Week 5 – Jss 1 Second Term Mathematics Lesson Notes

 
 WEEK 5             NUMBER SYSTEM

 5.1     Converting Numbers Base From One Baseto Other Bases
This is a 2-stage conversion process. The steps to do this are simple. First convert the given base to base ten and then convert the result to the required base.
Example5.1: Convert 2135₆ to a number in base two
Solution:
Step one: Converting to base ten
    2³1²3¹5⁰ = (2X6³) + (1×6²) + (3×6¹) + (5×6⁰)
        = (2x 216) + (1x 36) + (3x 6) + (5 x 1)
        = 432 + 36 + 18 + 5 = 491
Step two: Convert the answer in base ten to a number in base two
    2    491
    2    245 R 1
    2    122 R 1
    2    61 R 0
    2    30 R 1
    2    15 R 0
    2    7 R 1
    2    3 R 1
    2    1 R 1
        0 R 1
Therefore 2135₆ = 111101011₂

 
 
 
 Examples 5.2: Express 10111₂ to a number in base three
Solution:
1. 1⁴0³1²1¹1⁰= (1 X 2⁴) + (0 X 2³) + (1 X 2²) + (1 X 2¹) + (1 X 2⁰)
= 1 X 16 + 0 X 8 + 1 X 4 + 1 X 2 + 1 X 1
= 16 + 0 + 4 + 2 + 1
= 23₁₀
    3    23
    3    7 R 2
    3    2 R 1
        0 R 2
Therefore 10111₂ = 212₃

 
 5.2    Addition and Subtraction of Number Base

 Note the following STEPS in adding and subtracting.
To add in base two
0+0 = 0
1+ 0 = 1
1 + 1 = 10
1 + 1+ 1 = 11

 To subtract in base two
0 – 0 =0
1 – 0 = 1
10 – 1 = 1
11 – 1 = 10    

 Example 5.3: Add the following
1.    1101₂ and 1111₂
2.    11011₂, 10101₂, and 1001₂
Solution:
1.    1101 + 1111 = 1 1 0 1 [using the addition steps 1 + 1 = 10, write 0 down and move the 1 to the next number]
         + 1 1 1 1 [also take the 1 to next, and move them till you get to last one].
        1 1 1 0 0₂

 2.    11011 + 10101 + 1001 = 1 1 0 1 1₂
                1 0 1 0 1₂
                + 1 0 0 1₂
                1 1 1 0 0 1₂

 Example 5.4: Subtract (a) 1011₂ from 10101₂ (b) 11011₂ from 111010₂
Solution:
    1.    1 0 1 0 1[using the subtraction steps 1- 1 = 0, write 0 down and move to the next number]
        -1 0 1 1₂
        1 0 1 0₂
    2.    1 1 1 0 1 0₂
        -1 1 0 1 1₂
        1 1 1 1 1₂

 5.3    Multiplication of Binary
Example 5.5: Find the product of 1011₂ and 101₂
Solution:
        1 0 1 1₂
        x 1 0 1₂
        1 0 1 1
     0 0 0 0
    1 0 1 1
1 1 0 1 1 1₂
    
Example 5.6: multiply 1111 by 110
Solution:
        1 1 1 1₂
         1 1 0₂
        0 0 0 0
     1 1 1 1
     1 1 1 1
1 0 1 1 0 1 0₂

 
 EVALUATION: DO THESE
1.    Add the following (a) 1001₂ + 1111₂
            (b) 101110₂+10010₂
2.    Subtract 1011₂ from 1101101₂
3.    Multiply 110101₂ by 1011₂
4. find the value of the following binary numbers 1011+10-111

 ASSIGNMENT:
EXERCISE 9.5 page 107 NO 1(B,C,D), NO 2 A and B, NO 3 (E,F,G,H)
EXERCISE 9.6 NO 2,5,8 and 11. PAGE 108