Simple inter conversions of digital number systems.

2. Decimal Number system is composed of 10 numerals or symbols.
These numerals are 0 to 9. Using these symbols as digits we can
express any quantity. It is also called base-10 system. It is a
positional value system in which the value of a digit depends on
its position.
These digits can represent any value, for example: 754.
The value is formed by the sum of each digit, multiplied by the
base (in this case it is 10 because there are 10 digits in decimal
system) in power of digit position (counting from zero):

3. In Binary Number system there are only two digits i.e. 0 and 1. It
is base-2 system. It can be used to represent any quantity that
can be represented in decimal or other number system. It is a
positional value system, where each binary digit has its own value
or weight expressed as power of 2.
The following are some examples of binary numbers:
1011012 ,112 ,101102 .
Binary system is preferred more in digital systems and machines
than any other system because in it, there are only two symbols, 0
and 1. In its comparison, if we take the example of decimal
system, an electronic equipment can’t work with 10 different
voltage levels easily that are provided by it in the form of 10 digits.

4. # It has eight unique symbols i.e. 0 to 7.
# It has base of 8.
# Each octal digit has its own value or weight expressed as a
power of 8.
Example – (321.12)8 = (3x82) + (2x81) + (1x80) + (1x8-1 ) + (2x8-2)
# The sequence of these numbers goes 0, 1, 2, 3, 4, 5, 6, 7, 10,
11, 12, 13, 14, 15, 16, 17, 20, 21,…
i.e. – each successive number after 7 is a combination of 2 or
more unique symbols of octal system.

5. # The hexadecimal system uses base 16.
# It has 16 possible digit symbols. It uses the digits 0 to 9 plus the
letters A,B,C,D,E,F as the letter symbols.
# Each hexadecimal digit has its own value or weight expressed
as a power of 16.
# Example : 35616 = (3x162 ) + (5x161 ) + (6x160 )
= 768 + 80 + 6
= 85410

6. Decimal Binary Hexadecimal Octal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
10000
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
10
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
20

7. * Convert 45(10) to Χ(2)
Here, the number is continuously divided by 2 (as it’s the base for
binary system) until the quotient is 0.
Division Quotient Remainder
45 / 2 22 1
22 / 2 11 0
11 / 2 5 1
5 / 2 2 1
2 / 2 1 0
1 / 2 0 1
=> 45(10) =101101(2)

8. Here, the fractional part of a number is repeatedly multiplied by 2
(as it’s the base for binary system) until we get the fractional part as
0.
*Convert 0.182(10) to Χ(2)
Multiplication Product Integer value
0.182 * 2 0.364 0
0.364 * 2 0.728 0
0.728 * 2 1.456 1
0.456 * 2 0.912 0
0.912 * 2 1.824 1
0.824 * 2 1.648 1
0.648 * 2 1.296 1
=> 0.182(10) = 0.0010111(2)

9. *Convert 101101.0010111(2) to Χ(10)
150413120110.0-10-21-30-41-51-61-7
1 * 25 + 0 * 24 + 1 * 23 + 1 * 22 + 0 * 21 + 1 * 20 + 0 * 2-1 + 0 * 2-2 +
1 * 2-3 + 0 * 2-4 + 1 * 2-5 + 1 * 2-6 + 1 * 2-7
= 32 + 0 + 8 + 4 + 0 + 1 + 0 + 0 + 0.125 + 0 + 0.03125 + 0.015625
+ 0.007813
= 45.179688(10)

10. * Convert 45(10) to X(8)
Here, the number is repeatedly divided by 8 (as it’s the base for
octal system) until the quotient is 0. The remainders are combined
in bottom to top order to get the required octal number.
Division Quotient Remainder
45 / 8 5 5
5 / 8 0 5
=> 45(10) = 55(8)

11. * Convert 0.182(10) to Χ(8)
Here, the fractional part of the number is repeatedly multiplied by 8
(as it’s the base for octal system) until it becomes 0.
Multiplication Product Integer
0.182 * 8 1.456 1
0.456 * 8 3.648 3
0.648 * 8 5.184 5
0.184 * 8 1.472 1
0.472 * 8 3.776 3
0.776 * 8 6.208 6
=> 0.182(10) = 0.135136(8)

12. * Convert 55.135136(8) to Χ(10)
5150.1-13-25-31-43-56-6
5 * 81 + 5 * 80 + 1 * 8-1 + 3 * 8-2 + 5 * 8-3 + 1 * 8-4 + 3 * 8-5 + 6 * 8-6
=40 + 5 + 0.125 + 0.03125 + 0.009766 + 0.000244 + 0.0001 +
0.0000229
= 45.1663829(10)

13. Octal
Digit
0 1 2 3 4 5 6 7
Binary
No.
000 001 010 011 100 101 110 111

14. For this conversion, make groups of three digits from right to left
before decimal & left to right after decimal.
Then assign the specific octal value.
* Convert 110101000.101010(2) to X(8)
110 101 000 . 101 010
6 5 0 . 5 2
=> 110101000.101010(2) = 650.52(8)

15. * Convert 650.52(8) to X(2)
6 5 0 . 5 2
110 101 000 . 101 010
=> 650.52(8) = 110101000.101010(2)

16. * Convert 45(10) to X(16)
Here, the number is repeatedly divided by 16 (as it’s the base for
hexadecimal) until the quotient is 0. The remainder is converted to
its hexadecimal equivalent.
Division Quotient Remainder Hex No. (Χ)
45 / 16 2 13 D
2 / 16 0 2 2
=> 45(10) = 2D(16)

17. * Convert 0.182(10) to Χ(16)
Here, the fractional part is repeatedly multiplied by 16 (as it’s the
base for hexadecimal) until it becomes 0. The integer part obtained
in product is converted to its hexadecimal equivalent.
Multiplication Product Integer Hexadecimal No.
0.182 * 16 2.912 2 2
0.912 * 16 14.592 14 Ε
0.592 * 16 9.472 9 9
0.472 * 16 7.552 7 7
0.552 * 16 8.832 8 8
0.832 * 16 13.312 13 D
=> 0.182(10) = 0.2E978D(16)

18. * Convert 2D.2E978D(16) to Χ(10)
21130.2-114-29-37-48-513-6
2 * 161 + 13 * 160 + 2 * 16-1 + 14 * 16-2 + 9 * 16-3 + 7 * 16-4 + 8 *
16-5 + 13 * 16-6
= 32 + 13 + 0.125 + 0.0546875 + 0.00219727 + 0.00010681 +
0.00000762 + 0.00000077
= 45.18199997(10)

19. For this conversion, make groups of four digits from right to left
before decimal & left to right after decimal.
Then assign the specific Hexadecimal value.
* Convert 110101000.101010(2) to X(16)
0001 1010 1000 .1010 1000
1 Α 8 . Α 8
=> 110101000(2) = 1Α8.Α8(16)

20. * Convert 1Α8.Α8(16) to X(2)
Here, the binary equivalents of hexadecimals are written to get the
required number.
1 Α 8 . Α 8
0001 1010 1000 . 1010 1000