*What* "seems to be the default"?
-s
On Thu, Nov 10, 2011 at 13:15, cheater cheater <cheater00 at gmail.com> wrote:
> That seems to be the default then, doesn't it?
>
> Cheers
>
> On Thu, Nov 10, 2011 at 19:04, Richard Fateman
> <fateman at eecs.berkeley.edu> wrote:
> > On 11/10/2011 9:30 AM, cheater cheater wrote:
> >>
> >> Hi Rene,
> >> they do:
> >>
> >> (%i1) sqrt(0.1)-sqrt(1/10);
> >> (%o1) .3162277660168379 - 1/sqrt(10)
> >> (%i2) ev(sqrt(0.1)-sqrt(1/10),numer);
> >> (%o2) 0.0
> >>
> >> You just have to invoke numeric evaluation.
> >>
> >> Hope this helps
> >
> > Actually, 0.1 is not equal to 1/10.
> >
> > 0.1 is a floating point number which is converted to double-float
> binary,
> > in which internal form it is
> > exactly equal to 3602879701896397/2^55.
> >
> > This is NOT exactly equal to 1/10. They differ by 1/180143985094819840 or
> > 1/(5*2^55).
> >
> > The only way 0.1 - 1/10 is computed as zero is if the user forces exact
> > numbers be converted to nearby floating-point numbers. In this case the
> two
> > values coincide.
> >
> >
> >
> >
> >
> >
> >> On Thu, Nov 10, 2011 at 17:51, rene kaelin<renekaelin at gmx.ch> wrote:
> >>>
> >>> Dear maxima users
> >>> Why don't both expressions simplify to zero?
> >>> (%i1) sqrt(0.1)-sqrt(1/10);
> >>> (%o1) .3162277660168379-1/sqrt(10)
> >>> (%i2) 0.1^0.5-(1/10)^0.5;
> >>> (%o2) 0.0
> >>> Manual:
> >>> Function: SQRT (X)the square root of X. It is represented internally by
> >>> X^(1/2). Also see ROOTSCONTRACT. RADEXPAND[TRUE] - if TRUE will cause
> nth
> >>> roots of factors of a product which are powers of n to be pulled
> outside
> >>> of
> >>> the radical, e.g. SQRT(16*X^2) will become 4*X only if RADEXPAND is
> TRUE.
> >>> Maxima version: 5.21.1Maxima build date: 16:28 5/16/2010Host type:
> >>> i686-apple-darwin10.3.0Lisp implementation type: SBCLLisp
> implementation
> >>> version: 1.0.38
> >>>
> >>>
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> >>>
> >>>
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> >
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