Answer:
| Decimal | Binary so far | |
|---|---|---|
| Start | 0.750 | 0. |
| ×2 | 1.50 | 0.1 |
| erase | .50 | 0.1 |
| ×2 | 1.00 | 0.11 |
| erase | .00 | 0.11 |
| Result | 0.11 |
So 0.7510 = 0.112.
To check this, go in the other direction:
0.112 = 2-1 + 2-2 = 0.5 + 0.25 = 0.75
| Decimal | Binary so far | |
|---|---|---|
| Start | 0.750 | 0. |
| ×2 | 1.50 | 0.1 |
| erase | .50 | 0.1 |
| ×2 | 1.00 | 0.11 |
| erase | .00 | 0.11 |
| Result | 0.11 |
So 0.7510 = 0.112.
To check this, go in the other direction:
0.112 = 2-1 + 2-2 = 0.5 + 0.25 = 0.75
| Decimal | Binary so far | |
|---|---|---|
| Start | 0.1 | 0. |
| ×2 | 0.2 | 0.0 |
| ×2 | 0.4 | 0.00 |
| ×2 | 0.8 | 0.000 |
| ×2 | 1.6 | 0.0001 |
| .6 | 0.0001 | |
| ×2 | 1.2 | 0.00011 |
| 0.2 | 0.00011 | |
| ×2 | 0.4 | 0.000110 |
| ×2 | 0.8 | 0.0001100 |
| ×2 | 1.6 | 0.00011001 |
| .6 | 0.00011001 | |
| ×2 | 1.2 | 0.000110011 |
| 0.2 | 0.000110011 | |
| ×2 | 0.4 | 0.0001100110 |
| ×2 | 0.8 | 0.00011001100 |
| Result | 0.00011001100... |
At
right the algorithm is used to
convert 0.110
to binary.
(The "erase" steps are not shown when a 0 is copied to the "binary so far" column.)
The algorithm does not end.
After it has started up,
the same pattern 0.2, 0.4, 0.8, 1.6, 0.6, 1.2, 0.2
repeats endlessly.
The pattern 0011 is appended to the
growing binary fraction for each repitition.
Unexpected Fact: The value "one tenth" cannot be represented precisely using a binary fraction.
This is true in the base two positional notation used here, and also in floating point representation used in programming languages. This is sometimes an important consideration when high accuracy is needed.
Can "one third" be represented accurately in decimal?