 # Chapter - 3 Number System and their Conversion

1.       Define Number System.
The system concerned with the number and represented by sequence of digits is called number system. It plays vital role in computing and electronics. Number system also refers to the digits, its arrangement, positional value and base of number system.
2.       What is base or radix of number system?
Ø  The total number of digits used by the particular number system is called base or radix of that number system. For Example: Base of Binary number  system is 2 because it uses two digits 0 and 1 only.
3.       List out the different types of number system with their bases.
Ø  There are 4 types of number system and its base are given below:
 Number System Base Binary Number System 2 Octal Number System 8 Decimal Number System 10 Hexadecimal Number System 16

4.       Define Binary number system.
Ø  The number system having base 2 and consists of digits: 0 and 1 is called Binary number system.
5.       Define Octal number system.
Ø  The number system having base 8 and consists of digits: 0, 1, 2, 3, 4, 5, 6 and 7 is called Octal number system.
6.       Define Decimal (denary) number system.
Ø  The number system having base 10 and consists of digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 is called Decimal (Denary) number system.

Ø  The number system having base 10 and consists of digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 , 9  and alphabets A, B, C, D, E and F is called Hexadecimal number system.
Where:
 A 10 B 11 C 12 D 13 E 14 F 15

8.       Why do digital computers use binary number system for their operations?
Ø  The number system having base 2 and consists of digits: 0 and 1 is called Binary number system. It has two bits 0 and 1. An electronic circuit has two states either ON state or OFF state. The bit 1 represents high voltage (ON state) and the bit 0 represents the low voltage (OFF state) of an electronic circuit. That’s why digital computer uses binary number for their operations.

According it the order of writing symbols for a number, number systems are divided into two types.
A)     Positional Number System:-
The position of a symbol in number has positional significance. Positional number is easy for arithmetic computations.  Different position of the same symbol signifies different meaning (values).Examples of positional number systems is decimal, binary etc.
Example: - 294=200+90+4 and 429=400+20+9. Therefore 294and 429 are different.
B)     Non-Positional Number System:-
The order of the symbols is the number does not have any significance. Examples of non-positional number systems are Roman numbers, Greek numbers etc. Example:- I,II,III,IV,V,VI,X,L,M etc.  Arithmetic with these numbers is difficult or complex arithmetic is almost impossible. In today’s world non-positional number systems are rarely used because of this complex nature and difficulty in arithmetic.
# CONVERSION OF NUMBER SYSTEM RULE (dfxfdGq)

(A) From other Number System to Decimal
We should multiply the each digit by its base.
For example:
i)        Convert (11111)2 to its decimal equivalent. [HISSAN 2069-D1)
Solution: Converting from Binary to decimal, we get
 Face Value 1 1 1 1 1 Place Value 24 23 22 21 20
Here,
=1x24+1x23+1x22+1x21+1x20
=1x16+1x8+1x4+1x2+1x1
= 16+8+4+2+1
= (31)10
ii)      (1011.01)2= (?)10
Solution: Converting from Binary to decimal, we get

 Face Value 1 0 1 1 . 0 1 Place Value 23 22 21 20 2-1 2-2
Here,
=1x23+0x22+1x21+1x20+0x2-1+1x2-2
=8+0+2+1+0+0.25
= (11.25)10
(1011.01)2 = (11.25)10

iii)    (1570.04) 8= (?)10
Solution: Converting from Octal to decimal, we get

 Face Value 1 5 7 0 0 4 Place Value 83 82 81 80 8-1 8-2
Here,

=1x83+5x82+7x81+0x80+0x8-1+4x8-2
=512+320+56+0+0+0.0625
= (888.0625)10
(1570.04)8 = (888.0625)10
iv)    (CAB) 16= (?)10
Here,
 Face Value C A B Place Value 162 161 160
Here,
=Cx162+Ax161+Bx160
=12x256+10x16+11x1=3072+160+11
= (3243)10
(CAB)16 = (3243)10

(B) From Decimal to other System

i) Decimal to Binary
(456)10= (?)2
Solution:
Converting from decimal to binary, we get
(456)10 = (111001000)2   Ans
(45.125)10 = ( ? )2
 2 2 2 2 2 45        Rem 22 1 0 1 1 0 11 5 2 1
(0.125)10 = (?)2
0.125 x 2 = 0.25 integer part = 0
0.25 x 2= 0.5      integer part = 0
0.5 x 2 = 1           integer part = 1
So, (0.125)10=(0.001)2
Finally we can write,
(45.125)10 = (101101.001)2
ii) Decimal to Octal
(827)10= (?)8

 8       8       8 827     Rem 103 3 7 4 12 1
So, (827)10=(1473)16
(983)10= (?)16
 16 16 16 983 61 7 13(D) 3 3 0
So, (983)10=(3D7)16
QN 1. Convert (BABA)16 to its equivalent octal number. [HSEB-2073-“D”]
Solution:       At first converting Hexadecimal to decimal, we get
 Face Value B B A Place Value 162 161 160

Here,
= B x 162 + B x 161 + A x 160
= 11 x 256 + 11 x 16 + 10 x 1
= (3002)10
Now, Converting decimal to octal, we get
 8 3002 Remainder Tips:     3002 / 8 = 375. 25 8 375 -          2 write 375; for remainder 8 46 -          7 3002 – 8 x 375 = 2 8 5 -          6 0 -          5

#   9’s and 10’s Complements Decimal Subtraction
By using the complement method, addition process can be used to perform subtraction. This reduces hardware of the computer. With the same hardware used for addition, computers can perform subtraction through the complement method.

There are two types of complement s for each base or radix ‘r’ system:
The r’s complement and the (r-1)’s complement
When the value of the base or radix ‘r’ is 2 then the two types are referred as the 2’s and 1’s complement for Binary Number System and the 10’s and 9’s Complement for the Decimal Number System.

By: r’s Complement method” 10’s Complement for Decimal number
1.       Let M=72532 and N= 13250 ,
10’s Complement of i.e. 13250 = (99999-13250) +1=86750
. . .   72532
+ 86750
1 59282
2.       If M>N, the sum will produce an end carry which is discarded and what is left is the result is M-N. Here M>N, so end carry 1 is discarded, so answer is 59282.
3.       If M<N , the sum does not produce an end carry and is equal to r’s complement of the sum and place negative sign in front.
Let M= 13250 and N=72532,
10’s complement of N = (99999-72532) +1=27468
13250+27468=40718
Since, M<N, so answer = 10’s complement of sum i.e. 40718
= (99999-40718) +1= - 59282

By (r-1)‘s Complement method for decimal number
1.       Let M=72532 and N=13250
Here, 9’s complement of N i.e. 13250= 99999-13250=86749

. . .   72532
+ 86749
1 59281
2.       If M>N, the sum will produce an end carry which is discarded and what is left is the result. Here, M>N, so end carry 1 is discarded and is added to the result to get the final result. So, answer = 59282+1=59282.
3.       If M<N, the sum does not produce an end carry and is equal to (r-1)’s complement of the sum and place negative sign in front.
Let M=13250 and N=72532,
So, 9’s Complement of N i.e. 72532 = 99999-72532 = 27467
. . .  13250
+ 27467
40717
Here, M>N, So answer = 9’s Complement of 40717
= 99999-40717= - 59282
As with decimal addition, to add two binary numbers we need to add two corresponding bits from the two numbers at a time. The result of addition of two bits from the two numbers is given in the table.

 X Y X+Y 0 0 0 0 1 1 1 0 1 1 1 0 (carry 1)

Workout Examples:

 a)       110+10 110              + 010                1000 b)      101101+1001  101101               + 001001                  110110 c)       1011+0111             1011          + 0111           10010

Binary Subtraction:
As with decimal subtraction, to subtract a binary numbers from other we need to subtract a correspond bit from other. The result of subtraction of a bit from the other is given in the table.

 X Y X-Y 0 0 0 0 1 1 (borrow) 1 0 1 1 1 0

Workout Examples:
 i) 1101-111     1101  -  0111     0110 ii) 10101-1010         10101     -   01010                         01011 iii) 1010-100       1010     - 0100       0110

1’s and 2’s Complement Methods of Binary Subtraction:
Complements are used in digital computers for simplifying the subtraction operation and for logical manipulation. There are two types of complements for each base or radix ‘r’ system.
1’s Complement
The 1’s complement of binary number is obtained by subtracting each digit from 1. Because of the subtraction of a binary digit from 1, the bit changes from 0 to 1 and 1 to 0. For example, the complement of 1’s of 1011001 is 01001110 and the 1’s complement of 000111 is 111000.
2’s Complement
2’s complement of binary number is obtained by adding 1 to 1’s complement value. The 2’s complement of 101100 is 010100 and is obtained by adding 1 to 1’s complement of 101100, i.e.
= (111111-101100) +1=010100
Some solved 1’s and 2’s complement problems:
1.     Subtract (1011)2 from (1100)2 by using 1’s and 2’s complement method.
Solution:     We have to find 1100-1011 = ?
1’s complement method
Let, P = 1100 and Q = 1011
Since, both numbers have equal bits. No need to add zero in either case in front to make them equal.
1’s Complement of Q = 1111 – 1011
= 0100
Now, adding 1’s complement of Q  with P, we get
1
0100
+ 1100
10000
Since, P > Q, there is an end carry 1. So, adding the end carry 1 at last, we get
= (0001)2

2’s complement method
Let, P = 1100 and Q = 1011
Since, both numbers have equal bits. No need to add zero in either case in front to make them equal.
2’s Complement of Q = (1111 – 1011) + 1
= 0100 + 1
= 0101
Now, adding 2’s complement of Q with P, we get
1
0101
+ 1100
10001
Since, P > Q, there is an end carry 1. In 2’s complement we have to discard the end carry 1 , then
2.     Subtract (111)2 from (1000)2 by using 1’s and 2’s complement method.
Solution:     We have to find 1000-111 = ?

1’s complement method
Let, P = 1000 and Q = 0111
Since, both numbers have not equal bits. We have to add one zero in 2nd number in order to make both equal bits.
1’s Complement of Q = 1111 – 0111
= 1000
Now, adding 1’s complement of Q  with P, we get
1000
+ 1000
10000
Since, P > Q, there is an end carry 1. So, adding the end carry 1 at last, we get
= (0001)2

2’s complement method
Let, P = 1000 and Q = 0111
Since, both numbers have not equal bits. We have to add one zero in 2nd number in order to make both equal bits.
2’s Complement of Q = (1111 – 0111) + 1
= 1000 + 1
= 1001
Now, adding 2’s complement of Q with P, we get
1001
+ 1000
10001
Since, P > Q, there is an end carry 1. In 2’s complement we have to discard the end carry 1 , then
3.     Subtract (1000)2 from (111)2 by using 1’s and 2’s complement method.
Solution:     We have to find 111-1000 = ?
1’s complement method
Let, P = 0111 and Q = 1000
Since, both numbers have not equal bits. We have to place one zero in front of 1st number to make them equal.
1’s Complement of Q = 1111 – 1000
= 0111
Now, adding 1’s complement of Q  with P, we get
111
0111
+ 0111
1110
Since, P < Q, there is no end carry 1. So, we have to find the 1’s complement of 1110 and place the minus sign in front of final answer, we get
1’s complement of 1110= 1111-1110
= - (0001)2
2’s complement method
Let, P = 0111 and Q = 1000
Since, both numbers have not equal bits. We have to place one zero in front of 1st number to make them equal.
2’s Complement of Q = (1111 – 1000)+1
= 0111+1 = 1000
Now, adding 2’s complement of Q  with P, we get
1000
+ 0111
1111
Since, P < Q, there is no end carry 1. So, we have to find the 2’s complement of 1111 and place the minus sign in front of final answer, we get
2’s complement of 1111= (1111-1110) + 1
= 0000 + 1
= - (0001)2

Homework [Unit – 3]
1.     List out the different number system with its base. Convert (456)10 to binary number system.
2.     Convert (BBA)16 to its octal number.
3.     Subtract (1001)2 from (1111)2 by using 1’s and  2’s complement method.
4.     Subtract (1000)2 from (1111)2 by using 1’s and 2’s complement method.
5.     Subtract (100000)2 from (111)2 by using 1’s and 2’s complement method.
6.     Perform the following binary operation:
 i.            10111+111 ii.            1111-1001 iii.            11001-1101 iv.            1111+1111
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