Different Types of Computer Number System and Their Conversion
Number system refers to the digits, its arrangements, positional weight and base of number system. Different types of number system are as follows:
a) Decimal number system: A number system having base or radix 10 is called decimal number system.
a. Examples: (789)10 or (789)D
b) Binary number system: A number system having base or radix 2 is called binary number system.
a. Examples: (110)2
c) Octal number system: A number system having base or radix 6 is called octal number system.
a. Examples: (1234)8
b.
d) Hexadecimal number system: A number system having base or radix 16 is called hexadecimal number system.
a. Examples: (C01F)16 or (C01F)H
The base or radix of a number system is the number of different symbols available to represent any digit within that system. For example, the decimal system (Base 10) has a radix of 10. Decimal uses different combinations of 10 symbols to represent any value (i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
Here,
(11011.011)2 = 24 x 1 + 23 x 1 + 22 x 0 + 21 x 1 + 20 x 1 + 2-1 x 0 + 2-2 x 1 + 2-3 x 1
= 16 + 8 + 0 + 2 + 1 + 0 + 0.25 + 0.125
= 27.375
A number system having base or radix 2 is called binary number system. It consists of 2 bits: 0 and 1. It is also known as Binary digits. It is specially used in internal processing of computer system.
An electronic circuit has two states either ON state or OFF state. The bit 1 represents the high voltage i.e. ON state and the bit 0 represents the low voltage i.e. OFF state of an electronic circuit. So it is used in computer system.
A number system having base or radix 2 is called binary number system. It consists of 2 bits: 0 and 1. It is also known as Binary digits.
Here, (520)10
|
16
|
520
|
8
|
|
16
|
32
|
0
|
|
16
|
2
|
2
|
|
0
|
Therefore, (520)10 = (208)16
A number system having base or radix 16 is called hexadecimal number system. It consists of 16 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. It is also used in computer basically in memory management.
Here, (111011)2
|
0011
|
1011
|
|
3
|
B
|
= (3B)16
A number system having base or radix 16 is called hexadecimal number system. It consists of 16 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. It is also used in computer basically in memory management.
Here, (B8C)16
|
B
|
8
|
C
|
|
1011
|
1000
|
1100
|
= (101110001100)2
Now,
|
101
|
110
|
001
|
100
|
|
5
|
6
|
1
|
4
|
= (5614)8
A compliment is process of representing the negative numbers or bits in digital computer system. Complements are used in digital computers for simplifying the subtraction operation and for logical manipulation. Using complements, all the arithmetic operators can be performed in the form of addition.
The process of subtraction using 2's complement is given below:
a) Make the number of digits equal in both minuend and subtrahend.
b) Calculate 2's complement of subtrahend.
c) Calculate sum of minuend and 2's complement of subtrahend.
d) Check the overflow digit(carry).
1. If there is overflow digit, discard it and the remaining bits would be final answer.
2. If there is no overflow bit then the result must be negative. So again, calculate 2's complement of the sum and that would be the final answer.
Here, (149)10
|
2
|
149
|
1
|
|
2
|
74
|
0
|
|
2
|
37
|
1
|
|
2
|
18
|
0
|
|
2
|
9
|
1
|
|
2
|
4
|
0
|
|
2
|
2
|
0
|
|
2
|
1
|
1
|
|
0
|
Therefore, (149)10 = (10010101)2
Here, (804)10
|
8
|
804
|
4
|
|
8
|
100
|
4
|
|
8
|
12
|
4
|
|
8
|
1
|
1
|
|
0
|
Therefore, (804)10 = (1444)8
Here, (1600)10
|
16
|
1600
|
0
|
|
16
|
100
|
4
|
|
16
|
6
|
6
|
|
0
|
Therefore, (1600)10 = (640)16
Here, (100100)2 = 25 x 1 + 24 x 0 + 23 x 0 + 22 x 1 + 21 x 0 + 20 x 0
= 32 + 0 + 0 + 4 + 0 + 0
= (36)10
Here, (2040)8 = 83 x 2 + 82 x 0 + 81 x 4 + 80 x 0
= 1024 + 32
= (1056)10
Here, (1E0D)16 = 163 x 1 + 162 x E + 161 x 0 + 160 x D
= 4096 + 256 x 14 + 0 + 1 x 13
= 4096 + 3584 + 13
= (7693)10
Here, (110111101)2
|
110
|
111
|
101
|
|
6
|
7
|
5
|
Therefore, (110111101)2 = (675)8
Here, (1001110111)2
|
0010
|
0111
|
0111
|
|
2
|
7
|
7
|
Therefore, (1001110111)2 = (277)16
Here, (375)8
|
3
|
7
|
5
|
|
011
|
111
|
101
|
Therefore, (375)8 = (11111101)2
Here, (ABC)16
|
A
|
B
|
C
|
|
1010
|
1011
|
1100
|
Therefore, (ABC)16 = (101010111100)2
Here, (555)8
|
5
|
5
|
5
|
|
101
|
101
|
101
|
= (101101101)2
|
0001
|
0110
|
1101
|
|
1
|
6
|
D
|
Therefore, (555)8 = (16D)16
Here, (BCA)16
|
B
|
C
|
A
|
|
1011
|
1100
|
1010
|
= (101111001010)16
|
101
|
111
|
001
|
010
|
|
5
|
7
|
1
|
2
|
Therefore, (BCA)16 = (5712)8
Here, (0.55)10
|
0.55 x 2 = 1.1
|
1
|
|
0.1 x 2 = 0.2
|
0
|
|
0.2 x 2 = 0.4
|
0
|
|
0.4 x 2 = 0.8
|
0
|
|
0.8 x 2 = 1.6
|
1
|
|
0.6 x 2 = 1.2
|
1
|
Therefore, (0.55)10 = (0.100011)2
Here, (234.997)10
|
8
|
234
|
2
|
|
8
|
29
|
5
|
|
8
|
3
|
3
|
|
0
|
Also
|
0.997 x 8 = 7.976
|
7
|
|
0.976 x 8 = 7.808
|
7
|
|
0.808 x 8 = 6.464
|
6
|
|
0.464 x 8 = 3.712
|
3
|
|
0.712 x 8 = 5.696
|
5
|
|
0.696 x 8 = 5.568
|
5
|
Therefore, (234.997)10 = (352.776355)2
Here, (689.336)10
|
16
|
689
|
1
|
|
16
|
43
|
11 = B
|
|
16
|
2
|
2
|
|
0
|
Also
|
0.336 x 16 = 5.376
|
5
|
|
0.376 x 16 = 6.016
|
6
|
|
0.016 x 16 = 0.256
|
0
|
|
0.256 x 16 = 4.096
|
4
|
|
0.096 x 16 = 1.536
|
1
|
|
0.536 x 16 = 8.576
|
8
|
Therefore, (689.336)10 = (2B1.560418)16
Here, (101.1101)2 = 22 x 1 + 21 x 0 + 20 x 1 + 2-1 x 1 + 2-2 x 1 + 2-3 x 0 + 2-4 x 1
= 4 + 0 + 1 + 0.5 + 0.25 + 0 + 0.0625
= (5.8125)10
Here, (0.1042)8 = 8-1 x 1 + 8-2 x 0 + 8-3 x 4 + 8-4 x 2
= 0.125 + 0 + 0.0078125 + 0.00048828125
= (0.1333)10
Here, (FA.AEF)16 = 161 x 15 + 160 x 10 + 16-1 x 10 + 16-2 x 14 + 16-3 x 15
= 240 + 10 + 0.625 + 0.0546875 + 0.00366211
= 250.68335
Here, (101010.110111)2
|
101
|
010
|
110
|
111
|
|
5
|
2
|
6
|
7
|
Therefore, (101010.110111)2 = (52.67)8
Here, (10101.11011)2
|
0001
|
0101
|
1101
|
1000
|
|
1
|
5
|
D
|
8
|
Therefore, (10101.11011)2 = (15.D8)16
Here, (77.226)8
|
7
|
7
|
2
|
2
|
6
|
|
111
|
111
|
010
|
010
|
110
|
Therefore, (77.226)8 = (111111.010010110)2
Here, (0.376)8
|
3
|
7
|
6
|
|
011
|
111
|
110
|
= (011111110)2
|
0111
|
1111
|
|
7
|
F
|
Therefore, (0.376)8 = (0.7F)16
Here, (0.5AB)16
|
5
|
A
|
B
|
|
0101
|
1010
|
1011
|
Therefore, (0.5AB)16 = (0.010110101011)2
Here, (0.226)16
|
2
|
2
|
6
|
|
0010
|
0010
|
0110
|
= (0.001000100110)2
|
001
|
000
|
100
|
110
|
|
1
|
0
|
4
|
6
|
Therefore, (0.226)16 = (0.1046)8
Here, 11111 + 10001
|
11111
|
|
|
+
|
10001
|
|
110000
|
Here, 1111 + 1111
|
1111
|
|
|
+
|
1111
|
|
11110
|
Here, 1100 - 11
|
1100
|
|
|
-
|
0011
|
|
1001
|
Here, 101010 - 1001
|
101010
|
|
|
-
|
001001
|
|
100001
|
Here, 1010 x 1010
|
1010
|
|
|
x
|
1010
|
|
0000
|
|
|
1010x
|
|
|
0000xx
|
|
|
1010xxx
|
|
|
1100100
|
Here, 1111 x 1111
|
1111
|
|
|
x
|
1111
|
|
1111
|
|
|
1111x
|
|
|
1111xx
|
|
|
1111xxx
|
|
|
11100001
|
Here, 110100 / 110
|
110)
|
110100 (1000
|
|
|
110
|
||
|
________
|
||
|
100
|
||
|
000
|
||
|
_______
|
||
|
100
|
||
Therefore, Quotient = 1000
And Remainder = 100
Here, 11010001 / 1001
|
1001)
|
11010001 (10111
|
|||||
|
1001
|
||||||
|
________
|
||||||
|
1000
|
||||||
|
0000
|
||||||
|
_______
|
||||||
|
10000
|
||||||
|
01001
|
||||||
|
_______
|
||||||
|
1110
|
||||||
|
1001
|
||||||
|
_____
|
||||||
|
1011
|
||||||
|
1001
|
||||||
|
______
|
||||||
|
10
|
||||||
Therefore, Quotient = 10111
And Remainder = 10
Here, 1101 - 110
Using 1's complement method:
i. Make the number of bits equal as 1101 and 0110.
ii. 1's complement of 0110 is 1001.
iii. Add both the bits:
|
1101
|
|
|
+
|
1001
|
|
1
|
0110
|
Here, we got overflow bit so discard it and add the remaining part.
0110 + 1 = 0111
Thus, 1101 - 110 = 111
Using 2's complement method:
i. Make the number of bits equal as 1101 and 0110.
ii. 2's complement of 0110 is 1001 + 1 = 1010.
iii. Add both numbers:
|
1101
|
|
|
+
|
1010
|
|
1
|
0111
|
Here, we got overflow bit which is discarded and the remaining part is our answer.
Thus, 1101 - 110 = 111
b.
Here, 1100 - 1111
Using 1's complement method:
i. 1's complement of 1111 is 0000.
ii. Add both the bits:
|
1100
|
|
|
+
|
0000
|
|
1100
|
Here, we did not get overflow bit so we again calculate 1's complement of 1100 i.e. 0011 and put minus sign before it.
Thus, 1100 - 1111 = -11
Using 2's complement method:
i. 2's complement of 1111 is 0000 + 1 = 0001.
ii. Add both numbers:
|
1100
|
|
|
+
|
0001
|
|
1101
|
Here, we did not get overflow bit so, we again calculate 2's complement of 1101 and put minus sign before it.
i.e. 0010 + 1 =0011
Thus, 1101 - 110 = -11
Here, 1234 - 123
Using 9's complement method:
i. Making the numbers equal in both minuend and subtrahend as 1234 and 0123.
ii. 9's complement of 0123 is 9876.
iii. Adding both numbers:
|
1234
|
|
|
+
|
9876
|
|
1
|
1110
|
Here, we got overflow digit, so we discard it and add it to the remaining part.
Thus, 1110 + 1 = 1111
Using 10's complement method:
i. Making the numbers equal in both minuend and subtrahend as 1234 and 0123.
ii. 10's complement of 0123 is 9877.
iii. Adding both numbers:
|
1234
|
|
|
+
|
9877
|
|
1
|
1111
|
Here, we got overflow digit, so we discard it and remaining will the answer.
Thus, 1234 - 123 = 1111
Here, 567 - 4567
Using 9's complement method:
i. Making the numbers equal in both minuend and subtrahend as 0567 and 4567.
ii. 9's complement of 4567 is 5432.
iii. Adding both numbers:
|
0567
|
|
|
+
|
5432
|
|
5999
|
Here, we did not get overflow digit, so we again calculate 9's complement of it.
i.e. 5999 becomes 4000
Thus, 567 - 4567 = 4000
Using 10's complement method:
i. Making the numbers equal in both minuend and subtrahend as 0567 and 4567.
ii. 10's complement of 4567 is 5433.
iv. Adding both numbers:
|
0567
|
|
|
+
|
5433
|
|
6000
|
Here, we did not get overflow digit, so we again calculate 10's complement of it.
i.e. 3999 + 1 = 4000
Thus, 567 - 4567 = 4000
