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

From: Paul <pbrazier_at_cosmos-uk.co.uk>
Date: 27 Feb 2003 05:56:21 -0800

Steve Kass <skass_at_drew.edu> wrote in message news:<b3k8ks\$kca\$1_at_slb5.atl.mindspring.net>...
> I'm lost on who is countering whose argument here, but you have
> described one type that represents the 9 points on and at the
> center of the 2 x 2 square at the origin of Z x Z, and another type
> that represents 9 points on and at the center of a circle (I think you
> may have meant to allow -0.75PI, ..., 0.75PI, 1 PI), instead of twice
> these values).
>
> These certainly represent very different subsets of cartesian space,
> so what do you mean by saying these are equivalent? Once you stray
> from points of the cartesian form (N,0) and (0,N), I don't think you'll
> have much luck finding equivalent polar and cartesian representations.

If you stick to representations of the form R x T for some sets R and T then it would be impossible because to express cartesian (1,1) in polar you need r=sqrt(2). Thus your polar domain would also contain polar (sqrt(2),0) which is (the same coincidentally) cartesian (sqrt(2),0) which isn't in the original cartesian domain.

You could have your polar domain defined as the subset of R x [0,2*pi) (R being the reals here) where r*sin(th) and r*cos(th) are both integers. But then for exactness you'd need to introduce symbols for sqrt(2), sqrt(3), and all your angles (maybe these are all of the form pi*n/m though?), etc. so you make things massively complicated and may as well stick to cartesians anyway.

Paul. Received on Thu Feb 27 2003 - 14:56:21 CET

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