Unit 2 Number System and Conversion Boolean Logic
NEB Class 11 Computer Chapter 2 Number System and Boolean Algebra Notes, Exercise Solution 2080.
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Computer Number System
The number system plays a vital role in computer calculations. Number system is an organized and systematic way of representing numbers. Number systems are basically of two types: nonpositional and positional number systems.
 NonPositional Number System
The nonpositional number system is a number system in which each symbol represents the same value, regardless of its position in the number. The symbols are simply added to find out the value of a particular number. The most common nonpositional number system is the Roman Number System. It is a system of representing numbers devised by the ancient Romans. It is based on certain letters which are given values as numerals.
 Positional Number System
Positional number system can be represented by a few symbols called digits, which represent different values depending on the position that they occupy. The value of each digit in such a number is determined by the digits itself, the position of the digit in the number, and by the base of the number system. The main positional number systems used in computer are decimal, binary, octal and hexadecimal.
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Different types of number system are as follows:
 Decimal number system: A number system having base or radix 10 is called decimal number system.
Examples: (789)_{10} or (789)_{D}
_{}
 Binary number system: A number system having base or radix 2 is called binary number system.
Examples: (110)_{2}
_{}
 Octal number system: A number system having base or radix 6 is called octal number system.
Examples: (1234)_{8}
_{}
 Hexadecimal number system: A number system having base or radix 16 is called hexadecimal number system. Examples: (C01F)_{16} or (C01F)_{H}
 Decimal Number System
The base or radix of a number system is the number of different symbols available to represent any digit within that system. For example, the decimal system (Base 10) has a radix of 10. Decimal uses different combinations of 10 symbols to represent any value (i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
 Binary Number System
A number system having base or radix 2 is called binary number system. It consists of 2 bits: 0 and 1. It is also known as Binary digits. It is specially used in internal processing of computer system.
An electronic circuit has two states either ON state or OFF state. The bit 1 represents the high voltage i.e. ON state and the bit 0 represents the low voltage i.e. OFF state of an electronic circuit. So it is used in computer system.
 Octal Number System
A number system having base or radix 8 is called octal number system. It consists of 8 digits: 0, 1, 2, 3, 4, 5, 6, 7). It is also known as octonary number system. The octal system is used in computing as a simple means of expressing binary quantities.
 Hexadecimal Number System
A number system having base or radix 16 is called hexadecimal number system. It consists of 16 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. It is also used in computer basically in memory management.
Check: Class 11 All Chapter Important Question For NEB Exams.
Number Conversion:
Rules
 Decimal Number System to Others (Binary, Octal, HexaDecimal) Number System = Divide (2,8,16)
 Others (Binary, Octal, HexaDecimal) Number System to Decimal Number System= Multiply (2,8,16)
 Binary Number system to Octal Number system and Vice versa = 3 bit binary Grouping Method (421)
 Binary Number System to HexaDecimal Number System and Vice versa= 4 bit binary Grouping Method ( 8421)
 Octal Number System to Hexadecimal Number System and Vice versa= 3 bit & 4 bit groping Method (421 & 8421)
Number Conversion: Rules in short  
1  Decimal N.S._{(10) }To Others N.S._{(2,8,16) }= _{ }Divide (2,8,16)

2  Others N.S._{ (2,8,16) }To Decimal N.S._{(10) }= Multiply (2,8,16) x

3 
3 bit binary grouping method (421) Octal _{(8)}
Binary _{(2)}
4 bit binary grouping method (8421) Hexa Decimal _{(16)}

Number Conversion examples
 Decimal to Binary
Here, (149)_{10}
2  149  1 
2  74  0 
2  37  1 
2  18  0 
2  9  1 
2  4  0 
2  2  0 
2  1  1 
0 
Therefore, (149)_{10} = (10010101)_{2}
 Decimal to Octal
Here, (804)_{10}
8  804  4 
8  100  4 
8  12  4 
8  1  1 
0 
Therefore, (804)_{10} = (1444)_{8}
 Decimal to Hexa decimal
Here, (1600)_{10}
16  1600  0 
16  100  4 
16  6  6 
0 
Therefore, (1600)_{10} = (640)_{16}
 Binary to Decimal
Here, (100100)_{2} = 2^{5} x 1 + 2^{4} x 0 + 2^{3} x 0 + 2^{2} x 1 + 2^{1} x 0 + 2^{0} x 0
= 32 + 0 + 0 + 4 + 0 + 0
= (36)_{10}
 Octal to Decimal
Here, (2040)_{8} = 8^{3} x 2 + 8^{2} x 0 + 8^{1} x 4 + 8^{0} x 0
= 1024 + 32
= (1056)_{10}
 Hexadecimal to Decimal
Here, (1E0D)_{16} = 16^{3} x 1 + 16^{2} x E + 16^{1} x 0 + 16^{0} x D
= 4096 + 256 x 14 + 0 + 1 x 13
= 4096 + 3584 + 13
= (7693)_{10}
 Binary to Octal
Here, (110111101)_{2}
110  111  101  (: 3 bit binary grouping method 
6  7  5 

Therefore, (110111101)_{2} = (675)_{8}
_{}
 Binary to Hexadecimal
Here, (1001110111)_{2}
0010  0111  0111 
2  7  7 
Therefore, (1001110111)_{2} = (277)_{16}
 Octal to Binary
Here, (375)_{8}
3  7  5 
011  111  101 
Therefore, (375)_{8 }= (11111101)_{2}
 Hexadecimal to Binary
Here, (ABC)_{16}
A  B  C 
1010  1011  1100 
Therefore, (ABC)_{16} = (101010111100)_{2}
 Octal to Hexadecimal
Here, (555)_{8}
5  5  5 
101  101  101 
= (101101101)_{2}
0001  0110  1101 
1  6  D 
Therefore, (555)_{8} = (16D)_{16}
 Hexadecimal to Octal
Here, (BCA)_{16}
B  C  A 
1011  1100  1010 
= (101111001010)_{16}
101  111  001  010 
5  7  1  2 
Therefore, (BCA)_{16} = (5712)_{8}
_{}
Fractional Number System Conversion
 Decimal to Binary
Here, (0.55)_{10}
0.55 x 2 = 1.1  1 
0.1 x 2 = 0.2  0 
0.2 x 2 = 0.4  0 
0.4 x 2 = 0.8  0 
0.8 x 2 = 1.6  1 
0.6 x 2 = 1.2  1 
Therefore, (0.55)_{10} = (0.100011)_{2}
 Decimal to Octal
Here, (234.997)_{10}
8  234  2 
8  29  5 
8  3  3 
0 
Also
0.997 x 8 = 7.976  7 
0.976 x 8 = 7.808  7 
0.808 x 8 = 6.464  6 
0.464 x 8 = 3.712  3 
0.712 x 8 = 5.696  5 
0.696 x 8 = 5.568  5 
Therefore, (234.997)_{10} = (352.776355)_{2}
 Decimal to Hexadecimal
Here, (689.336)_{10}
16  689  1 
16  43  11 = B 
16  2  2 
0 
Also
0.336 x 16 = 5.376  5 
0.376 x 16 = 6.016  6 
0.016 x 16 = 0.256  0 
0.256 x 16 = 4.096  4 
0.096 x 16 = 1.536  1 
0.536 x 16 = 8.576  8 
Therefore, (689.336)_{10} = (2B1.560418)_{16}
 Binary to Decimal
Here, (101.1101)_{2} = 2^{2} x 1 + 2^{1} x 0 + 2^{0} x 1 + 2^{1} x 1 + 2^{2} x 1 + 2^{3} x 0 + 2^{4} x 1
= 4 + 0 + 1 + 0.5 + 0.25 + 0 + 0.0625
= (5.8125)_{10}
 Octal to Decimal
Here, (0.1042)_{8} = 8^{1} x 1 + 8^{2} x 0 + 8^{3} x 4 + 8^{4} x 2
= 0.125 + 0 + 0.0078125 + 0.00048828125
= (0.1333)_{10}
 Hexadecimal to Decimal
Here, (FA.AEF)_{16 } = 16^{1} x 15 + 16^{0} x 10 + 16^{1} x 10 + 16^{2} x 14 + 16^{3} x 15
= 240 + 10 + 0.625 + 0.0546875 + 0.00366211
= 250.68335
 Binary to Octal
Here, (101010.110111)_{2}
101  010  110  111 
5  2  6  7 
Therefore, (101010.110111)_{2} = (52.67)_{8}
 Binary to Hexadecimal
Here, (10101.11011)_{2}
0001  0101  1101  1000 
1  5  D  8 
Therefore, (10101.11011)_{2} = (15.D8)_{16}
 Octal to Binary
Here, (77.226)_{8}
7  7  2  2  6 
111  111  010  010  110 
Therefore, (77.226)_{8} = (111111.010010110)_{2}
 Octal to Hexadecimal
Here, (0.376)_{8}
3  7  6 
011  111  110 
= (011111110)_{2}
0111  1111 
7  F 
Therefore, (0.376)_{8} = (0.7F)_{16}
_{}
 Hexadecimal to Binary
Here, (0.5AB)_{16}
5  A  B 
0101  1010  1011 
Therefore, (0.5AB)_{16} = (0.010110101011)_{2}
 Hexadecimal to Octal
Here, (0.226)_{16}
2  2  6 
0010  0010  0110 
= (0.001000100110)_{2}
001  000  100  110 
1  0  4  6 
Therefore, (0.226)_{16} = (0.1046)_{8}
Binary Arithmetic
Rule for addition0+0=0 0+1=1 1+0=1 1+1=10 (0 with carry over 1) 1+1+1=11 (1 with carry over 1) 1+1+1+1=100 (0 with carry over 10) 1+1+1+1+1=101 (0 with carry over 10)

 Binary Addition
Here, 11111 + 10001
11111  
+  10001 
110000 
Here, 1111 + 1111
1111  
+  1111 
11110 
 Binary Subtraction
Rule for Subtraction00=0 10=1 11=0 01=1 (with borrowing 1 from left side Then it becomes 10 i.e. 101=1)

Here, 1100 – 11
1100  
–  0011 
1001 
Here, 101010 – 1001
101010  
–  001001 
100001 
 Binary Multiplication
Rule for Multiplication0x0=0 1×0=0 0x1=0 1×1=1

Here, 1010 x 1010
1010  
x  1010 
0000  
1010x  
0000xx  
1010xxx  
1100100 
Here, 1111 x 1111
1111  
x  1111 
1111  
1111x  
1111xx  
1111xxx  
11100001 
 Binary Division
Here, 110100 / 110
110) 
110100 (1000  
110  
________  
100  
000  
_______  
100 
Therefore, Quotient = 1000 and Remainder = 100
Here, 11010001 / 1001
1001)  11010001 (10111  
1001  
________  
1000  
0000  
_______  
10000  
01001  
_______  
1110  
1001  
_____  
1011  
1001  
______  
10  
Therefore, Quotient = 10111 and Remainder = 10
1’s and 2’s Binary Subtraction Method
A compliment is process of representing the negative numbers or bits in digital computer system. Complements are used in digital computers for simplifying the subtraction operation and for logical manipulation. Using complements, all the arithmetic operators can be performed in the form of addition.
The process of subtraction using 2’s complement is given below:
 Make the number of digits equal in both minuend and subtrahend.
 Calculate 2’s complement of subtrahend.
 Calculate sum of minuend and 2’s complement of subtrahend.
 Check the overflow digit (carry).
 If there is overflow digit, discard it and the remaining bits would be final answer.
 If there is no overflow bit then the result must be negative. So again, calculate 2’s complement of the sum and that would be the final answer.
1’S AND 2’S COMPLEMENT BINARY SUBTRACTION METHOD
 Subtract 110 from 1101 using 1’s complement binary subtraction method.
Using 1’s complement method:
Here Main value= 1101
Second Value= 110
 Make the number of bits equal as 1101 and 0110.
 1’s complement of second value 0110 is 1001.
 Add both the bits:
1101  
+  1001 
1  0110 
Here, we got overflow bit so discard it and add the remaining part.
0110 + 1 = 0111
Thus, 1101 – 110 = 111
Using 2’s complement method:
Here Main value= 1101
Second Value= 110
 Make the number of bits equal as 1101 and 0110.
 1’s complement of second value 0110 is 1001.
 2’s Complement of 1001 + 1 = 1010.
 Add both numbers:
1101  
+  1010 
1  0111 
Here, we got overflow bit which is discarded and the remaining part is our answer.
Thus, 1101 – 110 = 111
 Subtract 1111 from 1100 using 1’s and 2’s complement binary subtraction method.
Here, Main value =1100
Second value= 1111
Both are equal digit.
Using 1’s complement method:
 1’s complement of second value 1111 is 0000.
 Add both the bits:
1100  
+  0000 
1100 
Here, we did not get overflow bit so we again calculate 1’s complement of 1100 i.e. 0011 and put minus sign before it.
Thus, 1100 – 1111 = 11
Using 2’s complement method:
 1’s complement of second value 1111 is 0000.
 2’s complement of 0000 + 1 = 0001.
 Add both numbers:
1100  
+  0001 
1101 
Here, we did not get overflow bit so, we again calculate 2’s complement of 1101 and put minus sign before it.
i.e. 0010 + 1 =0011
Thus, 1101 – 110 = 11
9’S AND 10’S COMPLEMENT DECIMAL SUBTRACTION METHOD
 Subtract 123 from 1234 using 9’s and 10’s decimal subtraction method.
Using 9’s complement method:
Main value=1234
Second value=123
 Making the numbers equal in both minuend and subtrahend as 1234 and 0123.
 9’s complement of 0123 is (99990123) = 9876.
 Adding both numbers:
1234  
+  9876 
1  1110 
Here, we got overflow digit, so we discard it and add it to the remaining part.
Thus, 1110 + 1 = 1111
Using 10’s complement method:
Main value=1234
Second value=123
 Making the numbers equal in both minuend and subtrahend as 1234 and 0123.
 9’s complement of 0123 is (99990123) = 9876.
 10’s complement of 0123 is (9876+1) = 9877.
 Adding both numbers:
1234  
+  9877 
1  1111 
Here, we got overflow digit, so we discard it and remaining will the answer.
Thus, 1234 – 123 = 1111
 Subtract 4567 from 567 using 9’s and 10’s decimal subtraction method.
Using 9’s complement method:
Main value= 567
Second value= 4567
Using 9’s complement method:
 Making the numbers equal in both minuend and subtrahend as 0567 and 4567.
 9’s complement of 4567 (99994567) is 5432.
 Adding both numbers:
0567  
+  5432 
5999 
Here, we did not get overflow digit, so we again calculate 9’s complement of it.
i.e. 99995999 becomes 4000
Thus, 567 – 4567 = 4000
Using 10’s complement method:
 Making the numbers equal in both minuend and subtrahend as 0567 and 4567.
 9’s complement of 4567 (99994567) is 5432.
 10’s complement of 4567 (5432+1) is 5433.
 Adding both numbers:
0567  
+  5433 
6000 
Here, we did not get overflow digit, so we again calculate 10’s complement of it.
i.e. (99996000)= 3999 + 1 = 4000
Thus, 567 – 4567 = 4000
Boolean algebra
Boolean algebra:
Boolean algebra is a study of mathematical operations performed on certain variables (called binary variables) that can have only two values: true (represented by 1) or false (represented by 0).
Logic Function (Boolean Function):
A logic function is an expression algebraically with binary variables, logical operation symbols, parenthesis and equal sign, is known as Boolean function. Example, F= A.B.C’+A.B
Truth Table:
Truth table is a table which represents all the possible values of logical variables/statements along with all the possible results of the given combinations of values. For example, following logical statements can have only one of the two values (TRUE (YES) or FALSE (NO)).
Logic gates:
Logic gates perform basic logical functions and are the fundamental building blocks of digital integrated circuits. Most logic gates take an input of two binary values, and output a single value of a 1 or 0.
AND Gate:
Logical Symbol
Truth Table
Input  Output  
A  B  F=A.B 
0  0  0 
0  1  0 
1  0  0 
1  1  1 
OR Gate:
Output. It is denoted by (+) operator and graphically represented by:
Logical Symbol
Truth Table
Input  Output  
A  B  F=A+B 
0  0  0 
0  1  1 
1  0  1 
1  1  1 
NOT Gate:
Logical Symbol
Truth Table
Input  Output 
A  F=A’ 
0  1 
1  0 
NAND Gate:
NAND gate is the combination of AND and NOT gate. NAND gate generates true (1) output if at least any of the input is false otherwise, it generates false output. Graphically it is represented by:
Logical Symbol
Truth table
Input  Output  
A  B  A.B  F=(A.B)’ 
0  0  0  1 
0  1  0  1 
1  0  0  1 
1  1  1  0 
NOR Gate:
Logical Symbol:
The truthtable of NOR gate is:
Input  Output  
A  B  (A+B)  F=(A+B)’ 
0  0  0  1 
0  1  1  0 
1  0  1  0 
1  1  1  0 
Exclusive (XOR)Gates:
The XOR gate produces false output (0) when both the inputs are same otherwise, the output will be true (1). It can also have two or more inputs which produces only one output.
Logical Symbol:
The truth table of XOR gate is given below:
Inputs 



 Output  
A  B  A’  B’  A’.B  A.B’  F=A’.B+A.B’ 
0  0  1  1  0  0  0 
0  1  1  0  1  0  1 
1  0  0  1  0  1  1 
1  1  0  0  0  0  0 
ExclusiveNOR(XNOR) Gate:
The XNOR (exclusiveNOR) gate is a combination XOR gate followed by an inverter. Its output is “true” if the inputs are the same and output is “false” if the inputs are different.
The XNOR gate symbol is given below:
The truth table of XNOR gate is given below:
Input 



 Output  
A  B  A’  B’  A’.B’  A.B  F=A.B+A’.B’ 
0  0  1  1  1  0  1 
0  1  1  0  0  0  0 
1  0  0  1  0  0  0 
1  1  0  0  0  1  1 
There are 2 De Morgan’s laws or theorems:
Theorem 1: The complement of a sum of variables is equal to the product of the complement of each variables.
(A+B)’ = A’.B’
A  B  A’  B’  A+B  (A+B)’  A’.B’ 
0  0  1  1  0  1  1 
0  1  1  0  1  0  0 
1  0  0  1  1  0  0 
1  1  0  0  1  0  0 
Here, (A+B)’ = A’.B’ thus proved.
Theorem 2: The complement of a product of variables is equal to the sum of the complement of each variables.
(A.B)’ = A’ + B’
A  B  A’  B’  A.B  (A.B)’  A’+B’ 
0  0  1  1  0  1  1 
0  1  1  0  0  1  1 
1  0  0  1  0  1  1 
1  1  0  0  1  0  0 
Here, (A.B)’ = A’ + B’ thus proved.
Duality principle:
Duality principle state can be obtained by replacing AND (.) with OR (+) and vice versa, 1 with 0 and vice versa keeping variables and complements and variables are unchanged.
For example, duality of the expression A.B’ = A+B’ and A’.B+C = A’+B.C
Laws of Boolean algebra
The Boolean laws are the rules for manipulating Boolean variables by using Boolean operations. This set of rules states that how variables and values assigned with the Boolean operators reacts. These laws are used to circuit minimization. The process of shorting the length and using minimum components to construct same functioning circuit is called circuit minimization. The Boolean laws are as follows.
7.
a.
Here, A.(B +C) = A+(B.C)
b.
Here, A.B + C.1 + 0.1 = A + B.C + 0.1 + 0
c.
Here, C.D + A.0 + 1 = C + D.A + 1.0
d.
Here, 1.0 + A + C.1 = 0 + 1.A.C + 0
8.
Associative law states that when ORing or ANDing more than two variables, the result is the same regardless of the grouping of the variables.
(a) (A + B) + C = A + (B + C)
(b) (A B) C = A (B C)
Proof:
A  B  C  A+B  B+C  (A+B)+C  A+(B+C) 
0  0  0  0  0  0  0 
0  0  1  0  1  1  1 
0  1  0  1  1  1  1 
0  1  1  1  1  1  1 
1  0  0  1  0  1  1 
1  0  1  1  1  1  1 
1  1  0  1  1  1  1 
1  1  1  1  1  1  1 
Here, (A + B) + C = A + (B + C) thus proved.
Distributive law states that ORing/ANDing two or more variables and then ANDing/ORing the result with a single variable is equivalent to ANDing/ORing the single variable with each of the two or more variables and then ORing/ANDing the products/sums.
(a) A (B + C) = A.B + A . C
(b) A + (B . C) = (A + B) . (A + C)
Proof:
A  B  C  B+C  A.B  A.C  A.(B+C)  A.B+A.C 
0  0  0  0  0  0  0  0 
0  0  1  0  0  0  0  0 
0  1  0  1  0  0  0  0 
0  1  1  1  0  0  0  0 
1  0  0  1  0  0  0  0 
1  0  1  1  0  1  1  1 
1  1  0  1  1  0  1  1 
1  1  1  1  1  1  1  1 
Here, A.(B+C) = A.B + A.C thus proved.
9.
a.
Here,
AB + A’BC + BC
= AB + BC(A’+1)
=AB + BC
=B(A + C)
The truth table is:
A  B  C  A+C  B(A+C) 
0  0  0  0  0 
0  0  1  1  0 
0  1  0  0  0 
0  1  1  1  1 
1  0  0  1  0 
1  0  1  1  0 
1  1  0  1  1 
1  1  1  1  1 
b.
Here,
PQ’ + Q(P + Q) + P(P’ + Q)
= PQ’ + PQ + QQ + PP’ + PQ
= PQ’ + PQ + Q + 0
= P(Q’+Q) + Q
= P + Q
The truth table is:
P  Q  P+Q 
0  0  0 
0  1  1 
1  0  1 
1  1  1 
c.
Here,
(X + Y)(XY’Z + XYZ + XY’Z’)
= XY’Z + XYZ + XY’Z’ + XYY’Z + XYZ + XYY’Z’
= XY’Z + XYZ + XY’Z’ + 0 + 0
=XY’ (Z + Z’) + XYZ
= XY’ + XYZ
= X(Y’+YZ)
= X(Y’+ Z)
The truth table is:
X  Y  Z  Y’  Y’+Z  X(Y’+Z) 
0  0  0  1  1  0 
0  0  1  1  1  0 
0  1  0  0  0  0 
0  1  1  0  1  1 
1  0  0  1  1  1 
1  0  1  1  1  1 
1  1  0  0  0  0 
1  1  1  0  1  1 
10.
Here,
a = A.B.C
b = A’ + B + C
c = A’.B’.C
Now, a + b + c = A.B.C + A’+ B + C + A’.B’.C
= B + A’ + C + A’B’C
= A’ + B’ + C
Principle of duality
 Changing to symbol OR (+) operation from AND (.) operation or AND (.) operation from OR (+) operation and digits 0 and 1 that is called principle of duality.
 Example, (x+y).z=(x.y)+z
Uses of gates
 All the gates are used to design digital electronic circuit.
 XOR and XNOR gates are used for circuit minimization.
 The NAND and NOR gates are used for construction of flash memory. Flash memories are used in various devices like PDA, Laptops, mobiles phones etc.
 Flash memory are constructed by using NOR are gates and external memories constructor by using NAND gates. For example, mini SD, Micro SD, Memory Cards.
Uses of Boolean algebra in computer science
 Logical function are useful not only to the hardware designer in implementing circuits but also to software designer in making decision, performing arithmetic, recognizing characters and patterns, checking for errors, formatting output and assembling and disassembling data.
Check: Chapter 1: Computer System Complete note
Check: Class 11 All Chapter Important Question For NEB Exams.