Second Term Lesson Note for Week Four
Class : JSS 1
Subject : Mathematics
Topic : Number Bases (Binary numbers)
Duration : 80 Minutes
Period : Double Period
Reference Book :
- Essential Mathematics for Junior Secondary School, Book 1
- Lagos State Unified schemes of work for Junior Secondary School, JSS 1 – 3.
- Online Resources
Instructional Material : Chart showing bases of numbers.
Learning Objectives : By the end of the lesson learners will be able to :
i. Identify the basic number Bases.
ii. Convert from decimal(base ten) to binary (base two) and vice versa.
iii. Change from base ten to other bases
Content :
NUMBER BASE
4.0 Introduction :
The usual system of counting in our days is called the decimal or denary system. The denary or decimal system is also called base ten. This system enables us to be able to write small or large numbers using the combination of the digits i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
4.1 Expansion and Conversion to Base Ten
Expanded Notation
Example 4.1 : Write the following in expanded form :
(a) 1101100 2
(b) 21356
(c) 45678
Solution :
11011002 = (1 x 26) + (1 x 25) + (0 x 24) + (1 x 23) + (1 x 22) + (0 x 21) + (0 x 20).
The base is used to expand it and the power for each expansion.
(b) 2 1 3 5 6 = (2 X 6³) + (1 x 6²) + (3 x 6¹) + (5 x 60)
(c) 4 5 6 7 = (4 x 8³) + (5 x 8²) + (6 x 8¹) + (7 x 80).
4.2 Conversion to Base Ten
Example 4.2 : convert the numbers in example 1 above to Base ten
Solution :
In other to do this we simply continue from the expanded notation, evaluate and get our answers.
1.) 1101100 = (1 x 26) + (1 x 25) + (0 x 24) + (1 x 23) + (1 x 22) + (0 x 21) + (0 x 20).
= (1 x 64) + (1 x 32) + (0 x 16) + (1 x 8) + (1x 4) + (0 x 2) + (0 x 1)
= 64 + 32 + 0 + 8 + 4 + 0 + 0
= 108 ten
2.) 2135 = (2 x 63) + (1 x 62) + (3 x 61) + (5 x 60)
= (2 x 216) + (1 x 36) + (3 x 6) + (5 x 1)
= 432 + 36 + 18 + 5
= 491
3.) 4567 = (4 x 83) + (5 x 82) + (6 x 81) + (7 x 80)
= (4 x 512) + (5 x 64) + (6 x 8) + (7 x 1)
= 2048 + 320 + 48 + 7
= 2423
4.3 Binary System (Base two number system)
In binary system, the greatest digit used is 1, so the two digits available in binary system are 0 and 1. Remember that each digit in a binary number has a place value.
Converting Number in Base ten to Numbers in Base two
Examples 4.3 :
(a) Convert 29ten to base 2.
(b) Convert 79ten to base 2
(c) Convert 145ten to base 2
Solution :
(a) Convert 29ten to base 2.
To convert 29 base 10 to a number in base 2, we divide through by 2 record the remainder.
| 2 | 29 | |
| 2 | 14 | R 1 |
| 2 | 7 | R 0 |
| 2 | 3 | R 1 |
| 2 | 1 | R 1 |
| 0 | R 1 ⇑ |
Hence, 29 ten = 11101 two
b.) Convert 79ten to base 2
| 2 | 79 | |
| 2 | 39 | R 1 |
| 2 | 19 | R 1 |
| 2 | 9 | R 1 |
| 2 | 4 | R 1 |
| 2 | 2 | R 0 |
| 2 | 1 | R 0 |
| 0 | R 1 ⇑ |
Hence, 79 ten = 1 0 0 1 1 1 1 two
C. Convert 145ten to base 2
EVALUATION:
(1) Expand the following with their bases.
(a) 35318
(b) 101010²
(c) 1110110242
(2) Convert the following number to base 2
1.) 35610
2.) 4710
3.) 21810
ASSIGNMENT :
PAGE 105 EXERCISE 9.2 : NO 3 (e,f,g,h) , NO 5 (a,b,d)