Page 41 - Computer Science Class 11 Without Functions
P. 41
Given two non-negative numbers n and m (m ≠ 0), using the division algorithm, we can write n = m.q + r, where
0<=r<m, for a suitable choice of q and r. Thus, q is the quotient obtained by dividing n by m, and r is the remainder.
Now, let us rewrite the above computations using the division algorithm, which reveals more clearly why the above
method of converting a number from base 10 to base 2 works.
70
= 2 × 35 + 0
0
= 2 × 35 + 2 × 0
0
= 2 × (2 × 17 + 1) + 2 × 0
0
2
1
= 2 × 17 + 2 × 1 + 2 × 0
2
1
0
= 2 × (2 × 8 + 1) + 2 × 1 +2 × 0
3
2
1
0
= 2 × 8 + 2 × 1 + 2 × 1 + 2 × 0
2
1
0
3
= 2 × (2 × 4 + 0) + 2 × 1 + 2 × 1 + 2 × 0
2
4
1
= 2 × 4 +2 × 0 + 2 × 1 + 2 × 1 + 2 × 0
3
0
3
1
0
2
= 2 × (2 × 2 +0) + 2 × 0 +2 × 1 + 2 × 1 + 2 × 0
4
3
2
4
0
5
1
= 2 × 2 + 2 × 0 + 2 × 0 + 2 × 1 + 2 × 1 + 2 × 0
2
5
1
4
3
0
= 2 × (2 × 1 + 0) + 2 × 0 +2 × 0 + 2 × 1 + 2 × 1 + 2 × 0
3
5
1
2
6
0
4
= 2 × 1 +2 × 0 + 2 × 0 +2 × 0 +2 × 1 + 2 × 1 + 2 × 0
Thus, we have expressed 70 in terms of a sum of the powers of 2:
4
1
5
6
70 = 2 × 1 +2 × 0 +2 × 0 +2 × 0 +2 × 1 + 2 × 1 + 2 × 0
0
2
3
So, we can write binary representation of 70, as a sequence of coefficients of powers of 2 (beginning the coefficient of
the highest power of 2) in the above sum as follows:
(70 ) = (1000110)
10 2
Find binary equivalents of 5, 8, 16.
Binary to Decimal Conversion
Let us convert a binary number to decimal. For this, we use the place values with a base value of 2. The conversion
process involves the following steps:
1. Write the binary number
2. Below each digit, write the place value of the digit, represented as a power of 2.
3. Multiply the face value of each digit with its place value.
4. Add all values obtained in step 3 to get the decimal number.
In example 1, (1000110) is the binary representation of the decimal number (70) . So, if we express the binary
2 10
number (1000110) as a decimal number, we must get the original number (70) shown below:
2 10
Find decimal equivalents of 1001, 1100, and 110.
Number Systems and Encoding Schemes 39

