Re: Mixing OO and DB

From: Dmitry A. Kazakov <mailbox_at_dmitry-kazakov.de>
Date: Fri, 22 Feb 2008 21:47:42 +0100
Message-ID: <q4h80hdtuw90.urp4kl1xvext$.dlg_at_40tude.net>


On Fri, 22 Feb 2008 12:20:02 -0800 (PST), Marshall wrote:

> On Feb 22, 11:36 am, "Dmitry A. Kazakov" <mail..._at_dmitry-kazakov.de>
> wrote:

>> On Fri, 22 Feb 2008 08:13:10 -0800 (PST), Marshall wrote:
>>
>>> Earlier your threw around the word "uncountable"
>>> a few times so maybe you have some related meaning in
>>> mind. But it doesn't matter; if we limit our context to what
>>> is computable, then mathematical relationships don't somehow
>>> vanish;
>>
>> Of course they do. For example this vanishes:
>>
>>    forall x, circle exist y, circle twice as big

>
> Does this mean that you are claiming that given a
> computable specification of a circle, it is impossible
> to determine a computable specification of a
> circle with twice the radius? Or maybe area?

No, you claimed that a set of computable circles retain the properties of the set of all circles. This is wrong, as my example shows.

> It appears you are somehow claiming that multiplication
> is not defined on computable numbers.

Sure. Multiplication (addition, subtraction, division) is incomputable and thus cannot be defined.

>>> a computable circle is still an ellipse, just as much
>>> as an uncomputable circle is. Also: every countable set of
>>> circles is still a subset of the set of all ellipses.
>>
>> So what?

>
> So what?! It's what we're discussing. So you agree then?
>
> What did you think we were discussing?

Whether a set of circle values can have the structure of the set of circles. In order to prove that you have to show an isomorphism, which trivially does not exist. So far your single argument was that you could find a circle for each circle value. Great, but that alone does not constitute isomorphism.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de
Received on Fri Feb 22 2008 - 21:47:42 CET

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