Higher Engineering Mathematics, Sixth Edition

Binary, octal and hexadecimal 91

Table 10.1

Octal digit Natural

binary number

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

The ‘0’ on the extreme left does not signify anything,
thus 26. 358 =10 110.011 101 2
Conversion of decimal to binary via octal is demon-
strated in the following worked problems.

Problem 11. Convert 3714 10 to a binary number,

via octal.

Dividing repeatedly by 8, and noting the remainder
gives:

Remainder

7202

8 3714

8 464

858

87

0

2

0

2

7

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

i.e. 371410 =111 010 000 010 2

Problem 12. Convert 0.59375 10 to a binary

number, via octal.

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

.4 6

0.75 3 8 5

0.59375 3 8 5

6.00

4.75

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 13. Convert 5613.90625 10 to a binary
number, via octal.
The integer part is repeatedly divided by 8, noting the
remainder, giving:
8 5613
8 701
887
810
81
0
1 2755
Remainder
5
5
7
2
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:
.7 2
0.25 3 8 5
0.90625 3 8 5
2.00
7.25
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 14. Convert 11 110 011.100 01 2 to a
decimal number via octal.
Grouping the binary number in three’s from the binary
point gives: 011 110 011.100 010 2