Re: JSH: Operator ambiguity, Escultura

From: Rupert (rupertmccallum_at_yahoo.com)
Date: 11/28/04 

Date: 27 Nov 2004 19:01:59 -0800

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411271359.ca44e69@posting.google.com>...
> I would like to pull out and highlight something interesting that E.
> E. Escultura posted a few days ago, which I'd guess he's probably
> talked about many times before, but I just noticed it and think it's
> neat.
>
> First some more preamble as *by convention* as has been noted when I
> brought up the subject of operator ambiguity before, sqrt(x) is taken
> to be positive.
>
> So, by the convention, sqrt(4) = 2, and that's good as, -2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0,
> which is not good.
>
> Naively then, you may believe that you can just say, take the positive
> of the square root but as Escultura showed, that doesn't work:
>
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
>

If we make the convention that sqrt(-1)=i, then it's not true that
sqrt(1/-1)=sqrt(1)/sqrt(-1).

> You see, the ambiguity in the square root operator still remains,
> despite the convention.
>
> It doesn't work to just try and always take the positive as
> Escultura's example shows so clearly.
>
> Who has the resolution? I'm curious as to whether or not any of you
> think you can answer.
>
>
> James Harris

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