|
|
| Author |
Message |
| James Harris |
 Posted: Sat Nov 27, 2004 4:59 pm |
|
|
|
|
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
|
|
| Back to top |
|
| N. Silver |
| Posted: Sat Nov 27, 2004 8:25 pm |
|
|
|
|
James Harris wrote:
[quote:c12fd46985]i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
Who has the resolution? I'm curious as to whether
or not any of you think you can answer.
[/quote:c12fd46985]
As this example shows, in the domain of complex numbers,
the square root of the product is not equal to the product of the
square roots and the same for quotients.
What does work is DeMoivre's Theorem.
Since -1 = cos(pi) + i*sin(pi), we have
r1 = [cos(pi)+i*sin(pi)]^(1/2)
= cos(pi/2)+i*sin(pi/2) = i
r2 = [cos(3pi)+i*sin(3pi)]^(1/2)
= cos(3pi/2)+i*sin(3pi/2) = -i
As a consequence of the Fundamental Theorem of Algebra,
every complex number has exactly two complex square roots.
So, i and -i are the only square roots of -1.
|
|
| Back to top |
|
| N. Silver |
| Posted: Sat Nov 27, 2004 8:36 pm |
|
|
|
|
James Harris wrote:
[quote:e327f6c85a]I would like to pull out and highlight something interesting
at 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.
[/quote:e327f6c85a]
You just ache to show mathematics is not on firm foundations,
don't you?! Because you are stumped at times does not mean
that mathematics is contradictory. On the contrary, it means
you are not working hard enough, do not know enough, or
are not mathematically sophisticated enough to overcome
your mathematical difficulties.
This elementary stuff has been worked over many, many
times before by those much smarter than you or I.
It would be prudent to give up the ghost now. You have
not a snowball's chance of ever winning in this forum.
|
|
| Back to top |
|
| darrenn |
| Posted: Sat Nov 27, 2004 8:55 pm |
|
|
|
|
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...
[quote:30571e416d]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[/quote:30571e416d]
|
|
| Back to top |
|
| Rupert |
| Posted: Sat Nov 27, 2004 10:01 pm |
|
|
|
|
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411271359.ca44e69@posting.google.com>...
[quote:713dfcdf4a]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.
[/quote:713dfcdf4a]
If we make the convention that sqrt(-1)=i, then it's not true that
sqrt(1/-1)=sqrt(1)/sqrt(-1).
[quote:713dfcdf4a]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[/quote:713dfcdf4a]
|
|
| Back to top |
|
| José Carlos Santos |
| Posted: Sun Nov 28, 2004 2:42 am |
|
|
|
|
N. Silver wrote:
[quote:d3fab4a3c5]As this example shows, in the domain of complex numbers,
the square root of the product is not equal to the product of the
square roots and the same for quotients.
[/quote:d3fab4a3c5]
And how do you define "the square root" in the domain of complex
numbers?
[quote:d3fab4a3c5]As a consequence of the Fundamental Theorem of Algebra,
every complex number has exactly two complex square roots.
[/quote:d3fab4a3c5]
In any field, every element has at most two square roots. Besides,
you don't need the Fundamental Theorem of Algebra in order to prove
that every complex complex number has a square root.
Best regards,
Jose Carlos Santos
|
|
| Back to top |
|
| N. Silver |
| Posted: Sun Nov 28, 2004 3:00 am |
|
|
|
|
José Carlos Santos wrote:
[quote:0cb41c2912]N. Silver wrote:
As this example shows, in the domain of complex numbers,
the square root of the product is not equal to the product of the
square roots and the same for quotients.
And how do you define "the square root" in the domain of complex
numbers?
[/quote:0cb41c2912]
A square root of a complex number a+bi is a solution of the equation
z^2 = a+ib.
[quote:0cb41c2912]As a consequence of the Fundamental Theorem of Algebra,
every complex number has exactly two complex square roots.
In any field, every element has at most two square roots. Besides,
you don't need the Fundamental Theorem of Algebra in order to prove
that every complex complex number has a square root.[/quote:0cb41c2912]
|
|
| Back to top |
|
| José Carlos Santos |
| Posted: Sun Nov 28, 2004 3:13 am |
|
|
|
|
N. Silver wrote:
[quote:b836f6b6dc]As this example shows, in the domain of complex numbers,
the square root of the product is not equal to the product of the
square roots and the same for quotients.
And how do you define "the square root" in the domain of complex
numbers?
A square root of a complex number a+bi is a solution of the equation
z^2 = a+ib.
[/quote:b836f6b6dc]
That's not an answer. In your reply to James Harris you wrote (and it's
reproduced above) something about "the square root". Now, I know very
well what _a_ square root is, thank you very much, but when you use the
expression "the square root" you are making a choice: you are choosing
one among the two square roots of a complex number (different from 0).
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?
Best regards,
Jose Carlos Santos
|
|
| Back to top |
|
| David C. Ullrich |
| Posted: Sun Nov 28, 2004 6:13 am |
|
|
|
|
On 27 Nov 2004 13:59:51 -0800, jstevh@msn.com (James Harris) wrote:
[quote:9bc3c612f9]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.
[/quote:9bc3c612f9]
Of course you can't take the positive square root of a negative
number - it doesn't _have_ a positive square root! You seem to
be under the impression that there is such a convention - there
is not, the convention about the meaning of sqrt(x) is only
for x > 0.
You also seem to be under the impression that i is positive,
which is kind of funny - it's not. It's only real numbers
that are positive or negative.
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).
Except for totally missing the main points it was an
excellent post.
[quote:9bc3c612f9]Who has the resolution? I'm curious as to whether or not any of you
think you can answer.
[/quote:9bc3c612f9]
Guffaw. There's no contradiction to be resolved, because the
calculuation above is simply wrong, using properties of sqrt
that nobody ever claimed were valid.
[quote:9bc3c612f9]James Harris
[/quote:9bc3c612f9]
************************
David C. Ullrich
|
|
| Back to top |
|
| George Cox |
| Posted: Sun Nov 28, 2004 7:16 am |
|
|
|
|
James Harris wrote:
[quote:65b3a262c3]
...
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:65b3a262c3]
You, or Escultura, whomever, thereby proves that
sqrt(x/y) != sqrt(x)/sqrt(y)
|
|
| Back to top |
|
| James Harris |
| Posted: Sun Nov 28, 2004 9:52 am |
|
|
|
|
rupertmccallum@yahoo.com (Rupert) wrote in message news:<d6af759.0411271901.4a62a25@posting.google.com>...
[quote:1c4d0b25f8]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).
[/quote:1c4d0b25f8]
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.
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.
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
|
|
| Back to top |
|
| Guest |
| Posted: Sun Nov 28, 2004 11:30 am |
|
|
|
|
James Harris wrote:
[quote:ca94c54815]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:ca94c54815]
sqrt(a/b) = sqrt(a)/sqrt(b) is not true for every real a, b: a <> b
[quote:ca94c54815]The ambiguity extends to other operators like the cuberoot operator
or
an infinity of other operators where you get more than one answer.
Can you show few examples with the other operators?[/quote:ca94c54815]
|
|
| Back to top |
|
| C. Bond |
| Posted: Sun Nov 28, 2004 1:25 pm |
|
|
|
|
James Harris wrote:
[snip]
[quote:b7d4f80b29]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:b7d4f80b29]
Sure it can. Consider a square surface with an area of 4 square units. Answer the question: What is the length
of one side? Are you saying that -2 units is a correct answer?
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
|
|
| Back to top |
|
| William Hughes |
| Posted: Sun Nov 28, 2004 1:47 pm |
|
|
|
|
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411271359.ca44e69@posting.google.com>...
[quote:4aab4adbbb]
It doesn't work to just try and always take the positive as
Escultura's example shows so clearly.
[/quote:4aab4adbbb]
In general a complex number has two square roots, both complex.
So it doesn't make sense to try to "take the positive".
However, there are two cases
i) x^2 - a = 0 has two roots, one with positive real
part and one with negative real part.
ii) x^2 - a = 0 has two roots both with a real part of zero
So to get a unique square root.
i) take the root with positive real part
ii) take the root with positive imaginary part
Now x^2 +1 =0 has two roots, i and -i. Both have 0 real part. So
by ii) we choose i and say the (principlal value of) the square root
of -1 is i.
[quote:4aab4adbbb]Who has the resolution?
[/quote:4aab4adbbb]
Me, No one else. Nobody ever though about this before me.
I deserve an Acadamy Award.
[quote:4aab4adbbb]I'm curious as to whether or not any of you
think you can answer.
James Harris
[/quote:4aab4adbbb]
Be curious no longer.
-"William Hughes"
|
|
| Back to top |
|
| Christian Bau |
| Posted: Sun Nov 28, 2004 2:13 pm |
|
|
|
|
In article <3c65f87.0411271359.ca44e69@posting.google.com>,
jstevh@msn.com (James Harris) wrote:
[quote:ceea15d82c]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.
[/quote:ceea15d82c]
Typical mistake of a beginner: In the complex numbers, sqrt (a/b) = sqrt
(a) / sqrt (b) is not usually true. The convention decides about the
sign of the result. This decision cannot possible made to be consistent
in all cases for sqrt (a), sqrt (b) and sqrt (a/b).
Convert your complex numbers into polar coordinates, look at what the
convention does to the phase angle, and it should be quite obvious where
your mistake is.
|
|
| Back to top |
|
|
|
All times are GMT - 5 Hours
The time now is Wed May 16, 2012 6:01 pm
|
|
