On Wed, 2009-09-23 at 13:10 -0400, Bradley Lucier wrote:
> Ray Dillinger scripsit:
> As an aside, I'm interested in how this function is
> implemented.
> To answer Ray, if you have (expt x p/q) and x is a positive fixnum,
> bignum, or ratnum, it takes the qth root of x and sees if it's exact; if
> so, it computes (expt (expt x 1/q) p).
Okay, I got that far. But as I said, my problem is getting an exact
result while taking a qth root. That's the part I don't know how to
do, except by getting lucky in interval division.
Bear
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