Re: infinities reformulated Thomas Bushnell BSG 02 Jun 2005 15:23 UTC

Aubrey Jaffer <xxxxxx@alum.mit.edu> writes:

> So in a Scheme implementation which has "arbitrarily big" precision,
> how many digits is (sqrt 2)?  How many digits is (sin 7/5)?

There is no function "precision-of", so there is no need for an
answer.  Arbitrarily big precision arithmetic (generally) works pretty
well; you carry around symbolic representations and operate on them.