Re: JSH: Operator ambiguity, Escultura

From: Jim Ferry (corklebath_at_hotmail.com)
Date: 11/28/04 

Date: 28 Nov 2004 13:33:24 -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.

Wait, you're not suggesting that sqrt(0) is positive are you? Just kidding.
Of course you aren't. As you well know, the convention that sqrt(x) is
positive applies only to positive numbers x.

> 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.

Hey! I remember thinking this was really cool. I did it with 5 = -5, but
it was the same idea. I showed it to my teacher (this was in sixth grade),
and he said I couldn't do that, but he wasn't able to give me a good reason
why not.

The reason it appealed to me so much was that once you have a result like
that, you can prove anything. Wow! All you need to prove anything is a
fuzzy definition like "sqrt(n) is the thing that gives you n when you square
it." Provided you're willing to put your fingers in your ears and say "la
la la la I am not lis en ing" when mathematicians try to explain why such a
definition is invalid.

> 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.

Oh, this is the more sophisticated version. Usually it's shown with a
few more steps:

i = sqrt(-1) = sqrt(1/-1) = sqrt(1)/sqrt(-1) = 1/i = -i.
                          ^
                        ERROR!

> You see, the ambiguity in the square root operator still remains,
> despite the convention.

Umm, no.

> 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.

You ask this question like it's at the forefront of mathematical knowledge
because (a) it's at the frontier of your mathematical knowledge, and (b)
you believe that you are blazing new trails in mathematics.

Assumption (b) is what's holding you back from learning math. Before you
can learn, you have to realize that you don't know.

> James Harris

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