y = -2
Find value of
C=2  r
at
Solution:
C=2  r
C = 2(3.14)(14)
C = 87.9
r = 14
= x + 2xy - y
= (3) 2 + 2(3)(-2) – (-2) 2
= 9 – 12 – (4)
= 9 – 12 – 4
= -7
Solving Linear Equations
There are some steps to solve linear equations in one variable:
1. Remove small brackets
2. Remove fractions
3. Collect the like terms
4. Solve like terms and find value of given variable
Example1
Example2
Solve x + 7 = 4
Solve 4x - 3 = 9
Solution:
Solution:
x+7=4
4x - 3 = 9
x=4–7
4x = 9 + 3
x = -3
4x = 12
4 x 12

4
4
x=3
Example3
Example4
Solve 6(x + 3) = 9
Solve 13x + 15 = 5x - 9
Solution:
Solution:
6(x + 3) = 9
13x + 15 = 5x - 9
6x + 18 = 9
13x – 5x = -9 -15
6x = 9 - 18
8x = -24
6x = -9
8 x  24

6x  9
3
8
8

or x =
x = -3
6
6
2
OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Example4
Example5
Solve 4x - 3 = 4 + 3x
Solution:
Solve
x + 4 = 15
Solution:
4x - 3 = 4 + 3x
4x – 3x = 4 + 3
x=7
x + 4 = 15
x = 15 - 4
x = 11
Taking square on both sides
( x ) 2 = (11) 2
x = 121
Example6
Example7
Solve
x3 = 6
Solve
5x  4 + 3 = 1
Solution:
Solution:
x3 = 6
Taking square on both sides
( x  3 ) 2 = (6)
x + 3 = 36
x = 36 – 3
x = 33
5x  4 + 3 = 1
5x  4 = 1 – 3
5 x  4 = -2
Taking square on both sides
( 5 x  4 ) 2 = (-2)
5x + 4 = 4
5x = 4 – 4
5x = 0
5x 0

5
5
x=0
2
Example8
Example9
Solve
2
x3 = 6
5
Solve
3
x4 = 4
2
Solution:
Solution:
2
x3 = 6
5
Taking square on both sides
2
(
x  3 ) 2 = (6)
5
2
x  3 = 36
5
2
x = 36 – 3
5
2
x = 33
5
2x = 33 × 5
2x = 165
2 x 165

2
2
x = 82.5
3
x4 = 4
2
Taking square on both sides
3
(
x  4 ) 2 = (4)
2
3
x  4 = 16
2
3
x = 16 + 4
2
3
x = 20
2
3x = 20 × 2
3x = 40
3 x 40

3
3
x = 13.3
2
2
2
OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Example10
2x  3 x  2

Solve
5
2
Solution:
2x  3 x  2

5
2
Multiplying both sides with 10
2x  3 x  2
10 

 10
5
2
2(2x - 3) = 5(x-2)
4x – 6 = 5x – 10
4x – 5x = -10 + 6
- x = -4
x=4
Example12
Solve
2x x 1
 
5 4 5
Solution:
2x x 1
 
5 4 5
Multiplying both sides with 20
2x
x 1
20 
 20    20
5
4 5
4(2x) + 5(x) = 4(1)
8x + 5x = 4
13x = 4
13 x 4

13 13
x = 0.3
Example14
Solve
Solution:
2x 4 x  2

5
3
2
2x 4 x  2

5
3
2
Multiplying both sides with 6
2x
4x 2
6
 6
 5 6
3
2
2(2x) – 3(4x - 2) = 30
4x – 12x + 6 = 30
4x – 12x = 30 – 6
-8x = 24
 8 x 24

8 8
x = -3
Example11
Solve
Solution:
2x  5 x  3

6
4
2x  5 x  3

6
4
Multiplying both sides with 24
2x  5 x  3
24 

 24
6
4
4(2x - 5) = 6(x-3)
8x – 20 = 6x – 18
8x – 6x = -18 + 20
2x = 2
2x 2

2
2
x=1
Example13
2x  1 x  2

3
Solve
3
2
Solution:
2x  1 x  2

3
3
2
Multiplying both sides with 20
2x  1
x 2
6
 6
 3 6
3
2
2(2x - 1) + 3(x + 2) = 18
4x – 2 + 3x + 6 = 18
7x + 4 = 18
7x = 18 – 4
7x = 14
7 x 14

7
7
x=2
Example15
x 1 2 x 1

3
Solve
4
6
Solution:
x 1 2 x 1

3
4
6
Multiplying both sides with 24
x 1
2 x 1
24 
 24 
 3  24
4
6
6(x-1) – 4(2x - 1) = 72
6x - 6 – 8x + 4 = 72
- 2x – 2 = 72
-2x = 72 + 2
-2x = 74
 2 x 74

2
2
x = -37
OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Modeling with Linear Equations
This topic will enable the students to translate worded problems into mathematical expressions.
Example1
Example2
When 8 is added to a number, the result is
When 67 is subtracted from 2 times of a
equal to 21. Find the number
number, the result is equal to 33. 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
Solution:
Let x is the required number:
According to the given condition
2x - 67 = 33
2x = 33 + 67
2x = 100
2 x 100

2
2
x = 50
so the required number is 50
Example3
Example4
5 is subtracted from 4 times a number and
What is the number which when
the result is doubled. If the answer is equal to 6, multiplied by 2 and added to 8 gives same result as
what is the number?
when it is divided by 2 and added to 32
Solution:
Let x is the required number:
According to the given condition
2(4x – 5) = 6
8x – 10 = 6
8x = 6 + 10
8x = 16
8 x 16

8
8
x=2
so the required number is 2
Solution:
Let x is the required number:
According to the given condition
x
2x + 8 =  32
2
x
2  2 x  2  8  2   2  32
2
4x + 16 = x + 64
4x – x = 64 – 16
3x = 48
3 x 48

3
3
x = 16
so the required number is 16
Example5
Example6
The sum of three consecutive natural
The sum of three consecutive odd
numbers is 192. Find the numbers.
numbers is 279. Find the numbers.
(Hint: three consecutive natural numbers are x - 1, x, x + 1)
(Hint: three consecutive natural numbers are x - 2, x, x + 2)
Solution:
According to the given condition
(x - 1) + (x) + (x + 1) = 192
x – 1 + x + x + 1 = 192
3x = 192
3x 192

3
3
x = 64
so the required numbers are 63, 64 and 65
Solution:
According to the given condition
(x - 2) + (x) + (x + 2) = 279
x – 2 + x + x + 2 = 279
3x = 279
3 x 279

3
3
x = 93
so the required numbers are 91, 93 and 95
OMAN COLLEGE OF MANAGMENT AND TECHONOLGY
Solving Inequalities
An open sentence containing the symbol < or > is called inequality or inequation.
Example1
Find solution set of 4n  1  7
Example2
( n W )
Find solution set of 7  5a  32
Solution:
Solution:
Example3
Example4
4n  1  7
4n  7  1
4n  8
4n 8

4 4
n2
so the solution set is { n | n  W and n  2 }
Find solution set of
Solution:
x  5 25  4 x

10
5
x  5 25  4 x

10
5
x  5 25  4 x
50 

 50
10
5
5( x  5)  10(25  4 x)
5x  25  250  40x
5x  40x  250  25
45x  225
45 x 225

45
45
x5
so the solution set is { x | x  N and x  5 }
( aZ )
7  5a  32
5a  32  7
5a  25
5a 25

5
5
a5
so the solution set is { a | a  Z and a  5 }
( xN )
Find solution set of
Solution:
7  5x 1  x

3
2
7  5x 1  x

3
2
7  5x 1  x
6

6
3
2
2(7  5 x)  3(1  x)
14  10x  3  3x
10x  3x  3  14
13x  11
13 x  11

13
13
 11
x
13
so the solution set is { x | x  R and x 
( xR)
 11
}
13