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Re: infinities reformulated



Aubrey Jaffer <agj@xxxxxxxxxxxx> 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.