This topic introduces the numbering systems: decimal, binary, octal and hexadecimal. The topic covers the conversion between numbering systems, binary arithmetic, one's complement, two's complement, signed number and coding system. This topic also covers the digital logic components.

1. Chapter 2:
DATA REPRESENTATION
IN COMPUTER MEMORY
SUMMARY: This topic introduces the numbering systems: decimal, binary, octal
and hexadecimal. The topic covers the conversion between numbering systems,
binary arithmetic, one's complement, two's complement, signed number and
coding system. This topic also covers the digital logic components.
CLO 2:apply appropriate method to solve arithmetic problem in numbering
system (C3).
RTA: (08 : 08)

2. 2.1 Understand data representation
on CPU.
2.1.1 Define decimal, binary, octal and
hexadecimal number.
2.1.2 Perform arithmetic operation (addition
and subtraction) in different number
bases.
2.1.3 Convert decimal, binary, octal and
hexadecimal numbers to different bases and
vice-versa

3. INTRODUCTION
 The binary system and decimal system is
most important in digital system.
 Decimal - Universally used to represent
quantities outside a digital system.
 Its means, there will be situations decimal
values must be converted to binary values
before entered to digital system.
 Example : Calculator / Computer

4. DECIMAL NUMBERING SYSTEM
 Decimal system is composed of 10 numerals
or symbols.
 These 10 sysmbols are 0, 1, 2, 3, 4, 5, 6,
7, 8, 9.
 Using these symbols as digits of a
number, it can express any quantity.

5.  Base number = 10
 Basic number = 0,1,2,3,4,5,6,7,8,9
23410
Basic number
Base number

6. Positional Values (weights)
2 7 4 6 . 2
102
103
101
100
10-1
Positional values
(weights)
MSD LSD
2746.210 is from calculation below:
2746.2 = (2x103
) + (7x102
) + (4x101
) + (6x100
) + (2x10-1
)
= 2000 + 700 + 40 +6 +0.2
= 2746.2
Most significant digit Least significant digit

7. BINARY NUMBERING SYSTEM
 Define Binary numbers
 Binary numbers representing number in which only
digits 0 or 1.
 ADDITION BINARY NUMBERS
Basic binary addition rule :
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10
1 + 1 + 1 = 11
 Example : 101 + 101 = 1010
1011 + 1011 = ?

8.  Ex 1:
110112 + 100012 = 1011002
 Ex 2:
 101112 + 1112 = ________
Exercise

9. Subtraction
Four conditions in binary subtraction
 0 - 0 = 0
 0 - 1 = 1 borrow 1
 1 - 0 = 1
 1 - 1 = 0
 10 - 1 = 1
 If a 10 being borrow a 1, what’s left with
that 10 is a 1

10.  Ex 1:
10012 – 102 = 1112
 Ex 2:
 1010112 – 11112 =__________

11. Conversions of Binary Numbers
Binary to Decimal conversions
 Example : 1 1 0 1 12
 24
+ 23
+ 22
+ 21
+ 20
= 16 + 8 + 2 + 1
= 2710

12. Decimal to Binary conversions
 Convert 2510 to binary number
Exercise: Convert 3010 to binary number

13. OCTAL NUMBERING SYSTEM
 The octal number system has a base of eight,
meaning that it has eight possible digits:
0,1,2,3,4,5,6 and 7.
 The digit positions in an octal number have
weights as follows :
84
83
82
81
80 •. 8-1
8-2
8-3

14.  Ex:
5248 – 1678 = 3358
1678 – 248 = _________
Octal number - (Subtraction -
Pengurangan)

15. Octal –to-decimal conversion
Convert 3728 to decimal number
3728 = 3 x (8
2
) + 7 x (8
1
) + 2 x (8
0
)
= (3 x 64) + (7x 8) + (2 x 1)
= 25010

16. Octal number - Addition (Penambahan)
 Ex:
1238
+ 3218
4448
 Ex:
4578
+ 2458

17.  Decimal integer can be converted to octal by
using the same repeated-division method
with a division factor of 8.
Decimal-to-Octal Conversion

18. Octal –to- Binary conversion

19. Binary to Octal conversion

20.  The hexadecimal number system uses base
16.
 It has 16 possible digit symbols.
 It uses the digits 0 through 9 plus the letters
A, B, C, D, E and F as the 16 digit symbols.
HEXADECIMAL NUMBERING
SYSTEM
7A16
Basic number
Base number

21. Hexadecimal number - Addition
(Penambahan)
 Ex:
3 316
 + 4 716
 Ex:
2 0 D 316
+ 1 2 B C16
7 A16

22. Hexadecimal number - Subtraction
(Pengurangan)
 Ex:
4 416
- 1 716
 Ex:
3 2 5 516
- 3 1 8 216
2 D16

23. Hexadecimal-To-Decimal Conversion
 A hexadecimal number can be converted to
its decimal equivalent by using the fact that
each hex digit position has a weight that is a
power of 16.

24.  Ex 1:
1416 = (1 x 161
) + (4 x 160
)
= 16 + 4
= 2010

25.  Ex 2:
ABC16 = (10 x 162
) + (11 x 161
)
= (12 x 160
)
= 2560 + 176 + 12
= 274810

26.  Decimal to hex conversion can be done using
repeated division by 16.
 Ex: Convert 2010 to hex
Decimal-To-Hexadecimal Conversion
16 20
16 1 4
2010 = 1416

27.  Like the octal number system, the
hexadecimal number system is used
primarily as a “shorthand” method for
representing binary numbers.
 It is a relatively simple matter to convert a
hex number to binary .
 Each hex digit is converted to its four-bit
binary equivalent.
Hexadecimal-to-Binary Conversion

28.  Ex:
111010012 = __________
00101101001111002 =___________

29.  The binary number is grouped into groups
of four bits, and each group is converted
to its equivalent hex digit.
 Zero are added, as needed to complete a
four-bit group.
 Ex:
1012 = 0101
= 516
Binary-to-Hexadecimal Conversion

30. Summary
Hexadecimal Decimal Binary
0 0 0000
1 1 0001
2 2 0010
3 3 0011
4 4 0100
5 5 0101
6 6 0110
7 7 0111

31. Summary
Hexadecimal Decimal Binary
8 8 1000
9 9 1001
A 10 1010
B 11 1011
C 12 1100
D 13 1101
E 14 1110
F 15 1111

32. 2.1.4 Describe the coding system
a. Sign and magnitude
b. 1’s Complement and 2’s Complement
c. Binary Coded Decimal (BCD system)
d. ASCII and EBCDIC

33. Describe the coding system
Sign and Magnitude
Positive sign, ( 0 )
or
Negative sign ( 1 )
Magnitude =
Size
Or value

34. Describe the coding system
Sign and Magnitude
Example 1 : Represent -9 in sign
and magnitude.
-9
sign magnitude
1 1001
11001
-9 in sign &magnitude value is 11001

35. Describe the coding system
Sign and Magnitude
Example 2 : Represent 25 in sign
and magnitude.
+25
sign magnitude
0 11001
011001
25 in sign & magnitude value is 011001

36. One’s Complements and Two’s
Complements
 One’s Complements
 One’s complements is used in binary number.
 The one’s complement of a binary number is
obtained by changing each 0 to 1 and 1 to a 0.
 Only change negative number
 In other words, change each bit in the number to
its complement.

37.  Exp:
 10011001 – original binary number
 01100110 – complement each bit to
form 1’s complement
 Thus, we say that the 1’s complement of
10011001 is 01100110.

38.  Exp:
 Convert -2710 to 1’s complement
a) -2710 = 001002
 Convert -4510 to 1’s complement
-------------

39.  Two’s Complement
 The 2’s complement of a binary number is formed
by taking the 1’s complement of the number and
adding 1 to the least-significant-bit (LSB)
position.
 Exp:
101101 binary equivalent of 45
010010 complement each bit to form 1’s
complement
+ 1 add 1
010011 2’s complement of original binary
number
Two’s complement = One’s Complement + 1

40. Exercise
1. Convert the number below to
1’s complement 2’s
complement
------------------- -------------------
-101110012 0100 0110 0100 0111
-5768 01000 0001 01000 0010
-124516
-45

41. Addition in 1’s complement
 Exp 1 : 810 + (-310)
1000 8 change to binary number
+ 1100 -3 change to 1’s complement
10100
1 add carry to LSB
0101
 Exercise 2 : 510 + (-210)
----------------

42.  Exp 1: 810 + (-310)
1100 -3 change to 1’s complement
+ 1
1101 -3 change to 2’s complement
+ 1000 8 change to binary number
10101
Addition in 2’s complement
This carry is disregarded, the result is 0101 (sum=5)

43.  Exp 2: -810 + 310
0111 -8 change to 1’s complement
+ 1
1000 -8 change to 2’s complement
+ 0011 3 change to binary number
1011 negative sign bit

44.  Only binary number which have –ve sign need to
change to 1’s complement. If the number is decimal
number, change the number to binary number.
 The –ve number that already change to 1’s
complement, means that the number already
change to +ve. So that, the subtraction process
have change to addition process.
 Overflow bit in addition process need to carry to LSB
and add with the number.
Subtraction in 1’s complement

45.  Exp 1: 2510 – 1310
Step 1 : Convert 2510 and -1310 to binary number.
2510 = 110012 1310 = 11012
Step 2 : Change 1310 = 1101 to 1’s complement
1310 = 11012 change to 1’s complement = 10010
11001 25 change to binary number
+ 10010 -13 change to 1’s complement
101011
+ 1 add 1
01100 total=12

46.  Change the number are given to binary
number.
 For each –ve binary number, we must change
to 1’s complement (change 0 to 1 and 1 to 0).
 Then, add the number with 1.
Subtraction in 2’s complement

47.  Exp 1: 2510 – 1310
Step 1 : Convert 2510 and -1310 to binary number.
2510 = 110012 1310 = 11012
Step 2 : Change -1310 to 2’s complement
-1310 = 011012 change to 1’s complement = 10010
10010 change to 2’s complement = 10011
11001 25 change to binary
+ 10010 -13 change to 2’s complement
101100 total=12
Carry disregard

48.  Solve this arithmetic with 2’s complement.
i. 428 – 158
ii. 101110 - 56910
Exercise

49. Signed Number
 Addition of Signed Number
 Addition number same sign
 Exp: +4 + (+8) = +12
+4 00000100
+8 00001000
+12 00001100

50.  Addition number different sign
 Exp 1:
(-4) + (+8) = +4
-4 11111100
+ (+8) 00001000
+4 1 00000100
This carry is disregarded
00000100 +4
11111011 1’s complements
+ 1
11111100 2’s complements

51.  Exp 1:
(-12) + (+5) = -7
-12 11110100
+ +5 00000101
-7 11111001
00001100 +12
11110011 1’s complements
+ 1
11110100 2’s complements
Negatif sign bit

52. Subtraction of Signed Number
Positive Number (-) Negative Number
 Exp:
+10 – (-5) = +10 + (+5)
= + 15
+10 00001010
+ 5 00000101
+15 00001111

53. Negative Number (-) Positive Number
 Exp:
-10 – (+5) = - 15
- 10 11110110 11110110
- (+5) 00000101 + 11111011
-15 111110001

54. Negative Number (-) Negative Number
 Exp:
-10 – (-5) = -10 + (5)
= - 5
-10 11110110
- (- 5) + 00000101
- 5 11111011

55. BCD Code
 BCD – Binary Coded Decimal
 It’s contain BCD 8421, 2421, 3321, 5421, 5311,
4221 and etc.
 The common used is BCD 8421 and BCD 2421.

56. 4 bit BCD Code
Desimal 5421 5311 4221 3321 2421 8421 7421
0 0000 0000 0000 0000 0000 0000 0000
1 0001 0001 0001 0001 0001 0001 0001
2 0010 0011 0010 0010 0010 0010 0010
3 0011 0100 0011 0011 0011 0011 0011
4 0100 0101 1000 0101 0100 0100 0100
5 1000 1000 0111 1010 1011 0101 0101
6 1001 1001 1100 1100 1100 0110 0110
7 1010 1011 1101 1101 1101 0111 1000
8 1011 1100 1110 1110 1110 1000 1001
9 1100 1101 1111 1111 1111 1001 1010

57. Binary-Coded-Decimal Code
 If each digit of a decimal number is represented by
its binary equivalent, the result is a code called
binary-coded-decimal.
 Decimal digit can be as large as 9, four bits are
required to code each digit (the binary code for 9 is
1001)

58.  Exp: 87410
8 7 4 (decimal)
1000 0111 0100 (BCD)

59. BCD 8421 Code – to – Binary Number
 Exp:
Convert 1001 0110BCD 8421 to binary number.
Step 1: Change BCD 8421 code to decimal
number.
1001 0110
9 6
Step 2 : Change decimal number to binary
number.
1001 0110BCD 8421 = 11000002

60.  Exp:
Convert 10010102 to BCD 8421
code.
Step 1: Change binary number to decimal
number.
10010102 = 7410
Step 2: Change decimal number to BCD
8421 code.
10010102 = 01110111BCD 8421
Binary Number – to – BCD 8421 Code

61.  The most widely used alphanumeric code is the
American Standard Code for Information
Interchange (ASCII).
 The ASCII code is a seven-bit code and so it has 27
= 128 possible code groups.
ASCII Code

62. MSB
LSB
Binary 000 001 010 011 100 101 110 111
Binary Hex 0 1 2 3 4 5 6 7
0000 0 Nul Del sp 0 @ P p
0001 1 Soh Dc1 1 1 A Q a q
0010 2 Stx Dc2 “ 2 B R b r
0011 3 Etx Dc3 # 3 C S c s
0100 4 Eot Dc4 $ 4 D T d t
0101 5 End Nak % 5 E U e u
0110 6 Ack Syn & 6 F V f v
0111 7 Bel Etb ‘ 7 G W g w
1000 8 Bs Can ( 8 H X h x
1001 9 HT Em ) 9 I Y i y
1010 A LF Sub . : J Z j z
1011 B VT Esc + ; K k
1100 C FF FS , < L l
1101 D CR GS - = M m
1110 E SO RS . > N n
1111 F SI US / ? O o
ASCII code

63.  Exp:
An operator is typing in a BASIC program
at the keyboard of a certain
microcomputer. The computer converts
each keystroke into its ASCII code and
stores the code as a byte in memory.
Determine the binary strings that will be
entered into memory when operator types
in the following BASIC statement:
GOTO 25

64.  Solution:
Locate each character (including the
space) and record ASCII code.
G 01000111
O 01001111
T 01010100
O 01001111
(space) 00100000
2 00110010
5 00110101
*0 was added to the leftmost bit of each ASCII code because the
Codes must be stored as bytes (eight bits).

65. Exercise :
1.The following message encode in ASCII
code. What the meaning of this code ?
 a) 54 4F 4C 4F 45 47
 b) 48 45 4C 4C 4F
 c) 41 50 41 4B 48 41 42 41 52

66. EBCDIC
 Stand for Extended Binary Coded Decimal
Interchange Code.
 was first used on the IBM 360 computer,
which was presented to the market in 1964.
 Used with large computer such as
mainframe.