>http://www.jal.cc.il.us/math/faculty/jim.html
Would anyone have any insight into the following:
Assuming a logorithmic increase in volume, how would I determine the
proper decibel value for any point between two decibel values? example
- At 0 sec, the signal is attenuated to -145 db, then at 1 minute,
the dignal is 0 decibels (unity gain). What would the volume in
decibels be at 30sec? Does anyone know of a mathematical function that
solves for this? Any help would be appreciated.
Depends on what you mean by a logarithmic increase
in volume. Our ears are logarithmic, so the perceived
volume goes as the log of the actual sound intensity.
So probably you mean a linearly-increasing intensity.
Convert your dB values to linear intensity values
(arbitrary units):
I(0) = 10^(-145/10) = 10^(-14.5) = 3.16*10^(-15)
I(1 minute) = 10^(0/10) = 1
The first value is essential 0. At 30 seconds the
intensity is halfway between I(0) and I(1 minute)
which is pretty close to 0.5, or -3 dB.
(dB value = 10*log10(I) = 10*log10(0.5)
= -3.01 dB).
Some other values:
1 second: 10*log10(1/60) = -17.8 dB
5 seconds: 10*log10(5/60) = -10.8 dB
10 seconds: 10*log10(10/60) = -7.8 dB
Does that seem like the kind of trend you were thinking
of as logarithmic?
- Randy
====
---------------------------------------------------------------------
--------------------------------------------------------------------------
x^2 - 2
need a superscript 2 for ^2 but pretty good.
Herc
How does this look?
ok, I can read yours as html, has a vertical bar and different font, I
set this back
to html send to try again.
also had to do this first..
To edit the HTML source tags, click the View menu and make sure a check
mark appears next to Source Edit. Then, select the Source tab and start
editing.
test : ±
thats a +- symbol if you only have text reader.
yep, it generates verbose html but it works, its for this site
http://www.alanwood.net/demos/symbol.html
there's a lot of sci.math posts complaining about no standard for
posting
maths symbols so I'll do a html post of all the available ones, a lot
of
them use an add in program instead.
Herc
Mind if I try it out before making a silly of myself - test -
±
What I suspect is happening though is that OE is buggering with the
post after we press send which is why I am being cautious and testing :o)
If that worked however then you need to insert your code snippet (
± ) directly into the source. You are
using Outlook Express it seems. To edit the source you need to select
Edit-->Source Edit and then use the Source tab that appears at the bottom of
the screen.
Obviously the person reading the message needs an HTML enabled reader
- I hope that your problem does not lie with a non-gifted reader at the
other end of your messages ;-)
this is a html post but the code wont display
±
test
Herc
--
www.winternet.com/~mikelr/flame76.html
Christos Dimitrakakis
====
>
>>Using Mozilla, I can.
> Using a killfile on Herc, I can't ;-)
Dirk Vdm
Most of the time I just manually ignore. I guess I should consider
killfiles one of these days :)
--
Will Twentyman
====
Christos Dimitrakakis scribbled the following:
>> Where did you find this list?
>
> Latin-1 Supplement <../unicode/latin_1_supplement.html> : Unicode
> I guess the unicode spec.
> I prefer people using tex-like notation rather than MATHML, as it is
> almost human-readable. MATHML is not. Also, if you use emacs (gasp) as
> your newsreader you might be able to view tex with the preview-latex
> mode, though I have not tried it.
I would choose pseudo-TeX over pseudo-{HT,MATH}ML over any day. Why?
Which of these do you find more readable on an ASCII newsreader?
The circumference of a circle is pi times as long as its diameter.
The circumference of a circle is p
times as long as its diameter.
--
/-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------
| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
----------------------------------------- Finland rules! ------------/
No, Maggie, not Aztec, Olmec! Ol-mec!
- Lisa Simpson
====
> Christos Dimitrakakis scribbled the following:
> I would choose pseudo-TeX over pseudo-{HT,MATH}ML over any day. Why?
> Which of these do you find more readable on an ASCII newsreader?
The circumference of a circle is pi times as long as its diameter.
The circumference of a circle is p
Mozilla. Apparently it is incorrect in some way, and Mozilla refuses to
support it. It was the things like π that I was interested in (as these
did
render properly in Mozilla). Where do I find a list of these? What I am
particularly interested in is things like $sum; and $int;, if such things
exists.
--
Stephen Montgomery-Smith
stephen@math.missouri.edu
http://www.math.missouri.edu/~stephen
====
Stephen Montgomery-Smith scribbled the
following:
>> Christos Dimitrakakis scribbled the following:
>> I would choose pseudo-TeX over pseudo-{HT,MATH}ML over any day. Why?
>> Which of these do you find more readable on an ASCII newsreader?
>>
>> The circumference of a circle is pi times as long as its diameter.
>>
>> The circumference of a circle is p times as long as its diameter.
> Mozilla. Apparently it is incorrect in some way, and Mozilla refuses
to
> support it. It was the things like π that I was interested in (as
these did
> render properly in Mozilla). Where do I find a list of these? What I am
> particularly interested in is things like $sum; and $int;, if such things
exists.
by a human.
--
/-- Joona Palaste (palaste@cc.helsinki.fi) ---------------------------
| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
----------------------------------------- Finland rules! ------------/
B-but Angus! You're a dragon!
- Mickey Mouse
====
---------------------------------------------------------------------
å± å« åü
Ìù åÏ G D Q L X P S F Y W a
b g d e h q l x p r s f y w åÛ å.b9
åÇ å å¨
åø åÀ Ì[YAcute]
Ì[Thorn] Ìÿ
Ìõ ÌÄ
$ ' * - @ ^ ~ å£ å´
åÒ å[Micro]
http://www.alanwood.net/demos/symbol.html
Here's the main ones, and you just copy and paste them into your equations.
x å± 3W
Bet none of you knew list constructors are derived as pure functions!
CONS = (la. lb. lf. f a b)
HEAD = (lc. c (la. lb. a)
TAIL = (lc. c (la. lb. b)
HEAD (CONS p q)
= lc.c (la. lb. a)) (CONS p q)
å¨ CONS p q (la. lb. a)
= (la. lb. lf. f a b) p q (la. lb. a)
å¨ (lb. lf. f p b) q (la. lb. a)
å¨ (lf. f p q) (la. lb. a)
å¨ (la. lb. a) p q
å¨ (lb. p) q
å¨ p
Even has a single character for ... å.b9
I assume the majority of people can read it, and other text readers are
supposed to get the
html as an attachment they can use a browser for.
Tex is no good just because its a dual platform, majority see html and only
minority use Tex,
mainly Maths institutions making it an intranet protocol, and if there is a
reason to
use symbols in the first place then /pi wont be appropriate.
Herc
id QQowza09331
====
I'm mucking about with an AVR/PIC uC and want to choose a suitable
accelerometer
(probably from Analog Devices - but please let me know of any suitable
alternatives!)
to give me good accuracy to calculate 0-62mph times, max lateral G force,
braking G's
etc
What I do need is for someone to point me in the right direction with
I can calculate such things using only the accelerometer and a known car
weight?
TIA!
Adam
====
>Is there any clear definition of quartiles?
Yes, but unfortunately there's more than one.
>I understand 2nd quartile is the same as median,
No, it isn't quite the same.
The k'th quartile is the (k/4)'th quantile.
The definition of the q'th quantile Q(q) for a random variable is the
following:
Q(q) = inf { x: F(x) >= q }
where F(x) is the cdf of the random variable. As far as I know everybody
agrees on that. However, for data, there seems to be no general agreement
on the correct definition of a quantile (which would be called the sample
or empirical quantile).
The most official definition is perhaps the following, according to the
Encyclopedia of Statistical Sciences:
If there are n data points X[1] <= X[2] <= ... <= X[n],
Q(q) = X[j] where (j-1)/n < q <= j/n.
No interpolation between data points here!
So e.g. for four data points [1,2,3,4], the 1st quartile would be
1 because (1-1)/4 < 1/4 <= 1/4, the 2nd quartile would be 2 because
(2-1)/4 < 2/4 <= 2/4, and the 3rd quartile would be 3 because
(3-1)/4 < 3/4 <= 3/4. For five points [1,2,3,4,5] the 1st quartile
would be 2, the 2nd would be 3 and the 3rd would be 4.
Parzen, J. Amer. Stat. Assoc. 74 (1979) 105-121, lists two other,
piecewise linear, versions. The first is
Q(q) = (j - n q) X[j-1] + (n q - j + 1) X[j] where (j-1)/n <= q <= j/n
(or X[1] if j = 1) which seems to be the definition used by Maple.
According to this, for [1,2,3,4,5] the 1st, 2nd and 3rd quartiles
would be 1.25, 2.5 and 3.75.
The second definition is
Q(q) = (j - n q + 1/2) X[j] + (n q - j + 1/2) X[j+1] where
(j-1/2)/n <= q <= (j+1/2)/n
(undefined if q < 1/(2 n) or q > 1 - 1/(2 n))
According to this, for [1,2,3,4,5] the 1st, 2nd and 3rd quartiles
would be 1.75, 3 and 4.25
MINITAB defines the 1st quartile as
(j - n/4 + 3/4) X[j] + (n/4 - j + 1/4) X[j+1] where j <= (n+1)/4 <= j+1
and the 3rd as
(j - 3n/4 + 1/4) X[j] + (3n/4 - j + 3/4) X[j+1] where j <= 3(n+1)/4 <= j+1
which (if it had quantiles in general, which I don't think it does) would
generalize to
Q(q) = (j - q (n+1) + 1) X[j] + (q (n+1) - j) X[j+1] where
j <= q (n+1) <= j+1.
According to this, for [1,2,3,4,5] the 1st, 2nd and 3rd quartiles
would be 1.5, 3 and 4.5.
I'm not sure what Excel uses; it seems to be different from all of these.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
====
It's the various definitions that made me confused.
It's much clearer now.
--
ES Kim
====
i got it nowYou've been very helpfull.
--
V.8anligen
Konrad
---------------------------------------------------
phone #1: (+46/0) 708 - 70 73 92
phone #2: (+46/0) 704 - 79 96 95
url: http://konrads.webbsida.com
-----------------------------------
Sleep - thing used by ineffective people
as a substitute for coffee
Ambition - a poor excuse for not having
enough sence to be lazy
---------------------------------------------------
====
...
>>on two texts, and of them I second seems most promising so far.
>>The readings I've found are:
>>1. 'Quaternionic and Clifford Calculus for Physicists and Engineers', K.
>>Gurlebeck & W. Sprossig, Wiley 1997
>>2. 'Rotations, Quaternions, and Double Groups', S. Altmann, Oxford 1986
>
>>MK
> I forgot to include the general category of direction cosine matrices.
> These are 3x3 matrices whose elements are dot products of orthogonal
> unit vectors in the rotated frame and unrotated frame. Sequences of
> rotations are represented by products of these matrices.
representations. This last one seems I think to be covered by
quaternions since there is a such dot product in the pure quaternion
component.
> John E. Prussing
> University of Illinois at Urbana-Champaign
> Department of Aerospace Engineering
> http://www.uiuc.edu/~prussing
Just based on the Altmann book I think I should look carefully in the
quaternion direction, although he gives ample warning that it's
perilous,although many of the perils can be avoided if one adheres to
all the suggested conventions, of which there are many...
I'm still not sure if, at the end of this, I will discover that
isomorphism betwen the notation and rotation sequence. I do know that
there are simplifications in my application that might help along the
way. Principally, that the piecewise paths in the SO(3) rotations I am
looking at have to be orthogonal.
MK.
====
You might want to look up Ken Shoemake in Google.
Dick Palais
> Does anyone know of a notation or a scheme for describing
> rotational paths (locus of a point on a sphere)?
> In three-dimensions a single Pi/2 rotation is nicely captured by a
> 3x3 orthogonal matrix, and that one can compose a sequence of them by
> multiplying the appropriate matrices in the same order.
> The question is, if one were to compose a sequence of these
> rotations, is there a notational scheme that's been devised to annotate
> each type of rotation (with a symbol) and to then describe such composed
> paths as a word, made up of the symbols?
> Would it be possible to associate one word with a unique path, or
> would there sometimes be multiple distinct paths satisfying a particular
> word?
> Anyone know of a good reference for this type of problem?
MK
>
====
> You might want to look up Ken Shoemake in Google.
Dick Palais
site. I notice that his context seems to be the rendering of rotations
as seen through the eye of a camera.
Going through a refernce I cited earlier (Altmann) I am continuing to
see promise in the SU(2) covering of SO(3) since my rotational segments
are orthogonal, and the four unit angular momentum matrices seem to be
good starting point for developing the symbols for discovering what I am
begining to learn is a 'homotopy' for the rotational paths of interest.
MK
====
Can anyone prove this?
====
> Can anyone prove this?
Yes, for instance, I can.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
The League of Gentlemen
====
in message <7fed4c14.0307112212.3161e585@posting.google.com>:
> Can anyone prove this?
I hope you yourself can do the |G/Z(G)| = 1 case. ;-)
If |G/Z(G)| = p, then G/Z(G) is cyclic and so is generated by a
single element, say xZ. So every element of G is of the form
{x^n}Z and so commutes with every other element.
--
Jim Heckman
====
> Can anyone prove this?
Here is a more complete proof:
If |G/Z(G)| = 1, then |G| = |Z(G)| ,thus G = Z(G). DUH
If |G/Z(G)| = p, p prime, then G/Z(G) is cyclic, and can be written in
the form:
G/Z(G) = { Z(G), aZ(G), (a^2)Z(G),...(a^n)Z(G)}, for some a in G
Now, let x, y be elements of G, and say, x is in (a^i)Z(G), y in
(a^j)Z(G), then: x = (a^i)g, y = (a^j)h, where g and h are in Z(G)
So, xy = (a^i)g(a^j)h = a^(j+i)gh = (a^j)h(a^i)g = yx, since g,h are
in Z(G).
There ya go.
====
Can anyone prove this?
====
> Can anyone prove this?
Yes, for instance, I can.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
The League of Gentlemen
====
in message <7fed4c14.0307112214.669f4561@posting.google.com>:
> Can anyone prove this?
is a normal subgroup of Aut(G).
--
Jim Heckman
====
> Can anyone prove this?
Yes, for instance, I can.
This guy is sooo cool. Lucky I live in the other hemisphere. :)
I just started learning about groups, what about making naive guesses
for start? Sometimes later we can reach a stage where the wise man
(Dr. Chapman, that is) can no longer keep quiet (only because seeing
us doing it so wrong must irritate him so much) and that he (I am
sure) will give away the clues.
Prove: If Aut(G) has a prime order, then G/Z(G) has order p or 1.
Looks creepy. What do we know about group orders? If Aut(G) is of
prime order, then any single element eta belongs to Aut(G) generates
Aut(G) [shoot me if I'm wrong]. That means Aut(G) is abelian.
Z(G) is always a normal subgroup isn't it. Let me see. f((g-)zg) =
f((g-)gz) = f(z), what was I thinking. Let E be an endomorphism with
kernel Z(G) [oops, cannot do that]. Then Ok, E is a homomorphism. Can
I safely say G/Z(G) = E? I think so if I spell the equal sign as 'is
isomorphic to'. Oh yeah, the First Theorem.
Interesting. The theorem to prove actually claims that G/Z(G) is
abelian!
The |G/Z(G)| = 1 case makes me think. Surely it says G is abelian [or
else that angry professor shoots me again]. I think |Aut(G)| is prime
says you get to make your decision only once. It may be that when
|G/Z(G)| = 1, G is of order p, namely, isomorphic to Z_p. If not, you
can just write 'no' here. That will make me behave more cautiously
next time.
The |G/Z(G)| = p case seems just incredible. How can that be? I see no
way [with my naive eyes] other than implying that Z(G) = { 1 }. You
can just write 'no' also if you like [provided that's true, of course
:)]
====
> is a normal subgroup of Aut(G).
====
Most likely I do not fully understand something, but I have a question
about
imaginary numbers and infinity. If there are positive and negative numbers
that represents plus and minus measure, and the exponents n ^ > 1 and n ^ <
1 represents an increase or decrease in dimension, then wouldn't i (-1^.5)
lie in an inverted topographical dimension to our number system? A
dimension
smaller than a negative measure? Wouldn't these number systems intersect at
infinity (inf) and zero?
If i were merely a representation of a unit of 1 smaller (I don't mean
less) than zero, then would zero then be the intersection of:
lim(1 -> 1/inf) and lim(1 -> inf/1)
Then would:
0 = inf
due to
(+/-)i = inf - 1
and
0i = inf
and
0 = inf + 1
and
i*(inf-i) = lim(1 -> inf/1)
The two parallel number lines would look like this:
o-----------.-----------.-----------o
0 1 lim(1 -> inf/1) inf
i*inf i*(inf-i) i 0i
o-----------.-----------.-----------o
I realize that using infinity in this way is not correct mathematically,
but
if infinity is 0i, then would (inf + 1) be a definition of zero? I would
expect there being a (Hadamard?) gap of 1 between 0 and infinity, but this
gap would exists due to there being no fractional portion to infinity.
Infinity would have to be a whole number, in this case, due to no number
existing that is larger than infinity; therefore infinity is still an
abstraction but no fraction could be added to infinity (i.e.,
inf = .nnn + inf
).
Without going into it here, there seems to be support for this in the QM
waveform definition (Euler formula) and special relativity (Lorenz
transformation). I realize that mathematics should never take direction for
applied science, but mathematics was created by physical systems (i.e.
humans).
If there is some obscure area of number theory that I seem to be attempting
to describe, please let me know. I've tried reading some on theoretic
recursive number theory, but that is some tough reading.
====
> Most likely I do not fully understand something, but I have a question
about
> imaginary numbers and infinity. If there are positive and negative
numbers
> that represents plus and minus measure, and the exponents n ^ > 1 and n ^
<
> 1 represents an increase or decrease in dimension, then wouldn't i
(-1^.5)
> lie in an inverted topographical dimension to our number system? A
dimension
> smaller than a negative measure? Wouldn't these number systems intersect
at
> infinity (inf) and zero?
>
None of that makes much sense. Complex numbers are thought of as being a
two dimensional plane; z = x + yi, where z is complex and x,y real. z is
plotted as tho it's the point (x,y) in the xy plane. A complex number
goes to infinity when |z| = sqr(x^2 + y^2) goes to infinity. Unlike the
real line that has a positive infinity, oo at one end and a negative
infinity -oo at the other, the complex plane has one infinity, as far as
the eye can see in any direction, like the polar point of an infinite
sphere.
====
Most likely I do not fully understand something, but I have a question
about
> imaginary numbers and infinity. If there are positive and negative
numbers
> that represents plus and minus measure, and the exponents n ^ > 1 and n
^ <
> 1 represents an increase or decrease in dimension, then wouldn't i
(-1^.5)
> lie in an inverted topographical dimension to our number system? A
dimension
> smaller than a negative measure? Wouldn't these number systems
intersect
at
> infinity (inf) and zero?
None of that makes much sense. Complex numbers are thought of as being a
> two dimensional plane; z = x + yi, where z is complex and x,y real. z
is
> plotted as tho it's the point (x,y) in the xy plane. A complex number
> goes to infinity when |z| = sqr(x^2 + y^2) goes to infinity. Unlike the
> real line that has a positive infinity, oo at one end and a negative
> infinity -oo at the other, the complex plane has one infinity, as far as
> the eye can see in any direction, like the polar point of an infinite
> sphere.
>
Sorry, I reread my post, and a crappy job phrasing it (e.g., our number
====
> Most likely I do not fully understand something, but I have a
question
> about
> imaginary numbers and infinity. If there are positive and negative
> numbers
> that represents plus and minus measure, and the exponents n ^ > 1 and
n
> ^ <
> 1 represents an increase or decrease in dimension, then wouldn't i
> (-1^.5)
> lie in an inverted topographical dimension to our number system? A
> dimension
> smaller than a negative measure? Wouldn't these number systems
intersect
> at
> infinity (inf) and zero?
> None of that makes much sense. Complex numbers are thought of as being
a
> two dimensional plane; z = x + yi, where z is complex and x,y real. z
is
> plotted as tho it's the point (x,y) in the xy plane. A complex number
> goes to infinity when |z| = sqr(x^2 + y^2) goes to infinity. Unlike
the
> real line that has a positive infinity, oo at one end and a negative
> infinity -oo at the other, the complex plane has one infinity, as far
as
> the eye can see in any direction, like the polar point of an infinite
> sphere.
>
Sorry, I reread my post, and a crappy job phrasing it (e.g., our
number
But, it would be fantastic for *me* if you would take the pain to
clarify what you said, I have heard it before, but didn't understand.
Real number is a line, having +inf at one end, -inf at the other.
Cool. Intuitive. Euclidean plane has so many infinities that it would
be nice to say it has infinite number of infinities, in every possible
direction. But what makes complex plane so different that it is
considered to be topologically equivalent to a sphere? I mean, the
stereographic projection works, but I don't see what is the reason
that people don't project the Euclidean plane that way.
And I would appreciate very much if you give me some hints about what
implication it has on calculus. It must be fascinating, or else people
wouldn't care. What integral theorems follow?
It would be WONDERFUL if you could illustrate it in one concrete
example, that is, something that gives different result for the
Euclidean plane. It must be fun to do that, if the topology thingies
are worth it.
somewhat fundamental. I hold that real numbers are to be preferred,
because I can write any complex number in the form (x+iy) that makes
them 'constructed from' the real numbers. Of course the mathematicians
would call these 'emotions' as they are essentially numbers, helping
us in calculations, not to be get emotional about. I just want to know
what gives the complex numbers such a unique place, apart from the
fact that they are algebraically closed.
====
Doe any one know the problems of the first
[Micro]*o
====
http://www.mathlinks.go.ro/
On the forum...
--
Julien Santini,
CMI Technop.99le de Ch.89teau-Gombert, France
Home page: http://www.analgebra.com
====
http://www22.brinkster.com/lordgandalf/imo/imo.html
Gandalf the Grey.
Note: Someone posted these problems on a forum but thats taking a long
time to load. You can find those problems here ::
http://www.mathlinks.go.ro/
====
In Herstein's 'Topics in Algebra', there is a question, to find an
integral domain which has an infinite number of elements, yet is of
finite characteristic.
I can't seem to find one.
Any help, anyone?
====
Andrew Duncan schrieb im Newsbeitrag
> In Herstein's 'Topics in Algebra', there is a question, to find an
> integral domain which has an infinite number of elements, yet is of
> finite characteristic.
I can't seem to find one.
Any help, anyone?
For example the polynomial ring over a finite field Z/(p)
(Infinite number of elements, characteristic p>0)
====
>In Herstein's 'Topics in Algebra', there is a question, to find an
>integral domain which has an infinite number of elements, yet is of
>finite characteristic.
I can't seem to find one.
Any help, anyone?
Think about polynomials.
************************
David C. Ullrich
====
I would like to know of any theoretical methods for solving an equation
of this kind, where a, b, c, d are positive integers:
a + b + c + d = 8
where a >=1, b>=2, c >=1 d>=2
I solved it by inspection and, if I am not mistaken, there are 10
possible sets of solutions for a, b, c, d. This is fairly easy, but my
constants could be in the 1 - 13 range, and a + b + c + d =10 for
example, would take quite some time by inspection.
I could write my own algorithm , but I'd rather not reinvent the
wheel...
Pierre
====
Reinvent the wheel, that's how you learn mathematics.
> I would like to know of any theoretical methods for solving an equation
> of this kind, where a, b, c, d are positive integers:
a + b + c + d = 8
where a >=1, b>=2, c >=1 d>=2
I solved it by inspection and, if I am not mistaken, there are 10
> possible sets of solutions for a, b, c, d. This is fairly easy, but my
> constants could be in the 1 - 13 range, and a + b + c + d =10 for
> example, would take quite some time by inspection.
> I could write my own algorithm , but I'd rather not reinvent the
> wheel...
Pierre
>
====
> I would like to know of any theoretical methods for solving an equation
> of this kind, where a, b, c, d are positive integers:
a + b + c + d = 8
where a >=1, b>=2, c >=1 d>=2
I solved it by inspection and, if I am not mistaken, there are 10
> possible sets of solutions for a, b, c, d. This is fairly easy, but my
> constants could be in the 1 - 13 range, and a + b + c + d =10 for
> example, would take quite some time by inspection.
> I could write my own algorithm , but I'd rather not reinvent the
> wheel...
Pierre
Reinvent the wheel, that's how you learn mathematics.
In books, this is known as a partition problem.
For your original problem, I would proceed like this: Those
inequalities are inconvenient, so let a=A+1, b=B+2, c=C+1, d=D+2. Then
you want to solve A+B+C+D=2 in nonnegative integers. That is what you
find in the partition formula.
====
Suppose we have a1 + a2 + ... + an = k. The number of solutions to
this equation, where the variables are non-negative integers (0 is ok)
is the number of length (n-1+k) bitstrings with exactly n-1 ones:
for example: a1 + a2 + a3 + a4 = 8 can be represented as:
00110001000, with a1 = 2, a2 = 0, a3 = 3, a4 = 3. You are placing 3
(=n-1) 1s at any of the (n-1+k) positions, so the solution is
C(n-1+k,n-1).
Now if you have constraints such as a1 >=1, a2>=2, a3 >=1, a4>=2, then
you know for sure that your bitstring has 6 zeroes already taken care
of (the sum of the constraints), so all you need to do now is find the
number of solutions to
a1 + a2 + a3 + a4 = 8-6 = 2, which is just C(4-1+2,4-1) = C(5,3) = 10
(as you stated).
Another example: solve a1 + a2 + a3 + a4 + a5 = 126, with ai>=i.
Sum of constraints: 1+2+3+4+5=15. So the number of solutions is
C(5-1+126-15,5-1)=C(115,4).
> I would like to know of any theoretical methods for solving an equation
> of this kind, where a, b, c, d are positive integers:
a + b + c + d = 8
where a >=1, b>=2, c >=1 d>=2
I solved it by inspection and, if I am not mistaken, there are 10
> possible sets of solutions for a, b, c, d. This is fairly easy, but my
> constants could be in the 1 - 13 range, and a + b + c + d =10 for
> example, would take quite some time by inspection.
> I could write my own algorithm , but I'd rather not reinvent the
> wheel...
Pierre
====
If r(h,i) describes a closed 3d shape, where h and i are angles and r
describes a radius for angles h and i, how can I calculate it's volume
?
In other words, what is
Integral(0If r(h,i) describes a closed 3d shape, where h and i are angles and r
>describes a radius for angles h and i, how can I calculate it's volume
>?
>In other words, what is
>Integral(0in terms of x,y and z ?
I don't know what angles your h and i are. The usual angles in spherical
coordinates are longitude theta (0 <= theta < 2 pi) and colatitude phi
(0 <= phi <= pi). The element of volume is then r^2 sin(phi) dr dphi
dtheta, so the volume of your object (0 <= r <= r(theta, phi)) would be
1/3 int_0^{2 pi} dtheta int_0^pi dphi r(theta,phi)^3 sin(phi)
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
====
> I am asking:
> if uncountably infinite is valid, then in the same way we require
> a (countably) infinite number of decimal places even if the
> _value_ of a number can be expressed in a finite number of digits,
> why don't we require an uncountably infinite number of decimal
> places even though any number's _value_ will be expressed in
> a countably infinite amount of places? (allow calculations
> to continue as the new size permits....)
We don't do that because a countable number of digits is sufficient to
specify each real number.
There's also the fact that what you are proposing wouldn't make sense for
the real numbers. We know what a digit d_n in the n'th decimal place
means: it contributes a value of d_n * 10^(-n) to the value of the
number.
What would a digit d_(alpha) in the alpha-th place mean, if alpha is a
transfinite ordinal? It would have to represent a real value smaller
than 10^(-n) for any finite n, and that would violate the Archimedean
property of the real numbers: given any real number epsilon > 0, there
exists an integer N > 0 such that 1/N < epsilon.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
>In Cantor's Diagonal Argument reals are listed by their decimal
>representations. However, if the digits in a decimal representation
>of a real are denoted d_1, d_2...d_n, then the set of numbers that
>comprise the subscripts of d are integers, and therefore are
>countably infinite. Assuming the validity of uncountablty infinite,
>a countably infinite set of subscript numbers is not comprehensive;
>there must exist decimal places greater than the nth place where
n
>is an integer.
Uh, right. Just like if the idea of infinite sets is valid then every
> set must be infinite.
Hmm.
Well, a number such as 1.0 _could_ be expressed in a finite number of
> places, but because of the validity of carrying out calculations past
> the zero (infinitely so), this number is thought to have an endless
> trail of zeros.
Focusing only on those reals which have finite decimal representation
is misleading because they constitute a minority of all reals; and
are in fact a countable subset.
> I am asking:
if uncountably infinite is valid, then in the same way we require
> a (countably) infinite number of decimal places even if the
> _value_ of a number can be expressed in a finite number of digits,
> why don't we require an uncountably infinite number of decimal
> places even though any number's _value_ will be expressed in
> a countably infinite amount of places? (allow calculations
> to continue as the new size permits....)
There are many real numbers which can't be expressed by
anything but an infinite length sequence, the requirement to
represent them with an infinite sequence isn't an arbitrary
choice.
For any one real that can be expressed with a finite length string,
there are an infinite* number of reals based on that finite
string as a prefix...
I think your concept of a real number is off.
Any conceivable infinite length string (of decimal digits in
this case) can represent a single real number, important point:
it doesn't have to end... ever.
* not countably infinite.
dg
>As the value of a given real is completed in a
>countably infinite number of digits, any remaining digits must be
>zeros. If all reals have some point after which the remaining digits
>are zeros, diagonalization no longer works.
************************
David C. Ullrich
====
> I am asking:
>
> if uncountably infinite is valid, then in the same way we require
> a (countably) infinite number of decimal places even if the
> _value_ of a number can be expressed in a finite number of digits,
> why don't we require an uncountably infinite number of decimal
> places even though any number's _value_ will be expressed in
> a countably infinite amount of places? (allow calculations
> to continue as the new size permits....)
We don't do that because a countable number of digits is sufficient to
> specify each real number.
There's also the fact that what you are proposing wouldn't make sense for
> the real numbers. We know what a digit d_n in the n'th decimal place
> means: it contributes a value of d_n * 10^(-n) to the value of the
> number.
What would a digit d_(alpha) in the alpha-th place mean, if alpha is a
> transfinite ordinal? It would have to represent a real value smaller
> than 10^(-n) for any finite n, and that would violate the Archimedean
> property of the real numbers: given any real number epsilon > 0, there
> exists an integer N > 0 such that 1/N < epsilon.
Agreed, to the extent that any number contributed by digit d_(alpha)
cannot be in the form of an integer * 1/n. Obviously, if a digit
d_(alpha) is to exist the only value that can occupy that digit is 0.
As I am in agreement with the above, this zero cannot be arrived at by
having digit d_(alpha)contribute 0 * 1/n for any Integer n. Lets say a
digit d_(alpha) in the alpha-th place contributes a value of d_(alpha)
* 0, or 0. Is this a problem?
====
Dave Seaman Wrote:
>> I am asking:
> if uncountably infinite is valid, then in the same way we require
>> a (countably) infinite number of decimal places even if the
>> _value_ of a number can be expressed in a finite number of digits,
>> why don't we require an uncountably infinite number of decimal
>> places even though any number's _value_ will be expressed in
>> a countably infinite amount of places? (allow calculations
>> to continue as the new size permits....)
We don't do that because a countable number of digits is sufficient to
>specify each real number.
>
Exactly. I wonder if there is *any* representation of the real numbers
that
requires an uncountable number of digits? I've never seen one, but such a
critter might be kind of interesting.
Rich Burge
====
>> What would a digit d_(alpha) in the alpha-th place mean, if alpha is a
>> transfinite ordinal? It would have to represent a real value smaller
>> than 10^(-n) for any finite n, and that would violate the Archimedean
>> property of the real numbers: given any real number epsilon > 0, there
>> exists an integer N > 0 such that 1/N < epsilon.
> Agreed, to the extent that any number contributed by digit d_(alpha)
> cannot be in the form of an integer * 1/n. Obviously, if a digit
> d_(alpha) is to exist the only value that can occupy that digit is 0.
> As I am in agreement with the above, this zero cannot be arrived at by
> having digit d_(alpha)contribute 0 * 1/n for any Integer n. Lets say a
> digit d_(alpha) in the alpha-th place contributes a value of d_(alpha)
> * 0, or 0. Is this a problem?
You tell me. You are the one who is claiming there is a problem in the
Cantor diagonal argument, but since the Cantor diagonal argument does not
mention any such thing as a d_alpha for transfinite alpha, I fail to see
what your argument has to do with Cantor's.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
>
> What would a digit d_(alpha) in the alpha-th place mean, if alpha is a
> transfinite ordinal? It would have to represent a real value smaller
> than 10^(-n) for any finite n, and that would violate the Archimedean
> property of the real numbers: given any real number epsilon > 0,
there
> exists an integer N > 0 such that 1/N < epsilon.
>
> Agreed, to the extent that any number contributed by digit d_(alpha)
> cannot be in the form of an integer * 1/n. Obviously, if a digit
> d_(alpha) is to exist the only value that can occupy that digit is 0.
> As I am in agreement with the above, this zero cannot be arrived at by
> having digit d_(alpha)contribute 0 * 1/n for any Integer n. Lets say a
> digit d_(alpha) in the alpha-th place contributes a value of d_(alpha)
> * 0, or 0. Is this a problem?
You tell me. You are the one who is claiming there is a problem in the
> Cantor diagonal argument, but since the Cantor diagonal argument does not
> mention any such thing as a d_alpha for transfinite alpha, I fail to see
> what your argument has to do with Cantor's.
I am being too free with my wording· I am not asking you for help
in
disproving something. Believe it or not (I don't care either way) I
have no vested interest in disproving anything. However, I do have
questions about what is claimed to have been proven that, at least to
me, are unsettling.
I am not claiming, I am asking.
Now onto addressing the issue.
Cantor's diagonal argument establishes the existence of transfinite.
Any transfinite amount can be split into two sections
1) the portion that can be mapped 1 to 1 with the integers
2) the remaining portion
Call section 1) A and section 2) B.
I would like to note that I am **going to change the suggested
representation**
from what I had previously suggested, but not the idea.
_Make_ a representation for all real numbers that _requires_ a
transfinite amount of digits to be filled. For the section of that
representation that falls into A let the digits be generated in
their normal fashion, that is, the nth digit = d_n * 10^(-n). For the
section of that representation that falls into B ,(might as well use
the borrowed term) , alpha-th digit = either 1 or 0; the chosen
number being 1 if a zero can still be added to the value of the real
without exceeding the desired value, or 0 if it cannot be. Being that
an uncountably infinite number of zeros can be added to any number and
not interfere with the value of that number, every d_(alpha) would =
1.
I am stating rather then asking, and I don't like doing that as I feel
it makes me look confident about what I am stating. But since I have
stated, and am not confident, please tell me if any of these steps
violates rules I don't know about or haven't thought of.
dg
====
>>
>>In Cantor's Diagonal Argument reals are listed by their decimal
>>representations. However, if the digits in a decimal representation
>>of a real are denoted d_1, d_2...d_n, then the set of numbers that
>>comprise the subscripts of d are integers, and therefore are
>>countably infinite. Assuming the validity of uncountablty infinite,
>>a countably infinite set of subscript numbers is not comprehensive;
>>there must exist decimal places greater than the nth place where
n
>>is an integer.
>>
>> Uh, right. Just like if the idea of infinite sets is valid then every
>> set must be infinite.
>>
>> Hmm.
Well, a number such as 1.0 _could_ be expressed in a finite number of
>places, but because of the validity of carrying out calculations past
>the zero (infinitely so), this number is thought to have an endless
>trail of zeros.
I am asking:
if uncountably infinite is valid, then in the same way we require
>a (countably) infinite number of decimal places even if the
>_value_ of a number can be expressed in a finite number of digits,
>why don't we require an uncountably infinite number of decimal
>places even though any number's _value_ will be expressed in
>a countably infinite amount of places? (allow calculations
>to continue as the new size permits....)
No, you didn't ask that, you _asserted_ that it must be done
that way.
The reason we don't do it that way, to answer your question,
is that (as long as we're just talking about real numbers) it
would be stupid - what would be the point to adding
uncounatbly many zeroes to the decimal representation
of a real number?
>dg
>>As the value of a given real is completed in a
>>countably infinite number of digits, any remaining digits must be
>>zeros. If all reals have some point after which the remaining digits
>>are zeros, diagonalization no longer works.
>>
>> ************************
>>
>> David C. Ullrich
************************
David C. Ullrich
====
>> You tell me. You are the one who is claiming there is a problem in the
>> Cantor diagonal argument, but since the Cantor diagonal argument does
not
>> mention any such thing as a d_alpha for transfinite alpha, I fail to see
>> what your argument has to do with Cantor's.
> I am being too free with my wording· I am not asking you for
help in
> disproving something. Believe it or not (I don't care either way) I
> have no vested interest in disproving anything. However, I do have
> questions about what is claimed to have been proven that, at least to
> me, are unsettling.
> I am not claiming, I am asking.
> Now onto addressing the issue.
> Cantor's diagonal argument establishes the existence of transfinite.
> Any transfinite amount can be split into two sections
> 1) the portion that can be mapped 1 to 1 with the integers
> 2) the remaining portion
> Call section 1) A and section 2) B.
> I would like to note that I am **going to change the suggested
> representation**
> from what I had previously suggested, but not the idea.
The representation you choose doesn't really matter. The reason for that
is that there are only two possibilities:
(A) Your proposed representation produces a field isomorphic
to the reals, or
(B) it doesn't.
In case (A), it follows from the uncountability of the reals that your
representation is likewise uncountable (in particular, an isomorphism is
also a bijection). In case (B), nothing about your representation can
possibly contradict the fact that the reals are uncountable. Either way,
your proposed representation is not relevant to the uncountability of the
reals.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
>In Cantor's Diagonal Argument reals are listed by their decimal
>representations. However, if the digits in a decimal representation
>of a real are denoted d_1, d_2...d_n, then the set of numbers that
>comprise the subscripts of d are integers, and therefore are
>countably infinite. Assuming the validity of uncountablty infinite,
>a countably infinite set of subscript numbers is not comprehensive;
>there must exist decimal places greater than the nth place where
n
>is an integer.
Uh, right. Just like if the idea of infinite sets is valid then every
> set must be infinite.
Hmm.
Well, a number such as 1.0 _could_ be expressed in a finite number of
>places, but because of the validity of carrying out calculations past
>the zero (infinitely so), this number is thought to have an endless
>trail of zeros.
I am asking:
if uncountably infinite is valid, then in the same way we require
>a (countably) infinite number of decimal places even if the
>_value_ of a number can be expressed in a finite number of digits,
>why don't we require an uncountably infinite number of decimal
>places even though any number's _value_ will be expressed in
>a countably infinite amount of places? (allow calculations
>to continue as the new size permits....)
No, you didn't ask that, you _asserted_ that it must be done
> that way.
The reason we don't do it that way, to answer your question,
> is that (as long as we're just talking about real numbers) it
> would be stupid - what would be the point to adding
> uncounatbly many zeroes to the decimal representation
> of a real number?
I agree it is stupid.
Is it valid? My point is not that numbers _should_ be represented this
way, it is that if they _can_ be it would pose a problem (all reals
would have digits in which they do not differ from each other).
Dave Seaman posted that my proposed representation would not be
allowable because it would produce a field isomorphic to the reals and
would therefore itself be uncountable. This is what i've been
wondering all along but obviously havent stated clearly enough:
First,
can a transfinite (uncountable) amount be split into a countable and
uncountable section:
a) the amount that maps 1-1 with the reals
b) the remaining portion
Second,
If the divide is allowable would it matter how we order section b) of
the representation considering that every digit would have to be the
same: There doesn't have to be a next digit in b) but there have to
be _digits_ in b) *and* b) must be _next_ after a)?
I assume there are problems with one or both of these ideas, because
the answer I got when I posted this was it follows from the
uncountability of the reals that your representation is likewise
uncountable. However that doesn't give a reason uncountable can not
be divided as suggested above, or a reason that if the division is
acceptable the structure of b) is not.
(to be fair though, this isn't the fault of the person responding, I
didn't say anything about a structure or order of b when I posted.)
If you or anyone has a reason for the explicit problems created by
accepting what is suggested in either of these two questions I would
greatly appreciate it.
I really do not wish to have this thread go on forever, or to bother
people with repetitive arguments. If I could have an answer to those
two questions I would be more than satisfied.
thanks,
dg
>dg
>As the value of a given real is completed in a
>countably infinite number of digits, any remaining digits must be
>zeros. If all reals have some point after which the remaining digits
>are zeros, diagonalization no longer works.
************************
David C. Ullrich
************************
David C. Ullrich
====
>>I am asking:
>>if uncountably infinite is valid, then in the same way we require
>>a (countably) infinite number of decimal places even if the
>>_value_ of a number can be expressed in a finite number of digits,
>>why don't we require an uncountably infinite number of decimal
>>places even though any number's _value_ will be expressed in
>>a countably infinite amount of places? (allow calculations
>>to continue as the new size permits....)
>> No, you didn't ask that, you _asserted_ that it must be done
>> that way.
>> The reason we don't do it that way, to answer your question,
>> is that (as long as we're just talking about real numbers) it
>> would be stupid - what would be the point to adding
>> uncounatbly many zeroes to the decimal representation
>> of a real number?
> I agree it is stupid.
> Is it valid? My point is not that numbers _should_ be represented this
> way, it is that if they _can_ be it would pose a problem (all reals
> would have digits in which they do not differ from each other).
In what way would that be a problem? There are proofs of the
uncountability of the reals that do not mention digits at all. Such
proofs obviously would be unaffected by the chosen representation.
> Dave Seaman posted that my proposed representation would not be
> allowable because it would produce a field isomorphic to the reals and
> would therefore itself be uncountable. This is what i've been
> wondering all along but obviously havent stated clearly enough:
That is not what I said. If your representation produces a field
isomorphic to the reals, then it is allowable, but the existence of an
isomorphism demonstrates that your version of the reals has the same
cardinality as the ordinary reals. That is, both are uncountable.
> First,
> can a transfinite (uncountable) amount be split into a countable and
> uncountable section:
No countable set can have an uncountable subset. An uncountable set can
be shown to have a countable subset if we assume the axiom of choice.
> a) the amount that maps 1-1 with the reals
> b) the remaining portion
Not every infinite set has a subset that maps 1-1 with the reals. If
such a subset exists, then the original set must be uncountable.
I get the impression that you didn't say what you actually meant in part
a). Rather than a set that maps 1-1 with the reals, you seem to be
talking about a set that maps 1-1 with the natural numbers (i.e., a
countable set), such as the set of digit positions in the representation
of a real number.
> Second,
> If the divide is allowable would it matter how we order section b) of
> the representation considering that every digit would have to be the
> same: There doesn't have to be a next digit in b) but there have to
> be _digits_ in b) *and* b) must be _next_ after a)?
I can't say whether it matters or not how you order things, because
I have no idea what you are trying to do.
As best I can make out, you seem determined to transform the Cantor
diagonal proof into some entirely different argument and then attack the
transformed argument instead of the original. This is known as the
straw man fallacy. If you want to understand what the diagonal
argument means, you should study what the argument actually says instead
of trying to force it to say something different.
> I assume there are problems with one or both of these ideas, because
> the answer I got when I posted this was it follows from the
> uncountability of the reals that your representation is likewise
> uncountable. However that doesn't give a reason uncountable can not
> be divided as suggested above, or a reason that if the division is
> acceptable the structure of b) is not.
You are correct; it doesn't give a reason that sets can't be divided.
Instead, it gives a reason why it makes no difference to the
uncountability of the reals whether sets can be divided or not.
The real numbers can be divided into the rationals and the irrationals.
The irrationals are uncountable and have the same cardinality as the
reals. Does that answer your question about whether sets can be divided?
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
>I am asking:
>
>if uncountably infinite is valid, then in the same way we require
>a (countably) infinite number of decimal places even if the
>_value_ of a number can be expressed in a finite number of digits,
>why don't we require an uncountably infinite number of decimal
>places even though any number's _value_ will be expressed in
>a countably infinite amount of places? (allow calculations
>to continue as the new size permits....)
>
> No, you didn't ask that, you _asserted_ that it must be done
> that way.
>
> The reason we don't do it that way, to answer your question,
> is that (as long as we're just talking about real numbers) it
> would be stupid - what would be the point to adding
> uncounatbly many zeroes to the decimal representation
> of a real number?
> I agree it is stupid.
>
> Is it valid? My point is not that numbers _should_ be represented this
> way, it is that if they _can_ be it would pose a problem (all reals
> would have digits in which they do not differ from each other).
In what way would that be a problem? There are proofs of the
> uncountability of the reals that do not mention digits at all. Such
> proofs obviously would be unaffected by the chosen representation.
>
Yes. I have seen the generalized proof. It relies on definitions of
sets I am uncomfortable with. If the definitions are acceptable then
so is the proof, but for this post, I would like to stay away from it
altogether.
> Dave Seaman posted that my proposed representation would not be
> allowable because it would produce a field isomorphic to the reals and
> would therefore itself be uncountable. This is what i've been
> wondering all along but obviously havent stated clearly enough:
That is not what I said. If your representation produces a field
> isomorphic to the reals, then it is allowable, but the existence of
an
> isomorphism demonstrates that your version of the reals has the same
> cardinality as the ordinary reals. That is, both are uncountable.
> First,
>
> can a transfinite (uncountable) amount be split into a countable and
> uncountable section:
No countable set can have an uncountable subset.
The proposed representation would be uncountable, this is irrelevant.
> An uncountable set can be shown to have a countable subset if we assume
the >axiom of choice.
>
*This* is **exactly** what I was wondering.
(In no way do I think it is your fault for not understanding my lack
of clarity or knowledge in posts. Actually, I appreciate your
continued responses allowing me to try to clarify what should have
been clear originally.)
Now , assuming the Axiom of Choice, could a real number have the
following representation:
The countably infinite subset A of an uncountably infinite set B
contains the traditional digits of a real number, and the part of the
set that is not within that subset A contains 0's, each present 0
standing for a valid addition of itself to the real without exceeding
the desired value.
[cut]
> As best I can make out, you seem determined to transform the Cantor
> diagonal proof into some entirely different argument and then attack the
> transformed argument instead of the original. This is known as the
> straw man fallacy. If you want to understand what the diagonal
> argument means, you should study what the argument actually says instead
> of trying to force it to say something different.
>
No. I am only trying to change one thing in the argument (Specifically
the Diagonal Argument concerning rationals and reals, not the
generalized version):
the decimal representation.
> I assume there are problems with one or both of these ideas,
because
> the answer I got when I posted this was it follows from the
> uncountability of the reals that your representation is likewise
> uncountable. However that doesn't give a reason uncountable can not
> be divided as suggested above, or a reason that if the division is
> acceptable the structure of b) is not.
You are correct; it doesn't give a reason that sets can't be divided.
> Instead, it gives a reason why it makes no difference to the
> uncountability of the reals whether sets can be divided or not.
The real numbers can be divided into the rationals and the irrationals.
> The irrationals are uncountable and have the same cardinality as the
> reals. Does that answer your question about whether sets can be divided?
Your answer involving the axiom of choice fully answered my question
about the particular division of uncountable sets I was wondering
about.
If I am (hopefully) stating things clearly enough now, and an answer
of why or why not my proposed representation is valid/invalid is
possible, along with whether or not it would affect the diagonal
argument to use such representations in place of traditional decimal
representations,
everything will be completely cleared up for me.
dg
====
>> Is it valid? My point is not that numbers _should_ be represented this
>> way, it is that if they _can_ be it would pose a problem (all reals
>> would have digits in which they do not differ from each other).
>> In what way would that be a problem? There are proofs of the
>> uncountability of the reals that do not mention digits at all. Such
>> proofs obviously would be unaffected by the chosen representation.
> Yes. I have seen the generalized proof. It relies on definitions of
> sets I am uncomfortable with. If the definitions are acceptable then
> so is the proof, but for this post, I would like to stay away from it
> altogether.
*The* generalized proof? Which one is that?
For example, there is the proof from measure theory: Lebesque measure is
countably additive, and points have zero measure. Yet, the unit interval
[0,1] has Lebesgue measure 1. Therefore, it follows that [0,1] must be
uncountable.
Or, there is the Baire Category Theorem: A complete metric space cannot
be a countable union of nowhere-dense sets. The reals are a complete
metric space. Single points on the real line are nowhere dense.
Therefore, the reals are uncountable.
>> Dave Seaman posted that my proposed representation would not be
>> allowable because it would produce a field isomorphic to the reals and
>> would therefore itself be uncountable. This is what i've been
>> wondering all along but obviously havent stated clearly enough:
>> That is not what I said. If your representation produces a field
>> isomorphic to the reals, then it is allowable, but the existence of
an
>> isomorphism demonstrates that your version of the reals has the same
>> cardinality as the ordinary reals. That is, both are uncountable.
>> First,
>> can a transfinite (uncountable) amount be split into a countable and
>> uncountable section:
>> No countable set can have an uncountable subset.
> The proposed representation would be uncountable, this is irrelevant.
The set of digit positions used in the representation of a real number is
countable. You are trying to add on more digit positions to make it
uncountable, but what you get is not in any way a subset of what you
started with.
>> An uncountable set can be shown to have a countable subset if we assume
the >axiom of choice.
> *This* is **exactly** what I was wondering.
> (In no way do I think it is your fault for not understanding my lack
> of clarity or knowledge in posts. Actually, I appreciate your
> continued responses allowing me to try to clarify what should have
> been clear originally.)
> Now , assuming the Axiom of Choice, could a real number have the
> following representation:
> The countably infinite subset A of an uncountably infinite set B
> contains the traditional digits of a real number, and the part of the
> set that is not within that subset A contains 0's, each present 0
> standing for a valid addition of itself to the real without exceeding
> the desired value.
In other words, you are appending an uncountable number of zeroes to the
decimal representation of each real numbers. Yes, there is an obvious
isomorphism between your representation and the standard decimal
representation.
By the way, we don't need the axiom of choice to assert that every set
has an uncountable superset. Let X be any set at all, and let Y = X
union R (the reals). Then Y is a superset of X, and Y is uncountable.
This is entirely different from the question you asked previously.
> [cut]
>> As best I can make out, you seem determined to transform the Cantor
>> diagonal proof into some entirely different argument and then attack the
>> transformed argument instead of the original. This is known as the
>> straw man fallacy. If you want to understand what the diagonal
>> argument means, you should study what the argument actually says instead
>> of trying to force it to say something different.
> No. I am only trying to change one thing in the argument (Specifically
> the Diagonal Argument concerning rationals and reals, not the
> generalized version):
> the decimal representation.
As I said, you are trying to change the argument. You are committing the
straw man fallacy.
As it happens, it doesn't really make any difference, because the
diagonal argument still works when applied to your representation. We
only diagonalize over the countable digit positions and ignore the
uncountably many zeroes that you have appended.
But if things had turned out differently and somehow the diagonal
argument didn't work for your representation, it still wouldn't make any
difference. The argument works for the standard representation, and
yours is isomorphic to the standard representation. That's all we need
to show that your representation is likewise uncountable.
>> The real numbers can be divided into the rationals and the irrationals.
>> The irrationals are uncountable and have the same cardinality as the
>> reals. Does that answer your question about whether sets can be
divided?
> Your answer involving the axiom of choice fully answered my question
> about the particular division of uncountable sets I was wondering
> about.
No, it didn't. My answer involving the axiom of choice was for a
question entirely unrelated to the question of whether every set has an
uncountable superset. That is not the question you originally asked.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
>
> Is it valid? My point is not that numbers _should_ be represented
this
> way, it is that if they _can_ be it would pose a problem (all reals
> would have digits in which they do not differ from each other).
>
> In what way would that be a problem? There are proofs of the
> uncountability of the reals that do not mention digits at all.
Such
> proofs obviously would be unaffected by the chosen representation.
> Yes. I have seen the generalized proof. It relies on definitions of
> sets I am uncomfortable with. If the definitions are acceptable then
> so is the proof, but for this post, I would like to stay away from it
> altogether.
*The* generalized proof? Which one is that?
For example, there is the proof from measure theory: Lebesque measure is
> countably additive, and points have zero measure. Yet, the unit interval
> [0,1] has Lebesgue measure 1. Therefore, it follows that [0,1] must be
> uncountable.
Or, there is the Baire Category Theorem: A complete metric space cannot
> be a countable union of nowhere-dense sets. The reals are a complete
> metric space. Single points on the real line are nowhere dense.
> Therefore, the reals are uncountable.
> Dave Seaman posted that my proposed representation would not be
> allowable because it would produce a field isomorphic to the reals
and
> would therefore itself be uncountable. This is what i've been
> wondering all along but obviously havent stated clearly enough:
>
> That is not what I said. If your representation produces a field
> isomorphic to the reals, then it is allowable, but the existence of
an
> isomorphism demonstrates that your version of the reals has the same
> cardinality as the ordinary reals. That is, both are uncountable.
>
> First,
>
> can a transfinite (uncountable) amount be split into a countable and
> uncountable section:
>
> No countable set can have an uncountable subset.
>
> The proposed representation would be uncountable, this is irrelevant.
The set of digit positions used in the representation of a real number is
> countable. You are trying to add on more digit positions to make it
> uncountable, but what you get is not in any way a subset of what you
> started with.
An uncountable set can be shown to have a countable subset if we assume
the >axiom of choice.
> *This* is **exactly** what I was wondering.
>
> (In no way do I think it is your fault for not understanding my lack
> of clarity or knowledge in posts. Actually, I appreciate your
> continued responses allowing me to try to clarify what should have
> been clear originally.)
>
> Now , assuming the Axiom of Choice, could a real number have the
> following representation:
> The countably infinite subset A of an uncountably infinite set
B
> contains the traditional digits of a real number, and the part of the
> set that is not within that subset A contains 0's, each present 0
> standing for a valid addition of itself to the real without exceeding
> the desired value.
In other words, you are appending an uncountable number of zeroes to the
> decimal representation of each real numbers. Yes, there is an obvious
> isomorphism between your representation and the standard decimal
> representation.
By the way, we don't need the axiom of choice to assert that every set
> has an uncountable superset. Let X be any set at all, and let Y = X
> union R (the reals). Then Y is a superset of X, and Y is uncountable.
> This is entirely different from the question you asked previously.
> [cut]
> As best I can make out, you seem determined to transform the Cantor
> diagonal proof into some entirely different argument and then attack
the
> transformed argument instead of the original. This is known as the
> straw man fallacy. If you want to understand what the diagonal
> argument means, you should study what the argument actually says
instead
> of trying to force it to say something different.
> No. I am only trying to change one thing in the argument (Specifically
> the Diagonal Argument concerning rationals and reals, not the
> generalized version):
>
> the decimal representation.
As I said, you are trying to change the argument. You are committing the
> straw man fallacy.
As it happens, it doesn't really make any difference, because the
> diagonal argument still works when applied to your representation. We
> only diagonalize over the countable digit positions and ignore the
> uncountably many zeroes that you have appended.
But if things had turned out differently and somehow the diagonal
> argument didn't work for your representation, it still wouldn't make any
> difference. The argument works for the standard representation, and
> yours is isomorphic to the standard representation. That's all we need
> to show that your representation is likewise uncountable.
> The real numbers can be divided into the rationals and the
irrationals.
> The irrationals are uncountable and have the same cardinality as the
> reals. Does that answer your question about whether sets can be
divided?
>
> Your answer involving the axiom of choice fully answered my question
> about the particular division of uncountable sets I was wondering
> about.
No, it didn't. My answer involving the axiom of choice was for a
> question entirely unrelated to the question of whether every set has an
> uncountable superset. That is not the question you originally asked.
Ok.
dg
====
Cantor's diagonal argument establishes the existence of transfinite.
>
I always thought that transfinite was an adjective.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
The League of Gentlemen
====
> Cantor's diagonal argument establishes the existence of transfinite.
>
> I always thought that transfinite was an adjective.
it is. transfinite [numbers].
in any case, i now realize the problem with the argument.
so, as confused as i was, i guess the post helped me after all.
dg
====
>
>>
>> Cantor's diagonal argument establishes the existence of transfinite.
I always thought that transfinite was an adjective.
OED lists it as an adjective and noun. There are uses of it like,
...into the transfinite and so forth in the literature. This
particular
example is by Goedel and listed under the OED entry for incompleteness.
The word is new enough that there doesn't seem to be established rules for
its usage. I'm not sure if Cantor used transfinite as a noun. I have
a
vague recollection he may have.
====
>The reason we don't do it that way, to answer your question,
>is that (as long as we're just talking about real numbers) it
>would be stupid - what would be the point to adding
>uncounatbly many zeroes to the decimal representation
>of a real number?
I have a bigger problem with this concept: Since the digits are discrete
items, in a linear ordering, I don't see how you could meaningfully claim
to even define such a critter. Doesn't the mere fact that you have a
sequential list of items kind of imply that the cardinality of that list
*must* be Aleph null?
If you could come up with a meaning for an uncountable list of digits,
wouldn't that be just what the folks who try to claim 1.0 != 0.999999...
want to do -- put stuff after the last digit in 0.99999...?
--
Michael F. Stemper
#include
The FAQ for rec.arts.sf.written is at:
http://www.geocities.com/evelynleeper/sf-written.htm
Please read it before posting.
====
>>The reason we don't do it that way, to answer your question,
>>is that (as long as we're just talking about real numbers) it
>>would be stupid - what would be the point to adding
>>uncounatbly many zeroes to the decimal representation
>>of a real number?
> I have a bigger problem with this concept: Since the digits are discrete
> items, in a linear ordering, I don't see how you could meaningfully claim
> to even define such a critter. Doesn't the mere fact that you have a
> sequential list of items kind of imply that the cardinality of that list
> *must* be Aleph null?
Consider omega_1, the first uncountable ordinal. The members of omega_1
are discrete. Each one has an immediate successor. Yet it makes
perfectly good sense to regard a function defined on omega_1 as a
sequence (in an extended sense).
I don't think the OP specified which uncountable infinity he was
considering, but if we accept the axiom of choice, or if we accept the
continuum hypothesis, then the real numbers can be well-ordered, and
therefore a similar answer applies.
> If you could come up with a meaning for an uncountable list of
digits,
> wouldn't that be just what the folks who try to claim 1.0 != 0.999999...
> want to do -- put stuff after the last digit in 0.99999...?
A sequence of digits is not the same thing as a real number. There is no
problem in defining what it means to have an uncountable sequence of
digits. The problem is in trying to interpret such a sequence as a real
number. It's not possible for all such sequences to represent distinct
real numbers, because there are simply too many of them.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
Dave Seaman Wrote:
>A sequence of digits is not the same thing as a real number. There is no
>problem in defining what it means to have an uncountable sequence of
>digits. The problem is in trying to interpret such a sequence as a real
>number.
I was thinking this afternoon that another problem would be how to compute
such
a sequence, given some real number one wanted to find a representation for.
It
could take some time to get a good approximation.
As for an example of a representation of the reals that uses an uncountable
number of digits, if there are an uncountable number of ways of
representing
the reals with a countable number of digits then perhaps one could find a
nice
uncountable representation by selecting a few digits from each of the
countable
representations.
Rich Burge
> It's not possible for all such sequences to represent distinct
>real numbers, because there are simply too many of them.
--
>Dave Seaman
>Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
>
====
> Dave Seaman Wrote:
>>A sequence of digits is not the same thing as a real number. There is no
>>problem in defining what it means to have an uncountable sequence of
>>digits. The problem is in trying to interpret such a sequence as a real
>>number.
> I was thinking this afternoon that another problem would be how to compute
such
> a sequence, given some real number one wanted to find a representation
for. It
> could take some time to get a good approximation.
The usual decimal representation expresses each real number in [0,1) in
the form of a countable sum
x = sum_k d_k * 10^(-k)
where the sum is over integers k > 0 and d_k is the digit in position k.
We could consider generalizing this to an uncountable sum of the form
x = sum_a d_a * v(a)
where the index a is allowed to range over some uncountable set, d_a is
the digit in position a, and v(a) is the value of place a in the
representation, taking the place of 10^(-a) (which doesn't make sense as
a real number if a is infinite).
If v(a) = 0 for a particular a, then the value of d_a doesn't matter and
can be chosen arbitarily. Therefore, we may as well concentrate on the
set
A = { a : v(a) != 0 }.
However, if the expression for x is to converge, this set A must be
countable. Otherwise, the set of a's such that abs(v(a)) > 1/n would
have to be infinite for some n.
As I see it, that leaves just two possibilities:
(1) Ignore the a's except for the countable set A, since
the rest have no effect on the representation, or
(2) Abandon the idea of each place in the string having a
fixed value v(a) associated with it. Each number woould
have to have its own individual set of v(a)'s. That makes
it pretty difficult to describe a useful mapping between
uncountable digit strings and real numbers.
> As for an example of a representation of the reals that uses an
uncountable
> number of digits, if there are an uncountable number of ways of
representing
> the reals with a countable number of digits then perhaps one could find a
nice
> uncountable representation by selecting a few digits from each of the
countable
> representations.
There is not an uncountable number of ways of representing any real with
a countable number of digits. In fact, the maximum number of ways is 2.
For example, 1.000... and 0.999... represent the same number.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
====
> Is this a new fundamental discovery?
In my work with computers,
I believe that I have come across a new fundamental function
that relates basic math constants of nature and the lower integers.
----
Most remarkable is that the function is of the form
Y = mX + b or a straight line
that resulted from curve fitting eight points below,
where the point has significant integers or
important constants as its components.
---
Even more interesting is the slope 'm' is an important
constant from the area of Energy; the Ryberg Constant
This constant is related to radiated energy observations
and Planck's constant..
---------- points of Proof
Here are the eight points; try plotting therm yourself.
# X Y (to a resolution of four significant digits )
1 pi/2 pi/2
2 P^2 e
3 3 pi
4 4 P^3
5 5 16/3
6 8 69/8
7 -4/4 -5/4
8 -9 -10
Readers are reminded P is the golden mean,
and that P^2 = P + 1 and 4 = P^3 - P^-3.
-------------------------------------------
I get the following curve from these 8 points:
Y = Rbc * X - 0.153 where Rbc is the Ryberg constant of 1.0973
It is as if, value 'X' has weight 'Y'
I imagine it as a plane with a Ryberg slope;
which I call it Jod's Plane.
------------
I have never found any reference to the above
so I believe I can therefore claim to be its discoverer!
What do you think?
--
====
> Is this a new fundamental discovery?
> In my work with computers,
> I believe that I have come across a new fundamental function
> that relates basic math constants of nature and the lower integers.
----
Most remarkable is that the function is of the form
> Y = mX + b or a straight line
that resulted from curve fitting eight points below,
where the point has significant integers or
> important constants as its components.
---
Even more interesting is the slope 'm' is an important
> constant from the area of Energy; the Ryberg Constant
This constant is related to radiated energy observations
> and Planck's constant..
---------- points of Proof
Here are the eight points; try plotting therm yourself.
# X Y (to a resolution of four significant digits )
1 pi/2 pi/2
2 P^2 e
3 3 pi
4 4 P^3
5 5 16/3
6 8 69/8
7 -4/4 -5/4
8 -9 -10
Readers are reminded P is the golden mean,
> and that P^2 = P + 1 and 4 = P^3 - P^-3.
-------------------------------------------
I get the following curve from these 8 points:
Y = Rbc * X - 0.153 where Rbc is the Ryberg constant of 1.0973
It is as if, value 'X' has weight 'Y'
I imagine it as a plane with a Ryberg slope;
> which I call it Jod's Plane.
------------
I have never found any reference to the above
> so I believe I can therefore claim to be its discoverer!
What do you think?
You discovered the equation of a straight line? I don't get it.
====
Rydberg Constant
http://scienceworld.wolfram.com/physics/RydbergConstant.html
====
>I believe that I have come across a new fundamental function
>that relates basic math constants of nature and the lower integers.
>Even more interesting is the slope 'm' is an important
>constant from the area of Energy; the Ryberg Constant
> Y = Rbc * X - 0.153 where Rbc is the Ryberg constant of 1.0973
The Rydberg constant is not a pure number. It has units of 1/length:
approximately 1.09737 * 10^7 m^(-1). If some other unit of length was
used, the number would be different. The fact that you get a number
approximately 1.0973 (without any units) is therefore just a coincidence.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
====
>I believe that I have come across a new fundamental function
>that relates basic math constants of nature and the lower integers.
>Even more interesting is the slope 'm' is an important
>constant from the area of Energy; the Ryberg Constant
> Y = Rbc * X - 0.153 where Rbc is the Ryberg constant of 1.0973
The Rydberg constant is not a pure number. It has units of 1/length:
> approximately 1.09737 * 10^7 m^(-1). If some other unit of length was
> used, the number would be different. The fact that you get a number
> approximately 1.0973 (without any units) is therefore just a coincidence.
Well, either that, or proof positive that the metric system is
the right way to go.
--
====
>I believe that I have come across a new fundamental function
>that relates basic math constants of nature and the lower integers.
>Even more interesting is the slope 'm' is an important
>constant from the area of Energy; the Ryberg Constant
> Y = Rbc * X - 0.153 where Rbc is the Ryberg constant of 1.0973
The Rydberg constant is not a pure number. It has units of 1/length:
> approximately 1.09737 * 10^7 m^(-1). If some other unit of length was
> used, the number would be different. The fact that you get a number
> approximately 1.0973 (without any units) is therefore just a
coincidence.
Well, either that, or proof positive that the metric system is
> the right way to go.
I wonder how that formula would work
in furlong, fortnight, stone units?
--
Tom Potter http://tompotter.us
====
Consider G_tt = H(t)G_xx
Is this a standard, solvable PDE with a known solution?
If so, what is G(x,t) in terms of the known function H(t)?
G=G(x,t)
G_tt is the second partial derivative of G with respect to t.
G_xx is the second partial derivative of G with respect to x.
Eugene Shubert
http://www.everythingimportant.org/relativity
====
>Consider G_tt = H(t)G_xx
Is this a standard, solvable PDE with a known solution?
>If so, what is G(x,t) in terms of the known function H(t)?
G=G(x,t)
G_tt is the second partial derivative of G with respect to t.
>G_xx is the second partial derivative of G with respect to x.
>Eugene Shubert
>
I do believe you should try plugging in a solution of the form
G(x,t) = A(x)*B(t)
Because then you can get everything with t dependence on
one side, and everything with x dependence on the other
side... thus both sides must be equal to a constant.
And you've got yourself two nice little ODEs instead of
one mean little PDE.
====
> Consider G_tt = H(t)G_xx
We could call it a wave equation where the speed of light varies with
time.
> I do believe you should try plugging in a solution of the form
G(x,t) = A(x)*B(t)
A good idea ... BUT ... if you do, please keep in mind that not every
solution has that form.
====
> >Your reasons for this assertion are irrelevent and fail to explain the
> >mechanics behind point mass collisions.
>
> Point masses don't exist except as idealizations. Along with idealizing
> a point mass, you have to take what goes along with the fiction if you
> expect to get something other than non-sense. You've already proved that
> when you got complete non-sense by ignoring the physics.
Idiot. If you have a continuous space and a mass then that mass is the
integral sum of its containing point masses.
[...]
> >> That's your mistake. If the collision occurs, where did the energy
> >> go? It went into additional mass at the point of impact. or else some
> >> or all of it went into heat. Whatever energy is not dissipated as
heat,
> >> is the kinetic energy given to the masses as they separate.
> >The issue is not with dynamics but with mechanics - specifically,
> >Newtons THIRD law claiming that paired equal and opposite forces occur
> >at the point of collision.
>
> You're too busy firing off a load of crap and not busy enough reading
> so that the impulse is removed from the equations. Then the forces are
> always balanced in any interval dt as t-> 0.
Idiot. The dynamics of a collision are irrelevant if the collision is
logically impossible.
> >> Wrong. The kinetic energy is transformed into heat if the collision
> >> is perfectly inelastic. If you use forces, then you have to use
impulses.
> >This is an example of the irrelevancy your posts tend to contain.
>
> You not only don't know as much as you think you do, you don't know
> as much as the average college sophomore.
Idiot. This reminds me of the time you called my equation for
computing a real with the natural numbers as the mantissa wrong. How
many more lessons do I need to teach you before you learn?
> [...]
> >Mass density is not infinite, but continuous over space. You have no
> >evidence suggesting otherwise unless you invoke quantum mechanics
> >which is ultimately derived from Newtons work and inapplicable in this
> >very basic scenario.
>
> If you ever took a physics course, I suggest you get a refund and
> consider legal action against your advisor for negligence in failing
> to steer you into a field that requires less in the way of analytical
> skills. As a physicist you simply wouldn't be employed. As an engineer
> you'd be dangerous.
Idiot.
====
> John Schoenfeld: crackpot
> >> >Your reasons for this assertion are irrelevent and fail to explain
the
> >> >mechanics behind point mass collisions.
> >>
> >> Point masses don't exist except as idealizations. Along with
idealizing
> >> a point mass, you have to take what goes along with the fiction if
you
> >> expect to get something other than non-sense. You've already proved
that
> >> when you got complete non-sense by ignoring the physics.
> >Idiot. If you have a continuous space and a mass then that mass is the
> >integral sum of its containing point masses.
>
> No, it isn't. It's the integral of a mass density. Are stoned or just
> stupid?
Your mind, riddled with fallacies accepts the notion of continuous
mass density yet dismisses its consequence of point mass existence.
They are not mutually exclusive, crackpot.
JS
====
> No, it isn't. It's the integral of a mass density. Are stoned or
just
> stupid?
Your mind, riddled with fallacies accepts the notion of continuous
> mass density yet dismisses its consequence of point mass existence.
> They are not mutually exclusive, crackpot.
They are if density is finite everywhere.
- Randy
====
[snip detailed exposition of refutation of James' alleged proof]
> ...(if you don't agree), then show me (hey, don't worry about me, show
> your public!)
> where my error is. I've done all the multiplication; you can verify
> or refute all this very easily, given the ability to multiply
> or expand polynomial expressions.
Show how highly you value algebra: Do some.
It's easy: ordinary polynomials, ordinary polynomial multiplication,
> nothing up my sleeves, no salesman will call. You don't even need
> to solve any equations. Just multiply, combine terms, show me wrong!
Nice challenge, Dale. But James has shown repeatedly that he is unable to
muliply polynomials, even a small number of binomials, without screwing up
the exponents and/or signs. He has an amazing record of posts with
inexcusable errors in simple arithmetic.
--
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
====
> I created several threads today where I went ahead and used my ability
> to put in some actual numbers for symbols in some key expressions, and
> looking over the newsgroup, I see a lot of replies.
My suggestion is that you consider those replies and notice how few
> talked about the math presented by actually referencing it, and how
> many just asserted things--after deleting it out.
It's almost funny until you realize why there was so much frenetic
> energy, and why these people are so desperate to hide the math that
> they continually delete it out:
The mathematics is rather simple, the algebra is basic, but the
> conclusion is dramatic.
So let's say you were facing people who needed to keep people from
> looking closely, and every time you tried to simplify and show
> details, they'd jump in and hide details and make things more
> convoluted?
Well you might do what I've done today and stretch them out.
What I'm doing is not very complicated as I'm using the extra symbols
> to factor a polynomial into non-polynomial factors.
Now that's a fascinating idea for factoring polynomials that I guess
> is new to the math world.
And yes, bad apples can take advantage when there's something new.
What is a surprise though is that mathematical society can be taken in
> by people like Arturo Magidin or Nora Baron, when they're posting so
> desperately, so quickly when stretched out, that they can barely get
> to the math, or say things that are clearly false.
But what if they never really believed in mathematics?
What if for them it is a fashion show?
Remember mathematics can be about appearance for some people. They
> might have gotten a lot of mileage out of being able to put down math
> that looked good.
That is, you may have people who are cons in your midst who are used
> to playing a game upon other mathematicians, and now they're
> trapped--stretched out--by a lot of threads and math where I did the
> sudden move of putting in numbers where once there were symbols.
It seems to me that in mathematics there is a certain respect that is
> to be given to ideas, logic, and mathematical truth. Sure being a
> rather large society it's not surprising that corrupt people can slip
> in, and maybe get some stature, or get through a Ph.d program.
But in society when the corrupt people reveal themselves clearly by
> finally pushing beyond obvious limits--like denying basic algebra--it
> becomes time to clean house. Look over those threads, look at their
> desperation, and their contempt for algebra and your algebra
> knowledge, then please ask yourselves how it cannot now be that time.
> James Harris
I looked a little more closely at Advanced Polynomial Factorization.
It turns out that, in addition to basic conceptual problems, it also
has careless mistakes in algebra.
Start with
P(m) = f^2 *((m^3*f^4 - 3*m^2*f^4 + 3*m)*x^3
- 3*(-1 + m*f^2)*x*u^2 + u^3*f),
as you give early in Section 2 of APF. You note that P(m) has
a factor of f^2.
Nothing wrong with this particular formula. However, on the last
page, when you substitute in m = 1, f = sqrt(5), and u = 1, you do
NOT obtain
65*x^3 - 12*x + 1.
Instead you get
65*x^3 - 60*x + 5*sqrt(5).
Now if this is factored in the form
(a1*x + 1)*(a2*x + 1)*(a3*x + 1),
it obviously doesn't work: 1*1*1 is not equal to 5*sqrt(5).
Back on the second page, in a parenthetical note at
the bottom, you say: Note: the a's are roots of a monic
polynomial with algebraic integer coefficients so they
are algebraic integers. Not true at all. Note that
the polynomial P(m), as given above, if considered as a
polynomial in x, does not have constant term equal to 1
or -1. The constant term with respect to x is u^3*f^3.
Thus the polynomial of which the a's are roots is not monic.
I am not sure what you actually intended.
This is carelessly written, even independent
of the unsoundness of the ideas. Parts of your posts
from yesterday on this topic also contained similar
algebraic errors.
I think it was still your intention in APF however to
show that if
65*x^3 - 12*x + 1
were factored in the form
(a1*x + 1)*(a2*x + 1)*(a3*x + 1),
where a1, a2, and a3 are algebraic integers, then one
of a1, a2, or a3 is coprime to 5. No matter
how you cut it, that is still false, as shown by
the proofs given in other threads by me and W. Dale
Hall. Yes, you have started other several new threads
today, but you have explicitly avoided a direct response to
either me or Dale Hall on this.
Even if you are unable to perform the computations that
Dale proposes and are incapable of understanding the proof that
I presented, you cannot ignore the simple algebraic errors
in APF. Something there has to change.
Nora B.