# rational to floating point

## Thant Tessman <thant@acm.org>9 Mar 2003 17:34:36 -0500

From comp.compilers

Related articles
rational to floating point thant@acm.org (Thant Tessman) (2003-03-09)
Re: rational to floating point chase@theworld.com (David Chase) (2003-03-14)
Re: rational to floating point nmm1@cus.cam.ac.uk (2003-03-14)
Re: rational to floating point thant@acm.org (Thant Tessman) (2003-03-14)
Re: rational to floating point joachim_d@gmx.de (Joachim Durchholz) (2003-03-14)
Re: rational to floating point ajo@andrew.cmu.edu (Arthur J. O'Dwyer) (2003-03-14)
Re: rational to floating point gah@ugcs.caltech.edu (Glen Herrmannsfeldt) (2003-03-14)
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 From: Thant Tessman Newsgroups: comp.compilers Date: 9 Mar 2003 17:34:36 -0500 Organization: XMission http://www.xmission.com/ Keywords: arithmetic, question Posted-Date: 09 Mar 2003 17:34:36 EST

This is a math question, but one I hope is relevant to this newsgroup.

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.

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. 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.

Is my reasoning sound? Is there a simpler test?

For the curious, a description and source for my interpreter can be
found at:

http://www.standarddeviance.com

My gratitude goes to folks in this newsgroup who have helped improve my
understanding of concepts I put to use building this.

-thant

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