JSH: Operator ambiguity, Escultura

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JSH: Operator ambiguity, Escultura

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Jim Ferry

Posted: Sun Nov 28, 2004 4:33 pm

Guest

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411271359.ca44e69@posting.google.com>...
[quote:c19c788ebe]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.
[/quote:c19c788ebe]
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.

[quote:c19c788ebe]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.
[/quote:c19c788ebe]
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.

[quote:c19c788ebe]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.
[/quote:c19c788ebe]
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!

[quote:c19c788ebe]You see, the ambiguity in the square root operator still remains,
despite the convention.
[/quote:c19c788ebe]
Umm, no.

[quote:c19c788ebe]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.
[/quote:c19c788ebe]
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

Richard Tobin

Posted: Sun Nov 28, 2004 5:21 pm

Guest

In article <30tfl3F30gq59U1@uni-berlin.de>,
José Carlos Santos <jcsantos@fc.up.pt> wrote:

[quote:61e0e21a21]Only after that you can state that "in the domain of complex numbers,
the square root of the product is not equal to the product of the square
roots".

So, I repeat my question: how do you define "the square root" in the
domain of complex numbers?
[/quote:61e0e21a21]
Any definition of which root is *the* root will have the problem
stated, so the statement is true regardless of which definition you
use.

-- Richard

José Carlos Santos

Posted: Sun Nov 28, 2004 5:53 pm

Guest

Richard Tobin wrote:

[quote:adcaf42c41]Only after that you can state that "in the domain of complex numbers,
the square root of the product is not equal to the product of the square
roots".

So, I repeat my question: how do you define "the square root" in the
domain of complex numbers?

Any definition of which root is *the* root will have the problem
stated, so the statement is true regardless of which definition you
use.
[/quote:adcaf42c41]
I know that. I just wanted to know which definition was N. Silver
thinking about when he wrote that "in the domain of complex numbers,
the square root of the product is not equal to the product of the square
roots". Note that he did not add "regardless of which definition you
use".

Best regards,

Jose Carlos Santos

Rupert

Posted: Sun Nov 28, 2004 6:24 pm

Guest

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411280652.17b14c6@posting.google.com>...
[quote:db04eed91a]rupertmccallum@yahoo.com (Rupert) wrote in message news:<d6af759.0411271901.4a62a25@posting.google.com>...
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).

So your assertion is that the substitution 1/-1 = -1 cannot be made?

[/quote:db04eed91a]
No.

[quote:db04eed91a]The resolution is that the sqrt() operator gives TWO answers. It
always does, and convention can't force it to give only one answer.

[/quote:db04eed91a]
You can make a convention that it only gives one answer, but you may
have to sacrifice identities like sqrt(a/b)=sqrt(a)/sqrt(b).

<snip>

Dik T. Winter

Posted: Sun Nov 28, 2004 7:32 pm

Guest

In article <qfcjq0lu2gdkir58dnvsfoo24hs4pv5dvi@4ax.com> ullrich@math.okstate.edu writes:
[quote:c5ce314794]On 27 Nov 2004 13:59:51 -0800, jstevh@msn.com (James Harris) wrote:
....
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.
....
Finally, you seem to have missed the most important point
in the "contradiction" above, which is that it's impossible
to define sqrt(z) for complex z in such a way that
sqrt(zw) = sqrt(z) sqrt(w).
[/quote:c5ce314794]
He is missing one more point. That you can not algebraically distinguish
i and -i is also a result from Galois theory (you can not algebraically
distinguish the various roots of a primitive polynomial).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

KeithK

Posted: Mon Nov 29, 2004 3:27 am

Guest

"James Harris" <jstevh@msn.com> wrote in message
news:3c65f87.0411280652.17b14c6@posting.google.com...
[quote:79c0be907b]rupertmccallum@yahoo.com (Rupert) wrote in message
news:<d6af759.0411271901.4a62a25@posting.google.com>...
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).

So your assertion is that the substitution 1/-1 = -1 cannot be made?

The resolution is that the sqrt() operator gives TWO answers. It
always does, and convention can't force it to give only one answer.

No, you're confusing the square root of a number (which has both a positive[/quote:79c0be907b]
answer and it's negation) with the sqrt() operator, which is _defined_ to
yield the positive root.

This definition prohibits the implied step above: i = sqrt(1)/sqrt(-1),
because this equation is false since 1/sqrt(-1) = 1/i = -i.

[quote:79c0be907b]So i = +/-sqrt(-1), where the +/- in front is a nod to the reality
that the result of using the square root operator is two solutions.

That's the operator ambiguity.

No, i does not = +/-sqrt(-1). You're tilting at definitions again. The[/quote:79c0be907b]
_definition_ of 'i' is i = sqrt(-1).

KeithK

[quote:79c0be907b]Naively you may believe that if you simply say, take only ONE answer
from the square root, and try to figure out some way to do it that you
can succeed, but you will always have a contradiction lurking.

The ambiguity extends to other operators like the cuberoot operator or
an infinity of other operators where you get more than one answer.

James Harris[/quote:79c0be907b]

David C. Ullrich

Posted: Mon Nov 29, 2004 6:24 am

Guest

On Mon, 29 Nov 2004 00:32:21 GMT, "Dik T. Winter" <Dik.Winter@cwi.nl>
wrote:

[quote:e6a96cf15d]In article <qfcjq0lu2gdkir58dnvsfoo24hs4pv5dvi@4ax.com> ullrich@math.okstate.edu writes:
On 27 Nov 2004 13:59:51 -0800, jstevh@msn.com (James Harris) wrote:
...
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.
...
Finally, you seem to have missed the most important point
in the "contradiction" above, which is that it's impossible
to define sqrt(z) for complex z in such a way that
sqrt(zw) = sqrt(z) sqrt(w).

He is missing one more point. That you can not algebraically distinguish
i and -i is also a result from Galois theory (you can not algebraically
distinguish the various roots of a primitive polynomial).
[/quote:e6a96cf15d]
Well, yes, although that's at least related to one I did point out:
"You also seem to be under the impression that i is positive".

************************

David C. Ullrich

Dik T. Winter

Posted: Mon Nov 29, 2004 7:04 am

Guest

In article <ml1mq0hni2tolpakq5j1vk3tsi7mnacilk@4ax.com> ullrich@math.okstate.edu writes:
[quote:275d271402]On Mon, 29 Nov 2004 00:32:21 GMT, "Dik T. Winter" <Dik.Winter@cwi.nl
wrote:

In article <qfcjq0lu2gdkir58dnvsfoo24hs4pv5dvi@4ax.com> ullrich@math.okstate.edu writes:
On 27 Nov 2004 13:59:51 -0800, jstevh@msn.com (James Harris) wrote:
...
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.
...
Finally, you seem to have missed the most important point
in the "contradiction" above, which is that it's impossible
to define sqrt(z) for complex z in such a way that
sqrt(zw) = sqrt(z) sqrt(w).

He is missing one more point. That you can not algebraically distinguish
i and -i is also a result from Galois theory (you can not algebraically
distinguish the various roots of a primitive polynomial).

Well, yes, although that's at least related to one I did point out:
"You also seem to be under the impression that i is positive".
[/quote:275d271402]
But "being positive" is not an algebraic property.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

thdmuync

Posted: Mon Nov 29, 2004 7:32 am

Joined: 25 Nov 2004 Posts: 7

[quote:72c331cffd="James Harris"]

i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
[/quote:72c331cffd]

I don't think this is the right definition of the imaginary number i. According to met i is defined as i^2 = -1. The function f(x)= x^2 is well defined because if x=y then x^2=y^2. However, the inverse 'function' is not really a function, because it can have more than one image f^-1(4) can be both 2 and -2.
So, by [u:72c331cffd][b:72c331cffd]convention[/b:72c331cffd][/u:72c331cffd] the inverse function of f is defined from the positive orthonant of R to the positive orthonant of R, so we have sqrt(4) = 2. and sqrt(-1), [i:72c331cffd]sadly enough[/i:72c331cffd] is not defined. the correspondence that you define, call this sqrt*) is not a function that has an element as an image, but a whole set as an image (sqrt*(4)= {2,-2}).[/b]

George Cox

Posted: Mon Nov 29, 2004 2:22 pm

Guest

James Harris wrote:
[quote:7a23ab06f0]
...

i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction,

[/quote:7a23ab06f0]
This is a reductio proof that sqrt(x/y) <> sqrt(x)/sqrt(y) .

This is a well-known and not very exciting fact.

Gib Bogle

Posted: Mon Nov 29, 2004 2:26 pm

Guest

Jim Ferry wrote:

[quote:3547fa6a68]Assumption (b) is what's holding you back from learning math. Before you
can learn, you have to realize that you don't know.
[/quote:3547fa6a68]
A Fundamental Truth.

James Harris

Posted: Mon Nov 29, 2004 6:22 pm

Guest

Gib Bogle <bogle@ihug.too.much.spam.co.nz> wrote in message news:<coft74$ifr$1@lust.ihug.co.nz>...
[quote:c9db5c4193]Jim Ferry wrote:

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

A Fundamental Truth.
[/quote:c9db5c4193]

Oh please, the poster never answered the mathematical issue raised but
instead turned to personal attacks. You deleted out all of the
context and made a reply that had nothing to do with the issue.

For those who missed it, I used Escultura's example of

i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction,

to show that it's naive to think that you can remove the ambiguity
from the square root operator.

Rather than give a cogent answer the sci.math poster Jim Ferry babbled
about his childhood, and simply *claimed* as if that's all it takes
that there's no problem.

The regular sci.math poster Gib Bogle then came in to delete out all
of the context and make his own reply.

That's how sci.math'ers managed to paint me as a crank. They do this
consistently, and sci.math readers cheer them on!

The sci.math readership is remarkably stupid and gullible.

James Harris

Brian Quincy Hutchings

Posted: Mon Nov 29, 2004 8:17 pm

Guest

hey, then 0 = 5 - 5 = -5 - 5 = -10;
you should turn that into a tutorial on Zero Divisors ... and
watch out for those chaotic "rounding-off" errors!

corklebath@hotmail.com (Jim Ferry) wrote in message news:<80142a28.0411281333.461a4740@posting.google.com>...

[quote:df61cdb68d]Hey! I remember thinking this was really cool. I did it with 5 = -5, but
[/quote:df61cdb68d]
--Advice 0.05; Free, if wrong, again!
http://tarpley.net/bush6.htm

Dik T. Winter

Posted: Mon Nov 29, 2004 8:52 pm

Guest

In article <3c65f87.0411291522.63c4747c@posting.google.com> jstevh@msn.com (James Harris) writes:
....
[quote:eb628278ce]Oh please, the poster never answered the mathematical issue raised but
instead turned to personal attacks. You deleted out all of the
context and made a reply that had nothing to do with the issue.
[/quote:eb628278ce]
But some posters answered it.

[quote:eb628278ce]For those who missed it, I used Escultura's example of

i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction,
[/quote:eb628278ce]
The contradiction exists because the assumption is made that
sqrt(a/b) = sqrt(a)/sqrt(b), which is false when in the complex
numbers. Similarly log(a.b) = log(a) + log(b) is false when in
the complex numbers. Read a bit about Riemann surfaces and you
may understand.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Tim Smith

Posted: Tue Nov 30, 2004 4:39 am

Guest

In article <I7yyK2.M0D@cwi.nl>, Dik T. Winter wrote:
[quote:2a98776646]Read a bit about Riemann surfaces and you may understand.
[/quote:2a98776646]
He won't.

--
--Tim Smith

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