|rational to floating point email@example.com (Thant Tessman) (2003-03-09)|
|Re: rational to floating point firstname.lastname@example.org (David Chase) (2003-03-14)|
|Re: rational to floating point email@example.com (2003-03-14)|
|Re: rational to floating point firstname.lastname@example.org (Thant Tessman) (2003-03-14)|
|Re: rational to floating point email@example.com (Joachim Durchholz) (2003-03-14)|
|Re: rational to floating point firstname.lastname@example.org (Arthur J. O'Dwyer) (2003-03-14)|
|Re: rational to floating point email@example.com (Glen Herrmannsfeldt) (2003-03-14)|
|Re: rational to floating point firstname.lastname@example.org (2003-03-14)|
|Re: rational to floating point Peter-Lawrence.Montgomery@cwi.nl (2003-03-14)|
|Re: rational to floating point email@example.com (John Stracke) (2003-03-14)|
|From:||"Arthur J. O'Dwyer" <firstname.lastname@example.org>|
|Date:||14 Mar 2003 11:31:54 -0500|
|Organization:||Carnegie Mellon, Pittsburgh, PA|
|Posted-Date:||14 Mar 2003 11:31:54 EST|
On Sun, 9 Mar 2003, Thant Tessman wrote:
> This is a math question, but one I hope is relevant to this newsgroup.
Possibly also comp.programming. Cross-posted.
> I've been writing an interpreter for a functional programming language
> in C++. One of the datatypes it supports is an arbitrary-precision
> rational number type. Currently the interpreter displays rational
> numbers as the ratio of two integers. I'd like it to display rational
> numbers as floating-point numbers whenever the conversion to floating
> point won't produce an infinite (repeating) stream of digits.
> The question is: Under what conditions will a rational number produce
> an infinite stream of digits for a given base? What I've come up with
> is this:
> Converting a rational number to a floating point value is equivalent
> to multiplying the numerator and denominator by some number that
> converts the denominator to a whole-number power of the base. That is:
> b^n = c * d where 'b' is the base, 'd' is the denominator, and 'n' and
> 'c' are some whole numbers that satisfy the equation.
This is absolutely correct. But for interfacing to real computer
languages like C++, you also have the constraint that 'b' is 2 (or
FLT_RADIX) and 'n' is no larger than some fixed value (the number of bits
in the mantissa of a floating-point number; I forget what C++ calls it
But I see from your website that you probably want to implement your
own floating-point class also, and you don't care about sizeof(double).
So that constraint isn't as relevant as it might be.
> I think there is no solution to the above equation if the denominator
> of the original rational number and the base contain no prime factors
> in common.
More generally, there is a solution c to (b^n = c*d) *if and only
if* d has no prime factors which are not prime factors of b. (For
example, 1/12 is not terminating in base 4, because 3 does not divide
4. But 1/4 is terminating in base 12, because 2 divides 12.)
> And I think that this in turn implies that if and only if
> gcd(d,b) is 1 and 'd' is not 1, then the original rational number can
> only be represented by an infinite stream of digits.
I don't think that's quite equivalent to what I said above, but you
were on the right track.
> Is my reasoning sound? Is there a simpler test?
> For the curious, a description and source for my interpreter can be
> found at:
> My gratitude goes to folks in this newsgroup who have helped improve my
> understanding of concepts I put to use building this.
Hope this helps.
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