Number_System .pdf

Unit - III: Number System
Dr. Tushar Bhatt
Assistant Professor in Mathematics
Department of Science and Technology
Faculty of Engineering and Technology
Atmiya University
Rajkot - 360005
Dr. Tushar Bhatt Unit - III: Number System 1 / 52

Table of Content
1 Introduction
2 Decimal Number System (Base 10)
3 Binary Number System (Base 2)
4 Conversation - I: Decimal to Binary
5 Conversation - II: Decimal fraction to Binary
6 Conversation - III: Binary to Decimal
7 Conversation - IV: Binary fraction to Decimal
8 Octal Number System (Base 8)
9 Conversation - V: Octal to Decimal
10 Conversation - VI: Octal fraction to Decimal
11 Conversation - VII: Decimal to Octal
12 Conversation - VIII: Decimal fraction to Octal
13 Conversation - IX: Octal to Binary
14 Conversation - X: Octal fraction to Binary

1. Introduction
A digital computer manipulates discrete elements of data and that these
elements are represented in the binary forms. Operands used for calculations
can be expressed in the binary number system.
Other discrete elements including the decimal digits, are represented in bi-
nary codes. Data processing is carried out by means of binary logic elements
using binary signals. Quantities are stored in binary storage elements.
The purpose of this unit is to introduce the various binary concepts as a
frame of reference for further study in the succeeding units.
Mainly there are two types of number systems as given below:
1 Non-positional Number system.
2 Positional number system.

1.1 Non-positional Number system
In this number system
Use symbols such as I for 1, II for 2, III for 3, IIII for 4, IIIII for 5, etc...
Each symbol represents the same value regardless of its position in
the number.
The symbols are simply added to find out the value of a particular
number.
It is difficult to perform arithmetic with such a number system.

1.2 Positional number system
In this number system
Use only a few symbols called digits.
These symbols represent different values depending on the position
they occupy in the number.
The value of each digit is determined by:
1 The digit itself.
2 The position of the digit in the number.
3 The base of the number system.
Radix or base = total number of digits in the number system.
The maximum value of a single digit is always equal to one less than
the value of the base.

1.3 Categories of positional number systems
There are four categories of positional number system defined as follows:
1 Decimal Number System.
2 Binary Number System.
3 Octal Number System.
4 Hexadecimal Binary Number System.

2. Decimal Number System (Base 10)
Decimal number system have ten digits represented by
0,1,2,3,4,5,6,7,8 and 9. So, the base or radix of such system is 10.
In this system the successive position to the left of the decimal point
represent units, tens, hundreds, thousands etc.
For example, if we consider a decimal number 257, then the digit
representations are,
Position Position Position
Hundred Tens Ones
2 5 7
The weight of each digit of a number depends on its relative position within
the number.

2.1 Decimal Number System (Base 10) - Example
Ex - 1 : The weight of each digit of the decimal number 6472
6472 = 6000 + 400 + 70 + 2
= (6 × 103
) + (4 × 102
) + (7 × 101
) + (2 × 100
)
∴ The weight of digits from right hand side are
Weight of 1st digit = 2 × 100.
Weight of 2nd digit = 7 × 101.
Weight of 3rd digit = 4 × 102.
Weight of 4th digit = 6 × 103.
The above expressions can be written in general forms as the weight of nth
digit of the number from the right hand side
= nth
digit × 10n−1

3. Binary Number System (Base 2)
Only two digits 0 and 1 are used to represent a binary number system. So
the base or radix of binary system is two (2). The digits 0 and 1 are called
bits (Binary Digits). In this number system the value of the digit will be
two times greater than its predecessor. Thus the value of the places are
< ... < −32 < −16 < −8 < −4 < −2 < −1 < ...
The weight of each binary bit depends on its relative position within the
number. It is explained by the following example:
The weight of bits of the binary number 10110 is:
10110 = (1 × 24
) + (0 × 23
) + (1 × 22
) + (1 × 21
) + (0 × 20
)
= 16 + 0 + 4 + 2 + 0
= 22(decimal number)

3. Binary Number System (Base 2)
The weight of each bit of a binary no. depends on its relative pointer
within the number and explained from right hand side
∴ The weight of digits from right hand side are
Weight of 1st bit = 0 × 20.
Weight of 2nd bit = 1 × 21.
Weight of 3rd bit = 1 × 22.
Weight of 4th bit = 0 × 23.
Weight of 5th bit = 1 × 24.
The above expressions can be written in general forms as the weight of nth
digit of the number from the right hand side
= nth
bit × 2n−1

3.1 How to convert Decimal number to Binary number?
Divide the given decimal number by 2 and note down the remainder.
Now, divide the obtained quotient by 2, and note the remainder again.
Repeat the above steps until you get 0 as the quotient.
Now, write the remainders in such a way that the last remainder is
written first, followed by the rest in the reverse order.
This can also be understood in another way which states that the
Least Significant Bit (LSB) of the binary number is at the top and
the Most Significant Bit (MSB) is at the bottom. This number is the
binary value of the given decimal number.

3.2 Decimal to Binary Conversation of numbers 0 to 9
Decimal Number Binary Number
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001

3.3 Decimal to Binary Conversation - Example

Conversation - I
Decimal ⇒ Binary

3.3 Decimal to Binary Conversation - Examples
Ex - 1: Convert (75)10 to its binary equivalent.
Solution:
Remainder
2 75
2 37 1
2 18 1
2 9 0
2 4 1
2 2 0
1 0
1
Now arrange remainders upward ↑. That is binary representation of given
number 75.
i.e.
(75)10 = (1001011)2

3.4 Decimal to Binary Conversation - Examples
Ex - 2: Convert (174)10 to binary.
Solution:
Division by 2 Quotient Remainder
174 ÷ 2 87 0 (LSB)
87 ÷ 2 43 1
43 ÷ 2 21 1
21 ÷ 2 10 1
10 ÷ 2 5 0
5 ÷ 2 2 1
2 ÷ 2 1 0
1 ÷ 2 0 1 (MSB)
After noting the remainders, we write them in the reverse order such that
the Most Significant Bit (MSB) is written first, and the Least Significant
Bit is written in the end. Hence, the binary equivalent for the given
decimal number (174)10 is (10101110)2.

Conversation - II
Decimal fraction ⇒ Binary

3.5 Convert Decimal fraction to Binary number - Steps
For this conversations, we will consider following three steps:
1 Convert the integral part of decimal to binary equivalent.
Divide the decimal number by 2 and store remainders in array.
Divide the quotient by 2.
Repeat step 2 until we get the quotient equal to zero.
Equivalent binary number would be reverse of all remainders of step 1.
2 Convert the fractional part of decimal to binary equivalent
Multiply the fractional decimal number by 2
Integral part of resultant decimal number will be first digit of fraction
binary number.
Repeat step 1 using only fractional part of decimal number and then
step 2.
If any fraction will have more and more fractions after multiply by 2
then fix the step size according to given instruction.
In the fraction part must arrange the integer part in the downward
direction ↓.
3 Combine both integral and fractional part of binary number.

3.6 Convert Decimal fraction to Binary number - Examples
Ex - 1: Convert (17.75)10 to binary.
Solution:
Step - 1: Convert the integral part of decimal to binary equivalent.
17 ÷ 2 8 1 (LSB)
8 ÷ 2 4 0
4 ÷ 2 2 0
2 ÷ 2 1 0
1 ÷ 2 0 1 (MSB)
Step - 2: Convert the fractional part of decimal to binary equivalent.
Fractional part × 2 Result Integer part of result
0.75 × 2 1.50 1 (LSB)
0.50 × 2 1.00 1(MSB)
Step - 3: Combine both integral and fractional part of binary
number i.e. (10001.11)2

3.6 Convert Decimal fraction to Binary number - Examples
Ex - 2: Convert (89.25)10 to binary.
Solution: Step - 1: Convert the integral part
89 ÷ 2 44 1 (LSB)
44 ÷ 2 22 0
22 ÷ 2 11 0
11 ÷ 2 5 1
5 ÷ 2 2 1
2 ÷ 2 1 0
1 ÷ 2 0 1 (MSB)
Step - 2: Convert the fractional part of decimal to binary equivalent.
Fractional part × 2 Result Integer part of result
0.25 × 2 0.50 0 (LSB)
0.50 × 2 1.00 1(MSB)
Step - 3: Combine both i.e. (1011001.01)2 .

Conversation - III
Binary ⇒ Decimal

3.7 How to convert Binary number to Decimal number?
We know that the binary to decimal number system conversion is the
process of changing from a binary (base 2) to a decimal number
system (base 10 number system).
To convert a binary number system to a decimal number system,
follow the procedures below.
Step - 1: Multiply each digit of the specified binary number by the
exponents of the base starting with the right most digit (i.e.,
20, 21, 22, and so on).
Step - 2: As we move right to left, the exponents should increase by
one, such that the exponents begin with 0.
Step - 3: Simplify and find the sum of each of the product values
obtained in the previous steps.

3.8 Binary to Decimal conversation examples
Ex - 1: Convert (11110)2 into a decimal number system.
Solution:
11110 = (1 × 24
) + (1 × 23
) + (1 × 22
) + (1 × 21
) + (0 × 20
)
= (16) + (8) + (4) + (2) + (1)
= 30
∴ (11110)2 = (30)10.

Solution:
0110 = (0 × 23
) + (1 × 22
) + (1 × 21
) + (0 × 20
)
= (0) + (4) + (2) + (0)
= 6
∴ (0110)2 = (6)10.

Solution:
1110110 = (1 × 26
) + (1 × 25
) + (1 × 24
) + (0 × 23
)
+ (1 × 22
) + (1 × 21
) + (0 × 20
)
= (64) + (32) + (16) + (0) + (4) + (2) + (0)
= 118
∴ (1110110)2 = (118)10.

Ex - 4: The binary number (1111)2 is equal to (x)10.Then find the
value of x.
Solution:
1111 = (1 × 23
) + (1 × 22
) + (1 × 21
) + (1 × 20
)
= (8) + (4) + (2) + (1)
= 15
∴ (1111)2 = (15)10.
∴ x = 15.

Conversation - IV
Binary fraction ⇒ Decimal

3.9 Fraction Binary number to Decimal number
conversation
Ex - 5: Convert (1110.1010)2 into a decimal number system.
Solution:
Here first we have to separate the integral and fraction part
Integral Part: 1110
1110 = (1 × 23) + (1 × 22) + (1 × 21) + (0 × 20) = 14
Fractional Part: .1010
.1010 = (1 × 2−1) + (0 × 2−2) + (1 × 2−3) + (0 × 2−4) = 1
2 + 0 + 1
8 + 0 =
0.5 + 0.125 = 0.625
Now
(1110.1010)2 = Integral part + Fractional part = 14 + 0.625 = 14.625
.
∴
(1110.1010)2 = (14.625)10

4. Octal Number System (Base 8)
A commonly used positional number system is the Octal Number
System. This system has eight (8) digit representations as
0,1,2,3,4,5,6 and 7. The base or radix of this system is 8.
The maximum value of a single digit is 7 (one less than the value of
the base.
Each position of a digit represents a specific power of the base (8).
Since there are only 8 digits, 3 bits (23 = 8) are sufficient to represent
any octal number in binary.

Conversation - V
Octal ⇒ Decimal

4.1 Steps to Convert Octal to Decimal
Step - 1: Since an octal number only uses digits from 0 to 7, we first
arrange the octal number with the power of 8.
Step - 2: We evaluate all the power of 8 values such as 80 is 1, 81 is
8, etc., and write down the value of each octal number.
Step - 3: Once the value is obtained, we multiply each number.
Step - 4: Final step is to add the product of all the numbers to obtain
the decimal number.

4.2 Examples on Octal to Decimal Conversation
Ex - 1: Convert (140)8 to decimal number.
Solution:

Ex - 2: Convert (2057)8 to decimal number.
Solution:
(2057)8 = (2 × 83
) + (0 × 82
) + (5 × 81
) + (7 × 80
)
= 1024 + 0 + 40 + 7
= 1071
∴ (2057)8 = (1071)10

Conversation - VI
Octal fraction ⇒ Decimal

Ex - 3: Convert (246.28)8 to decimal number.
Solution:

Conversation - VII
Decimal ⇒ Octal

4.3 Decimal to Octal conversation

4.3 Examples on Decimal to Octal Conversation
Ex - 1: Convert (1792)10 into an octal number.
Solution:
Decimal Number Operation Quotient Remainder
1792 ÷8 224 0
224 ÷8 28 0
28 ÷8 3 4
3 ÷8 0 3
Now write off the remainder from bottom to top, we have
(1792)10 = (3400)8

Solution:
127 ÷8 15 7
15 ÷8 1 7
1 ÷8 0 1
(127)10 = (177)8

Solution:
100 ÷8 12 4
12 ÷8 1 4
1 ÷8 0 1
(100)10 = (144)8

Conversation - VIII
Decimal fraction ⇒ Octal

Steps to Convert Decimal fraction to Octal
1 First, we calculate the integer part of the decimal point by dividing
the octal base number i.e. 8 until the quotient is less than 8.
2 The second part is calculated on the fraction part of the decimal
number where the number is multiplied with the base number 8 until
the fractional part is equal to zero.
3 Here, once multiplied we keep the integer part separate and the
fractional part separate.
4 The final octal number is calculated by adding both the integer and
the fractional number.

4.3 Examples on Decimal fraction to Octal Conversation
Ex - 4: Convert decimal number 29.45 into octal form, follows
seven steps.
Solution:
Step 1: Separate the decimal number into two parts - the integer and the
fractional. So, 29.45 = 29 + 0.45.
Step 2: Convert the integer part of the number first. So, we begin with 29
first by dividing it by the base number 8 until the quotient is less than 8.
Decimal Number(IP) Operation Quotient Remainder
29 ÷8 3 5
3 ÷8 0 3
Hence, 29 is 35 in an octal number.
Step - 3: Once the integer octal number is obtained, we proceed to the
fractional part. So, 0.45 is multiplied by 8 (octal base number) where the
result is again divided into its integer part and fractional part. The number
is multiplied by 8 until the fractional part is equal to zero.

4.3 Examples on Decimal fraction to Octal conversation;
Ex - 4: Solu...
Decimal Number(FP) Operation Result Integer part Fractional part
0.45 ×8 3.6 3 0.6
0.6 ×8 4.8 4 0.8
0.8 ×8 6.4 6 0.4
0.4 ×8 3.2 3 0.2
0.2 ×8 1.6 1 0.6
0.6 ×8 4.8 4 0.8
0.8 ×8 6.4 6 0.4
Write all the integer part from top to bottom that derives the octal
number of the fractional number. Hence, 0.45 = 0.3463146.
Step 4: Add both the integer and the fractional part together to obtain
the octal number. Hence, 35 + 0.3463146 = 35.3463146.
∴ (29.45)10 = (35.3463146)8 .

Conversation - IX
Octal ⇒ Binary

Steps for Convert Octal to Binary
Octal ⇒ Decimal ⇒ Binary

Examples on Octal to Binary conversation
Ex - 1: Convert (41)8 to a binary number.
Solution:
Step - 1: Octal to Decimal
(41)8 = (4 × 81) + (1 × 80) = 32 + 1 = 33 = (33)10.
Step - 2: Decimal to Binary
33 ÷2 16 1
16 ÷2 8 0
8 ÷2 4 0
4 ÷2 2 0
2 ÷2 1 0
1 ÷2 0 1
Therefore, the equivalent binary number is (100001)2.
Hence, (41)8 = (100001)2 .

Examples on Octal to Binary conversation
Ex - 2: Convert (10)8 to a binary number.
Solution:
Step - 1: Octal to Decimal
(10)8 = (1 × 81) + (0 × 80) = 8 + 0 = 8 = (8)10.
Step - 2: Decimal to Binary
8 ÷2 4 0
4 ÷2 2 0
2 ÷2 1 0
1 ÷2 0 1
Therefore, the equivalent binary number is (1000)2.
Hence, (10)8 = (1000)2 .

Conversation - IX
Octal fraction ⇒ Binary

Examples on Octal fraction to Binary conversation
Ex - 1: Convert (35.346)8 into binary number, follows five steps
Solution:
Step - 1: Here first we convert (35.346)8in decimal numbers.
(36.346)8 = (3 × 81
) + (5 × 80
) + (3 × 8−1
) + (4 × 8−2
) + (6 × 8−3
)
= 24 + 5 + 0.375 + 0.062 + 0.012
= 29.449
= (29.449)10

Examples on Octal fraction to Binary conversation - Ex-1:
Solu...
Step - 2: Now convert (29.449)10 to binary number.
Integer part (29) ⇒ Binary
Decimal No.(IP) Operation Quotient Remainder
29 ÷2 14 1
14 ÷2 7 0
7 ÷2 3 1
3 ÷2 1 1
1 ÷2 0 1
∴ (29)10 = (11101)2.

Examples on Octal fraction to Binary conversation - Ex-1:
Solu...
Step - 2: Now convert (0.449)10 to binary number.
Fraction part (0.449) ⇒ binary
Step size Decimal No.(FP) Operation Result IP FP
1 0.449 ×2 0.898 0 0.898
2 0.898 ×2 1.796 1 0.796
3 0.796 ×2 1.592 1 0.592
4 0.592 ×2 1.184 1 0.184
5 0.184 ×2 0.368 0 0.368
∴ (0.449)8 = (01110...)2.
Now added results of integer part and fraction part, we get
(35.346)8 = (29.449)10 = (11101.01110)2 .

Number_System .pdf

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

Similaire à Number_System .pdf

Similaire à Number_System .pdf (20)

Plus de Dr. Tushar J Bhatt

Plus de Dr. Tushar J Bhatt (20)

Dernier

Dernier (20)

Number_System .pdf