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Re: Extending my question. Was: The relational model and relational algebra - why did SQL become the industry standard?

From: Jan Hidders <jan.hidders_at_REMOVE.THIS.ua.ac.be>
Date: 10 Mar 2003 00:52:40 +0100
Message-ID: <3e6bd3c8.0@news.ruca.ua.ac.be>


Bob Badour wrote:
>"Jan Hidders" <jan.hidders_at_REMOVE.THIS.ua.ac.be> wrote in message
>news:3e5cb9f9.0_at_news.ruca.ua.ac.be...
>>
>> Having said this, it would be very strange that the equality relationship
>> as defined over the representations would not be an equivalence relation,
>> since that is more or less the accepted fundamental property of any
>> equality operator. If you are going to convert coordinates and do some
>> rounding this might go wrong.
>
>This is true any time one uses floating point representations and does some
>rounding.

No. Not necessarily. Note that I meant "equivalence relation" in its mathematical sense, i.e., a reflexive, symmetrica and transitive binary relation. With rounding you can still have well-defined equivalence classes. But if you have two different representations it might happen that if you translate from one represenation to another and then back again, you end up with a different value. That would make the relation intransitive.

>Consider:
>
>Y := X^(0.5)
>Y := Y*Y*Y*Y
>X := X*X
>
>Would you necessarily expect the logical expression (X = Y) to evaluate to
>TRUE?
No.

Received on Sun Mar 09 2003 - 17:52:40 CST

Original text of this message

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