Lesson Notes By Weeks and Term - Junior Secondary School 1

SUBJECT: MATHEMATICS

CLASS:  JSS 1

DATE:

TERM: 2nd TERM

REFERENCE BOOKS

• New General Mathematics, Junior Secondary Schools Book 1
• Essential Mathematics for Junior Secondary Schools Book 1

 
WEEK FIVE                                    Date:……………….

TOPIC: BASE NUMBERS

Content

• Multiplication in Base Two

• Conversion from (i) other bases to base ten (ii) ten to other bases

Multiplication  in Base Two

 Some method used in carrying out the long multiplication is still the same method used here.Where the conventional method is in base ten the one we want to work out now is strictly to base two.  See the examples below:

Example 1

Findthe product of 1101 two X 111two

Solution

    1  1  0   1

x      1   1   1

          1   1   0    1

      1  1   0   1

1 1  0   1______

1 01 1   0   1    1

Ans : 1011011 two

Examples 2

Calculate the following  binary numbers.

1. a)   ( 110 two)2
2. b)  ( 1011 two) 3

Solution

1. a)  ( 110two)2

( 110 two)2  = 110 two X 110 two

        1  1   0

x    1  1   0

                       0  0   0

                    1 1  0

 +   1 1 0_______

             1 0  0 1 0  0

          _______________

:. ( 110 two) 2  = 100100 two

b ( 1011two)3

( 10 1 1 two)  = 1011two x 1011two X 1011two

            1 0  11

        x 1 0 1 1

                  1  0  1  1

                                1 0  1   1

                              0 0 0 0

    1 0  1 1

The result above will finally be multiplied by 1011 two

        1  1  1  1  0  0  1

x            1  0  1  1

        1  1  1  1  0  0  1

                      1 1 1  1  0  0  1

                   0 0 0 0  0  0  0

                1 1 1 1 0  0  1

 1 0 10 01 1  0  0  1  1

Ans :  10100110011two

Evaluation:

1. multiply 1110 two  by 111 two
2. Calculate the following binary numbers
3. a)  ( 10 1 two)2 (b) ( 111two)2

1. Conversion

A From other Bases to Base Ten

Here, expansion method is applied. Refer to the previous work of last weeks

Example 1

Convert 11011two  to base ten.

Solution 

11011 two ------ ten

= ( 1 x 24) +  ( 1 x 23) +  ( 0 x 22) + ( 1 x 21)  + ( 1 x 20)

= 1 x 24 + 1 x 23 + 0 x 22 +  1 x 21 + 1 x 1

= 1 x 16  + 1 x 8 + 0 x 4 +  1 x 2 + 1 x 1

= 16 + 8 + 0 + 2 + 1

= 16 + 11

= 27ten

:.11011two  = 27 ten

Example 2

convert 451 eight  to ten

Solution

451 eight  ------ ten

( 4x 82)  + ( 5 x 81)  +  ( 1 x 80)

 4 x 82  +  5 x 81  +  1 x 1

 4 x 64  + 5 x 8 + 1

256 + 40 + 1

= 297ten

Evaluation:

Convert the following to base ten

(a) 3032 four        (b) 30021five

From Base Ten To Other Bases

Here, we apply the division rule

Example 1

Convert 27 ten to a number in base two

Solution 

• 27

1. 13 r 1

2        6 r 1

2        3 r 0

2        1 r 1

        0 r 1

27 ten = 11011 two

Example 2

Convert 403 ten to a number in base two

Solution 

1. 403

1. 201 r 1

1. 100 r 1

1. 50 r 0

1. 25 r 0

1. 12 r 1

1. 6 r 0

1. 3 x r 0

1. 1  r 1

0 r 1

403 ten = 110010011two

Note: We have been converting from other bases to base ten and vice versa. Let us try to convert from other bases to other bases other than ten.

The rule is simple. First convert to base ten and then to the required base.

Example

Convert 134 eight to base five

Solution 

1 3  4 eight ________ ten

( 1 x 82 )   + ( 3 x 81)  + ( 4 x 8 0)

1. x 82  + 3 x 81 + 4 x 1

1x 64  + 3 x 8 + 4 x 1

1. + 24 + 4 = 92ten

Then convert 92 ten ____ five using division

92ten _____ five

5    92

                5          18 r 2

                                                5    3 r 3

                                                             0 r 3

:. 92 ten = 332 five

Evaluation

1. Calculate the following :
2. a) ( 111) 2       (b)   ( 100)2
3. Convert:
4. a) 4035   to ten 

b)145 ten to binary number

1. c) 256 eight to base two

Reading Assignment 

Basic operations, Addition and Subtraction of numbers based on their place value and the use of number line.

Weekend Assignment 

1.Change 321four to base eight  (a) 71    (b) 81    (c) 62    (d) 75.

2.Change 101110two to octal number  (a) 67 (b) 57    (c) 56    (d) 54

3. Change 35471 eight to base ten(a) 15097 (b) 16081    (c ) 17097    (d) 16097
4. Simplify in base two (1101)2 (a) 1011011 (b) 10101001    (c ) 1101101      (d ) 1110111
5. The missing number in the expansion below is:

   4983 ten = 4 x 103+ 9 x --- + 8 x 101 + 3 x 1(a) 104    (b) 103        (c) 102    (d) 101

Theory

1. Convert the following to binary number (a) 234five        (b) 403five
2. Calculate the following binary numbers

    (a) 10001  x 11

    (b) 110111 x 111