1. By: Jonathan O. Cabriana
Number System
By: Jonathan O. Cabriana
Ronald Tocci. Digital Systems 8th Edition. Prentice Hall

2. Number System Binary Number System (base 2)
> number system in which there are only two possible digit values, 0 and 1.
> most important in digital systems.
Decimal Number System (base 10)
> number system that uses 10 different digits (0 – 9) or symbols.
> used to represent quantities outside a digital system which is the outside world.

3. Number System Octal Number System
> number system that has a base of 8.
> Digits from 0 – 7 are used to express an octal number.
Hexadecimal Number System
> number system that has a base of 16.
> Digits from 0 through 9 plus letters A through F are used to express a hex number.

4. Number System Helpfulness of Hex and Octal
> often used in a digital system as a “shorthand” way to represent string of bits.
> relatively easy to convert back and forth between binary and either hex or octal.
Example:
( )2 = (6E67)16

5. . Number System Binary – Decimal
> Binary system is a positional system where each digit carries a weight based on its position relative to the LSB.
Ex: ( )2 = ( ___________ )10
2 4
2 3
2 2
2 1
2 0
2 -1
2 -2
2 -3
2 -4 .
Binary Point

6. Conversions = .{ (0625) } (19.0625)10 Solution for whole number:
(10011)2 = (1 x 24) + (0 x 23) + (0 x 22) + (1 x 21) + (1 x 20)
= (1 x 16) + (0 x 8) + (0 x 4) + (1 x 2) + (1 x 1) =
= (19)10
Solution for fractional part:
(.0001)2 = .{ (0 x 2-1) + (0 x 2-2) + (0 x 2-3) + (1 x 2-4) }
= .{ (0) + (0) + (0) + (1 x ¼ ) }
= .{ (0625) }
= :
( )10

7. Conversions (11001)2 Decimal – Binary Ex: (25.0625)10 = ___________2
Solution for whole number: Repeated Division 25 2
= 12 + remainder of 1 12 2
= 6 + remainder of 0
(11001)2 6 2
= 3 + remainder of 0 3 2
= 1 + remainder of 1
Read from bottom to top for the binary equivalent 1 2
= 0 + remainder of 1

8. Conversions
Solution for fractional part: Repeated Multiplication
x 2 = 0.125
Read from top to bottom for the binary equivalent
. 125 x 2 = 0.25
(.0001)2
. 25 x = 0.5
. 5 x = 1.0 1
( )2
. :

9. . Conversions Octal – Decimal
> Use positional weight of octal as shown in the diagram below:
Ex 1: (372)8 = _________10
Ex 2: (24.6)8 = _________10
8 4
8 3
8 2
8 1
8 0
8 -1
8 -2
8 -3
8 -4 .
Octal Point

10. Conversions Decimal – Octal
> Solution for whole number: Repeated Division
> Solution for fractional part: Repeated Multiplication
Ex 1: (266) = _________8
Ex 2: (0.75)10 = _________8

11. Conversions Binary – Octal
> The bits of the binary number are grouped into groups of three bits starting at the LSB, then use 421 rule in obtaining the octal equivalent.
Ex 1: ( )2 = ___________8
Solution:
( )8 4 2 1 4 2 1 4 2 1

12. Conversions Ex 2: (0.001101)2 = ___________8 Solution: . 001 101
. ( )8 4 2 1 4 2 1

13. Conversions Octal – Binary
> reverse the process of binary to octal conversion.

14. Conversions Hexadecimal – Decimal
> Use positional weight of hexadecimal.
Ex: (B27)16 = _________10

15. Conversions Decimal – Hexadecimal
> Solution for whole number: Repeated Division
> Solution for fractional part: Repeated Multiplication
Ex 1: (378) = _________16

16. Conversions Binary – Hexadecimal
> The bits of the binary number are grouped into groups of 4 bits starting at the LSB, then use 8421 rule in obtaining the octal equivalent.
Ex 1: ( )2 = ___________16

17. Conversions Hexadecimal – Binary
> reverse the process of binary to hex conversion.

18. Assignment (1417)10 = _______2 (255)10 = _______2
(255)10 = _______2
( )2 = _______10
( )2 = _______10
(2497)10 = _______8
(511)10 = _______8
(235)8 = _______10
(7A9)16 = _______10
(1600)10 = _______16
(A56.B3C)16 = _______2