the sqrt(-i)
Robert Simmons Jr.
Fermat's Last theorem, I have found my interest in math rekindled. I
find though that I am having to relearn all of what I did learn in high
school and college courses because I haventused it in a long time.
However, while reading I came upon an interesting development with
numbers that brought a question to my mind.
We start with the discovery of imaginary numbers and thus complex
numbers. One finds them through the sqrt(-1) which cannot be expressed
in real numbers. Because math must be consistant,it was stated that
these numbers represent a whole new fieldof numbers. Further, the
cration ofthe imaginary numbers adds another axis to the traditonal
number line. This, in effect, turns the number line into anumber plane.
My question is this. If we try to take the sqrt(-i) what is the result ?
Is it yet another dimension in our number line? If we do, nd we call the
new numbers j, then could it be argued that the sqrt(-j) produces yet
another dimension and so on into infinity? If so, could this be a proof
that numerical infinity is not just a planar idea, but in fact an adea
that numbers are infinitely more complex from the dimensional point of
view?
I would LOVE to hear coments on this idea. If it is nieve or otherwise,
the comments will be appreciated. If it is nieve, isthere a proof that
numbers occupy only planar dimensionality when it comes to the visual
representation of a number line. That proof will probably be ver my head
right now but I would live tosee it anyway.
-- rob
Keith Reckdahl
>One finds them through the sqrt(-1) which cannot be expressed
>in real numbers.
>My question is this. If we try to take the sqrt(-i) what is the result ?
>Is it yet another dimension in our number line?
There is no real number x such that x*x = -1, so real
numbers cannot describe sqrt(-1). However, since both
of the following complex numbers
x= .707*( 1-i)
and x= .707*(-1+i)
satisfy x*x = -i, complex numbers are sufficient to
describe sqrt(-i).
Keith
Sam Sirlin
> My question is this. If we try to take the sqrt(-i) what is the result ?
There is a complex number (actually two of course) that works. See if
you can find it.
> Is it yet another dimension in our number line? If we do, nd we call the
> new numbers j, then could it be argued that the sqrt(-j) produces yet
> another dimension and so on into infinity? If so, could this be a proof
No. Just adding j doesn't work. But you can add j and k, both sqrt(-1)
as well. This gives the familiar algebra of quaternions (only the name
is unfamiliar; almost everyone is familiar with i,j,k and the cross
and dot product). You can also have a total of 7 sqrt(-1)'s, giving
the (unfamiliar) algegra of octaves or octonions or Cayley
numbers. That's it. There is a proof, but I've only seen it in an old
russian (translated) book.
A net search yeilds http://galaxy.cau.edu/tsmith/Bardo.html, where
you'll find some nice planar and non-planar ideas.
--
Sam Sirlin
Email: s...@kalessin.jpl.nasa.gov
Ron Shepard
<gnu...@oneimage.com> wrote:
[...]
>My question is this. If we try to take the sqrt(-i) what is the result ?
Unlike the situation with real numbers, it is not necessary to extend the
idea of complex numbers in order to find sqrt(-i). There is a complex
number that satisfies that role. In fact, there are two of them, just as
there are two distinct numbers that satisfy any square root expression.
The easiest way to think about these square roots is to think in terms of
a polar representation (an angle, and a distance from the origin). The
complex number -i is one unit down on the y axis, so this is either -90
degrees or 270 degrees. The two square roots of -i are just half of these
angles, -45 degrees or 135 degrees, with unit distance from the origin.
>Is it yet another dimension in our number line?
In the case of real numbers, the system *had* to be extended to find
solutions to such equations, the solution simply does not exist within the
real numbers. In the case of complex numbers, this extension is not
necessary. That is not to say however, that some new extension could not
be dreamed up that would give new distinct solutions to the question.
>I would LOVE to hear coments on this idea. If it is nieve or otherwise,
>the comments will be appreciated. [...]
No, this gets to a deep and interesting set of questions. Is mathematics
"discovered" or is it "invented"? Is it an abstract endeavor that happens
to have practical applications, or is it fundamentally and inescapably a
practical endeavor?
$.02 -Ron Shepard
David Wilkinson
>Recently, while reading "Enigma" which is the story of the proof of
>Fermat's Last theorem, I have found my interest in math rekindled. I
>find though that I am having to relearn all of what I did learn in high
>school and college courses because I haventused it in a long time.
>However, while reading I came upon an interesting development with
>numbers that brought a question to my mind.
>
>We start with the discovery of imaginary numbers and thus complex
>numbers. One finds them through the sqrt(-1) which cannot be expressed
>in real numbers. Because math must be consistant,it was stated that
>these numbers represent a whole new fieldof numbers. Further, the
>cration ofthe imaginary numbers adds another axis to the traditonal
>number line. This, in effect, turns the number line into anumber plane.
>
>new numbers j, then could it be argued that the sqrt(-j) produces yet
>another dimension and so on into infinity? If so, could this be a proof
>that numbers are infinitely more complex from the dimensional point of
>view?
>
0,+/-1,+/-2,..., then (-i)^0.5 = exp(-Pi/4+n*Pi) = (1-i)/2^0.5 and
(-1+i)/2^0.5, two complex numbers. If you square these you will confirm
they give -i.
Complex numbers have two elements but you can, of course, have vectors
in 2, 3 or any number of dimensions.
>I would LOVE to hear coments on this idea. If it is nieve or otherwise,
>the comments will be appreciated. If it is nieve, isthere a proof that
>numbers occupy only planar dimensionality when it comes to the visual
>representation of a number line. That proof will probably be ver my head
>right now but I would live tosee it anyway.
>
>-- rob
>
--
David Wilkinson
Paul Schlyter
Robert Simmons Jr. <gnu...@oneimage.com> wrote:
> We start with the discovery of imaginary numbers and thus complex
> numbers. One finds them through the sqrt(-1) which cannot be expressed
> in real numbers. Because math must be consistant,it was stated that
> these numbers represent a whole new fieldof numbers. Further, the
> cration ofthe imaginary numbers adds another axis to the traditonal
> number line. This, in effect, turns the number line into anumber plane.
>
> My question is this. If we try to take the sqrt(-i) what is the result ?
> Is it yet another dimension in our number line?
all you'll get is another complex number, not a new kind of number.
sqrt(-i) = (1-i)/sqrt(2)
You can easily verify this by squaring (1-i)/sqrt(2) to get -i back:
( (1-i)/sqrt(2) )^2 =
(1-i)*(1-i)/2 =
( 1 + i^2 - 2*i ) / 2 =
( 1 + (-1) - 2*i ) / 2 =
-2*i/2 =
-i
--
----------------------------------------------------------------
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
e-mail: pau...@saaf.se paul.s...@ausys.se pa...@inorbit.com
WWW: New URL at http://193.12.249.96/pausch -- updated daily!
Paul Schlyter
Keith Reckdahl <reck...@am-octane2.Stanford.EDU> wrote:
>Robert Simmons Jr. <gnu...@oneimage.com> wrote:
>>One finds them through the sqrt(-1) which cannot be expressed
>>in real numbers.
>>My question is this. If we try to take the sqrt(-i) what is the result ?
>>Is it yet another dimension in our number line?
>
>numbers cannot describe sqrt(-1). However, since both
>of the following complex numbers
> x= .707*( 1-i)
>and x= .707*(-1+i)
>satisfy x*x = -i, complex numbers are sufficient to
>describe sqrt(-i).
Sorry, but you should write 1/sqrt(2) instead of 0.707
If you square 0.707*(1-i) you'll only get -0.999698*i ....
Andreas Jung
: My question is this. If we try to take the sqrt(-i) what is the result ?
: Is it yet another dimension in our number line? If we do, nd we call the
: new numbers j, then could it be argued that the sqrt(-j) produces yet
: another dimension and so on into infinity? If so, could this be a proof
: that numerical infinity is not just a planar idea, but in fact an adea
: that numbers are infinitely more complex from the dimensional point of
: view?
The fundamental theorem of algebra states that the set C of complex
numbers is algebraically closed, which means that every univariate
polynomial over C of degree n is factorizable into n linear factors,
i.e. the polynomial has n roots (if you count a root of multiplicity
k as k roots).
Especially, there are two different complex square roots, three different
complex cube roots, four different complex 4-th roots, and so on, of
every complex number z<>0. You can calculate them by Moivre's formula
z^(1/n) = r^(1/n) * exp(i*(s+2*k*pi)/n)
= r^(1/n) * (cos((s+2*k*pi)/n) + i*sin((s+2*k*pi)/n))
for k=0,...,n-1, where z=r*exp(i*s), i.e. r=|z| and s=atan2(Im(z),Re(z)).
For the square root, there is also a formula that doesn't need trigonometric
functions and that can easily be obtained by solving (x+y*i)^2 = a+b*i
for x and y.
Greetings
Andreas Jung.
--
Andreas Gisbert Jung DL9AAI Tel:0381/498-3364 Fax:0381/498-3366
Theoretische Informatik mailto:aj...@informatik.uni-rostock.de
Universitaet Rostock http://www.informatik.uni-rostock.de/~ajung/
PGP fingerprint = 8A 0B 05 CA EE AB 7B 01 D9 07 6A D0 84 38 BB 82
Adam Russell
angle or cos(3pi/4) + isin(3pi/4). A second solution is that angle plus pi
or cos(7pi/4) + isin(7pi/4). If you like it better in cartesian coordinates
then the two solutions are -sqrt(1/2) + i*sqrt(1/2) and also sqrt(1/2) -
i*sqrt(1/2). If you square them you will see they both come out to -i.
Robert Simmons Jr. wrote in message <360A986D...@oneimage.com>...
>Recently, while reading "Enigma" which is the story of the proof of
>Fermat's Last theorem, I have found my interest in math rekindled. I
>find though that I am having to relearn all of what I did learn in high
>school and college courses because I haventused it in a long time.
>However, while reading I came upon an interesting development with
>numbers that brought a question to my mind.
>
>We start with the discovery of imaginary numbers and thus complex
>numbers. One finds them through the sqrt(-1) which cannot be expressed
>in real numbers. Because math must be consistant,it was stated that
>these numbers represent a whole new fieldof numbers. Further, the
>cration ofthe imaginary numbers adds another axis to the traditonal
>number line. This, in effect, turns the number line into anumber plane.
>
>My question is this. If we try to take the sqrt(-i) what is the result ?
>Is it yet another dimension in our number line? If we do, nd we call the
>new numbers j, then could it be argued that the sqrt(-j) produces yet
>another dimension and so on into infinity? If so, could this be a proof
>that numerical infinity is not just a planar idea, but in fact an adea
>that numbers are infinitely more complex from the dimensional point of
>view?
>
Kjinnovatn
sqrt(-i) = (1-i)/sqrt(2)