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Re: algebraic numbers
Actually, in Mathematica one has to carefully distinguish between Real
and Reals, Rational and Rationals, Integer and Integers etc. They are
really very different entities
Mathematica Reals contain the Rationals.
Element[2/3, Reals]
True
So a Rational belongs to the Reals but is not of the type Real.Also
Element[2/3, Rationals]
True
Element[1.2, Rationals]
False
I understand that what you mean is something completely different -
because Mathematica approximate numbers have only finite digits they are
"rationals". But this is really only a matter of how you choose to
interpret them: Mathematica interprets them as the first digits of
irrational numbers whose remaining digits are unknown (but there are
always infinitely many of them and they do not repeat themselves).
This is quite a consistent approach and in fact the only one that
justifies using approximate numbers in continuous probability
distributions.
Philosophically speaking, we cannot be sure that that is not exactly
what we do in our own minds when we thing of "real numbers" - our brains
are quite possibly only finite state automata and our reals are also
approximations to "reals" most of whose digits always remain unknown. So
I am not convinced that there is in this respect any fundamental
difference between humans and computers.
And by the way, reals numbers such as Pi or Sqrt[2], for which you have
a method which computes their digits as far as you wish to do are also
countable and form a set of measure zero. Most of real numbers are not
computable, either by computers or by us.
On 3 Jan 2010, at 06:34, DrMajorBob wrote:
> Mathematica Reals may not be Rational, but computer reals certainly
are. (I shouldn't have capitalized "reals" in the second case.)
>
> Bobby
>
> On Sat, 02 Jan 2010 04:03:30 -0600, Andrzej Kozlowski
<akoz at mimuw.edu.pl> wrote:
>
>>
>> On 1 Jan 2010, at 19:36, DrMajorBob wrote:
>>
>>> "If so, you will indeed have recognized the number x as algebraic,
from
>>> its first N figures."
>>>
>>> No... you will have identified an algebraic number that agrees with
x, to
>>> N figures.
>>>
>>> OTOH, every computer Real is rational, so they're all algebraic.
>>>
>>> Bobby
>>
>> Well, actually Mathematica does not agree with your last assertion:
>>
>> Head[1.12]
>>
>> Real
>>
>> Element[1.12, Rationals]
>>
>> False
>>
>> Element[1.12, Reals]
>>
>> True
>>
>> In fact it seems clear that the designers of Mathematica have decided
to interpret all approximate numbers (with head Real) as approximations
to irrationals rather than as finite expansions of rationals. The only
rationals in Mathematica are indeed the ones that have the head
Rational, i.e. fractions.
>>
>> Andrzej Kozlowski
>>
>>
>>
>>>
>>> On Thu, 31 Dec 2009 02:17:22 -0600, Robert Coquereaux
>>> <robert.coquereaux at gmail.com> wrote:
>>>
>>>> "Impossible....Not at all"
>>>> I think that one should be more precise:
>>>> Assume that x algebraic, and suppose you know (only) its first 50
>>>> digits. Then consider y = x + Pi/10^100.
>>>> Clearly x and y have the same first 50 digits , though y is not
>>>> algebraic.
>>>> Therefore you cannot recognize y as algebraic from its first 50
digits !
>>>> The quoted comment was in relation with the question first asked by
>>>> hautot.
>>>> Now, it is clear that, while looking for a solution x of some
>>>> equation (or definite integral or...), one can use the answer
obtained
>>>> by applying RootApproximant (or another function based on similar
>>>> algorithms) to numerical approximations of x, and then show that
the
>>>> suggested algebraic number indeed solves exactly the initial
problem.
>>>> If so, you will indeed have recognized the number x as algebraic,
from
>>>> its first N figures.
>>>> But this does not seem to be the question first asked by hautot.
>>>> Also, if one is able to obtain information, for any N, on the first
N
>>>> digits of a real number x, this is a different story... and a
>>>> different question.
>>>>
>>>> Le 30 d=E9c. 2009 =E0 18:11, Daniel Lichtblau a =E9crit :
>>>>
>>>>>
>>>>>> To recognize a number x as algebraic, from its N first figures,
is
>>>>>> impossible.
>>>>>
>>>>> Not at all. There are polynomial factorization algorithms based on
>>>>> this notion (maybe you knew that).
>>>>>
>>>>> Daniel Lichtblau
>>>>> Wolfram Research
>>>>
>>>>
>>>
>>>
>>> --
>>> DrMajorBob at yahoo.com
>>>
>>
>>
>
>
> --
> DrMajorBob at yahoo.com
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