Number System JSS1 Mathematics Lesson Note

Converting Number Bases From One Base to Another Base

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.

Example: 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

Example 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

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 the 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 : 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