Higher Engineering Mathematics, Sixth Edition

88 Higher Engineering Mathematics

=

1

2

+

1

8

+

1

16

= 0. 5 + 0. 125 + 0. 0625

=0.6875 10

Problem 3. Convert 101.0101 2 to a decimal

number.

101. 01012 = 1 × 22 + 0 × 21 + 1 × 20 + 0 × 2 −^1

+ 1 × 2 −^2 + 0 × 2 −^3 + 1 × 2 −^4

= 4 + 0 + 1 + 0 + 0. 25 + 0 + 0. 0625

=5.3125 10

Now try the following exercise

Exercise 38 Further problems on

conversion of binary to decimal numbers

In Problems 1 to 5, convert the binary numbers

given to decimal numbers.

1. (a) 110 (b) 1011 (c) 1110 (d) 1001
   [(a) 6 10 (b) 11 10 (c) 14 10 (d) 9 10 ]


2. (a) 10101 (b) 11001 (c) 101101 (d) 110011
   [(a) 21 10 (b) 25 10 (c) 45 10 (d) 51 10 ]


3. (a) 101010 (b) 111000 (c) 1000001
   (d) 10111000
   [(a) 42 10 (b) 56 10 (c) 65 10 (d) 184 10 ]


4. (a) 0.1101 (b) 0.11001 (c) 0.00111
   (d) 0.01011[

(a) 0. 812510 (b) 0. (^7812510)
(c) 0. 2187510 (d) 0. (^3437510)
]

5. (a) 11010.11 (b) 10111.011 (c) 110101.0111
   (d) 11010101.10111[

(a) 26. 7510 (b) 23. (^37510)
(c) 53. 437510 (d) 213. (^7187510)
]
(b) Conversion of decimal to binary
An integer decimal number can be converted to a cor-
responding binary number by repeatedly dividing by 2
and noting the remainder at each stage, as shown below
for 39 10.
Remainder
0
(most significant bit) (least significant bit)
1 00111
239
219
29
24
22
21
1 1 1 0 0 1
The result is obtained by writing the top digit of the
remainder as the least significant bit, (a bit is abinary
digitand the least significant bit is the one on the right).
The bottom bit of the remainder is the most significant
bit, i.e. the bit on the left.
Thus 3910 = (^1001112)
Thefractional part of adecimal number can beconverted
to a binary number by repeatedly multiplying by 2, as
shown below for the fraction 0.625
(most significant bit) 0.1 1 (least significant bit)

1. 250


2. 500


3. 000

0.625 3 2 5

0.250 3 2 5

0.500 3 2 5

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 multiplicationby 2.

Thus 0.625 10 =0.101 2

Problem 4. Convert 47 10 to a binary number.

From above, repeatedly dividing by 2 and noting the

remainder gives:

Remainder

0

101111

247

223

211

25

22

21

1 1 1 1 0 1

Thus 4710 = (^1011112)