# Lesson Notes By Weeks and Term - Junior Secondary School 3

BINARY NUMBERS (BASE 2 NUMBERS)

SUBJECT: MATHEMATICS

CLASS:  JSS 3

DATE:

TERM: 1st TERM

REFERENCE BOOKS

• New General Mathematics by M. F Macrae et al bk 3
• Essential Maths by AJS OluwasanmiBk 3

WEEK THREE

TOPIC: BINARY NUMBERS (BASE 2 NUMBERS)

• Addition in base 2
• Subtraction in base 2
• Multiplication & Division in base 2

ADDITION IN BASE TWO

We can add binary numbers in the same way as we separate with ordinary base 10 numbers.

The identities to remember are:-

0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 10, 1 + 1 + 1 = 11, 1 + 1 + 1 + 1 = 100

Worked Examples

Simplify the following

1. 1110 + 1001 2.    1111 + 1101 + 101

Solutions:

1. 1110

+    1001

10111

1. 1111

+ 1101

101

100001

Note: 11 take 1 carry 1

10 take 0 carry1

100 take 0 carry 10

EVALUATION

1. Simplify the following 101 + 101 +111
2. 10101 + 111

ADDITION IN BICIMALS

In bicimals, the binary point are placed underneath each other exactly the same way like ordinary decimals.

Example:

1.     1.1011two +   10.1001two  +  10.01
2.     10.001two+   101.111

Solution:

1.     1  .  1011

10 .  1001

10  .0100

1. 1000two

1.   101.111

10.001

1000.000

SUBTRACTION IN BASE TWO

The identities to remember on subtraction are: 0 - 0 = 0, 1 - 0 = 1, 10 - 1 = 1, 11 - 1 = 10, 100 - 1 = 11

Worked Examples

Simplify the following:-

(a)    1110 - 1001    (b) 101010 - 111

Solutions:

(a)     1110

-   1001

101

(b)    101010

-       111

1110

SUBTRACTION IN BICIMAL

Example

101.101two – 11.011two

101.101

11.011

10.010two

EVALUATION

1. 10111÷110
2. 10001 x 11

READING ASSIGNMENT

New Gen Maths Book 3, chapter 1 Exercise 1e pg 18 Nos 1-12

Essential Mathematics for J.S.S.3 Pg 8-10

WEEKEND ASSIGNMENT

1. Express 3426 as a number in base 10. (a) 342        (b) 3420    (c) 134
2. Change the number 10010 to base 10 (a) 18      (b) 34         (c) 40
3. Express in base two the square of 11 (a) 1001  (b)  1010    (c)  1011
4. Find the value of (101)2 in base two        (a) 1010  (b) 1111    (c)  1001
5. Multiply 1000012 by 11                 (a) 1001  (b) 1100011  (c)  10111

THEORY

1. Calculate 1102 x (10112 + 10012 – 1012)
2. Convert 110111 to base five

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