Re: Operator ambiguity, Escultura

From: darrenn (naylod-nospam_at_juno.com)
Date: 11/28/04 

Date: Sat, 27 Nov 2004 17:55:20 -0800

I think that the error is in the following step:

sqrt(1/-1) = 1/i

Its a bit like saying

sqrt((-3)(-3))=-3

And then proclaiming that you have taken the positive square root.
Clearly this is incorrect.

"James Harris" <jstevh@msn.com> 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.
>
> 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

Relevant Pages

  • JSH: Operator ambiguity, Escultura
    ... First some more preamble as *by convention* as has been noted when I ... brought up the subject of operator ambiguity before, ... You see, the ambiguity in the square root operator still remains, ...
    (sci.math)
  • Re: JSH: Operator ambiguity, Escultura
    ... >> I would like to pull out and highlight something interesting that E. ... that the result of using the square root operator is two solutions. ... That's the operator ambiguity. ...
    (sci.math)
  • Re: JSH: Operator ambiguity, Escultura
    ... > The resolution is that the sqrt() operator gives TWO answers. ... you're confusing the square root of a number (which has both a positive ... yield the positive root. ... > That's the operator ambiguity. ...
    (sci.math)
  • Re: Sign conventions
    ... The square root operator returns that number which when multipled times ... The square root *function* returns the nonnegative real number whose ... Justin ...
    (sci.math)