Chapter 2 : Number System
2.1 Decimal, Binary, Octal and Hexadecimal
Numbers
2.2 Relation between binary number system
with other number system
2.3 Representation of integer, character and
floating point numbers in binary
2.4 Binary Arithmetic
2.5 Arithmetic Operations for Ones
Complement, Twos Complement, magnitude
and sign and floating point number
Decimal, Binary, Octal and
Hexadecimal Numbers
Most numbering system use positional
notation :
N = a
n
r
n
+ a
n-1
r
n-1
+ + a
1
r
1
+ a
0
r
0
Where:
N: an integer with n+1 digits
r: base
a
i
{0, 1, 2, , r-1}
Examples:
a) N = 278
r = 10 (base 10) => decimal numbers
symbol: 0, 1, 2, 3, 4, 5, 6,
7, 8, 9 (10 different symbols)
N = 278 => n = 2;
a
2
= 2; a
1
= 7; a
0
= 8
278 = (2 x 10
2
) + (7 x 10
1
) + (8 x 10
0
)
Hundreds Ones
Tens
N = a
n
r
n
+ a
n-1
r
n-1
+ + a
1
r
1
+ a
0
r
0
b) N = 1001
2
r = 2 (base-2) => binary numbers
symbol: 0, 1 (2 different symbols)
N = 1001
2
=> n = 3;
a
3
= 1; a
2
= 0; a
1
= 0; a
0
= 1
1001
2
= (1 x 2
3
)+(0 x 2
2
)+(0 x 2
1
)+(1 x 2
0
)
c) N = 263
8
r = 8 (base-8) => Octal numbers
symbol : 0, 1, 2, 3, 4, 5, 6, 7,
(8 different symbols)
N = 263
8
=> n = 2; a
2
= 2; a
1
= 6; a
0
= 3
263
8
= (2 x 8
2
) + (6 x 8
1
) + (3 x 8
0
)
N = a
n
r
n
+ a
n-1
r
n-1
+ + a
1
r
1
+ a
0
r
0
d) N = 263
16
r = 16 (base-16) => Hexadecimal
numbers
symbol : 0, 1, 2, 3, 4, 5, 6, 7, 8,
9, A, B, C, D, E, F
(16 different symbols)
N = 263
16
=> n = 2;
a
2
= 2; a
1
= 6; a
0
= 3
263
16
= (2 x 16
2
)+(6 x 16
1
)+(3 x 16
0
)
Decimal Binary Octal Hexadecimal
0
0
0
0
1
1
1
1
2
10
2
2
3
11
3
3
4
100
4
4
5
101
5
5
6
110
6
6
7
111
7
7
8
1000
10
8
9
1001
11
9
10
1010
12
A
11
1011
13
B
12
1100
14
C
13
1101
15
D
14
1110
16
E
15
1111
17
F
16
10000
20
10
There are also non-positional numbering systems.
Example: Roman Number System
1987 = MCMLXXXVII
Relation between binary number
system and others
Binary and Decimal
Converting a decimal number into
binary (decimal binary)
Divide the decimal number by 2 and
take its remainder
The process is repeated until it
produces the result of 0
The binary number is obtained by
taking the remainder from the bottom
to the top
Example: Decimal Binary
53
10
=> 53 / 2 = 26 remainder 1
26 / 2 = 13 remainder 0
13 / 2 = 6 remainder 1
6 / 2 = 3 remainder 0
3 / 2 = 1 remainder 1
1 / 2 = 0 remainder 1
= 110101
2
(6 bits)
= 00110101
2
(8 bits)
(note: bit = binary digit)
Read from
the bottom
to the top
0.81
10
binary???
0.81
10
=> 0.81 x 2 = 1.62
0.62 x 2 = 1.24
0.24 x 2 = 0.48
0.48 x 2 = 0.96
0.96 x 2 = 1.92
0.92 x 2 = 1.84
= 0.110011
2
(approximately)
0.110011
2
Converting a binary number into decimal
(binary decimal)
Multiply each bit in the binary number
with the weight (or position)
Add up all the results of the
multiplication performed
The desired decimal number is the
total of the multiplication results
performed
Example: Binary Decimal
a)111001
2
(6 bits)
(1x2
5
) + (1x2
4
) + (1x2
3
) + (0x2
2
) +
(0x2
1
) + (1x2
0
)
= 32 + 16 + 8 + 0 + 0 + 1
= 57
10
b)00011010
2
(8 bits)
= 2
4
+ 2
3
+2
1
= 16 + 8 + 2
= 26
10
Binary and Octal
Theorem
If base R
1
is the integer power of
other base, R
2
, i.e.
R
1
= R
2
d
e.g., 8 = 2
3
Every group of d digits in R
2
(e.g., 3 digits)is equivalent to 1
digit in the R
1
base
(Note: This theorem is used to convert
binary numbers to octal and hexadecimal
or the other way round)
From the theorem, assume that
R
1
= 8 (base-8) octal
R
2
= 2 (base-2) binary
From the theorem above,
R
1
= R
2
d
8 = 2
3
So, 3 digits in base-2
(binary) is equivalent to 1
digit in base-8 (octal)
From the stated theorem, the
following is a binary-octal
conversion table.
Binary
Octal
000
0
001
1
010
2
011
3
100
4
101
5
110
6
111
7
In a computer
system, the
conversion from
binary to octal or
otherwise is based
on the conversion
table above.
3 digits in base-2 (binary) is equivalent to 1 digit in base-8 (octal)
Example: Binary Octal
Convert these binary numbers into octal
numbers:
(a) 00101111
2
(8 bits) (b) 11110100
2
(8 bits)
Refer to the binary-octal
conversion table
000 101 111
= 57
8
0
5
7
Refer to the binary-octal
conversion table
011 110 100
= 364
8
3
6
4
The same method employed in binary-octal
conversion is used once again.
Assume that:
R
1
= 16 (hexadecimal)
R
2
= 2 (binary)
From the theorem: 16 = 2
4
Hence, 4 digits in a binary number is
equivalent to 1 digit in the hexadecimal
number system (and otherwise)
The following is the binary-hexadecimal
conversion table
Binary and Hexadecimal
Binary
Hexadecimal
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
8
1001
9
1010
A
1011
B
1100
C
1101
D
1110
E
1111
F
Example:
1. Convert the following
binary numbers into
hexadecimal numbers:
(a) 00101111
2
Refer to the binary-
hexadecimal conversion table
above
0010 1111
2
= 2F
16
2 F
Convert the following octal numbers
into hexadecimal numbers (16 bits)
(a) 65
8
(b) 123
8
Refer to the binary-octal conversion table
6
8
5
8
110 101
0000 0000 0011 0101
2
0 0 3 5
= 35
16
Refer to the binary-octal conversion table
1
8
2
8
3
8
001 010 011
0000 0000 0101 0011
2
0 0 5 3
= 53
16
Example: Octal Hexadecimal
octal binary hexadecimal
Convert the following hexadecimal
numbers into binary numbers
(a) 12B
16
(b) ABCD
16
Refer to the binary-hexadecimal
conversion table
1 2 B
16
0001 0010 1011
2
(12 bits)
= 000100101011
2
Refer to the binary-hexadecimal
conversion table
A B C D
16
1010 1011 1101 1110
2
= 1010101111011110
2
Example: Hexadecimal Binary
Exercise 1
Binary decimal
001100
11100.011
Decimal binary
145
34.75
Octal hexadecimal
5655
8
Solution 1
Binary decimal
001100 = 12
11100.011 = 28.375
Decimal binary
145 = 10010001
34.75 = 100010.11
Octal hexadecimal
octal binary decimal hexadecimal
5655
8
= BAD
0 x 2
5
+ 0 x 2
4
+ 1 x 2
3
+ 1 x 2
2
+ 0 x 2
1
+ 0 x 2
0
=
8 +4 = 12
145/2 = 72 (reminder 1); 72/2=36(r=0); 36/2=18(r=0);
18/2=9(r=0); 9/2=4(r=1); 4/2=2(r=0); 2/2=1(r=0); =0(r=1)
Octal binary
101:110:101:101
Binary hexadecimal
1011:1010:1101
B A D
Solution 1
Binary decimal
001100 = 12
11100.011 = 28.375
Decimal binary
145 = 10010001
34.75 = 100010.11
Octal hexadecimal
octal binary hexadecimal
5655
8
= BAD
0x2
-1
+1x2
-2
+1x2
-3
0.75 x 2 = 1.5
0.5 x 2 = 1.0
Exercise 2
Binary decimal
110011.10011
Decimal binary
25.25
Octal hexadecimal
12
8
B
11001.01
51.59375
Representation of integer, character and
floating point numbers in binary
Introduction
Machine instructions operate on data. The most
important general categories of data are:
1. Addresses unsigned integer
2. Numbers integer or fixed point, floating point
numbers and decimal (eg, BCD (Binary Coded Decimal))
3. Characters IRA (International Reference
Alphabet), EBCDIC (Extended Binary Coded Decimal
Interchange Code), ASCII (American Standard Code for
Information Interchange)
4. Logical Data
- Those commonly used by computer users/programmers: signed
integer, floating point numbers and characters
Integer Representation
-1101.0101
2
= -13.3125
10
Computer storage &
processing do not have
benefit of minus signs (-)
and periods.
Need to represent the integer
Signed Integer Representation
Signed integers are usually used
by programmers
Unsigned integers are used for
addressing purposes in the
computer (especially for
assembly language programmers)
Three representations of signed
integers:
1. Sign-and-Magnitude
2. Ones Complement
3. Twos Complement
Sign-and-Magnitude
The easiest representation
The leftmost bit in the
binary number represents the
sign of the number. 0 if
positive and 1 if negative
The balance bits represent
the magnitude of the number.
Examples:
i) 8 bits binary number
__ __ __ __ __ __ __ __
7 bits for magnitude (value)
a) +7 = 0 0 0 0 0 1 1 1
(7 = 10000111
2
)
b) 10 = 1 0 0 0 1 0 1 0
(+10 = 00001010
2
)
Sign bit
0 => +ve 1 => ve
ii) 6 bits binary number
__ __ __ __ __ __
5 bits for magnitude (value)
a) +7 = 0 0 0 1 1 1
(7 = 1 0 0 1 1 1
2
)
b) 10 = 1 0 1 0 1 0
(+10 = 0 0 1 0 1 0
2
)
Sign bit
0 => +ve 1 => ve
Ones Complement
In the ones complement representation,
positive numbers are same as that of
sign-and-magnitude
Example: +5 = 00000101 (8 bit)
as in sign-and-magnitude representation
Sign-and-magnitude and ones complement
use the same representation above for +5
with 8 bits and all positive numbers.
For negative numbers, their
representation are obtained by changing
bit 0 1 and 1 0 from their positive
numbers
Example:
Convert 5 into ones complement
representation (8 bit)
Solution:
First, obtain +5 representation
in 8 bits 00000101
Change every bit in the number
from 0 to 1 and vice-versa.
5
10
in ones complement is
11111010
2
Exercise:
Get the representation of ones
complement (6 bit) for the
following numbers:
i) +7
10
ii) 10
10
Solution:
(+7) = 000111
2
Solution:
(+10)
10
= 001010
2
So,
(-10)
10
= 110101
2
Twos complement
Similar to ones complement, its
positive number is same as sign-
and-magnitude
Representation of its negative
number is obtained by adding 1 to
the ones complement of the number.
Example:
Convert 5 into twos complement
representation and give the answer
in 8 bits.
Solution:
First, obtain +5 representation in 8
bits 00000101
2
Obtain ones complement for 5
11111010
2
Add 1 to the ones complement number:
11111010
2
+ 1
2
= 11111011
2
5 in twos complement is 11111011
2
Exercise:
Obtain representation of twos
complement (6 bit) for the
following numbers
i) +7
10
ii)10
10
Solution:
(+7) = 000111
2
(same as sign-magnitude)
Solution:
(+10)
10
= 001010
2
(-10)
10
= 110101
2
+ 1
2
= 110110
2
So, twos compliment
for 10 is 110110
2
Exercise:
Obtain representation for the following
numbers
Decimal Sign-magnitude Twos complement
+7
+6
-4
-6
-7
+18
-18
-13
4 bits
8 bits
Solution:
Obtain representation for the following
numbers
Decimal Sign-magnitude Twos complement
+7 0111 0111
+6 0110 0110
-4 1100 1100
-6 1110 1010
-7 1111 1001
+18 00010010 00010010
-18 10010010 11101110
-13 11110010 11110011
Character Representation
For character data type, its
representation uses codes
such as the ASCII, IRA or
EBCDIC.
Note: Students are encouraged
to obtain the codes
Floating point representation
In binary, floating point
numbers are represented in the
form of : +S x B
+E
and the number
can be stored in computer words
with 3 fields:
i) Sign (+ve, ve)
ii) Significant S
iii) Exponent E
and B is base is implicit and
need not be stored because it is
the same for all numbers (base-
2).
Binary Arithmetics
1. Addition ( + )
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10
1 + 1 + 1 = (1 + 1) + 1 = 10 + 1 = 11
2
Example:
i. 010111
2
+ 011110
2
= 110101
2
ii. 100011
2
+ 011100
2
= 111111
2
010111
011110 +
110101
2. Multiplication ( x )
0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1
3. Subtraction ( )
0 0 = 0
0 1 = 1 (borrow 1)
1 0 = 1
1 1 = 0
4. Division ( / )
0 / 1 = 0
1 / 1 = 1
Example:
i. 010111
2
- 001110
2
= 001001
2
ii. 100011
2
- 011100
2
= 000111
2
Exercise:
i. 1000100 010010
ii. 1010100 + 1100
iii. 110100 1001
iv. 11001 x 11
v. 110111 + 001101
vi. 111000 + 1100110
vii. 110100 x 10
viii. 11001 - 1110
Arithmetic Operations for Ones Complement,
Twos Complement, sign-and-magnitude and
floating point number
Addition and subtraction for
signed integers
Reminder: All subtraction
operations will be changed into
addition operations
Example: 8 5 = 8 + (5)
10 + 2 = (10) + 2
6 (3) = 6 + 3
Sign-and-Magnitude
Z = X + Y
There are a few possibilities:
i. If both numbers, X and Y are
positive
o Just perform the addition operation
Example:
5
10
+ 3
10
= 000101
2
+ 000011
2
= 001000
2
= 8
10
ii. If both numbers are negative
o Add |X| and |Y| and set the sign bit = 1
to the result, Z
Example: 3
10
4
10
= (3) + (4)
= 100011
2
+ 100100
2
Only add the magnitude, i.e.:
00011
2
+ 00100
2
= 00111
2
Set the sign bit of the result
(Z) to 1 (ve)
= 100111
2
= 7
10
iii. If signs of both number differ
o There will be 2 cases:
a) | +ve Number | > | ve Number |
Example: (2) + (+4), (+5) + (3)
Set the sign bit of the ve
number to 0 (+ve), so that both
numbers become +ve.
Subtract the number of smaller
magnitude from the number with a
bigger magnitude
Sample solution:
Change the sign bit of the ve number to
+ve
(2) + (+4) = 100010
2
+ 000100
2
= 000100
2
000010
2
= 000010
2
= 2
10
b) | ve Number | > | +ve Number |
Subtract the +ve number from the ve number
Example: (+3
10
) + (5
10
)
= 000011
2
+ 100101
2
= 100101
2
000011
2
= 100010
2
= 2
10
Ones complement
In ones complement, it is easier than sign-
and-magnitude
Change the numbers to its representation
and perform the addition operation
However a situation called Overflow might
occur when addition is performed on the
following categories:
1. If both are negative numbers
2. If both are in difference sign and
|+ve Number| > | ve Number|
Overflow => the addition result
exceeds the number of bits that was
fixed
1. Both are ve numbers
Example: 3
10
4
10
= (3
10
) + (4
10
)
Solution:
Convert 3
10
and 4
10
into ones
complement representation
+3
10
= 00000011
2
(8 bits)
3
10
= 11111100
2
+4
10
= 00000100
2
(8 bits)
4
10
= 11111011
2
Perform the addition operation
(3
10
) => 11111100 (8 bit)
+(4
10
) => 11111011 (8 bit)
7
10
111110111 (9 bit)
11110111
+ 1
11111000
2
= 7
10
Overflow occurs. This value is called EAC and needs to be
added to the rightmost bit.
the answer
2. | +ve Number| > |ve Number|
This case will also cause an
overflow
Example: (2) + 4 = (2) + (+4)
Solution:
Change both of the numbers above
into ones complement
representation
2 = 11111101
2
+4 = 00000100
2
Add both of the numbers
(2
10
) => 11111101 (8 bit)
+ (+4
10
) => 00000100 (8 bit)
There is an EAC
+2
10
100000001 (9 bit)
Add the EAC to the rightmost bit
00000001
+ 1
00000010
2
= +2
10
the answer
Note:
For cases other than 1 & 2 above, overflow does not occur
and there will be no EAC and the need to perform addition to
the rightmost bit does not arise
Twos Complement
Addition operation in twos
complement is same with that
of ones complement, i.e.
overflow occurs if:
1. If both are negative numbers
2. If both are in difference
and |+ve Number| > |ve
Number|
Both numbers are ve
Example: 3
10
4
10
= (3
10
) + (4
10
)
Solution:
Convert both numbers into twos
complement representation
+3
10
= 000011
2
(6 bit)
3
10
= 111100
2
(ones complement)
3
10
= 111101
2
(twos complement)
4
10
= 111011
2
(ones complement)
4
10
= 111100
2
(twos complement)
111100 (3
10
)
111011 (4
10
)
= 111001
2
(twos complement)
= 7
10
Perform addition operation on both
the numbers in twos complement
representation and ignore the EAC.
1111001
Ignore the
EAC
The answer
Note:
In twos complement, EAC is
ignored (do not need to be added
to the leftmost bit, like that of
ones complement)
2. | +ve Number| > |ve Number|
Example: (2) + 4 = (2) + (+4)
Solution:
Change both of the numbers above
into twos complement representation
2 = 111110
2
+4 = 000100
2
Perform addition operation on both
numbers
(2
10
) => 111110 (6 bit)
+ (+4
10
) => 000100 (6 bit)
+2
10
1000010
Ignore the EAC
The answer is 000010
2
= +2
10
Note: For cases other than 1 and 2
above, overflow does not occur.
Exercise:
Perform the following arithmetic
operations in ones complement and also
twos complement
1. (+2) + (+3) [6 bit]
2. (2) + (3) [6 bit]
3. (2) + (+3) [6 bit]
4. (+2) + (3) [6 bit]
Compare your answers with the stated
theory
