ABSTRACT
For fractions, the most significant bit of the result is the top bit obtained from the integer part of multiplication by 2. The least significant bit of the result is the bottom bit obtained from the integer part of multiplication by 2. Thus 0.62510 = 0.1012
Problem 4. Convert 4710 to a binary number
From above, repeatedly dividing by 2 and noting the remainder gives:
2 47 Remainder
2 23 1
2 11 1
2 5 1
2 2 1
2 1 0
0 1
1 0 1 1 1 1
Thus 4710 = 1011112
Problem 5. Convert 0.4062510 to a binary number
From above, repeatedly multiplying by 2 gives:
0.40625 × 2 = 0. 8125
0.8125 × 2 = 1. 625
0.625 × 2 = 1. 25
0.25 × 2 = 0. 5
0.5 × 2 = 1. 0
. 0 1 1 0 1
i.e. 0.4062510 = 0.011012
Problem 6. Convert 58.312510 to a binary number
The integer part is repeatedly divided by 2, giving:
2 58 Remainder 2 29 0 2 14 1 2 7 0 2 3 1 2 1 1
0 1
1 1 1 1 0 0
The fractional part is repeatedly multiplied by 2 giving:
0.3125 × 2 = 0.625 0.625 × 2 = 1.25 0.25 × 2 = 0.5 0.5 × 2 = 1.0
. 0 1 0 1
Thus 58.312510 = 111010.01012
Now try the following exercise
Exercise 10 Further problems on conversion of decimal to binary numbers
In Problems 1 to 4, convert the decimal numbers given to binary numbers.
