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 26.35 8 = 10 110.011 101 2
Conversion of decimal to binary via octal is demonstrated 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
7 2 0 2
8 3714
8 464
8 58
8
7
0
2
0
2
7
From Table 10.1, 7202 8 = 111 010 000 010 2
i.e.
3714 10 = 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
0.59375 10 = 0.46 8
From Table 10.1,
0.46 8 = 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
8 87
8 10
8
1
0
1 2 7 5 5
Remainder
5
5
7
2
1
This octal number is converted to a binary number,
(see Table 10.1).
12755 8 = 001 010 111 101 101 2
i.e.
5613 10 = 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).
0.72 8 = 0.111 010 2
i.e.
0.90625 10 = 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