Week 1 and 2 – Jss 3 First Term Mathematics Lesson Notes

SCHEME OF WORK FOR MATHEMATICS JSS 3

 REFERENCE MATERIALS
ESSENTIAL MATHEMATICS for junior secondary school, book 3 by A. J. S Oluwasanmi
EFFECTIVE MATHEMATICS for junior secondary school book 3 by M.K.Akinsola, M.C.Ejike and A.Tella

 WEEK 1
REVISION OF JS S 2 WORK

 WEEK TWO
BINARY NUMBERS
Numbers in base two are called binary numbers at is made up two digit is 0 and 1
Converting base 10 numbers to base two number
We do this by dividing the base ten number repeatedly by 2, writing down the remainder until we get to zero and reading the remainder upwards.
Example: (a) Write 8₁₀ to a number in base two
b)    Express 85 in a binary number
c)    Convert 107₁₀ to a number in the base two
d)    Convert 152_ten to a number in base two
e)    Convert 3/8_ten to a binary fraction (bicimal)
f)    Express 15.125₁₀ in binary notation

 SOLUTION
(a)    2    8
    2    4    R    0
    2    2    R    0
        0    R    1
                        8₁₀ = 1000₂

 (b)    2    85
    2    42    R    1
    2    21    R    0
    2    10    R    1
    2    5    R    0
    2    2    R    1
        1    R    0
        0    R    1
85₁₀ = 1010101_two

 
 (c) 2 107
    2    53    R    1
    2    26    R    1
    2    13    R    0
    2    6    R    1
    2    3    R    0
    2    1    R    1
        0    R    1
                        107₁₀ = 11010011₂
(d)    2    152
    2    76    R    0
    2    38    R    0
    2    19    R    0
    2    9    R    1
    2    4    R    1
    2    2    R    0
    2    1    R    0
        0    R    1
                        152_ten = 10011000₂
(e)    2    3
    2    1    R    1
        0    R    1
                        3₁₀ = 11₂
    
  2 8
    2    4    R    0
    2    2    R    0
    2    1    R    0
    2    0    R    1
8₁₀ = 1000_two

 First express 3 and 8 in binary, ₁₀ = 11₂/1000₂= 0.011₂
(f)    15.125 = 15 = 15= ₁₀
    
    2    121

     2    60    R    1
    2    30    R    0
    2    15    R    0
    2    7    R    1
    2    3    R    1
        1    R    1
        0    R    1
121₁₀ = 1111001₂
    2    8    R
    2    4    0
    2    2    0
    2    1    0
    2    0    1
8₁₀ = 1000₂

 ₍= ₂ = 1111.001₂

 Exercise: Convert the following binary numbers.
(a) 72        (b)      (c) 0.875        (d) 32
Converting Base Two Numbers to Base 10 Numbers
We express the given binary numbers as a sum of multiples of powers of two 2⁰, 2¹, 2², 2³ etc.
Example: Convert (i) 101_two (ii) 10.1001₂ (iii) 111₂
SOLUTION

1. 101₂ = 1×2² + 0x2¹ + 1×2⁰

   = 4 + 0 + 1
   = 5₁₀

2. 111₂ = 1×2² + 1×2¹ + 1×2⁰

   = 4 + 2 + 1
   = 9₁₀

3. 10.1001₂ = 1×2¹ + 0x2⁰ + 1 x 2 ^-1 + 0 x 2^-2 + 0 x 2^-3 + 1 x 2^-4

   = 2 + 0 + + 0 + 0 +

    =
    = 2₁₀

1. 10101₂ = 1 x 2⁴ + 0 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰

   = 16 + 0 + 4 + 0 + 1
   = 27₁₀

Exercise: If 110₂ = P₁₀. Find the value of P

 Assignment

1. Write 1-10 in binary numbers
2. Convert to base 10 (a) 11101₂ (b) 11.0101₂ (c) 1011001₂
3. Convert to binary number (a) 43_ten (b) 1280_ten (c) 176₁₀

ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION IN BINARY NUMBERS
Examples

1. Add 101111₂ and 10011₂
2. Subtract 10011₂ from 101111₂
3. 1101₂ x 111₂
4. 10010001₂ 101₂

 Solution
1. 101111₂            2.    101111_{2        3.}    1101₂

• 10011₂                +10011₂            x111₂

11100₂             1000010₂            1101
                                 1101
                                 1101
                                 1011011
4.Convert to base 10 to have 145₁₀ 5₁₀ = 29₁₀ = 11101₂ or
111
101₂      1001001₂
     101
     1000
101
101
     101
        0
Exercise: (1) add 100110₂, 101010₂ and 111011₂
(2) Multiply 10011₂ x 11₂
(3) 1011010₂ – 100111₂
(4) 1010111₂ 111₂