Class 11 Computer Science – Unit 2: Number System and Boolean Logic
This unit provides a comprehensive overview of Number Systems and Boolean Logic, covering fundamental concepts of different number systems, their conversions, binary arithmetic, Boolean algebra, and logic gates. Understanding these concepts is crucial for computer science applications and forms the foundation of digital electronics and computer architecture.
Unit Information
Unit 2: Number System and Boolean Logic (15 hours)
Subject: Computer Science, Grade 11, NEB
Description: This guide provides complete notes covering number systems, binary arithmetic, Boolean algebra, logic gates, and Boolean laws as per NEB syllabus.
Credit: Important Notes
Table of Contents
Detailed Unit Notes
2.1 Number System and Conversion
2.1.1 Decimal, Binary, Octal, Hexadecimal Number System & Conversion
A number system is a set of values used to represent different quantities and characters. It deals with numbers and their representation (base, exponent, and radix in different systems).
Categories of Number Systems:
- Positional number system – The value of a digit depends on its position in the number
- Non-positional number system – The value of a digit does not depend on its position
1. Decimal or Denary Number System
- Most popular numbering system used in daily life
- Contains ten digits (0, 1, 2, …, 9)
- Also called base 10 number system
- Subscripted by 10 or ‘D’
- Example: 1510 or 15D
2. Binary Number System
- Number system of computers – the “mother language” of computers
- Consists of two digits (0 and 1)
- Also called base 2 number system
- Subscripted by 2 or ‘B’
- Example: 10112 or 1011B
3. Octal Number System
- Consists of 8 digits (0-7)
- Also called base 8 number system
- Subscripted by 8 or ‘O’
- Example: 1268 or 126O
| Octal | Decimal | Binary (421) |
|---|---|---|
| 0 | 0 | 000 |
| 1 | 1 | 001 |
| 2 | 2 | 010 |
| 3 | 3 | 011 |
| 4 | 4 | 100 |
| 5 | 5 | 101 |
| 6 | 6 | 110 |
| 7 | 7 | 111 |
4. Hexadecimal Number System
- Consists of 16 symbols: 10 digits (0-9) and 6 letters (A-F)
- Also called base 16 number system
- Subscripted by 16 or ‘H’
- Example: 12AB16 or 12ABH
| 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 |
| 8 | 8 | 1000 |
| 9 | 9 | 1001 |
| A | 10 | 1010 |
| B | 11 | 1011 |
| C | 12 | 1100 |
| D | 13 | 1101 |
| E | 14 | 1110 |
| F | 15 | 1111 |
Conversion of Number Systems
1. Decimal to Binary:
1510 = ?2
2 | 15
2 | 7 -- 1
2 | 3 -- 1
| 1 -- 1
1510 = 11112
43.62510 = ?2
Integer part:
2 | 43
2 | 21 -- 1
2 | 10 -- 1
2 | 5 -- 0
2 | 2 -- 1
| 1 -- 0
Fractional part:
0.625 × 2 = 1.25 → 1
0.25 × 2 = 0.5 → 0
0.5 × 2 = 1.0 → 1
43.62510 = 101011.1012
2. Decimal to Octal:
12310 = ?8
8 | 123
8 | 15 -- 3
| 1 -- 7
12310 = 1738
3. Decimal to Hexadecimal:
12310 = ?16
16 | 123
| 7 -- 11 (B)
12310 = 7B16
4. Binary to Decimal:
10112 = ?10
= 1×23 + 0×22 + 1×21 + 1×20
= 8 + 0 + 2 + 1 = 1110
5. Binary to Octal (Group bits in 3s):
0101112 = ?8
010 | 111
2 | 7
0101112 = 278
6. Binary to Hexadecimal (Group bits in 4s):
11011101112 = ?16
0011 | 0111 | 0111
3 | 7 | 7
11011101112 = 37716
2.1.2 Calculation in Binary Addition and Subtraction
Binary Addition Rules:
| A | B | Sum | Carry |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
Examples of Binary Addition:
Example 1:
1 1 1 0 1 0 1
+ 1 0 1 0 1 1 0
----------------
1 1 0 0 1 0 1 1
Example 2:
1 0 1 0
+ 1 1 1 1
----------
1 1 0 0 1
Binary Subtraction Rules:
| A | B | Difference | Borrow |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 |
Examples of Binary Subtraction:
Example 1:
1 0 1 1 0 1
- 1 0 1 1 1
--------------
1 0 1 1 0
Example 2:
1 1 1 1 1
- 1 0 0 0 1
------------
0 1 1 1 0
2.1.3 One’s and Two’s Complement Methods of Binary Subtraction
In computer systems, subtraction is not performed directly as arithmetic subtraction. It is performed using the technique called complement, which is essentially a process of repeated addition.
Types of Complement:
- r’s complement (e.g., 2’s or 10’s complement)
- (r-1)’s complement (e.g., 1’s or 9’s complement)
1’s Complement:
The 1’s complement of a binary number is obtained by subtracting each bit from 1. We can get the 1’s complement by simply replacing 1s with 0s and 0s with 1s.
Example: 1’s complement of 1010 is 0101
2’s Complement:
The 2’s complement of a binary number is obtained by adding binary 1 to the 1’s complement of the number.
Example:
Binary number: 110110
1's complement: 001001
Add 1: + 1
2's complement: 001010
Binary Subtraction using 1’s Complement:
Steps:
- Make both numbers have the same number of bits
- Determine the 1’s complement of the subtrahend
- Add the 1’s complement to the minuend
-
- If there is an additional bit (carry) in the result after addition, remove it and add it to the result
- If there is no carry, determine the 1’s complement of the result and prefix it with a negative (-) sign
Example: Subtract 1100002 from 11000002
Minuend = 1100000
Subtrahend = 110000
Step 1: 1's complement of subtrahend: 001111
Step 2: Add minuend and 1's complement:
1100000
+ 0011111
---------
1111111
Step 3: No carry, so take 1's complement and add negative sign:
1111111 → 0000000 → -0000000
Binary Subtraction using 2’s Complement:
Steps:
- Make both numbers have the same number of bits
- Determine the 2’s complement of the subtrahend
- Add the 2’s complement to the minuend
-
- If there is an additional bit in the result after addition, neglect the carry and the remaining bits are the required answer
- If there is no additional bit, determine the 2’s complement of the result and prefix it with a negative (-) sign
Example: Subtract 1012 from 1102
Minuend = 110
Subtrahend = 101
Step 1: 2's complement of subtrahend:
101 → 1's comp: 010 → 2's comp: 011
Step 2: Add minuend and 2's complement:
110
+ 011
-----
1 001
Step 3: Carry occurs, so remove carry: 001
Result: 0012
2.2 Logic Function and Boolean Algebra
2.2.1 Introduction to Boolean Algebra
Boolean algebra is the algebra of logic that deals with binary variables and logical operations. It is one of the most basic methods for analysis of logic. It was developed by the English mathematician George Boole.
In Boolean algebra, the variables are permitted to have only two values: true and false, usually written as one (1) or zero (0) respectively.
Key Characteristics:
- Deals with binary variables (0 and 1)
- Uses logical operations (AND, OR, NOT)
- Forms the foundation of digital circuit design
- Essential for computer logic and programming
2.2.2 Boolean Values, Truth Table, Boolean Expression and Function
Boolean Values:
Boolean algebra uses only two values:
- 0 (False) – represents logical false, low voltage, off state
- 1 (True) – represents logical true, high voltage, on state
Truth Table:
A truth table represents the input-output relationship of a binary variable logic gate. It shows the relationship between all inputs and outputs in a tabular format.
The number of rows in a truth table is equal to 2n, where n is the number of input variables.
| Input A | Input B | Output C = A+B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Boolean Expression:
A Boolean expression is an algebraic expression formed using Boolean constants, Boolean variables, and Boolean operators.
Examples:
- F = A + B (OR operation)
- F = A · B (AND operation)
- F = A’ (NOT operation)
Boolean Function:
A Boolean function is a function that maps Boolean inputs to Boolean outputs. It can be represented using:
- Boolean expressions
- Truth tables
- Logic diagrams
2.2.3 Logic Gates – AND, OR, NOT, NAND, NOR, XOR and XNOR
A logic gate is an electronic circuit that operates on one or more input signals to produce an output. It is a building block of digital circuits.
Types of Logic Gates:
- Fundamental or Basic Gates
- OR gate
- AND gate
- NOT gate
- Universal Gates
- NOR gate
- NAND gate
- Other Gates
- Exclusive OR gate (X-OR)
- Exclusive NOR gate (X-NOR)
Logic Gates with Diagrams:
Produces a high (1) output only when all inputs are 1.
| A | B | C = A·B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Produces a high (1) output when any input is 1.
| A | B | C = A+B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Produces the complement (inverse) of the input.
| A | C = A’ |
|---|---|
| 0 | 1 |
| 1 | 0 |
Complement of AND gate. Produces 0 only when all inputs are 1.
| A | B | C = (A·B)’ |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Complement of OR gate. Produces 1 only when all inputs are 0.
| A | B | C = (A+B)’ |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
Produces 1 when inputs are different.
| A | B | C = A ⊕ B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Produces 1 when inputs are the same.
| A | B | C = (A ⊕ B)’ |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
2.2.4 Laws of Boolean Algebra
Boolean algebra follows specific laws that help in simplifying Boolean expressions and designing efficient digital circuits.
1. De Morgan’s Laws
- First Theorem: (x+y)’ = x’ · y’
- Second Theorem: (x·y)’ = x’ + y’
2. Associative Law
- (A+B)+C = A+(B+C)
- (A·B)·C = A·(B·C)
3. Distributive Law
- A·(B+C) = A·B + A·C
- A+(B·C) = (A+B)·(A+C)
4. Commutative Law
- A+B = B+A
- A·B = B·A
5. Complement Law
- A + Ā = 1
- A · Ā = 0
6. Identity Law
OR Operations:
- A + 1 = 1
- A + A = A
- A + Ā = 1
- A + 0 = A
AND Operations:
- A · 1 = A
- A · A = A
- A · Ā = 0
- A · 0 = 0
2.2.5 Statement and Verification of Laws of Boolean Algebra
Verification of De Morgan’s First Theorem: (x+y)’ = x’ · y’
| x | y | x+y | (x+y)’ | x’ | y’ | x’·y’ |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Since columns for (x+y)’ and x’·y’ are identical, the law is verified.
Verification of Distributive Law: A·(B+C) = A·B + A·C
| A | B | C | B+C | A·(B+C) | A·B | A·C | A·B + A·C |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Since columns for A·(B+C) and A·B + A·C are identical, the law is verified.
Practice Questions and Answers
Question 1: Convert the following decimal numbers into octal and hexadecimal:
a) 756
Solution:
Octal conversion:
8 | 756
8 | 94 -- 4
8 | 11 -- 6
| 1 -- 3
75610 = 13648
Hexadecimal conversion:
16 | 756
16 | 47 -- 4
| 2 -- 15 (F)
75610 = 2F416
Question 2: Convert 33310 into hexadecimal and then to binary
Solution:
Hexadecimal conversion:
16 | 333
16 | 20 -- 13 (D)
| 1 -- 4
33310 = 14D16
Binary conversion from hexadecimal:
1 | 4 | D
0001 | 0100 | 1101
14D16 = 1010011012
Question 3: Convert ABC16 into binary number system
Solution:
A | B | C
1010 | 1011 | 1100
ABC16 = 1010101111002
Question 4: Convert 1110112 into base 16
Solution:
Binary: 0011 1011
Hexadecimal: 3 B
1110112 = 3B16
Question 5: Convert A5B16 into decimal number
Solution:
A5B16 = A×162 + 5×161 + B×160
= 10×256 + 5×16 + 11×1
= 2560 + 80 + 11 = 2651
A5B16 = 265110
Question 6: Perform binary arithmetic operations
a) Addition: 1001 + 1100
1001
+ 1100
-------
10101
b) Subtraction: 1100 – 0111
1100
- 0111
-------
0101
c) Multiplication: 1100 × 11
1100
× 11
-------
1100
1100
-------
100100
Question 7: Verify De Morgan’s Second Theorem using truth table
Theorem: (x·y)’ = x’ + y’
| x | y | x·y | (x·y)’ | x’ | y’ | x’ + y’ |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Since columns for (x·y)’ and x’ + y’ are identical, the theorem is verified.
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