1.
Define
Number System.
The system concerned with the number and
represented by sequence of digits is called number system. It plays vital role
in computing and electronics. Number system also refers to the digits, its arrangement,
positional value and base of number system.
2. What
is base or radix of number system?
Ã˜ The
total number of digits used by the particular number system is called base or
radix of that number system. For Example: Base of Binary number system is 2 because it uses two digits 0 and
1 only.
3. List
out the different types of number system with their bases.
Ã˜ There
are 4 types of number system and its base are given below:
Number System

Base

Binary Number System

2

Octal Number System

8

Decimal
Number System

10

Hexadecimal Number
System

16

4.
Define
Binary number system.
Ã˜ The
number system having base 2 and consists of digits: 0 and 1 is called Binary
number system.
5.
Define
Octal number system.
Ã˜ The
number system having base 8 and consists of digits: 0, 1, 2, 3, 4, 5, 6 and 7
is called Octal number system.
6.
Define
Decimal (denary) number system.
Ã˜ The
number system having base 10 and consists of digits: 0, 1, 2, 3, 4, 5, 6, 7, 8
and 9 is called Decimal (Denary) number system.
7.
Define
Hexadecimal number system.
Ã˜ The
number system having base 10 and consists of digits: 0, 1, 2, 3, 4, 5, 6, 7, 8
, 9 and alphabets A, B, C, D, E and F is
called Hexadecimal number system.
Where:
A

10

B

11

C

12

D

13

E

14

F

15

8.
Why
do digital computers use binary number system for their operations?
Ã˜ The
number system having base 2 and consists of digits: 0 and 1 is called Binary
number system. It has two bits 0 and 1. An electronic circuit has two states
either ON state or OFF state. The bit 1 represents high voltage (ON state) and
the bit 0 represents the low voltage (OFF state) of an electronic circuit.
That’s why digital computer uses binary number for their operations.
According it the order of writing symbols
for a number, number systems are divided into two types.
A)
Positional
Number System:
The position of a
symbol in number has positional significance. Positional number is easy for
arithmetic computations. Different
position of the same symbol signifies different meaning (values).Examples of
positional number systems is decimal, binary etc.
Example: 
294=200+90+4 and 429=400+20+9. Therefore 294and 429 are different.
B)
NonPositional
Number System:
The
order of the symbols is the number does not have any significance. Examples of
nonpositional number systems are Roman numbers, Greek numbers etc. Example:
I,II,III,IV,V,VI,X,L,M etc. Arithmetic
with these numbers is difficult or complex arithmetic is almost impossible. In
today’s world nonpositional number systems are rarely used because of this
complex nature and difficulty in arithmetic.
# CONVERSION OF NUMBER SYSTEM RULE (dfxfdGq)
(A)
From other Number System to Decimal
We should multiply the each digit
by its base.
For example:
i)
Convert (11111)_{2} to its decimal
equivalent. [HISSAN 2069D1)
Solution:
Converting from Binary to decimal, we get
Face Value

1

1

1

1

1

Place
Value

2^{4}

2^{3}

2^{2}

2^{1}

2^{0}

Here,
=1x2^{4}+1x2^{3}+1x2^{2}+1x2^{1}+1x2^{0}
=1x16+1x8+1x4+1x2+1x1
=
16+8+4+2+1
= (31)_{10}
ii)
(1011.01)_{2}=
(?)_{10}
Solution:
Converting from Binary to decimal, we get
Face
Value

1

0

1

1

. 0

1

Place
Value

2^{3}

2^{2}

2^{1}

2^{0}

2^{1}

2^{2}

Here,
=1x2^{3}+0x2^{2}+1x2^{1}+1x2^{0}+0x2^{1}+1x2^{2}
=8+0+2+1+0+0.25
= (11.25)_{10}
iii)
(1570.04) _{8}=
(?)_{10}
Solution: Converting from Octal to decimal, we
get
Face
Value

1

5

7

0

0

4

Place
Value

8^{3}

8^{2}

8^{1}

8^{0}

8^{1}

8^{2}

Here,
=1x8^{3}+5x8^{2}+7x8^{1}+0x8^{0}+0x8^{1}+4x8^{2}
=512+320+56+0+0+0.0625
= (888.0625)_{10}
iv)
(CAB)_{ 16}= (?)_{10}
Here,
Face
Value

C

A

B

Place
Value

16^{2}

16^{1}

16^{0}

Here,
=Cx16^{2}+Ax16^{1}+Bx16^{0}
=12x256+10x16+11x1=3072+160+11
=
(3243)_{10}
(B)
From Decimal to other System
i)
Decimal to Binary
(456)_{10}=
(?)_{2}
Solution:
Converting from decimal to
binary, we get
(45.125)_{10 }=
( ? )_{2}
(0.125)_{10 }= (?)_{2}
0.125 x 2 = 0.25 integer part = 0
0.25 x 2= 0.5 integer part = 0
0.5 x 2 = 1 integer part = 1
So, (0.125)_{10}=(0.001)_{2}
Finally
we can write,
(45.125)_{10} = (101101.001)_{2}

ii)
Decimal to Octal
(827)_{10}= (?)_{8}
So, (827)_{10}=(1473)_{16}



QN
1. Convert (BABA)_{16} to its equivalent octal number. [HSEB2073“D”]
Solution: At first converting Hexadecimal to decimal,
we get
Face Value

B

B

A

Place Value

16^{2}

16^{1}

16^{0}

Here,
= B x 16^{2} + B x 16^{1}
+ A x 16^{0}
= 11 x 256 + 11 x 16 + 10 x 1
=
(3002)_{10}
Now, Converting decimal to octal, we get
8

3002

Tips: 3002 / 8 = 375. 25


8

375


2

write
375; for remainder

8

46


7

3002
– 8 x 375 = 2

8

5


6


0


5

# 9’s
and 10’s Complements Decimal Subtraction
By using the
complement method, addition process can be used to perform subtraction. This
reduces hardware of the computer. With the same hardware used for addition,
computers can perform subtraction through the complement method.
There are two
types of complement s for each base or radix ‘r’ system:
The r’s
complement and the (r1)’s complement
When the value of
the base or radix ‘r’ is 2 then the two types are referred as the 2’s and 1’s
complement for Binary Number System and the 10’s and 9’s Complement for the
Decimal Number System.
By:
r’s Complement method” 10’s Complement for Decimal number
1. Let
M=72532 and N= 13250 ,
10’s
Complement of i.e. 13250 = (9999913250) +1=86750
.
^{.} . 72532
+ 86750
1 59282
2. If
M>N, the sum will produce an end carry which is discarded and what is left
is the result is MN. Here M>N, so end carry 1 is discarded, so answer is
59282.
3. If
M<N , the sum does not produce an end carry and is equal to r’s complement
of the sum and place negative sign in front.
Let
M= 13250 and N=72532,
10’s
complement of N = (9999972532) +1=27468
13250+27468=40718
Since,
M<N, so answer = 10’s complement of sum i.e. 40718
=
(9999940718) +1=  59282
By
(r1)‘s Complement method for decimal number
1. Let
M=72532 and N=13250
Here, 9’s complement of N
i.e. 13250= 9999913250=86749
.
^{.} . 72532
+ 86749
1 59281
2. If
M>N, the sum will produce an end carry which is discarded and what is left
is the result. Here, M>N, so end carry 1 is discarded and is added to the
result to get the final result. So, answer = 59282+1=59282.
3. If
M<N, the sum does not produce an end carry and is equal to (r1)’s
complement of the sum and place negative sign in front.
Let M=13250 and N=72532,
So, 9’s Complement of N
i.e. 72532 = 9999972532 = 27467
.
^{.} . 13250
+ 27467
40717
Here, M>N, So answer =
9’s Complement of 40717
= 9999940717=  59282
Binary
Addition:
As with decimal addition, to add two binary numbers we need to add two corresponding bits from the two numbers at a time. The result of addition of two bits from the two numbers is given in the table.
As with decimal addition, to add two binary numbers we need to add two corresponding bits from the two numbers at a time. The result of addition of two bits from the two numbers is given in the table.
X

Y

X+Y

0

0

0

0

1

1

1

0

1

1

1

0
(carry 1)

Workout
Examples:
a)
110+10
110
+ 010
1000

b) 101101+1001
101101
+ 001001
110110

c)
1011+0111
1011
+ 0111
10010

Binary
Subtraction:
As with decimal subtraction, to subtract a
binary numbers from other we need to subtract a correspond bit from other. The
result of subtraction of a bit from the other is given in the table.
X

Y

XY

0

0

0

0

1

1 (borrow)

1

0

1

1

1

0

Workout
Examples:
i) 1101111
1101

0111
0110

ii) 101011010
10101

01010
01011

iii) 1010100
1010
 0100
0110

1’s
and 2’s Complement Methods of Binary Subtraction:
Complements are used in digital computers
for simplifying the subtraction operation and for logical manipulation. There
are two types of complements for each base or radix ‘r’ system.
1’s Complement
The 1’s complement
of binary number is obtained by subtracting each digit from 1. Because of the subtraction
of a binary digit from 1, the bit changes from 0 to 1 and 1 to 0. For example,
the complement of 1’s of 1011001 is 01001110 and the 1’s complement of 000111
is 111000.
2’s Complement
2’s complement of
binary number is obtained by adding 1 to 1’s complement value. The 2’s
complement of 101100 is 010100 and is obtained by adding 1 to 1’s complement of
101100, i.e.
= (111111101100)
+1=010100
Some solved 1’s and 2’s complement problems:
1.
Subtract
(1011)_{2} from (1100)_{2} by using 1’s and 2’s complement
method.
Solution: We
have to find 11001011 = ?
1’s complement method
Let, P = 1100 and Q =
1011
Since, both numbers have
equal bits. No need to add zero in either case in front to make them equal.
1’s Complement of Q = 1111
– 1011
= 0100
Now, adding 1’s
complement of Q with P, we get
1
0100
+ 1100
10000
Since, P > Q, there is an end carry 1.
So, adding the end carry 1 at last, we get
Answer
= 0000 + 1
= (0001)_{2}
2’s complement method
Let, P = 1100 and Q =
1011
Since, both numbers have
equal bits. No need to add zero in either case in front to make them equal.
2’s Complement of Q = (1111
– 1011) + 1
= 0100 + 1
= 0101
Now, adding 2’s
complement of Q with P, we get
1
0101
+ 1100
10001
Since, P > Q, there is an end carry 1.
In 2’s complement we have to discard the end carry 1 , then
Answer
= (0001)_{2}
2.
Subtract
(111)_{2} from (1000)_{2} by using 1’s and 2’s complement
method.
Solution:
We have to find 1000111 = ?
1’s complement method
Let, P = 1000 and Q = 0111
Since, both numbers have not equal bits. We
have to add one zero in 2^{nd} number in order to make both equal bits.
1’s Complement of Q = 1111 – 0111
= 1000
Now, adding 1’s complement of Q with P, we get
1000
+ 1000
10000
Since, P > Q, there is an end carry 1.
So, adding the end carry 1 at last, we get
Answer
= 0000 + 1
= (0001)_{2}
2’s complement method
Let, P = 1000 and Q = 0111
Since, both numbers have not equal bits.
We have to add one zero in 2^{nd} number in order to make both equal
bits.
2’s Complement of Q = (1111
– 0111) + 1
= 1000 + 1
= 1001
Now, adding 2’s
complement of Q with P, we get
1001
+ 1000
10001
Since, P > Q, there is an end carry 1.
In 2’s complement we have to discard the end carry 1 , then
Answer
= (0001)_{2}
3.
Subtract
(1000)_{2} from (111)_{2} by using 1’s and 2’s complement
method.
Solution: We
have to find 1111000 = ?
1’s complement method
Let, P = 0111 and Q = 1000
Since, both numbers have not
equal bits. We have to place one zero in front of 1^{st} number to make
them equal.
1’s Complement of Q =
1111 – 1000
= 0111
Now, adding 1’s
complement of Q with P, we get
111
0111
+ 0111
1110
Since, P < Q, there is no end carry 1.
So, we have to find the 1’s complement of 1110 and place the minus sign in
front of final answer, we get
1’s
complement of 1110= 11111110
=
 (0001)_{2}
2’s complement method
Let, P = 0111 and Q = 1000
Since, both numbers have not
equal bits. We have to place one zero in front of 1^{st} number to make
them equal.
2’s Complement of Q = (1111
– 1000)+1
= 0111+1 = 1000
Now, adding 2’s
complement of Q with P, we get
1000
+ 0111
1111
Since, P < Q, there is no end carry 1.
So, we have to find the 2’s complement of 1111 and place the minus sign in
front of final answer, we get
2’s
complement of 1111= (11111110) + 1
= 0000 + 1
=  (0001)_{2}
Homework [Unit – 3]
1.
List out the
different number system with its base. Convert (456)_{10} to binary
number system.
2.
Convert (BBA)_{16}
to its octal number.
3.
Subtract (1001)_{2}
from (1111)_{2} by using 1’s and
2’s complement method.
4.
Subtract (1000)_{2}
from (1111)_{2} by using 1’s and 2’s complement method.
5.
Subtract (100000)_{2}
from (111)_{2} by using 1’s and 2’s complement method.
6.
Perform the
following binary operation:
i.
10111+111

ii.
11111001

iii.
110011101

iv.
1111+1111

0
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