mm-197
Heres another way to do the problem.
(Looking only at the top nappe). The top is the circle x^2+y^2=
a^2 at z= a and has circumference 2pi a. The distance from the
vertex
to the edge has length sqrt(2)a. Cut a straight line from the
edge to
the vertex and ¤atten it out. It 'ts into a circle of
radius
sqrt(2)a. The cut portion opens up: the cone does not 'll the
entire
circle because that circle has circumference 2pi(sqrt(2)a)
while the
circumference of the (¤attened) cone is only 2pi a. The
proportion
of the circle that it 'lls is (2pi a)/(2pi sqrt(2))=
1/sqrt(2). The
entire circle of radius sqrt(2)a has area pi(sqrt(2)a)^2= 2 pi
a^2 so
the area of the cone is 1/sqrt(2) times that:
2/sqrt(2) pi a^2= sqrt(2) pi a^2.
The two nappes together have area 2sqrt(2) pi a^2.
===
Subject: Re: vector calculus clue
pretty
simple
when you think about it.
The lecturer is a tricky bloke, this question was amid loads
of true vector
integrals - where this doesnOt need it.
Kev
===
Subject: Math
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1CLY4K14783;
The odds against the yankees winning the pennant are 7:2. What
is the
probability that the Yankees will win the pennant?
===
Subject: Odds and probability (Re: Math)
Place try to use more speci'c subject lines -- it will
increase the
likelihood of your getting a useful response.
The odds against the yankees winning the pennant are 7:2.
What is the
>probability that the Yankees will win the pennant?
Assuming you copied it correctly--
This says there are 7 chances yes for every 2 chances no. How
many
chances are there in total? (Hint: What is 7+2?)
Now the probability of a win is 7 out of that total, and the
probability of a loss is 2 out of that total.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
An expense does not have to be required to be considered
necessary. -- IRS Form 1040 line 23 instructions
===
Subject: Re: Math
2/9
 The odds against the yankees winning the pennant are 7:2.
What is the
> probability that the Yankees will win the pennant?
>
===
Subject: Algebraic integers, easy way to show problem
On sci.math a poster called Rick Decker came up with a
quadratic which
he thought showed that I was wrong about a problem in core
with the
current understanding of algebraic integers, but it turns out
that it
being a quadratic means that thereOs an easy way for me to
show you
that thereOs a problem.
DeckerOs quadratic example (reference information at bottom)
is
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his aOs are roots of
a^2 - (x - 1)a + 7(x^2 + x).
Letting x=2, you have
a_1(2)^2 - a_1(2) + 42 = 0, which gives
a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
Now it turns out that it amazingly enough does not have a
non-unit
factor in common with 7 in the ring of algebraic integers
which is
simple enough to see by using another quadratic.
Now consider the quadratic
y^2 - by - 7 = 0, which has as one of its roots
(b + sqrt(b^2 + 28))/2.
Note that root is an algebraic integer factor of 7 for all
algebraic
integers y, and b.
So now I can put in any algebraic integer b, to shift through
all the
possible factors in common with 7 in the ring of algebraic
integers.
Now then consider
(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
which is
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z, and solving for z,
gives
z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
where the question is, can z be an algebraic integer?
If I try to get a polynomial with integer coef'cients, as I
eliminate
the square roots, IOll get an extra solution, so the
polynomial
without the square roots will be a quartic, and it is
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
Now, as you loop through algebraic integer values for b, for
irrational ones, the de'ning polynomial with integer
coef'cients
will be given by the degree of the monic polynomial with
integer
coef'cients for which b is a root.
For instance, if b=sqrt(c), where c is an integer, youOd
have
a degree
8 polynomial.
The problem is that polynomial will always be non-monic and
irreducible over Q for a simple reason, look back at
z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
and notice what happens if I let b=0, as then I get
z = (1 + sqrt(-167))/sqrt(28)
which is NOT an algebraic integer!
Therefore, no matter what polynomial with integer coef'cients
you
come up with for z, it will have to be non-monic, as one of its
solutions will always be
z = (1 + sqrt(-167))/sqrt(28).
So it turns out that my discovery of a problem with the ring of
algebraic integers could have been made at any time by someone
who
decided to try and 'nd the result of dividing off some factor
that
youOd have thought should be there.
WhatOs happened is that I found my result through a
different
logical
argument, which seems dif'cult for people to understand, and
some
mathematicians decided not to inform people, while others
decided to
spend effort on Usenet arguing against it.
For those wondering what the 4 solutions for the quartic are,
they are
z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))
z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))
z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))
as working to eliminate square roots from any one of them will
give
you
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
Mathematicians have a responsibility to act here as
unfortunately
there are various characters who have put up webpages
attacking me and
my work.
Failing to move to correct the situation would be treason to
the
discipline of mathematics.
James Harris
Decker Quadratic Source Information
---------------------
Recently Rick Decker, a professor at Hamilton College,
apparently
trying to refute my research came up with a quadratic example,
which I
like because itOs a quadratic, and easier to manipulate than
the
cubics IOve used before.
If you wish to see his original post here are some headers
which also
show that he posts from Hamilton College:
===
Subject: Re: Mathematical consistency, courage
Decker put forward the quadratic
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his aOs are roots of
a^2 - (x - 1)a + 7(x^2 + x).
===
Subject: Re: Algebraic integers, easy way to show problem
> On sci.math a poster called Rick Decker came up with a
quadratic which
> he thought showed that I was wrong about a problem in core
with the
> current understanding of algebraic integers, but it turns
out that it
> being a quadratic means that thereOs an easy way for me to
show you
> that thereOs a problem.
 DeckerOs quadratic example (reference information at
bottom)
is
 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
 where his aOs are roots of
 a^2 - (x - 1)a + 7(x^2 + x).
 Letting x=2, you have
 a_1(2)^2 - a_1(2) + 42 = 0, which gives
 a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
 Now it turns out that it amazingly enough does not have a
non-unit
> factor in common with 7 in the ring of algebraic integers
which is
> simple enough to see by using another quadratic.
 Now consider the quadratic
 y^2 - by - 7 = 0, which has as one of its roots
 (b + sqrt(b^2 + 28))/2.
 Note that root is an algebraic integer factor of 7 for all
algebraic
> integers y, and b.
 So now I can put in any algebraic integer b, to shift
through all the
> possible factors in common with 7 in the ring of algebraic
integers.
 Now then consider
 (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
 which is
 (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z, and solving for z,
gives
 z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
 where the question is, can z be an algebraic integer?
 If I try to get a polynomial with integer coef'cients, as I
eliminate
> the square roots, IOll get an extra solution, so the
polynomial
> without the square roots will be a quartic, and it is
 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
 Now, as you loop through algebraic integer values for b, for
> irrational ones, the de'ning polynomial with integer
coef'cients
> will be given by the degree of the monic polynomial with
integer
> coef'cients for which b is a root.
 For instance, if b=sqrt(c), where c is an integer, youOd
have a degree
> 8 polynomial.
 The problem is that polynomial will always be non-monic and
> irreducible over Q for a simple reason, look back at
 z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
 and notice what happens if I let b=0, as then I get
 z = (1 + sqrt(-167))/sqrt(28)
 which is NOT an algebraic integer!
 Therefore, no matter what polynomial with integer
coef'cients you
> come up with for z, it will have to be non-monic, as one of
its
> solutions will always be
 z = (1 + sqrt(-167))/sqrt(28).
 So it turns out that my discovery of a problem with the ring
of
> algebraic integers could have been made at any time by
someone who
> decided to try and 'nd the result of dividing off some
factor that
> youOd have thought should be there.
 WhatOs happened is that I found my result through a
different logical
> argument, which seems dif'cult for people to understand,
and
some
> mathematicians decided not to inform people, while others
decided to
> spend effort on Usenet arguing against it.
 For those wondering what the 4 solutions for the quartic
are, they are
 z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
 z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))
 z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))
 z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))
 as working to eliminate square roots from any one of them
will give
> you
 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
 Mathematicians have a responsibility to act here as
unfortunately
> there are various characters who have put up webpages
attacking me and
> my work.
 Failing to move to correct the situation would be treason to
the
> discipline of mathematics.
> James Harris
> Decker Quadratic Source Information
> ---------------------
> Recently Rick Decker, a professor at Hamilton College,
apparently
> trying to refute my research came up with a quadratic
example, which I
> like because itOs a quadratic, and easier to manipulate
than
the
> cubics IOve used before.
 If you wish to see his original post here are some headers
which also
> show that he posts from Hamilton College:
>
===
> Subject: Re: Mathematical consistency, courage
 Decker put forward the quadratic
 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
 where his aOs are roots of
 a^2 - (x - 1)a + 7(x^2 + x).
James, Shut up and learn some math. If you read your replies
more
carefully,
maybe youOd learn something. You are unwilling to learn.
There
is no core
error in mathematics and youOd realize that if you knew
something about
math. Seems like math is your life, not much of a life if you
ask me.
David Moran
===
Subject: Re: Algebraic integers, easy way to show problem
> Mathematicians have a responsibility to act here as
unfortunately
> there are various characters who have put up webpages
attacking me and
> my work.
 Failing to move to correct the situation would be treason to
the
> discipline of mathematics.
> James Harris
That raises the question whether the mathematicians should
correct the
situation by having JSH hanged or having him shot for his
treason to the
discipline of mathematics.
Since JSH seems to doubt mathematical marksmanship, perhaps he
would
prefer being hanged.
I am sure that the mathematical world can do that in an
workmanlike
manner.
Pray, Mr. Harris, let me persuade you to be hanged!
(with apologies to GBS)
===
Subject: Re: Can you help me solve this riddle?
>I build up castles, I tear down mountains, I make some men
blind, I
>help others to see.
Did someone change this group to alt.riddles while I wasnOt
looking?
We seem to have more riddles than math these days.
To the OP -- try rec.puzzles .
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
An expense does not have to be required to be considered
necessary. -- IRS Form 1040 line 23 instructions
===
Subject: Re: Can you help me solve this riddle?
Spoiler....
Spoiler....
Spoiler....
Spoiler....
Spoiler....
Spoiler....
SAND
www.brainbashers.com
===
Subject: Re: Resolution to Decker Quadratic Issue
>
<deleted>
HereOs an example of what I *hate* about some posters here
and
how
they get away with bull. Consider this claim of spurious
roots.
> > For example, if b = -6 then the quartic is reducible, as
>  > 7z^4 - 6z^3 + 299z^2 + 36z + 252 = (z^2 - z + 42)(7z^2 + z
+ 6),
>  > so its four zeros are (1 + sqrt(-167))/2, (1 -
sqrt(-167))/2,
> > (-1 + sqrt(-167))/14 and (-1 - sqrt(-167))/14.
>  > The 'rst of these is the value of z you started with; it
is an
> > algebraic integer. The other three zeros are spurious.
 Nope. On what basis do you claim theyOre spurious?
 Because you started by assuming that
> (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
> so, unless you are claiming equal roots, only the 'rst is
relevant. The
> others are artifacts of the process by which the square
roots were
> eliminated.
 This is a situation well known to those familiar with
algebra but often
> overlooked by putzers.
>
Now the reality can be demonstrated in a simple way:
P(x) = x^2 + bx + 4, so
x = (-b+/-sqrt(b^2 - 16))/2, so a root is
s = (-b + sqrt(b^2 - 16))/2, so you can solve for s.
2s = - b + sqrt(b^2 -16), so 2s + b = sqrt(b^2 - 16), s
4s^2 + 4bs + b^2 = b^2 - 16, so
4s^2 + 4bs + 16 = 0, so
s^2 + bs + 4 = 0.
ItOs not magic; itOs algebra.
There are no spurious roots here!
Jsut because you say solution
s = (-b + sqrt(b^2 - 16))/2
it does not mean that if you solve for s that the quadratic
you have
has one correct root and one spurious one!
It goes back to that ambiguity of the square root operator
thingy.
Math is hard. Insults and b.s. on a newsgroup is easy.
 The four roots zOs are as follows:
 z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
 z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))
 z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))
 z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))
 so in fact the other roots are NOT spurious!!!
 You can check with your software package to verify.
Mathematics is not a democracy. Just because some of you may
have
your panties in a bunch, and want to cry because you thought
these
other posters like Arturo Magidin were teaching you important
and
correct mathematics and they were wrong, the math DOES NOT
CHANGE!!!
You can be as childish and intellectually immature as you want
to be,
and the MATH WILL NOT CHANGE!!!!!!
People are so weird. Math is not a democracy. Why do so many
of you
clearly think it is?
James Harris
===
Subject: Re: Resolution to Decker Quadratic Issue
 It goes back to that ambiguity of the square root operator
thingy.
There is no ambiguity of the square root operator thingy.
> Math is hard. Insults and b.s. on a newsgroup is easy.
Yes, we can see itOs hard for you, Barbie.
--
Wayne Brown (HPCC #1104) | When your tailOs in a crack, you
improvise
fwbrown@bellsouth.net | if youOre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: Resolution to Decker Quadratic Issue
 <deleted
> HereOs an example of what I *hate* about some posters here
and how
> they get away with bull. Consider this claim of spurious
roots.
Rather that spurious a more common adjective is extraneous.
When one starts with an equation containing radicals, such as
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z, and operates to
eliminate
those radicals, as JSH has done, the 'nal equation is liable
to contain
solutions which are not valid in the original equation. These
new
solutions are standardly called extraneous solutions or
spurious
solutions meaning that they do not satisfy the _original_
equation.
JSH has claimed 4 solutions for z to a certain polynomial
equation in
z. I claim that only one of them can satisfy the original
equation
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z, and that the others,
are
_extraneous_ or _spurious_ or whatever adjective JSH will
accept as
describing the situation of not satisfying the original
equation, (1 +
sqrt(-167)) = (b + sqrt(b^2 + 28))z.
> > For example, if b = -6 then the quartic is reducible, as
>  > 7z^4 - 6z^3 + 299z^2 + 36z + 252 = (z^2 - z + 42)(7z^2 + z
+ 6),
>  > so its four zeros are (1 + sqrt(-167))/2, (1 -
sqrt(-167))/2,
> > (-1 + sqrt(-167))/14 and (-1 - sqrt(-167))/14.
>  > The 'rst of these is the value of z you started with; it
is an
> > algebraic integer. The other three zeros are spurious.
>  > Nope. On what basis do you claim theyOre spurious?
 Because you started by assuming that
> (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
> so, unless you are claiming equal roots, only the 'rst is
relevant. The
> others are artifacts of the process by which the square
roots were
> eliminated.
 This is a situation well known to those familiar with
algebra but often
> overlooked by putzers.
>
===
Subject: Re: Resolution to Decker Quadratic Issue
> Mathematics is not a democracy. Just because some of you may
have
> your panties in a bunch, and want to cry because you thought
these
> other posters like Arturo Magidin were teaching you
important and
> correct mathematics and they were wrong, the math DOES NOT
CHANGE!!!
Right! No matter how much you want it to change, it wonOt.
 You can be as childish and intellectually immature as you
want to be,
> and the MATH WILL NOT CHANGE!!!!!!
That is the nice thing about it, that nothing you can do can
corrupt it.
 People are so weird. Math is not a democracy. Why do so many
of you
> clearly think it is?
i donOt think so. Mathematics is an aristocracy, in the
original meaning
of the word, and JSH is incurably plebian, at least in things
mathematical.
===
Subject: Re: Resolution to Decker Quadratic Issue
> Math is hard. Insults and b.s. on a newsgroup is easy.
So is posting crap to the newsgroup -- as you admit you have
repeatedly
done. (Citations available on
request.)
> People are so weird. Math is not a democracy. Why do so many
of you
> clearly think it is?
You are so weird. There is *no* evidence that any poster
believes that math
is a democracy (post a reference
to serve as a counterexample, if you wish). Only you, however,
think it is a
dictatorship with you at the
helm.
> James You canOt get rid of me with reason. Harris
Your manipulation of quadratics is ¤awed, as is your
interpretation of the
Osquare root thingyO. Better get
back to the playpen until you grow up.
--
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
===
Subject: Re: Resolution to Decker Quadratic Issue
>Turns out thereOs a rather direct approach to showing a
problem with
>the old concepts about the ring of algebraic integers.
Decker, a sci.math poster, posted (reference information at
bottom)
>the quadratic
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his aOs are roots of
a^2 - (x - 1)a + 7(x^2 + x).
Letting x=2, you have
a_1(2)^2 - a_1(2) + 42 = 0, which gives
a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
Now consider the quadratic
y^2 - by - 7 = 0, which has as one of its roots
>
You are saying this is the general form of a monic
quadratic with 7 as the constant term. True enough,
though one might quibble that you need to consider
where the constant term might be +7 also.
But this is *not* the general form of monic polynomials
which have either 7 or 7^k as the constant term.
What you show below is suf'cient only to show that
if (1 + sqrt(-167))/2 has a factor of 7, then that
factor is not the root of a monic quadratic with
constant term -7.
But that in no way eliminates all of the in'nitely
many other possibilities.
>(b + sqrt(b^2 + 28))/2.
Note that root is an algebraic integer factor of 7 for all
algebraic
>integers y, and b.
Now consider
(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
which is
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
and working to eliminate square root terms gives
28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0
and working still further I get
196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0
and I can divide both sides by 28 to 'nally get
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
Importantly, for any integer b, such that it is irreducible
over Q,
>*none* of the solutions for z can be an algebraic integer!
First question may be, after starting out with
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
why do I have a quartic and what are the other roots?
Well if I used
(1 - sqrt(-167)) = (b + sqrt(b^2 + 28))z
IOd get the same result, as eliminating the square roots at
each
>points creates *two* possible solutions that will work. I
handled two
>square roots, so I have four solutions, which are
z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))
z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))
z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))
Now then, imagine that there exists some algebraic integer b
for which
>it is reducible over Q, then the root will be a fraction with
a 7 in
>the denominator.
Let c/7 be such a root, where c is then an integer, then
IOd
have
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7, or
(1 - sqrt(-167)) = (b + sqrt(b^2 + 28))c/7, or
(1 + sqrt(-167)) = (b - sqrt(b^2 + 28))c/7, or
(1 - sqrt(-167)) = (b - sqrt(b^2 + 28))c/7.
but for any of those possibilities, the result on the right
side has
>to be coprime to 7,
since (b + sqrt(b^2 + 28))/2 [(b - sqrt(b^2 + 28))/2] = -7.
If you go to bOs that are not integers, but are some
arbitrary
>algebraic integer, you would get a de'ning polynomial for z
of higher
>and higher degree, but itOd still be non-monic, and that
result would
>still follow.
The problem is that no matter how high you go, just like with
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0
>
No - as noted above, all you have proved is that
the common factor is not a root of a polynomial of
the form
b^2 - b*y - 7.
There are in'nitely many other higher-degree
polynomials whose roots are algebraic integer divisors
of 7. What you have done says nothing about them.
In particular of course, Keith Ramsay has already shown
that there is an 11th degree polynomial whose roots
't the requirement.
Again, you are thinking along the right lines. But
you are being too restrictive about where the factors
might come from. There is no reason to restrict to
quadratics or quartics.
Just to forestall your continuing to try this kind of
thing, consider the following.
Let r(x) = GCD(a_1(x), 7) and let s(x) = GCD(a_2(x), 7),
where a_1(x) and a_2(x) are your coef'cients in factoring
DeckerOs quadratic.
Then when x^2 + x <> 0 mod(7), it can be shown that:
1. r(x)*s(x) = 7
2. a_1(x)/r(x) and a_2(x)/s(x) are algebraic integers.
3. r(x) and s(x) are algebraic integers
4. (5 a_1(x)/r(x) + s(x))*(5 a_2(x)/s(x) + r(x)) is a
factorization of
25*x^2 + 30*x + 2.
5. Neither r(x) nor s(x) equals 7 or is a unit.
In particular, this applies for x = 1, 2, 3, 4, 5, 8, 9, ....
Now, the underlying theory does not immediately tell you
what the degree of the monic polynomials are of which r(x) and
s(x) are roots. There is a way to compute them but it is
dif'cult.
>your result would be a non-monic polynomial, so for it to
ever have a
>rational root, at least one rational root would have to be a
fraction
>with a 7 in the denominator, as remember 7 is prime.
Therefore, (1 + sqrt(-167)) has no non-unit factors in common
with 7
>in the ring of algebraic integers!!!
>
No, not proved at all.
You have already started I believe 3 new threads on this same
exact topic - hardly even waiting for a response - every one of
them wrong for exactly the same reason, and this *after* you
tried to brush off Dik WinterOs completely correct comment.
DonOt you ever think you might be a little too hasty in
starting
to brag about something you are going to have to retract
later???
Nora B.
>QED
>James Harris
===
Subject: Re: Resolution to Decker Quadratic Issue
Adjunct Assistant Professor at the University of Montana.
>>Turns out thereOs a rather direct approach to showing a
problem with
>>the old concepts about the ring of algebraic integers.
>>Decker, a sci.math poster, posted (reference information at
bottom)
>>the quadratic
>>(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
>>where his aOs are roots of
>>a^2 - (x - 1)a + 7(x^2 + x).
>>Letting x=2, you have
>>a_1(2)^2 - a_1(2) + 42 = 0, which gives
>>a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
>>Now consider the quadratic
>>y^2 - by - 7 = 0, which has as one of its roots
 You are saying this is the general form of a monic
>quadratic with 7 as the constant term. True enough,
>though one might quibble that you need to consider
>where the constant term might be +7 also.
 But this is *not* the general form of monic polynomials
>which have either 7 or 7^k as the constant term.
 What you show below is suf'cient only to show that
>if (1 + sqrt(-167))/2 has a factor of 7, then that
>factor is not the root of a monic quadratic with
>constant term -7.
Let f(x) = x^2 + ax + b be a quadratic polynomial with
algebraic
integer coef'cients, and let P be an algebraic integer that
divides
b.
Let r1 and r2 be the roots of f(x). Then P divides the product
r1*r2,
and therefore, there is a factorization of P in the form P =
u1*u2,
such that u1 and u2 are both algebraic integers, u1 divides
r1, and u2
divides r2 (in the ring of algebraic integers).
Now consider the polynomial g(y) = y^2 - (u1-u2)y - P. This
factors as
g(y)=(y-u1)(y+u2).
Therefore:
PROP. Let f(x) be a monic quadratic polynomial with algebraic
integer
coef'cients, and let P be an algebraic integer that divides
f(0). Let
r1 and r2 be the two roots of f(x). Then there is an algebraic
integer
b such that the roots of g(y) = y^2 - by - P are associates of
gcd(P,r1) and gcd(P,r2).
The real problem is that if you now write a big polynomial
whose
coef'cients are expressed in terms of b, then it would be
very
dif'cult to apply the theorem about roots of non-monic
irreducible
primitive polynomials: you would need to show that it is
irreducible
over its 'eld of de'nition, and that it is
primitive over the
ring
of integers of its 'eld of de'nition; or you
would need to
prove
that the coef'cients are actually integers and then show the
polynomial is primitive. Without knowing what b is, it seems
like it
would be very dif'cult to argue about it in general.
--
=
ItOs not denial. IOm just very selective
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
=
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Resolution to Decker Quadratic Issue
> Turns out thereOs another approach to prove a problem with
the old
> concepts about the ring of algebraic integers.
 Decker put forward the quadratic
 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
 where his aOs are roots of
 a^2 - (x - 1)a + 7(x^2 + x).
 Letting x=2, you have
 a_1(2)^2 - a_1(2) + 42 = 0, which gives
 a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
 Now consider the quadratic
 y^2 - by - 7 = 0, which has as one of its roots
 (b + sqrt(b^2 + 28))/2.
 Note that root is an algebraic integer factor of 7 for all
algebraic
> integers y, and b.
 Now consider
 (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
 which is
 (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
 and working to eliminate square root terms gives
 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0
 and working still further I get
 196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0
As noted by Jones, that should be
196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0
 and I can divide both sides by 4 to 'nally get
 49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0.
With the correction I can divide both sides by 28 to get
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
 Importantly, for any integer b, such that it is irreducible
over Q,
> *none* of the solutions for z can be an algebraic integer!
>
Here it might be easier if I explain how it works that I end
up with a
quartic, after starting out with
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
as, what are the other roots?
Well if I used
(1 - sqrt(-167)) = (b + sqrt(b^2 + 28))z
IOd get the same result, as eliminating the square roots at
each
points creates *two* possible solutions that will work. I
handled two
square roots, so I have four solutions, which are
z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))
z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))
z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))
which is how you can consider what happens with algebraic
integer b,
if you keep going from integers. like considering more
radicals.
For instance, if b = sqrt(c), where c is an integer, then you
will
have 4 more solutions ending up with a polynomial of degree 8.
For a ObO that is some arbitrary algebraic
integer you can go
up as
high as necessary as I noticed a reply from Dik Winter, where
he
claimed that Keith Ramsay had a solution for ObO
from a monic
polynomial with integer coef'cients of degree 11.
Now you know that claim must be false.
The problem is that the zOs can never be integers, which is
what would
be required for an algebraic integer solution, because
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0
is non-monic, and that 7 on the front wonOt just go away as
you go
higher. Since algebraic integers cannot be roots of non-monic
primitives irreducible over Q, that seals it.
So, in fact (1 + sqrt(-167))/2 does not have *any* non-unit
algebraic
integer factors in common with 7.
That refutes the claims of various posters with what should be
an easy
to understand counterexample.
Based on their responses you can see whether or not my
assessment that
various posters are *deliberately* trying to push false
mathematics to
your detriment is in fact true.
If I was wrong about them, they should quickly come forward
now.
James Harris
===
Subject: Re: Resolution to Decker Quadratic Issue
> Turns out thereOs another approach to prove a problem with
the old
> concepts about the ring of algebraic integers.
 Decker put forward the quadratic
 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
 where his aOs are roots of
 a^2 - (x - 1)a + 7(x^2 + x).
 Letting x=2, you have
 a_1(2)^2 - a_1(2) + 42 = 0, which gives
 a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
 Now consider the quadratic
 y^2 - by - 7 = 0, which has as one of its roots
 (b + sqrt(b^2 + 28))/2.
 Note that root is an algebraic integer factor of 7 for all
algebraic
> integers y, and b.
 Now consider
 (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
 which is
 (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
 and working to eliminate square root terms gives
 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0
 and working still further I get
 196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0
> As noted by Jones, that should be
 196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0
>
I still had a mistake as it should be
196 z^4 + 28b z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0.
James Harris
===
Subject: Re: Resolution to Decker Quadratic Issue
>> [...]
>> As noted by Jones, that should be
>> [...]
I still had a mistake as it should be
[...]
ItOs hard to understand why people donOt see
that youOre
right about all this...
>James Harris
************************
David C. Ullrich
===
Subject: Re: Resolution to Decker Quadratic Issue
...
> Here it might be easier if I explain how it works that I end
up with a
> quartic, after starting out with
 (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
 as, what are the other roots?
Yup, you are ending up with a quartic, which shows that you
are wrong.
In this case you should end up with a polynomial of degree 22.
> For a ObO that is some arbitrary algebraic
integer you can
go up as
> high as necessary as I noticed a reply from Dik Winter,
where he
> claimed that Keith Ramsay had a solution for
ObO from a monic
> polynomial with integer coef'cients of degree 11.
You did not pay attention James. What I gave was *not* a
ObO,
it was
an OyO.
What I gave was the factor in common between 7 and (1 +
sqrt(-167))/2,
which *is* the root of a monic 22-degree polynomial. Call it
OsO.
Now s^11 *is* the root of a monic quadratic polynomial, but s
itself
is not.
> The problem is that the zOs can never be integers, which
is
what would
> be required for an algebraic integer solution, because
[ I thought the zOs must be algebraic integers?]
Yes, that is true *when* you start with a quadratic. There is
*no*
reason to assume that you can start with a quadratic. To start
of again:
> a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
 Now consider the quadratic
 y^2 - by - 7 = 0, which has as one of its roots
 (b + sqrt(b^2 + 28))/2.
 Note that root is an algebraic integer factor of 7 for all
algebraic
> integers y, and b.
Are *all* algebraic integer factors of 7 of that form for some
b? And
the answer is, yes they are (although you have not shown
that). So
apparently you can start with a quadratic.
> Now consider
 (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
 which is
 (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
 and working to eliminate square root terms gives
 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0
 and working still further I get
 196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0
 As noted by Jones, that should be
 196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0
It should be:
196 z^4 + 28b z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0.
 With the correction I can divide both sides by 28 to get
 7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
And this becomes:
7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
And from this you conclude that z can not be an algebraic
integer.
But that is only valid *if* b is an integer, not if it is an
arbitrary algebraic integer, and also only if the polynomial is
irreducible! So the conclusion is not justi'ed.
Again, the cart before the horse. It is as follows:
1. De'nition: an algebraic integer is a root of a monic
polynomial
with integer coef'cients.
2. Theorem: a root of a non-monic, irreducible and primitive
polynomial with integer coef'cients can not be an algebraic
integer.
3. Theorem: a root of a monic polynomial with algebraic integer
coef'cients is an algebraic integer.
You are using a theorem of the form:
4. Theorem: a root of a non-monic polynomial with algebraic
integer
coef'cients can not be an algebraic integer.
I have never seen a proof of such a theorem.
Now set a = (44444 - 111.sqrt(-167))^(1/11) and aO its
complex
conjugate.
Set y = a, we get: b = a + aO.
Set z = ([(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2])^(1/11).
And see that it works.
This gets stranger and stranger. Explicit expression have been
shown
for a and b such that a.aO = 7^11 and a.b = [(1 +
sqrt(-167))/2]^11.
Veri'able and all. Neveretheless 7 and (1 + sqrt(-167))/2
have
no
factor in common. Pray, James, where does that factor in common
between 7^11 and [(1 + sqrt(-167))/2]^11 come from?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Resolution to Decker Quadratic Issue
>
... fatally ¤awed argument deleted ...
> So, in fact (1 + sqrt(-167))/2 does not have *any* non-unit
algebraic
> integer factors in common with 7.
>
No, your method is incorrect.
There is a non-unit algebraic integer that divides both
7 and (1 + sqrt(-167))/2
I can show this number to you, albeit indirectly. However, I
refuse to
do so until you ask nicely.
> That refutes the claims of various posters with what should
be an easy
> to understand counterexample.
 Based on their responses you can see whether or not my
assessment that
> various posters are *deliberately* trying to push false
mathematics to
> your detriment is in fact true.
 If I was wrong about them, they should quickly come forward
now.
>
Why should anyone respond to this nonsense? If and when I
produce the
requisite common factor, you will no doubt do one of the
following
things:
1. Claim that the following argument is circular:
There are non-unit algebraic integers p,q,r
that satisfy the following properties:
pq = 7
pr = (1 + sqrt(-167))/2
Therefore, p is a non-unit algebraic integer
that divides both 7 and (1 + sqrt(-167))/2
2. Continue to ignore direct, arithmetic evidence of
your folly, preferring to rely on a ¤awed argument.
3. Hey, I was wrong! But this hereOs an example....
 James Harris
Dale
===
Subject: Re: Resolution to Decker Quadratic Issue
>[...]
So, in fact (1 + sqrt(-167))/2 does not have *any* non-unit
algebraic
>integer factors in common with 7.
That refutes the claims of various posters with what should
be an easy
>to understand counterexample.
Based on their responses you can see whether or not my
assessment that
>various posters are *deliberately* trying to push false
mathematics to
>your detriment is in fact true.
How many times has it happened that youOve implied that the
people
discussing your work were lying, only to have it turn out that
even
you admitted they were right?
>If I was wrong about them, they should quickly come forward
now.
>James Harris
************************
David C. Ullrich
===
Subject: Re: Resolution to Decker Quadratic Issue
> So, in fact (1 + sqrt(-167))/2 does not have *any* non-unit
algebraic
> integer factors in common with 7.
 That refutes the claims of various posters with what should
be an easy
> to understand counterexample.
 Based on their responses you can see whether or not my
assessment that
> various posters are *deliberately* trying to push false
mathematics to
> your detriment is in fact true.
 If I was wrong about them, they should quickly come forward
now.
Too bad you wonOt as quickly acknowledge that you are wrong
about your
conclusion. You have failed, in
spite of your claim, to refute the arguments posed against
your position.
You are wrong! Dead wrong! Get a
life!
> James Proving me wrong will never stop me. Harris
--
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
===
Subject: integral
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1DDdVl22797;
integral(0-pi/2)(sqrt(4+5cos^2(t)))dt =?
pi/2
|
| SqRoot(4+5cos^2(t)) dt --???
|
0
pleasee help me how to solve this problem !!!
Hope answer!!
pleasee send me !
===
Subject: Re: integral
> integral(0-pi/2)(sqrt(4+5cos^2(t)))dt =?
 pi/2
> |
> | SqRoot(4+5cos^2(t)) dt --???
> |
> 0
> pleasee help me how to solve this problem !!!
ItOs a complete elliptic integral of the second kind.
De'nition: E(m) = integral(0-pi/2)(sqrt(1 - m sin^2(t)))dt
You should then be able to show, very easily, that the value
of your
particular integral is 3 E(5/9), which BTW is approximately
3.96636 .
David
===
Subject: Re: linear functional
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1DDdc623030;
>How about this: a_1 = |x1|, a_k=0 (k>1), x_k=0, k>1
>then for any scalars b_ :
> | b_1*a_1 + ... + b_n*a_n | <= 1000 || b_1*x_1 + ... +
b_n*x_n ||
>|F(x_1)| = a_1 = |x_1|
>but I really donOt see, how it would follow, that ||F|| =
1000
>?
I must confess this is pretty opaque to me also.
I take it that x_1 is a vector, a_1= ||x_1||, all other x_k,
k>1
are 0 vectors so a_k= ||a_k||= 0 for k> 1.
Then | b_1*a_1 + ... + b_n*a_n | = |b_1*a_1| which certainly is
<= 1000 ||b_1*x_1|= 1000 |b_1*a_1| (actually strictly less)
because 1<
1000!
As far as showing that ||F||= 1000 is concerned, I have no
idea since
you havenOt said what F is, only that ||F(x_1)||= a_1.
===
Subject: Help 'nd equation for pianistOs
possibilities of
sound?
IOm no math expert but IOd like a valid number
for the
possibilities of
sound from a piano. I started with this:
P = 10 x 88 x V x T
--
P = possibilities
10 = pianisitOs 10 'ngers
88 = number of keys on the piano
V = Velocity or volume (one could assign a number of 100
different
discernable volume levels)
T = tempo or time between notes (one could assign a number of
1000
different
discernable tempos)
--
I donOt know if my equation needs exponents though. And if
so,
where? One
can play 10 notes all at once or one at a time. And, by
holding the sustain
pedal a pianist can make all 88 keys ring at once.
The discussion that explains my thinking is below:
-----------
Just wanted to throw something your way. A discussion I had
after a recent
trip to see Gonzalo Rubalcaba, Cuban piano genius, I thought
might spark
your interest a bit. I was thinking about what it means to
create art
and
the term was troubling for me. It seemed to emphasize the ego.
It turns the
artist into creator. Rather, I found it better to view art
making as a
discovery process, much like astronomers and geologists to
some extent. Let
me start at the beginning of my thinking.
Driving home from an amazing and inspiring jazz piano trio
concert my
mother
the hauntingly gorgeous etched melodies carved out of the slow
moving space
of his ballads to the ferocious explosions of his rhythmic
up-tempo tunes.
I
wondered why his improvisations and compositions seemed so
fresh and new,
unlike any other jazz music IOd heard. It was even fresh
compared to the
last 30 times IOd seen him play.
I thought about all the choices he made that night. All the
possibilities
he
had before him on the 88 key piano. I saw it best as the
algebra equation:
P = 10 x 88 x V x T
Where P is the playing 'eld or all the possibilities of sound
from the
piano. 10 represents the artistOs 10 'ngers. 88
is the number
of keys on
the piano. V represents the velocity or volume at which the
keys are
struck.
And T is time or rather tempo.
Even if I were to assign a value of only 100 possible Velocity
levels and
say 1000 tempos (or time values between two notes), that still
makes our
equation 10 x 88 x 100 x 1000 = 88,000,000 unique
possibilities. Now my
local mathematics professor would say I need to alter my
equation to
include
exponents. Something like 88vt^10 (8.8 million to the power of
10 'ngers)
might be closer to what my equation should look like. Or maybe
10^88vt?
Regardless of my feeble attempts at algebra we know it would
take millions
of years and millions of piano players to play all the
possibilities of the
pianoOs sound. Ok, so now letOs calculate
possibility values
for the
acoustic bass and the drum set? .... Just kidding.
The equations I thought about reminded me of equations for 3-D
space. It
seems attractive to think about the piano as a 3 dimensional
space so vast
my mind can barely grasp itOs expansion. Hence, the pianist
becomes an
explorer on a journey through a God created world, 'rst
traveling through
the space others have trodden, and then 'nding new paths.
The pianistOs possibilities might better be portrayed as an
enormous solid
substance that stretches throughout the universe. It would be
suitable for
chiseling away at like a sculpture artist might work with
marble. Or during
a pianistOs jazz improvisation it might become soft clay
that
is more
easily
formed into shapes as spontaneously as the artist imagines
them. Always
discovering new shapes. Always 'nding new paths.
The focus is not on creating. Rather, 'nding and discovering.
When you
'nish a wood carving do you say This is a wood carving I
created? Or
do
you say This is a piece of a tree God created? Though it is
your work
indeed, the substance, the essence, the material itself is
GodOs.
The weight and pressure of creating something from nothing is
lifted. Now I
can just play with and explore throughout GodOs already
created material
until I 'nd things I like.
------------
Please email if you have any suggestions
hamm@i.am
-- Dennis
===
Subject: Re: Help 'nd equation for pianistOs
possibilities of
sound?
> IOm no math expert but IOd like a valid
number for the
possibilities of
> sound from a piano. I started with this:
 P = 10 x 88 x V x T
 --
> P = possibilities
 10 = pianisitOs 10 'ngers
 88 = number of keys on the piano
 V = Velocity or volume (one could assign a number of 100
different
> discernable volume levels)
 T = tempo or time between notes (one could assign a number
of 1000
different
> discernable tempos)
> --
 I donOt know if my equation needs exponents though. And if
so, where? One
> can play 10 notes all at once or one at a time. And, by
holding the
sustain
> pedal a pianist can make all 88 keys ring at once.
 The discussion that explains my thinking is below:
> -----------
> Just wanted to throw something your way. A discussion I had
after a
recent
> trip to see Gonzalo Rubalcaba, Cuban piano genius, I thought
might spark
> your interest a bit. I was thinking about what it means to
create art
and
> the term was troubling for me. It seemed to emphasize the
ego. It turns
the
> artist into creator. Rather, I found it better to view art
making as a
> discovery process, much like astronomers and geologists to
some extent.
Let
> me start at the beginning of my thinking.
 Driving home from an amazing and inspiring jazz piano trio
concert my
mother
> the hauntingly gorgeous etched melodies carved out of the
slow moving
space
> of his ballads to the ferocious explosions of his rhythmic
up-tempo tunes.
I
> wondered why his improvisations and compositions seemed so
fresh and new,
> unlike any other jazz music IOd heard. It was even fresh
compared to the
> last 30 times IOd seen him play.
 I thought about all the choices he made that night. All the
possibilities
he
> had before him on the 88 key piano. I saw it best as the
algebra
equation:
 P = 10 x 88 x V x T
 Where P is the playing 'eld or all the possibilities of
sound from the
> piano. 10 represents the artistOs 10 'ngers.
88 is the
number of keys on
> the piano. V represents the velocity or volume at which the
keys are
struck.
> And T is time or rather tempo.
 Even if I were to assign a value of only 100 possible
Velocity levels and
> say 1000 tempos (or time values between two notes), that
still makes our
> equation 10 x 88 x 100 x 1000 = 88,000,000 unique
possibilities. Now my
> local mathematics professor would say I need to alter my
equation to
include
> exponents. Something like 88vt^10 (8.8 million to the power
of 10
'ngers)
> might be closer to what my equation should look like. Or
maybe 10^88vt?
> Regardless of my feeble attempts at algebra we know it would
take
millions
> of years and millions of piano players to play all the
possibilities of
the
> pianoOs sound. Ok, so now letOs calculate
possibility values
for the
> acoustic bass and the drum set? .... Just kidding.
 The equations I thought about reminded me of equations for
3-D space. It
> seems attractive to think about the piano as a 3 dimensional
space so
vast
> my mind can barely grasp itOs expansion. Hence, the
pianist
becomes an
> explorer on a journey through a God created world, 'rst
traveling
through
> the space others have trodden, and then 'nding new paths.
 The pianistOs possibilities might better be portrayed as
an
enormous
solid
> substance that stretches throughout the universe. It would
be suitable
for
> chiseling away at like a sculpture artist might work with
marble. Or
during
> a pianistOs jazz improvisation it might become soft clay
that is more
easily
> formed into shapes as spontaneously as the artist imagines
them. Always
> discovering new shapes. Always 'nding new paths.
 The focus is not on creating. Rather, 'nding and
discovering. When you
> 'nish a wood carving do you say This is a wood carving I
created? Or
do
> you say This is a piece of a tree God created? Though it is
your work
> indeed, the substance, the essence, the material itself is
GodOs.
 The weight and pressure of creating something from nothing
is lifted. Now
I
> can just play with and explore throughout GodOs already
created material
> until I 'nd things I like.
> ------------
 Please email if you have any suggestions
> hamm@i.am
 -- Dennis
>
Not really. You have 88 choices for the 'rst key. 87 for the
2nd. So 88x87
for just the 'rst two. If order does not matter, divide that
by 2. After
that
you start to get into limits as to how far one can reach with
one hand.
Bill
===
Subject: combinations and permutations
I know my logic is wrong, but can someone please explain why?
This example
is explained in the Grimaldi book, but I 'nd his brief
explanation to be
of limited help.
How many arrangements of the letters of the word TALLAHASSEE
have no
adjacent AOs?
The step I donOt understand is how to determine the number
ways the
adjacent AOs can be arranged. HereOs how I
think it should be
done:
There are 9 places to put the AOs:
1-T-2-L-3-L-4-H-5-S-6-S-7-E-8-E-9.
There are 6 ways to arrange the AOs so 2 are together. There
are 6 ways to
arrange the AOs so all 3 are together. DoesnOt
this mean there
are (6 + 6)
x 9 = 108 ways of arranging the AOs so at least 2 are
adjacent?
The book says there are C(9,3) = 84 ways. I just donOt
understand why you
use the combination formula for this.