Higher Engineering Mathematics

COMPUTER NUMBERING SYSTEMS 89

A

Dividing repeatedly by 8, and noting the remainder
gives:

7 2 0 2

8 3 714 Remainder

8 464 2

8 5 8 0

8 7 2

0 7

From Table 10.1, 7202 8 =111 010 000 010 2

i.e. 371410 =111 010 000 010 2

Problem 8. Convert 0.59375 10 to a binary

number, via octal.

Multiplying repeatedly by 8, and noting the integer
values, gives:

6.00

0. 59375 × 8 =
   0.7 5 × 8 =
   . 4 6

4.7 5

Thus 0. 5937510 = 0. (^468)
From Table 10.1, 0. 468 = 0 .100 110 2
i.e. 0.59375 10 =0.100 11 2
Problem 9. Convert 5613.90625 10 to a binary
number, via octal.
The integer part is repeatedly divided by 8, noting
the remainder, giving:
1 2 7 5 5
8 5613 Remainder
8 701 5
8 87 5
8 10 7
8 1 2
0 1
This octal number is converted to a binary number,
(see Table 10.1).
127558 =001 010 111 101 101 2
i.e. 561310 =1 010 111 101 101 2
The fractional part is repeatedly multiplied by 8, and
noting the integer part, giving:
2.00
0.9062 5 × 8 =
0.2 5 × 8 =
.7 2
7.2 5
This octal fraction is converted to a binary number,
(see Table 10.1).
0. 728 = 0 .111 010 2
i.e. 0. 9062510 = 0 .111 01 2
Thus, 5613.90625 10 =1 010 111 101 101.111 01 2
Problem 10. Convert 11 110 011.100 01 2 to a
denary number via octal.
Grouping the binary number in three’s from the
binary point gives: 011 110 011.100 010 2
Using Table 10.1 to convert this binary number to
an octal number gives 363. 428 and 363. (^428)
= 3 × 82 + 6 × 81 + 3 × 80 + 4 × 8 −^1 + 2 × 8 −^2
= 192 + 48 + 3 + 0. 5 + 0. 03125
=243.53125 10
Now try the following exercise.
Exercise 44 Further problems on conver-
sion between denary and binary numbers via
octal
In Problems 1 to 3, convert the denary numbers
given to binary numbers, via octal.

1. (a) 343 (b) 572 (c) 1265
   [
   (a) 101010111 2 (b) 1000111100 2
   (c) 10011110001 2

]

2. (a) 0.46875 (b) 0.6875 (c) 0.71875
   [

(a) 0. 011112 (b) 0. (^10112)
(c) 0. (^101112)
]

3. (a) 247.09375 (b) 514.4375 (c) 1716.78125
   [

(a) 11110111. (^000112)
(b) 1000000010. (^01112)
(c) 11010110100. (^110012)
]