Binary Number System
LESSON – BINARY SYSTEM
TOPIC
- Introduction
- Conversion of Base 2 to Base 10
- Conversion of Base 10 to Base 2
- Addition and Subtraction of Binary Numbers
- Multiplication and Division Binary System
INTRODUCTION
Binary numbers are composed of only 0 and 1, while decimal numbers are composed of digits from 0 to 9.
CONVERSION OF BASE 10 TO BASE 2
Binary is a number system that represent a base 2 number system.
This means it only has two numbers: 0 and 1.
Relationship between decimal and binary is as follows:
0 = 0₂
1 = 1₂
2 = 10₂
3 = 11₂
4 = 100₂
5 = 101₂
6 = 110₂
7 = 111₂
8 = 1000₂
9 = 1001₂
10 = 1010₂
11 = 1011₂
12 = 1100₂
13 = 1101₂
14 = 1110₂
15 = 1111₂
16 = 10000₂
17 = 10001₂
18 = 10010₂
19 = 10011₂
20 = 10100₂
Illustration,
CLASS EXERCISE 1 – Convert the following to base 2:
1. 15
2. 23
3. 78
4. 137
5. 200
SOLUTIONS
1. 15
2 | 15
2 | 7 R 1
2 | 3 R 1
2 | 1 R 1
__ | 0 r 1 ↑
15 = 1111₂
2. 23
2 | 23
2 | 11 r 1
2 | 5 r 1
2 | 2 r 1
2 | 1 r 0
_ | 0 r 1 ↑
23 = 10111₂
3. 78
2 | 78
2 | 39 r 0
2 | 19 r 1
2 | 9 r 1
2 | 4 r 1
2 | 2 r 0
2 | 1 r 0
__| 0 r 1 ↑
78 = 1001110₂
4. 137
2 | 137
2 | 68 r 1
2 | 34 r 0
2 | 17 r 0
2 | 8 r 1
2 | 4 r 0
2 | 2 r 0
2 | 1 r 0
__| 0 r 1 ↑
137 = 10001001₂
5. 200
2 | 200
2 | 100 r 0
2 | 50 r 0
2 | 25 r 0
2 | 12 r 1
2 | 6 r 0
2 | 3 r 0
2 | 1 r 1
__| 0 r 1
200 = 11001000₂
CONVERSION OF BASE 2 TO BASE 10
Decimal system is a number system that represent a 0 – 9 number system.
Relationship between binary and Decimal is as follows:
0₂ = 0
1₂ = 1
10₂ = 2
11₂ = 3
100₂ = 4
101₂ = 5
110₂ = 6
111₂ = 7
1000₂ = 8
1001₂ = 9
1010₂ = 10
1011₂ = 11
1100₂ = 12
1101₂ = 13
1110₂ = 14
1111₂ = 15
10000₂ = 16
10001₂ = 17
10010₂ = 18
10011₂ = 19
10100₂ = 20
WORKING EXAMPLE 1
1011₂
= 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2º
= 1 x 8 + 0 x 4 + 1 x 2 + 1 x 1
= 8 + 0 + 2 + 1
= 11
Therefore, 1011₂ = 11
WORKING EXAMPLE 2
1100₂
= 1 x 2³ + 1 x 2² + 0 x 2¹ + 0 x 2º
= 1 x 8 + 1 x 4 + 0 x 2 + 0 x 1
= 8 + 4 + 0 + 0
= 12
Therefore, 1100₂ = 12
WORKING EXAMPLE 3
1101₂
= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2º
= 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1
= 8 + 4 + 0 + 1
= 13
Therefore, 1101₂ = 13
WORKING EXAMPLE 4
11001₂
= 1 x 2^4 + 1 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2º
= 1 x 16 + 1 x 8 + 0 x 4 + 0 x 2 + 1 x 1
= 16 + 8 + 0 + 0 + 1
= 25
Therefore, 11001₂ = 25
WORKING EXAMPLE 5
1010110₂
= 1 x 2^6 + 0 x 2^5 + 1 x 2^4 + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2º
= 1 x 64 + 0 x 32 + 1 x 16 + 0 x 8 + 1 x 4 + 1 x 2 + 0 x 1 = 64 + 0 + 16 + 0 + 4 + 2 + 0
= 86
Therefore, 1010110₂ = 86
CLASS EXERCISE 2 – Convert the following to base 10,
1. 1111₂
2. 10111₂
3. 1001110₂
4. 10001001₂
ADDITION AND SUBSTATION OF BINARY NUMBERS
Addition and subtraction are much like your normal everyday addition (decimal addition), except that it carries on a value of 2 instead of a value of 10.
For example,
8 + 2 = 10 (decimal number), corresponding value in binary is 1₂ + 1₂ = 10.
WORKING EXAMPLE 1
¹1 ¹0 ¹1 ¹ 1 1₂
+ 1 1 1 1₂
_____________
1 0 0 1 1 0₂
Therefore, 10111₂ + 1111₂ = 100110₂
WORKING EXAMPLE 2
10111₂ – 1111₂
1º 0² 1 1 1₂
– 1 1 1 1₂
_____________
1 0 0 0₂
Therefore, 10111₂ – 1111₂ = 1000₂
Note – 1 borrow from the next digit is equal to 2 just as in decimal, the one you borrow from the next digit is equal to 10.
CLASS EXERCISE 3 – Addition and Subtraction of Binary
1. 11011₂ + 10101₂
2. 1111₂ – 110011₂
3. 11011₂ – 10101₂
4. 11011₂ – 11010₂
ANSWER KEY
1. 11011₂ + 10101₂
1 1 0 1 1₂
+ 1 0 1 0 1₂
_____________
1 1 0 0 0 0₂
2. 1111₂ – 1011₂
1 1 1 1₂
– 1 0 1 1₂
_____________
1 0 0₂
3. 11011₂ – 10101₂
1 1 0 1 1₂
– 1 0 1 0 1₂
_____________
1 1 0₂
4. 11011₂ – 11010₂
11011₂
– 11010₂
_____________
1₂
MULTIPLICATION OF BINARY NUMBER
Multiplication is actually much similar and simpler to calculate than decimal multiplication.
MIND ON ACTIVITIES
0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1
WORKING EXAMPLE 1
10111₂ x 11₂
1 0 1 1 1₂
x 1 1₂
_____________
1 0 1 1 1₂
+ 1 0 1 1 1₂…
_____________
1 0 0 0 1 0 1₂
WORKING EXAMPLE 2
11101₂ x 101₂
1 1 1 0 1₂
x 1 0 1₂
_____________
1 1 1 0 1₂
0 0 0 0 0₂..
+ 1 1 1 0 1₂….
_____________
1 0 0 1 0 0 0 1₂
CLASS EXERCISE – Multiplication of Binary Numbers
1. 1010₂ x 10₂
2. 1011₂ x 11₂
3. 1111₂ x 101₂
ANSWER KEY
1. 1010₂ x 10₂
1 0 1 0₂
x 1 0₂
_____________
0 0 0 0₂
+ 1 0 1 0₂..
_____________
1 0 1 0 0₂
2. 1011₂ x 11₂
1011₂
x 11₂
_____________
1 0 1 1₂
+ 1 0 1 1₂..
_____________
1 0 0 0 0 1₂
3. 1111₂ x 101₂
1 1 1 1₂
x 1 0 1₂
_____________
1 1 1 1₂
0 0 0 0₂..
+ 1 1 1 1₂….
_____________
1 0 0 1 0 1 1₂
DIVISION OF BINARY NUMBERS (OPTIONAL)
ACTIVITY 1 – INTRODUCTION
The easy way to solve division of binary numbers is to,
- Step 1 – convert the two binary numbers to base 10.
- Step 2 – Divide the dividend by the divisor.
- Step 3 – Convert the quotient back to binary number.
WORKING EXAMPLE
11001₂ ÷ 101₂
SOLUTION
Step 1 – Convert to base 10
- 11001₂ = 25
- 101₂ = 5
Step 2 – Divide dividend by divisor
25 ÷ 5 = 5
Step 3 – Convert 5 to binary.
5 = 101₂
CLASS EXERCISE – Division of Binary Numbers
1. 10010₂ ÷ 10₂
2. 10000₂ ÷ 100₂
3. 1111₂ ÷ 11₂
4. 11010₂ ÷ 1101₂
5. 11110₂ ÷ 1111₂
ANSWER KEY
1. 10010₂ ÷ 10₂
10010₂ = 18
10₂ = 2
18 ÷ 2 = 9
Therefore, 10010₂ ÷ 10₂ = 1001₂
2. 10000₂ ÷ 100₂
10000₂ = 16
100₂ = 4
16 ÷ 4 = 4
10000₂ ÷ 100₂ = 100₂
3. 1111₂ ÷ 11₂
1111₂ = 15
11₂ = 3
15 ÷ 3 = 5
Therefore, 1111₂ ÷ 11₂ = 101₂
4. 11010₂ ÷ 1101₂
11010₂ = 26
1101₂ = 13
26 ÷ 13 = 2
Therefore, 11010₂ ÷ 1101₂ = 10₂
5. 11110₂ ÷ 1111₂
11110₂ = 30
1111₂ = 15
30 ÷ 15 = 2
Therefore, 11110₂ ÷ 1111₂ = 10₂
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