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*To*: mathematica@xxxxxxxxx*Subject*: Re: infinities reformulated*From*: Aubrey Jaffer <agj@xxxxxxxxxxxx>*Date*: Tue, 31 May 2005 19:48:05 -0400 (EDT)*Cc*: alexshinn@xxxxxxxxx, srfi-70@xxxxxxxxxxxxxxxxx*Delivered-to*: srfi-70@xxxxxxxxxxxxxxxxx*In-reply-to*: <20050531071658.EC10C1167@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> (mathematica@xxxxxxxxx)*References*: <20050531071658.EC10C1167@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>

| Date: Tue, 31 May 2005 15:16:37 +0800 | From: "Chongkai Zhu" <mathematica@xxxxxxxxx> | | I mentioned Mathematica, only for the "inexact number" part of it, | not the "symbolic manipluation" part of it. For example, if you | want to save the square root of 2 as an inexact number, you can | write: | | v1=1.414 | | the precision or the inexact number v1 is 4 (decimal digits); | | but you can also write | | v2=1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573 | | and v2 will get precision 100 (all these digits are saved into memory). So in a Scheme implementation which has "arbitrarily big" precision, how many digits is (sqrt 2)? How many digits is (sin 7/5)?

**Follow-Ups**:**Re: infinities reformulated***From:*Thomas Bushnell BSG

**References**:**Re: infinities reformulated***From:*Chongkai Zhu

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