Convert (0.675)10 to binary.

Worked Examples · Example 2.14

Q. Convert (0.675)10 to binary.

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Concept understanding — Octal To Decimal Conversion

What Does "Octal to Decimal" Even Mean?

You already know how to count in decimal — the system with digits 0 through 9. When you see the number 342, you instinctively read it as "three hundred and forty-two." But why? Because each position in a decimal number stands for a power of 10:

$342 = 3 \times 10^2 + 4 \times 10^1 + 2 \times 10^0$

Now imagine a world where you only have eight fingers. You'd count 0, 1, 2, 3, 4, 5, 6, 7 — and then what? You'd run out of digits. So the next number would be 10 (which means one eight and zero ones, not ten). That's octal: base 8.

Octal uses only the digits 0 through 7. The number 342 in octal does not mean three hundred and forty-two. It means something completely different — and to find out what, you convert it to decimal.

The Core Idea: Positional Value in Base 8

Every number system works the same way: the rightmost digit is the "units" place, the next digit to the left is the "base" place, then "base squared," and so on.

For octal (base 8):

Rightmost digit: $8^0 = 1$ place
Next left: $8^1 = 8$ place
Next: $8^2 = 64$ place
Next: $8^3 = 512$ place
... and so on

So to convert an octal number to decimal, you multiply each digit by its corresponding power of 8 and add everything up.

(d_n d_{n-1} \dots d_1 d_0)_8 = d_n \times 8^n + d_{n-1} \times 8^{n-1} + \dots + d_1 \times 8^1 + d_0 \times 8^0

Worked Example: Convert $(342)_8$ to Decimal

Write the digits with their place values:

Digit	Place value (power of 8)	Contribution
3	$8^2 = 64$	$3 \times 64 = 192$
4	$8^1 = 8$	$4 \times 8 = 32$
2	$8^0 = 1$	$2 \times 1 = 2$

Now add: $192 + 32 + 2 = 226$

So $(342)_8 = 226$ in decimal.

Watch out

Never read an octal number like a decimal number. $(342)_8$ is not 342 — it's 226. The subscript 8 is your reminder.

Another Example: $(107)_8$

Here the digit 7 is fine (it's less than 8), but notice the digit 0. Zero still contributes nothing, but it holds the place.

Digit	Place value	Contribution
1	$8^2 = 64$	$1 \times 64 = 64$
0	$8^1 = 8$	$0 \times 8 = 0$
7	$8^0 = 1$	$7 \times 1 = 7$

Sum: $64 + 0 + 7 = 71$ …

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Multiply repeatedly by 2, taking the integer part as the next bit:

Step	Multiplication	Bit	Carry fraction
1	0.675 x 2 = 1.35	1	0.35
2	0.35 x 2 = 0.70	0	0.70
3	0.70 x 2 = 1.40	1	0.40
4	0.40 x 2 = 0.80	0	0.80
5	0.80 x 2 = 1.60	1	0.60
6	0.60 x 2 = 1.20	1	0.20
7	0.20 x 2 = 0.40	0	0.40 (seen at step 3 -> repeats)

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Repeated multiplication by 2 gives the bits after the point (top-down) — but 0.675 never reduces to 0: its binary expansion is NON-TERMINATING, 0.101 followed by the block 0110 repeating.

Why this works. Each multiplication by 2 moves the binary point one place right; the digit entering the integer position is the next bit of the fraction. We stop either when the fraction hits 0 (terminating) or when a fraction value REPEATS (recurring expansion).

Working for (0.675)10:

Step	Multiplication	Integer part (bit)	Fractional part carried on
1	0.675 x 2 = 1.35	1	0.35
2	0.35 x 2 = 0.70	0	0.70
3	0.70 x 2 = 1.40	1	0.40
4	0.40 x 2 = 0.80	0	0.80
5	0.80 x 2 = 1.60	1	0.60
6	0.60 x 2 = 1.20	1	0.20
7	0.20 x 2 = 0.40	0	0.40

At step 7 the carried fraction 0.40 is one we already had after step 3 — so steps 4-7 (bits 0, 1, 1, 0) repeat forever:

(0.675)10 = (0.101 0110 0110 0110 ...)2
          =  0.101 with the block 0110 recurring
          ~  (0.1010110)2 to 7 bits  (= 0.671875, error < 0.004)

Confirm the first bits in Python:

x, bits = 0.675, "" …

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Q.Convert (0.675)10 to binary.

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