Kurt Foster on Wed, 10 Jun 2009 17:41:52 +0200

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 Re: Is There a Way to Rationalize a Decimal in Pari/GP?

```On Jun 5, 2009, at 8:07 AM, Rick Regan wrote:

```
I didn't mention that I am only looking at this from a limited point of view. I want to convert dyadic decimal fractions -- those that represent double-precision floating-point binary values.
```
```
This suggests the following problem: Assuming the input x can be represented as a normalized double-precision value, find that value (or one such value). Then, determine the simplest fraction which is consistent with that value.
```
```
Since a normalized double-precision value has 52 significant bits with an implied leading 1, the default precision (either 28 or 38 decimal digits) should be plenty to determine a normalized double-precision representation of x, assuming it has (at least) one.
```
```
Assuming x!=0, e can let k = 52 - floor(log(x)/log(2)), then N = round(x*2^k) satisfies 2^52 < x <= 2^53. Assuming N < 2^53, the problem then becomes that of determining the simplest fraction which lies in the open interval
```
](N - 1/2)*2^(-k), (N + 1/2)*2^(-k)[

```
There are many details to deal with, of course. Among other things, if x is such that x*2^k is exactly midway between two integers, N isn't uniquely determined. Or what of N = 2^53? Or x == 0? And, of course, there's checking that the exponent k is consistent with a normalized double-precision value. But apart from these annoyances, there will be a unique answer. There's a way to use simple continued fractions to find it, but (as with the double-precision representation) I'm too lazy to work out the details.
```
```
I did use the above formulation on some examples. The sample input from the thread
```
0.1000000000000000055511151231257827021181583404541015626

gives the fraction 1/10,

0.33333333333333333333333333333333333333333

gives the fraction 1/3, and

0.759765625

gives the fraction 389/512.

```