Number Base System SS1 Mathematics Lesson Note

INTRODUCTION 

People count in twos, fives, twenties etc. Also, the days of the week can be counted as 24 hours. Generally, people count in tens. The digits 0,1,2,3,4,5,6,7,8,9 are used to represent numbers. The place value of the digits is shown in the number. 

Example: 395:- 3 Hundreds, 9 Tens and 5 Units. i.e.

                 39510      = 3 x102 + 9 x 101 +5 x 100.

Since the above number is based on the powers of ten, it is called the base ten number system i.e.

                               = 300 + 90 + 5. 

 Also 4075 = 4 Thousand 0 Hundred 7 Tens 5 Units i.e. 4 x 103 + 0 x 102 + 7 x 101 + 5 x 100 Other Number systems are sometimes used. 

Example: The base 8 system is based on the power of 8. 

For example: Expand 6477, 265237, 1011012,

(a)    6457       = 6 x 72 + 4 x 71 + 5 X 70 = 6 x 49 + 4 x 7 + 5 x 1

(b)    265237   = 2 x 74 + 6 x73 + 5 x 72 + 2 x 71 + 3 x 70

(c)    1011012 = 1 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x21 + 1 x 20

 CLASSWORK 

Expand The Following

1. 7358   2.    10100112

CONVERSION TO DENARY SCALE (BASE TEN)

When converting from other bases to base ten the number must be raised to the base and added.

Worked Examples:

Convert the following to base 10

(a)    278    (b)    110112

Solutions:

(a)    278 = 2 x 81 + 7 x 80 = 2 x 8 + 7 x 1 = 16 + 7 = 23

(b)    110112 = 1 x 24 + 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 1 x 16 + 1 x 8 + 0 x 4 + 1 x 2 + 1 x 1

                         = 16 + 8 + 0 + 2 + 1 = 27

 CLASSWORK

Convert The Following To Base Ten:

(a)    1010112        (b) 21203

CONVERSION FROM BASE TEN TO OTHER BASES

To change a number from base ten to another base

Divide the base ten numbers by the new base number;

Continue dividing until zero is reached;

Write down the remainder each time; 

Start at the last remainder and read upwards to get the answer.

CONVERSION OF NUMBERS FROM ONE BASE TO ANOTHER

A number given in one base other than base ten can be converted to another base via base ten. 

 Example 1 

Convert:    (a) 1534six to base eight  

Solution 

1534six to base eight

First, convert 1534six to base ten.

1534six    =    1  63 + 5  62 + 3  61 + 4  60

        =    216 + 180 + 18 + 4

        =    418ten

Now convert 418ten to base eight. 

8    418    Remainders 

8    52    2

8    6    4

    0    6                i.e. 418ten = 642eight

Thus, 1534six = 642eight

Example 2

Determine the number bases x and y in the following simultaneous equations: 

32x – 12y = 9 ten and 23x – 21y = 4ten

Solution

    32x – 12y = 9ten            (1)

    23x – 21y = 4ten            (2)

Change equation (1) to base ten as follows: 

(3  x1 + 2  x0) – (1  y1 + 2 y0) = 9 

3x + 2 – y – 2 = 9

3x – y = 9                     (1a)

Similarly, change equation (2) to base ten: 

i.e.    x – y = 1                 (2a)

subtracting equations (2a) from (1a):

    2x = 8

    X = 4 

Substituting x = 4 in (2a)

    4 – y = 1 

    4 – 1 = y 

    y = 3 

Thus, x = 4 and y = 3. 

CLASSWORK

If x represents a base number in the following equations, what is the value of x? 

315x – 223x = 72x

405x + 43eight = 184ten

Convert each of the following to the base indicated:

10401.11seven to base eight

1. 4836 sixteen to base twelve 

ASSIGNMENT

Choose the correct answer from the letter a – c

1. Express 3426 as number in base 10         (a) 134        (b) 341        (c) 143

2. Change the number 100102 to base 10    (a) 1001      (b) 40          (c) 18

3. Express in base 2, 10010                         (a) 100100  (b) 1100100 (c) 11001

4. Convert 120 base 10 to base 3                (a) 111103  (b) 12103      (c) 121103

5. Convert 25 base 10 to base 2                  (a) 110012   (b) 10012     (c) 11002

6. Convert 23647 to base 10

7. Convert 10510 to 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

Example 1

Simplify the following

1110 + 1001           2.      1111 + 1101 + 101

Solutions:

1.         1110

        +  1001

        = 10111

2.         1111

         + 1101

          + 101

=    100001

Note: 11 take 1 carry 1 

         10 take 0 carry1

         100 take 0 carry 10

CLASSWORK 

Simplify the following; 

a)1001 + 101 + 1111

1. b) 10101 + 111

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

MULTIPLICATION AND DIVISION IN BASE TWO

In multiplication, 0 x 0 = 0, 1 x 0 = 0, 1 x 1 = 1.

When there is a long multiplication of binary numbers, the principle of addition can be used to derive the answer. Under division, the principle of subtraction can be used.

Worked Examples:

1110 x 111    2. 110 ÷ 10

Solution:

1110                                 2.              110

x   110                                               10   110

            0000                                                       10

          1110                                                           10

1110      10

= 1010100                         

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

6. Convert the following to a binary number 
7. a) 10ten
8. b) (10ten)2

7. Calculate 1102 x (10112 + 10012 – 1012)

8. Multiply 345 by 225.