Page 36 - Computer Science Class 11 Without Functions
P. 36
We have seen that while representing the whole numbers in the positional system, the values of digits increase from
right to left in the powers of the base. Similarly, while representing the fractional part, the values of digits decrease in
powers of the base to the right of the radix point. Note that the rightmost digit remains the least significant digit. In the
decimal system, the radix point is called the decimal point. Thus, in the decimal system, the values of digits increase in
powers of ten on the left and decrease in powers of ten on the right of the decimal point. For example, consider the
number 56.91. having 56 as the integer part and 0.91 as the fraction part. The value represented by different digits in
the number 56.91 is described below:
1
● Digit 5 represents 5 tens (5 × 10 = 5 × 10)
0
● Digit 6 represents 6 units (6 × 10 = 6 × 1)
● Digit 9 represents 9 tenths 9 × 10 -1 = 9 × 1
10
-2
● Digit 1 represents 1 hundredth 1 × 10 = 1 × 1
100
The value of the number (56.91) can be calculated as follows:
10
(56.91) 10
0
1
= 5 × 10 + 6 × 10 + 9 × 10 + 1 × 10 -2
-1
= 5 × 10 + 6 × 1 + 9 × 1 + 1 × 1
10 100
Table 2.2 shows the digit's place value, face value, and value represented by the decimal number (56.91) .
10
Table 2.2: Digit's place value, face value, value represented by (56.91) .
10
Digit's place value ten (10) unit (1) radix point tenth 1 hundredth 1
10 100
Digit 5 6 , 9 1
Face value 5 6 . 9 1
Value represented 5 × 10 1 6 × 10 0 9 × 10 –1 1 × 10 –2
by digit
A number is the sum of products of its digits' place values and corresponding face values.
2.2 Binary Number System (Base-2)
Table 2.3: Binary Equivalent value of Digits of Decimal number system
Decimal Number Binary Equivalent
0 (0) = 0
2
1 (1) = 1
2
1
0
2 (10) = 1 × 2 + 0 × 2 = 2 + 0 = 2
2
1
0
3 (11) = 1 × 2 + 1 × 2 = 2 + 1 = 3
2
2
1
0
4 (100) = 1 × 2 + 0 × 2 + 0 × 2 = 4 + 0 + 0 = 4
2
2
0
1
5 (101) = 1 × 2 + 0 × 2 + 1 × 2 = 4 + 0 + 1 = 5
2
0
1
2
6 (110) = 1 × 2 + 1 × 2 + 0 × 2 = 4 + 2 + 0 = 6
2
0
1
7 (111) = 1 × 2 + 1 × 2 + 1 × 2 = 4 + 2 + 1 = 7
2
2
0
1
2
3
8 (1000) = 1 × 2 + 0 × 2 + 0 × 2 + 0 × 2 = 8 + 0 + 0 + 0 = 8
2
0
1
2
3
9 (1001) = 1 × 2 + 0 × 2 + 0 × 2 + 1 × 2 = 8 + 0 + 0 + 1 = 9
2
34 Touchpad Computer Science-XI

