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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>