mm-1033
===
Subject: Re: Another Number-Theory /Taylor-Expansion Theorem
> Let {a(k)} be any sequence of integers where the sum below
converges.
> Let {b(j)} be a sequence such that:
>
> sum{j=0 to oo} b(j) x^j /j! =
>
> exp(A_n(x)),
>
> where A_n(x) =
>
> sum{j>=2, GCD(j-1,n)=1} a(j) x^j /j!,
>
> where this sum is over all integers j, j >= 2, where
> n is coprime to (j-1), n = any fixed positive integer.
>
> Then:
>
> b(j) is an integer sequence, and (the main result...):
>
> n divides b(n+1).
> |A generalization:
> |m also divides b(m+1),
> |for ALL positive integers m where every prime which
divides m also
divides n.
> |(obvious, but noteworthy)
> I like this one. I dont know how you prove it, though. It
doesnt seem
all
> that obvious to me. Perhaps Im missing something simple.
> By computing a few of them, it seems that b_k is the sum
over all of the
> ways of partitioning k distinct elements of the product
over the
elements*
> of the partition of a_i where i is the number of elements
in it.
[*Theres
> got to be a more descriptive term than element. I was about
to write
> partitions in the partition but thats worse. The 
partition
corresponds
> to an equivalence relation, and I mean the individual
equivalence
classes.
> There should be a good term for them.]
> It looks like thats not so hard to prove by induction. 
The
value of the
> k-th derivative of e^A at x=0 gives b_k. The k-th
derivative itself is
the
> sum over the equivalence relations on a k element set of
> e^A * (d/dx)^{i1}(A) * ... * (d/dx)^{it}(A)
> where i1,i2,...,it are the numbers of elements in the
equivalence
classes.
> Taking the derivative of that term gives us t+1 terms which
correspond to
> the t+1 partitions of a k+1 element set that restrict to
the given
> partition on the first k elements. The term we get by taking
the
derivative
> of (d/dx)^{ij}(A) corresponds to adding the k+1-st element
to the j-th
> equivalence class. The term we get by taking the derivative
of e^A (which
> gives Ae^A) corresponds to creating an additional
partition to put the
> k+1-st element into. And of course the value of the ij-th
derivative of
> A at x=0 is a_ij, while A(0)=0 and so e^A(0)=1.
> It seems that the result is true because the coefficient of
each term
> is separately divisible by m. The partitions of m+1
elements can be
> transformed by cyclic permutation of the first m elements.
This preserves
> the number of elements in the equivalence classes. So the
number of
> equivalence relations having i1,i2,...,it elements in their
equivalence
> classes is a multiple of m unless at least one of them is
fixed by a
> cyclic permutation of the first m elements.
> The only ones which are fixed are ones where the equivalence
class that
> contains the m+1-st element is fixed, since the k+1-st
element is fixed.
> So if there are i_j elements in that equivalence class, for
it to be
fixed
> there must be a common factor between i_j-1 and m. The
orbits of a
> nontrivial cyclic permutation of m elements have equal
order, all
dividing
> m. But since A_n(x) is a sum only over the j where
(j-1,n)=1, that
amounts
> to setting a_ij=0 when (j-1,n)>1. So for the nonzero terms,
the number of
> associated equivalence relations is actually divisible by m.
> Im curious to know in what terms youve 
been thinking
about it.
> Keith Ramsay
Your reply makes my theorem look MUCH more impressive than it
is.
:)
Your analysis reminds me of Robert Israels in his reply to
another
recent sci.math post of mine.
As with the other similar theorem, I used a recursive means to
calculate the b-sequence; and with the recursion, the main
theorem is
easily proved.
As with the other result, I assumed that:
b[0] = 1 (which is so when a[0] = 0);
and that
b[j+1] = sum{k=0 to j} binomial(j,k) b[j-k] a[k+1],
where (here) any a[k+1] is zero if GCD(n,k) > 1.
(Then {b} is obviously an integer sequence if {a} is as well.}
Since binomial(j,k) is divisible by j if (for
1 <= k <= j-1) GCD(j,k) = 1,
the result easily follows.
Leroy Quet
===
Subject: Re: Attacking my own math proof, fun
> a^3 + 3va^2 - (v^3 + 1) = 0
> is also true as an equation within the system.
> You *have* to consider the entire system, rather than try
to pull
> pieces out of it, and claim a single piece is the entire
thing.
> the entire P(m).
> Thats a false statement, which I can easily show to be
false, by
> putting in some numbers. Using f=1, x=2, u=1, with
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f)
> I have
> P(m) = 8m^3 - 24m^2 + 24m + 6 - 6m + 1 so
> P(m) = 8m^3 - 24m^2 + 18m + 7.
> But
> a^3 + 3va^2 - (v^3 + 1) = 0 is
> a^3 + 3(-1+m) -((-1+m)^3+1) = 0.
> Now then, would you mind explaining how you can use the
second
> expression to get P(m)?
Oh, OK. No problem.
v = -1 +mf^2
a^3 + 3va^2 - (v^3 + 1) = 0
Let the three roots of this be a_1, a_2, a_3 considering it
as a polynomial
in
a.
a^3 + 3va^2 - (v^3 + 1) =(a - a_1)(a - a_2)(a- a_3)
Define P(m) = (a_1x +uf)(a_2x + uf)(a_3x +uf)
Solve the former equation for a_1,a_2,a_3 and substitute into
the latter.
Explanation complete.
> A point to expressing it this way is to show that the m=0
case reduces
> it to.
> a^3 -3a^2 = 0
> You use two distinct arguments to conclude that a_3 is
coprime to f.
> One is that to show that a_3 is coprime to f at m=0, and to
conclude
> that therefore a_3 is coprime to f at m<>0. You dont use
any
mathematical
> argument to do this. By expressing the as directly, 
its
clear that
your
> conclusion is invalid.
> Why?
Because v = -1+mf^2
a^3 + 3va^2 - (v^3 + 1) = 0, so a is clearly a function of m.
Knowing their
values at m=0 doesnt make them independent of m for m<>0
> You put in a lot into one paragraph and I think you just
hoped to toss
> out a conclusion which *you* want readers to believe,
possibly hoping
> that youd not have to elaborate. I want you to explain
yourself in
> detail.
I think the detail above is sufficient. You need to prove that
theyre
somehow
not functions of m if you wish to persuade that knowing their
value at m=0
is
sufficient.
> What I will do is point out what youre trying to deny,
which is that
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f)
> is an expression that represents a complex system that you
cant
> reduce to a particular polynomial
> Even
> a^3 + 3va^2 - (v^3 + 1)
> is still not a polynomial because you have variable
coefficients.
> What I do is take
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f)
> and use m=0 to find the constant coefficient.
Which you havent proved is a mathematically sound approach.
> Now theres nothing improper or invalid about that, and of
course, the
> constant term of the expression viewed as a polynomial with
respect to
> m, will be given by m=0.
> That gives me P(0) = u^2 f^2(3x + uf).
> Now you can also multiply the expression out, group terms
with m, and
> eventually come upon the same expression as your last
coefficient, but
> you can also just use m=0.
Ooh, Maybe youd better show this. Perhaps this is the key.
So far you
havent
done that. Show the grouped terms with m, so we can see
whether
the a_s are coprime to f at m<>0.
> The other way is by claiming that b_1*b_2*a_3, which is an
algebraic
integer
> coprime to f, implies that a_3 is coprime to f. This is only
necessarily
true
> when b_1 and b_2 are algebraic integers. Furthermore you do
this
> without any mathematical arguments.
> Heres the mathematical argument.
> Consider, in the ring of algebraic integers,
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f).
> Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
> where w_1 w_2 w_3 = f, and
> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),
> and at m=0
> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),
> so two of the bs must equal 0, which means
> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)
> which is
> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)
> proving that w_1 w_2 must be coprime to 3, if f is coprime
to 3, which
> leaves b_3 = 3, or a unit multiple of 3.
Strictly speaking it shows that w_1w_2b_3 = 3. When m=0.
> Essentially objections to how f^2 divides off now come down
to
> claiming that the ws are functions of m, but consider 
that
w_1 w_2 is
> coprime to f, when m=0, if f is coprime to 3.
> But that was an arbitrary choice, so let f=3.
> Now it can be shown that w_1 w_2 is coprime to 3 without
regard to m.
Stricty speaking you can show that w_1w_2w_3 = 3, in this
case. Without
regard to m. If you like you can choose w_1w_2 to be a value
coprime to 3.
Without regard to m.
> That is, the ws can now be shown to all be constant with
regard to m,
> so they have the same value no matter what the value of m
is, so they
> are also constant with f coprime to 3.
Yes. No problem. Just define w_1 = 1, w_2 = 1 for all f
coprime to 3
for all m. So they are constant no matter what the value of
m. Also define
w_1=1 w_2=1 for all f not coprime to 3 as well, if you wish.
Or define them
to be something else entirely. No problem.
> Introducing a_1, a_2, and a_3, that is seen by considering
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f) =
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> with f=3, as then you have
> P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 -
> (-1+mf^2 )x u^2 + u^3 =
> (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u)
Slight typo there.
> so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3
divides
> off as a constant so it is without a dependency on m.
Fine.
> Now looking at
> P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
> I can again check at m=0, to see
> P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u),
> which forces w_3 to have a factor that is 3.
Slight error there. It should be:
P(0)/3^2 = u^2w_1w_2(b_3x + uw_3)
=u^2(b_3w_1w_2x + uw_1w_2w_3)=u^2(3x + 3u)
Which forces w_1w_2w_3 = 3. If you assume w_1=1, w_2=1, the
error
doesnt matter.
> Therefore, the factorization is
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f =
> (b_1 x + u)(b_2 x + u)(b_3 x + uf)
> where youll notice that the bs are 
algebraic integers
with m=1,
> f=sqrt(2), but thats a special case as generally they are
not, which
> shows a problem with the ring of algebraic integers.
The algebra is fine. It doesnt however show any 
problem with
the ring
of algebraic integers yet. Ill keep looking though.
> This is all notwithstanding the numerous claims you are
being shown in
other
> threads that none of the as are coprime to 5.
> Im not interested in claims. Im interested 
in proofs.
> And those problems arent dependent on on how f^2 divides
off. There is
no
> problem with your algebra in dividing the f^2 off the way
you do, and I
have
not
> any any point argued with it. I agree that your m and f are
independent,
> just most interesting when f is coprime to 3 and m.
> Its just your conclusions that need proving, in the face
of countless
alleged
> disproofs that are straightforward to verify, but so far
not refuted.
> I have refuted them.
Ill go back and reread them then.
> Why wont you at least admit the m dependency your
position would
> require on how f^2 divides off?
> Repeatedly youve claimed youre not 
asserting such a
dependency,
> which goes against the math.
> No I wont and no it doesnt.
> Well then, do you or do you not believe that m=0 is a
special case?
You get an answer at m=0, You might get other answers at m<>0.
You havent made the case at m<>0 yet. It 
doesnt take a very
strong
belief system to wonder, given that a is a function of m,
that it might
have
different values at different m. If that makes m=0 special,
sobeit.
> Note that your entire Advance Polynomial Factorization
paper hinges
> on the unproven assertion:
> Given that P(m) has a factor f^2 that separates off, two of
the gs
> should have a factor of f which would force two of the as
to have a
> factor that is f.
> Therefore, that leaves one factor coprime to f
> This is directly relevant to my arguments. Its in your
interests to
fix
this up
> properly and show the mathematics before you can expect to
make headway.
You
> wont be published with that statement in the paper.
> Phil Nicholson
> Well, as usual readers, you see a poster working to
convince, and in
> this case its with something tossed at you out of the 
blue.
Im not trying to convince. Im inviting you 
to convince by
showing you
some unproven assertions that youve made.
> In my paper Advanced Polynomial Factorizaton, you have
> P(m) = g_1 g_2 g_3
> g_1 = (a_1 x + uf), g_2 = (a_2 x + uf), g_3 =(a_3 x + uf)
> and I prove that one of the gs must be coprime to f, 
which
*should*
> force two of the gs to each have a factor that is f, for
reasons
> explained in detail, but the ring of algebraic integers has
this
> weird, esoteric problem, which means that youre forced 
out
of that
> ring.
No, you just state that two of the gs *should* have a 
factor
that is f.
The proof is missing. Check your paper, the statement stands
alone.
The above quote is identical to your erroneous b_1*b_2*a_3
proof, which
you
didnt present this time.
In the work just presented here, you didnt attempt to prove
anything
beyond showing that you get consistent behaviour if you just
let w_1=1,
w_2=1, w_3=f. Which begs the question as to why they were
introduced
in the first place.
> James Harris
===
Subject: General solution of AX=Y with rank constraints.
Given a matrix system of equations AX=Y, (A,X,Y are matrices)
the general
solution is
X=A*Y+(I-A*A)B, where A* is the pseudo-inverse of A and B is
any matrix
with
the same dimension as X.
Does anybody knows a generalization of the previous equation
when X is
constrained to be rank r?
===
Subject: General solution of AX=Y with rank constraints
Given a matrix system of equations AX=Y, (A,X,Y are matrices)
the general
solution is
X=A*Y+(I-A*A)B, where A* is the pseudo-inverse of A and B is
any matrix
with
the same dimension as X.
Does anybody knows a generalization of the previous equation
when X is
constrained to be rank r?
===
Subject: Re: How I know, linchpin of my FLT proof
<3F38305E.4050008@farir.com> <87r83r9qtj.fsf@phiwumbda.org>
Cc: jesse@phiwumbda.org (Jesse F. Hughes)
A8<L{-iZAH|#`e@d1Dd.&:lu`[O}2^rhg@6(x2l2qK1>EwTYfhf*u~,Eu,
tf6$HN*MY&)u0G
=N
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
J6m5.EN?>Zh
Xh;Y
V|,x(jsJfq02joVpj|#x
>> your useage of the English language is
>> substandard, even for a southerner.
> Dont be an ass.
> At least, not in the same sentence in which you misspell
usage and
> fail to capitalize Southerner.
> (Now, wheres my requisite typo in this spelling 
ßame?)
Oh, I see it. No quotes around usage.
How come I cant see it *before* the post?
--
If you *still* believe that [my proof is wrong], then I have
to think
that your mind is limited [...], and it may be the case that
not
everyone *can* achieve that, as the mental wiring may not be
there for
the task. -- James Harris, on faculties needed to accept his
proof.
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
><deleted>> The factor that has been provided,
>
> r(a) = 8 a^2 - 4 a - 45
>
> is
> (1) an algebraic integer for any algebraic integer a,
> (2) a divisor of both a and 5 for
> a = -(any of the ais in the above
> factorization of 65x^3 - 12x + 1)
>
> The factorizations have already been given, but since
youre so adept
> at ignoring the evidence, I will provide them once again:
>
> q(a) = 8 a^2 - 76 a - 185
> r(a) = 8 a^2 - 4 a - 45
> s(a) = 4 a^2 - 37 a - 104
>
> Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the
*negative*
> of any of the ais of the above factorization of 65x^3 -
12x + 1),
the
> following factorizations hold:
>
> q(a) r(a) = 5
> r(a) s(a) = a.
>>Hmmm...the poster has tried to dupe *you* the reader, and
it all has
>>to do with the word unit. See below...
>>
> In fact, the minimal polynomial of this number (r(a) for -a
= any of
> the above ais) is given as:
>
> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
>
> The above facts prove that this a has a non-unit algebraic
integer
> as a factor.
>>No it doesnt.
>> Why not? There is an algebraic integer, called here r(a),
with the
>> following three properties:
>> (1) It is not a unit in the ring of all algebraic
integers; you
>> apparently agree with that conclusion, but then engage in
>> sophistry about a supposed problem with it.
>It is true that r(a) is not a unit in the ring of algebraic
integers.
>Now what are the actual expressions?
>They are
> q(a) r(a) = 5
> r(a) s(a) = a.
Which means that r(a) is a nonunit common factor of 5 and a.
Which is exactly what you claimed was impossible to exhibit.
Thats it.
>where I simply note that r(a) may not have non-unit factors
in common
>with 5,
r(a) IS a non-unit factor OF 5.
>which *should* make q(a) have a factor of 5 ***in the ring of
>algebraic integers***,
q(a) IS a non-unit factor of 5.
> but the ring of algebraic integers is screwed
>up.
Nonsense.
>In trying to prove that r(a) must have a non-unit factor in
common
>with 5, you rely on the fact that it cant be a unit ***in
the ring of
>algebraic integers***!!!
r(a) IS a nonunit, and a factor of 5: there it is above.
There is
something that multiplied by r(a) equals 5. Thats the
DEFINITION of
being a factor. To do that, I do not use that r(a) is not a
unit in
the ring.
Only AFTER proving that r(a) is a common factor of 5 and a do
we show
that it is not a unit.
>The trick is to act as if proving the negative, proves that
it must
>have some non-unit factor in common with 5, but theres a
fascinating
>error with the ring of algebraic integers.
Apparently, the trick is to confuse yourself to a point where
you
dont know what you are saying... At least, that seems to be
what you
are doing.
>> (2) There is an algebraic integer, called s(a), such that
>> r(a)*s(a)=a. So r(a) is a factor of a in the ring of
algebraic
>> integers.
>Well, what if Ôa doesnt share 
non-unit factors in common
with 5?
Are you blind? r(a) is a factor of a, and it is a factor of
5, and it
is not a unit. There is no if.
>Then both r(a) and s(a) dont either.
r(a) is a factor of a, it is a factor of 5, and it is not a
unit.
>That *should* leave q(a) with a factor of 5, but the ring is
screwed
>up.
r(a) is a factor of 5, it is a factor of 5, and it is not a
unit.
>In order to try and prove that it does have a factor of 5,
you have to
>try and claim that it must because otherwise r(a) would be a
unit.
Nonsense. I have shown that r(a) is a factor of a; that it is
a factor
of 5, and that it is not a unit. That is the supposed thing
that
nobody could do: exhibit an algebraic integer which is a
factor of 5,
a factor of a, and not a unit.
>But the ring of algebraic integers is screwed up, which is
my point.
>> (3) There is an algebraic integer, called q(a), such that
>> r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of
algebraic
>> integers.
>However, how do you know that r(a) must share non-unit
factors in
>common with 5?
r(a) ITSELF is a nonunit factor of 5.
>Ultimately your claim must be that its because 
its not a
unit in the
>ring of algebraic integers!!!
r(a) is a factor of 5, and it is not a unit. You are
delusional.
>> So: r(a) is (1) not a unit; (2) a factor of a; and (3) a
factor of 5.
>> Why is it not the case that r(a) is a non-unit algebraic
integer which
>> is a common factor of a and 5?
>For readers, these posters are working to *convince* not get
to the
>bottom of the problem.
Answer the question: r(a) is (1) not a unit; (2) a factor of
a; and
(3) a factor of 5.
Why is it not the case that r(a) is a non-unit algebraic
integer which
is a common factor of a and 5?
>If they cared about the truth, then they only need consider
my proof
>of the problem with algebraic integers.
Answer the question: r(a) is (1) not a unit; (2) a factor of
a; and
(3) a factor of 5.
Why is it not the case that r(a) is a non-unit algebraic
integer which
is a common factor of a and 5?
[.snip.]
>> [.snip.]
>>What you have is
>> x^3 - 969 x^2 + 315 x + 5
>>and what you want to do is convince readers that none of
its roots can
>>be coprime to 5.
>> There is no convincing that needs to be done. NONE Of the
roots are
>> units, and ALL the roots are factors of 5. Youre done.
>No, you have a gap. The gap is that now you need to prove
that the
>roots share non-unit factors in common with 5 in the ring of
algebraic
>integers.
The roots ARE non-unit factors of 5 in the ring of algebraic
integers. There is no gap.
>>But you cannot do that,
>> Correction: James Harris cannot do that because he does
not know what
>> coprime means.
>Which is the semantic argument.
Do you know what semantic means? I dont think so.
>> and instead have maintained that proving that
>>none of the roots can be a unit proves that one of them
must have a
>>non-unit factor in common with 5 in the ring of algebraic
integers.
>> They are each FACTORS of 5, and EACH non-units.
>Which does NOT prove that they each have a non-unit factor
in common
>with 5, because the ring of algebraic integers is ßawed.
(1) You are assuming what you are claiming.
(2) Yes, it DOES prove that they each have a non-unit factor
in common
with 5: THEMSELVES.
>> Do you agree or disagree with (all happening in a
commutative ring
>> WITH 1):
>> (a) If a is a factor of b, then a and b have a common
factor
>> (namely, a).
>> (b) If a is a factor of b and is not a unit, then a is a
non-unit
>> common factor of a and b.
>> (c) If is a factor of b and is not a unit, then a and b
have a
>> non-unit common factor (namely, a).
>Hmmm...I can see that youre dedicated at working to
convince the
>audience, so Ill give a demonstration to show them *how*
youre
>working.
Answer the question. Do you agree or disagree with (a)? With
(b)? With
(c)?
You are trying really hard to avoid any direct question, with
specific
numbers, and instead rely on confusion. Why?
>Consider abc = 5, where a = sqrt(5),
b=sqrt(5)(-1+sqrt(-3))/2,
>c=(-1-sqrt(-3))/2, but imagine that youre talking to
someone who
>doesnt know about radicals, as all they know about are
integers.
>Now neither Ôa nor 
Ôb is an integer, but this person is
using d=ab,
>and they now see
> cd = 5
>and tell you that Ôc and 
Ôd must be factors of 5 ***in the
ring of
>integers***.
>You say nope.
That would be because neither of them are integers.
>Well they come back, and argue, and point out that neither
is a unit
>***in the ring of integers***.
>You say, yup, that is correct.
>Then they come back and say, well, cd = 5, neither is a unit
in the
>ring, so they must be factors of 5.
>And you say, nope, they are not.
This example is completely inapplicable. Because we agree
that:
r(a), s(a), and q(a) are ALL algebraic integers (which was
not the
case in your example);
r(a) is an algebraic integer which is not a unit in the ring
of
algebraic integers (which was not the case in your example);
r(a)*q(a)=5 and r(a)*s(a)=a.
And so, you agree that r(a) is an algebraic integer factor of
5 in the
ring of algebraic integers; that it is an algebraic integer
factor of
a in the ring of algebraic integers; and that it is an
algebraic
integer which is not a unit in the ring of algebraic integers.
So, why is it not the case that r(a) is an algebraic integer
common
factor of 5 and a which is not a unit?
No more nonsense about the ring being screwed up. Why is it
not a
nonunit common factor?
>But the ring of algebraic integers is VERY screwed up, so
that only
>gives you an idea.
No, you are screwed up. Nothing you have said even points to
any
problems in the ring of algebraic integers, only in your
understanding
of what a ring is, what factor means, and what unit means.
[.snip.]
>>The truth is that youre relying on the negative, which is
that NONE
>>of the roots of that expression can be units, to try and
prove the
>>positive, which is that then they all have a non-unit
factor in common
>>with 5 in the ring of algebraic integers.
>> You are completely, totally, utterly lost and confused.
>> The roots themselves are factors of 5. Each root is a
factor of
>> itself, and a factor of 5. Therefore, each root is a
common factor of
>> itself and 5. And since each of them is not a unit, then
we have
>> EXHIBITED a non-unit common factor of each with 5. The
roots are not
>> the original numbers we were interested in (the as), 
they
are the
>> common factors that have been produced.
>>
>> [.snip.]
>And again the trick is that use of the word unit as in fact
the
>factors are NOT units ***in the ring of algebraic
integers***.
This is again nonsense.
>In a higher, more complete ring, one of them IS a unit,
while two of
>them have a factor that is sqrt(5).
You have never defined complete, but never mind.
Yes, there are plenty of LARGER rings where one is a unit and
two have
a factor of sqrt(5). One example of such is Z[1/a1, 1/a2,
1/a3].
But you have still not answered the question:
Since r(a) is a factor of 5, a factor of a, and not a unit,
why is it
not a nonunit common factor of 5 and a?
[.snip.]
>For readers who dont realize how this ring business can be
confusing,
>consider my example with the ring of evens. That ring
doesnt have 1
>in it, as 1 is odd, so 2 and 6 do not share factors with
each other.
The example is inapplicable. The ring of algebraic integers
DOES have
a 1, so every element is a factor of itself.
[.snip.]
>Rather than be fascinated by it, and accept the mathematics,
you can
>see posters like Arturo Magidin instead working to convince
YOU.
>Why does he have to keep replying to me in posts?
I realize that you have absolutely no inkling about what
could drive a
man of principle, but that is exactly why I reply to you:
because I
have principles. And one of them is not to let mathematical
nonsense
stand unchallenged when it can be refuted.
>Because Im right. If he goes away, and isnt 
around to
confuse
>readers, he probably realizes that eventually I might
convince some of
>you to follow the mathematical logic.
I realize that someone like you cannot think of any other
reason,
because I am not gaining fame, money, or women, by my
actions, and
that is the stick by which you measure everything.
Nonetheless, that
is not the case.
>That explains why posters like Arturo Magidin are so
dedicated.
In you fantasy world only.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
>What you have is
> x^3 - 969 x^2 + 315 x + 5
>and what you want to do is convince readers that none of
its roots can
>be coprime to 5.
> There is no convincing that needs to be done. NONE Of the
roots are
> units, and ALL the roots are factors of 5. Youre done.
> No, you have a gap. The gap is that now you need to prove
that the
> roots share non-unit factors in common with 5 in the ring
of algebraic
> integers.
James, it has been pointed out many times that if 
Ôa divides
5 in the ring
of algebraic integers (as do
all roots of monic polynomials with integer coefficients and
constant term
+/- 5), then the square roots,
cube roots, etc. of Ôa are non-unit algebraic 
integers which
also divide 5
and clearly also divide Ôa.
Hence, if Ôa divides 5 in the ring of algebraic 
integers,
then there are an
*infinite* number of
algebraic integers which are factors in common with 
Ôa and
5. This is
elementary algebra. Is it beyond
you? Cant you see that your conclusion is false and that 
the
error is in
*your* argument?
--
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: How I know, linchpin of my FLT proof
<deleted >
> No it doesnt. What you can prove is that if 
Ôa is coprime
to 5 it
> cant be a unit in the ring of algebraic integers, but as
Ive said
> the ring of algebraic integers is screwed up.
>
> No, no one can prove that. After all, a unit *is* coprime
to every
> number!
Thats true, but it doesnt prove that in a 
given ring two
numbers
that dont share unit factors *in the ring* are coprime,
given your
definition of coprime.
Its not complicated.
Maybe it would help if you consider the ring of evens, where
2 and 6
dont share non-unit factors in the ring because 1 and 3 are
not in
the ring of evens, but by *your* definition they are NOT
coprime.
Think about it.
> Let u be a unit algebraic integer, and let p be any
algebraic integer.
> Then there is an algebraic integer v for which uv = 1, and
we get:
> u v + 0 p = 1,
> where 0 is zero.
Yeah, where youre saying that it IS a unit.
Remember, youve tried to use examples where numbers have 
been
determined to NOT be units.
Going from there, youve jumped to concluding that they then
must
share non-unit factors *in the ring of algebraic integers*
which you
have not proven.
Rather than actually work to figure out the math, 
youve
instead
engaged in chasing your tail, publicly through posts on these
forums.
> This establishes the fact that a UNIT algebraic integer is
coprime to
> every algebraic integer, in the ring of algebraic integers.
A very
> slight rewording proves the result for a unit of any
commutative
> ring.
It is true that a unit algebraic integer is coprime to every
algebraic
integer.
> No wonder you think the ring of algebraic integers is
screwed up!
Youre chasing your tail.
> Lets get the language straight. I have been clear in this
that Ôa
> refers to (-1) times any one of the coefficients ai in the
factorization
> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
> I claim that *each* of those numbers a has a non-unit
algebraic integer
> factor in common with 5.
But you do not prove it. Ill point out the gap in what you
have that
follows.
> PROOF: The above expressions q(a),r(a),s(a) all yield
algebraic integers
> for any algebraic integer a. When, in addition, a is a root
of the
> polynomial
> x^3 - 12 x^2 + 65
> then the products are as I mentioned above:
> q(a) r(a) = 5
> r(a) s(a) = a.
> Your claim:
> No it doesnt. What you can prove is that if 
Ôa is
> coprime to 5 it cant be a unit in the ring of algebraic
> integers, but as Ive said the ring of algebraic 
integers
> is screwed up.
Yup.
> amounts to this:
> PARAPHRASE:
> It doesnt matter whether anyone provides algebraic 
integers
> q,r, and s for which
> 5 = q*r
> a = r*s
> it doesnt prove that a and 5 are not coprime.
That is correct, and in fact, for ONE of the three values for
Ôa they
are in fact coprime.
> However, it *does* prove that. Assume otherwise, that a is
coprime to 5.
> Then there exist algebraic integers u and v for which
> a u + 5 v = 1
> and the above factorizations show this:
> rs u + qr v = 1
> so
> r( su + qv ) = 1,
> and we deduce that r is a unit.
You do not deduce you claim, but you cannot prove.
In fact, you *can* prove that it is not a unit.
If you assume that the ring of algebraic integers is ok, then
you
think youre done.
But youre chasing your tail, as the ring of algebraic
integers is NOT
ok, and in fact, in that situation where r *should* be a
unit, its
NOT in the ring of algebraic integers, as there is a
fascinating
problem with the ring.
>However, its not one of your fanciful
> Unit with no inverse hallucinations. In fact, the above
equation
> provides us with an inverse for r. Based on its
construction, it
> *must be* an algebraic integer: u and v are specified to be
algebraic
> integers, and q,r,s were already specified as algebraic
integers.
<deleted>
Why dont you show that inverse.
Now thats bizarre as the poster has just attacked his own
argument.
Now I dont claim that r is a unit, as its 
not *in the ring
of
algebraic integers*.
It amazes me how often posters just throw out false
statements.
Here the posters claim is that a non-monic primitive
irreducible over
Q, which is what gives you the inverse for r in context, is an
algebraic integer.
Dont see it?
Well consider the statements:
> Then there exist algebraic integers u and v for which
> a u + 5 v = 1
> and the above factorizations show this:
> rs u + qr v = 1
> so
> r( su + qv ) = 1,
> and we deduce that r is a unit.
The poster wants to claim that r must share a non-unit factor
in
common with 5 by showing a contradiction by assuming its a
unit,
which actually only shows that it cant be a unit in the 
ring
of
algebraic integers.
So he should have just ended up chasing his tale.
But he screwed up and managed to bite himself!
James Harris
===
Subject: Re: How I know, linchpin of my FLT proof
... stuff deleted ...
>> q(a) r(a) = 5
>> r(a) s(a) = a.
... stuff deleted ...
>>The above facts prove that this a has a non-unit algebraic
integer
>>as a factor.
>No it doesnt.
>>Why not? There is an algebraic integer, called here r(a),
with the
>>following three properties:
>> (1) It is not a unit in the ring of all algebraic
integers; you
>> apparently agree with that conclusion, but then engage in
>> sophistry about a supposed problem with it.
> It is true that r(a) is not a unit in the ring of algebraic
integers.
> Now what are the actual expressions?
> They are
> q(a) r(a) = 5
> r(a) s(a) = a.
> where I simply note that r(a) may not have non-unit factors
in common
> with 5, which *should* make q(a) have a factor of 5 ***in
the ring of
> algebraic integers***, but the ring of algebraic integers
is screwed
> up.
Apparently, you have some inkling of what it means for two
numbers to
share a factor. Allow me to propose a definition:
Ill assume Im dealing solely with integral 
domains for
the present discussion.
In the ring R, an element s is called a factor of the element
r
if there is an element u in R, for which
r = su
The element s is a *common factor* of the elements r and r,
if there are elements u and u of R, for which
r = s, and r = su
Would you agree with these definitions, at least to the point
that you
understand them?
Ill suppose so, and continue.
I have shown, and you appear to accept, these several facts:
r(a) is a factor of 5.
r(a) is a factor of a.
r(a) is a non-unit algebraic integer.
Now, you are claiming that the equations:
q r = 5
r s = a
may occur with r having *no* non-unit common factors with 5
in the ring of algebraic integers.
I claim that the number r *itself* serves as a non-unit factor
that r has in common with 5. It also serves as a non-unit
factor
that a has in common with 5. This latter was the point of the
exercise.
Proof of claim:
The (above) definition of common factor:
s is a common factor of r and r if there are
elements u, u in R for which the following
hold:
r = s u
r = s u
In our case, the claim is that r(a) is a
common factor of r(a) and 5. By definition,
we need to find the algebraic integers u
and u so that
r(a) = r(a)u
5 = r(a)u
However, if u = 1, and u = s(a), we have this:
r(a) = r(a)*1
5 = r(a)*s(a).
we already know that 1 and s(a) are algebraic integers,
so the result follows: r(a) is a common factor of
r(a) and 5.
> In trying to prove that r(a) must have a non-unit factor in
common
> with 5, you rely on the fact that it cant be a unit ***in
the ring of
> algebraic integers***!!!
Well, Ive shown that r(a) is not a unit, and you have
already agreed
to that. Whats the problem?
> The trick is to act as if proving the negative, proves that
it must
> have some non-unit factor in common with 5, but theres a
fascinating
> error with the ring of algebraic integers.
That is your claim. You have not provided any evidence, and
so may
not use it as a *basis* for your argument.
>> (2) There is an algebraic integer, called s(a), such that
>> r(a)*s(a)=a. So r(a) is a factor of a in the ring of
algebraic
>> integers.
> Well, what if Ôa doesnt share 
non-unit factors in common
with 5?
> Then both r(a) and s(a) dont either.
What? What do you think r(a) is? Its an algebraic integer,
but not
a unit! Well, another proof:
r(a) is a common factor of a and 5 in the ring of
algebraic integers.
The above definition
s is a factor of r and r if there exist
elements u and u for which the following
equations are satisfied:
r = s u
r= s u
In the present case, we need to find u and u
for which
a = r(a) u
5 = r(a) u
but we have the values q(a), s(a), and know
that
a = r(a) s(a)
5 = r(a) q(a).
Thus r(a) is a common factor of a and 5 in the
ring of algebraic integers. Since r(a) is
not a unit, r(a) is a *non-unit* factor that
a and 5 share in the ring of algebraic integers.
> That *should* leave q(a) with a factor of 5, but the ring
is screwed
> up.
No, r(a) is the common factor. Youre just throwing out
meaningless
objections to cover the fact that your argument is vacuous.
> In order to try and prove that it does have a factor of 5,
you have to
> try and claim that it must because otherwise r(a) would be
a unit.
NO ONE says that it has a factor of 5. 5 does NOT divide that
number,
rather r(a) divides 5. Get the division relation straight, or
at least
the language of whats a factor of what, before you start in
making
such irrelevant claims.
> But the ring of algebraic integers is screwed up, which is
my point.
Again, your argument is that my above derivations are
incorrect due to
this screwed up nature of the algebraic integers. However,
thats what
you are attempting to prove! It is not legitimate to assume
the truth of
ones conclusion in the course of proof.
That is, you may not refer to this defect of the ring of
algebraic
integers as a justification, during any argument that is
attempting
to establish the existence of such a defect.
Further, I have *shown* the common factor, you have *agreed*
that it
is a common factor, and you have also *agreed* that the
common factor
is (1) an algebraic integer, and (2) not a unit.
In short, you agree to the particulars of the argument, yet
you deny
the conclusion of the argument. Please explain how that
tactic is a
legitimate one.
>> (3) There is an algebraic integer, called q(a), such that
>> r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of
algebraic
>> integers.
> However, how do you know that r(a) must share non-unit
factors in
> common with 5?
I showed you.
> Ultimately your claim must be that its because 
its not a
unit in the
> ring of algebraic integers!!!
So what? I showed that also.
>>So: r(a) is (1) not a unit; (2) a factor of a; and (3) a
factor of 5.
>>Why is it not the case that r(a) is a non-unit algebraic
integer which
>>is a common factor of a and 5?
> For readers, these posters are working to *convince* not
get to the
> bottom of the problem.
No, you refuse to admit to the conclusion of the argument,
while
agreeing to the hypotheses as well as the logic of the
argument.
You are abusing the process of argument.
> If they cared about the truth, then they only need consider
my proof
> of the problem with algebraic integers.
Plus, you cast aspersions as to the ethics of those who argue
against
you. If you had a mathematical argument, you wouldnt do 
that.
> Proofs dont duel. If theyre correct, then 
theyd be able
to find an
> error with my argument, but it has no error as the ring of
algebraic
> integers IS screwed up, so instead they work to convince
YOU the
> reader.
I have found the arithmetic that you say does not exist.
Perhaps you
arent aware, but it is completely standard in rebuttal of
arguments
to display that the argument produces erroneous results.
I have shown that your claim (that the as and 5 are 
coprime)
is
incorrect. You, on the other hand, need to deny the facts, by
denying
the fact that a divisor that is not a unit must then be a
non-unit
divisor. The very nature of the contortions you apparently
need to
go through exposes the poverty of your case.
>> [.snip.]
>What you have is
> x^3 - 969 x^2 + 315 x + 5
>and what you want to do is convince readers that none of its
roots can
>be coprime to 5.
>>There is no convincing that needs to be done. NONE Of the
roots are
>>units, and ALL the roots are factors of 5. Youre done.
> No, you have a gap. The gap is that now you need to prove
that the
> roots share non-unit factors in common with 5 in the ring
of algebraic
> integers.
Duh.
5 = q(a) r(a)
a = r(a) s(a).
Its the polynomial
x^3 - 12 x^2 + 65
for which roots we are to demonstrate commonality with 5, not
the
cubic youre displaying here. That cubic is used to prove
that r(a)
is not a unit. However, youve seized on the word unit and
cant
quite grasp what relation that has with the original problem.
>But you cannot do that,
>>Correction: James Harris cannot do that because he does not
know what
>>coprime means.
> Which is the semantic argument.
Right. In mathematics, it is not incorrect to call into play
the very
definitions of what youre talking about. That 
way, a person
can stay
on track, rather than meandering all over the place.
Put another way, the definition (which youve 
always
pooh-poohed as
being some form of chicanery or circular argument) is an
essential
ingredient in any proof. The definition tells what it *is*
that youre
talking about, and a person ignores definitions at his peril.
>and instead have maintained that proving that
>none of the roots can be a unit proves that one of them must
have a
>non-unit factor in common with 5 in the ring of algebraic
integers.
>>They are each FACTORS of 5, and EACH non-units.
> Which does NOT prove that they each have a non-unit factor
in common
> with 5, because the ring of algebraic integers is ßawed.
No, that is just what it does prove, and your use of your
intended goal
the ring of algebraic integers is ßawed
as a justification (i.e., ... because the ring is ßawed ) 
is an
illegitimate tactic: you assume what has not been proven.
>>Do you agree or disagree with (all happening in a
commutative ring
>>WITH 1):
>> (a) If a is a factor of b, then a and b have a common
factor
>> (namely, a).
>> (b) If a is a factor of b and is not a unit, then a is a
non-unit
>> common factor of a and b.
>> (c) If is a factor of b and is not a unit, then a and b
have a
>> non-unit common factor (namely, a).
> Hmmm...I can see that youre dedicated at working to
convince the
> audience, so Ill give a demonstration to show them *how*
youre
> working.
> Consider abc = 5, where a = sqrt(5),
b=sqrt(5)(-1+sqrt(-3))/2,
> c=(-1-sqrt(-3))/2, but imagine that youre talking to
someone who
> doesnt know about radicals, as all they know about are
integers.
> Now neither Ôa nor 
Ôb is an integer, but this person is
using d=ab,
> and they now see
> cd = 5
> and tell you that Ôc and 
Ôd must be factors of 5 ***in
the ring of
> integers***.
> You say nope.
This is a false comparison. Note how none of the numbers in
your example
are actually in the ring youre discussing.
Are you claiming that none of the numbers q(a), r(a), and
s(a) is an
algebraic integer? I didnt notice you doing so before.
> Well they come back, and argue, and point out that neither
is a unit
> ***in the ring of integers***.
> You say, yup, that is correct.
> Then they come back and say, well, cd = 5, neither is a
unit in the
> ring, so they must be factors of 5.
> And you say, nope, they are not.
> But the ring of algebraic integers is VERY screwed up, so
that only
> gives you an idea.
Again, its your belief of this screwed up business that is
at the
same time your intended goal AND your justification for
denying these
straightforward arguments.
>Ive just shot down that little trick, but 
youre squirming.
>>I do see a lot of squirming, but its not coming from 
Dale.
> What mathematicians can do, because its such an odd 
error,
is keep
> casting doubt, and running away from the proof of the error.
You havent found an error. Rather, you are straining at the
concept
that a number could be a factor of itself.
> Probably for most of you the idea that you could have abc=5
where
> neither Ôa, Ôb, 
nor Ôc is a factor of 5, in the ring of
algebraic
> integers, seems nonsensical.
Um, I showed that q(a), r(a), and s(a) are algebraic
integers, unlike
your strawman argument where NONE of the elements belonged to
the ring
under discussion. Not a legitimate comparison, unless youre
claiming
that q,r,s are NOT in the ring.
> However, its an esoteric problem in an esoteric branch of
> mathematics, which has been there for over a *hundred*
years.
Yeah, back to the *esoteric* nature of arithmetic. Multiply
the
polynomials, compare coefficients. Anyone can see I told the
truth.
You still want to deny it.
> If mathematicians want to confuse most of you about it,
they can.
Right. Pointing out SPECIFIC polynomials, providing the
manipulations
to do, THAT is confusing, but talking about UBER polynomials,
that is
just regular stuff.
>> [.snip.]
>The truth is that youre relying on the negative, which is
that NONE
>of the roots of that expression can be units, to try and
prove the
>positive, which is that then they all have a non-unit factor
in common
>with 5 in the ring of algebraic integers.
>>You are completely, totally, utterly lost and confused.
>>The roots themselves are factors of 5. Each root is a
factor of
>>itself, and a factor of 5. Therefore, each root is a common
factor of
>>itself and 5. And since each of them is not a unit, then we
have
>>EXHIBITED a non-unit common factor of each with 5. The
roots are not
>>the original numbers we were interested in (the as), they
are the
>>common factors that have been produced.
>> [.snip.]
> And again the trick is that use of the word unit as in fact
the
> factors are NOT units ***in the ring of algebraic
integers***.
Yes, apparently it has you confused.
> In a higher, more complete ring, one of them IS a unit,
while two of
> them have a factor that is sqrt(5).
Oh, now I need to go to a ring where one of the roots is
invertible?
Why? I found common factors without inverting any of the
roots.
You have not proven what you claim, and are struggling in
vain to
give it some sense.
>>As I mentioned above, the factor r that I provided *is* an
algebraic
>>integer, and in the case that a is (-1) times any of the
ais in
the
>>factorization cited, I have provided its minimal
polynomial. It is
>>simple to verify from that minimal polynomial that its an
algebraic
>>integer and that its not a unit in the ring of algebraic
integers.
>See readers? You can catch the trick here as notice the
poster said
>not a unit, and what Im telling you is that these posters
are
>working hard to convince YOU the reader.
>>Are you claiming that non-unit and not a unit are not the
same
>>thing?
>> [.snip.]
> Nope.
> For readers who dont realize how this ring business can 
be
confusing,
> consider my example with the ring of evens. That ring
doesnt have 1
> in it, as 1 is odd, so 2 and 6 do not share factors with
each other.
> However, you go to the higher ring--integers--and they do.
Oh, where some element of 2Z becomes invertible? I dont
think so.
> Mathematicians have this problem where they didnt think
there was
> one, until I pushed it into the open.
Baloney
> Rather than be fascinated by it, and accept the
mathematics, you can
> see posters like Arturo Magidin instead working to convince
YOU.
Fascination does not do mathematics. Its a child looking at 
a
bubble, thinking its magic. Mathematics must be built, and
you
are not doing any building.
> Why does he have to keep replying to me in posts?
Oh, its back to the heads I win, tails you lose argument:
Dig this, folks:
1. Magidin responds to JSH, and that means:
> Because Im right. If he goes away, and 
isnt around to
confuse
> readers, he probably realizes that eventually I might
convince some of
> you to follow the mathematical logic.
> That explains why posters like Arturo Magidin are so
dedicated.
> They need to hang around to keep YOU confused.
2. Magidin doesnt respond, and that means:
See, I won!
> James Harris
JSH is back to his denial of direct arguments, accusing his
critics of
dishonesty, fraud, and intent to confuse the reader. No one
can utter
a contrary argument to JSH, unless he is evil.
Dale.
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
[.snip.]
[.newsgroups trimmed.]
>Why not? There is an algebraic integer, called here r(a),
with the
>following three properties:
> (1) It is not a unit in the ring of all algebraic integers;
you
> apparently agree with that conclusion, but then engage in
> sophistry about a supposed problem with it.
>> It is true that r(a) is not a unit in the ring of
algebraic integers.
>> Now what are the actual expressions?
>> They are
>> q(a) r(a) = 5
>> r(a) s(a) = a.
>> where I simply note that r(a) may not have non-unit
factors in common
>> with 5, which *should* make q(a) have a factor of 5 ***in
the ring of
>> algebraic integers***, but the ring of algebraic integers
is screwed
>> up.
>Apparently, you have some inkling of what it means for two
numbers to
>share a factor. Allow me to propose a definition:
>Ill assume Im dealing solely with integral 
domains for
>the present discussion.
> In the ring R, an element s is called a factor of the
element r
> if there is an element u in R, for which
> r = su
> The element s is a *common factor* of the elements r and 
r,
> if there are elements u and u of R, for which
> r = s, and r = su
Presumably, you meant r=su.
>Would you agree with these definitions, at least to the point
that you
>understand them?
>Ill suppose so, and continue.
>I have shown, and you appear to accept, these several facts:
> r(a) is a factor of 5.
> r(a) is a factor of a.
> r(a) is a non-unit algebraic integer.
>Now, you are claiming that the equations:
> q r = 5
> r s = a
>may occur with r having *no* non-unit common factors with 5
>in the ring of algebraic integers.
Perhaps an integer example would clarify Jamess confusion?
Suppose you want to show that the number 15 and 65 have a
nonunit
integer factor in common. There are many ways to do so, but
surely
James would agree that the following process accomplishes the
result:
(1) 3, 5, and 13 are all integers;
(2) 13*5 = 65, so 5 is a factor of 15;
(3) 3*5 = 15, so 5 is a factor of 65;
(4) 5 is not a unit in the integers.
Now, this would seem to me to accomplish the objective: we
have
explicitly exhibited 5 as a factor of both 15 and 65 in the
integers
(by exhibiting other integers that multiplied by 5 yield the
desired
numbers), and so it is a common factor.
James now is saying that we need to show that 5 shares a
non-unit
integer factor with 15 (or with 65). But he is confused. 5
itself was
exhibited as the nonunit common factor of 15 and 65. Thats
it.
[.snip.]
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
>> your useage of the English language is
>> substandard, even for a southerner.
> Dont be an ass.
> At least, not in the same sentence in which you misspell
usage and
> fail to capitalize Southerner.
> (Now, wheres my requisite typo in this spelling 
ßame?)
Well, you caught me there.
My bad.
Dale.
===
Subject: Re: How I know, linchpin of my FLT proof
> <deleted>>No it doesnt. What you can prove is that if 
Ôa
is coprime to 5 it
>cant be a unit in the ring of algebraic integers, but as
Ive said
>the ring of algebraic integers is screwed up.
>>No, no one can prove that. After all, a unit *is* coprime
to every
>>number!
> Thats true, but it doesnt prove that in a 
given ring two
numbers
> that dont share unit factors *in the ring* are coprime,
given your
> definition of coprime.
So, why did you state that one *could* prove this, if it is
false?
Actually, every element of a ring shares ALL unit factors
with EVERY
OTHER element of the ring. Why would a person even talk about
numbers
that DONT share unit factors? That is literally talking
about nothing!
> Its not complicated.
> Maybe it would help if you consider the ring of evens,
where 2 and 6
> dont share non-unit factors in the ring because 1 and 3
are not in
> the ring of evens, but by *your* definition they are NOT
coprime.
> Think about it.
Yes, I see that the rings Ive been working with have a
multiplicative
identity. Your ring doesnt, does it?
>>Let u be a unit algebraic integer, and let p be any
algebraic integer.
>>Then there is an algebraic integer v for which uv = 1, and
we get:
>> u v + 0 p = 1,
>>where 0 is zero.
> Yeah, where youre saying that it IS a unit.
Im showing that a unit is coprime to everything. Duh.
> Remember, youve tried to use examples where numbers have
been
> determined to NOT be units.
Im showing that if you insist that the following setup:
5 = q(a)*r(a)
a = r(a)*s(a)
with Ôa and 5 being coprime, then it is *YOU* 
who is
insisting
that r(a) is a unit.
> Going from there, youve jumped to concluding that they
then must
> share non-unit factors *in the ring of algebraic integers*
which you
> have not proven.
I have shown the non-unit factor: r(a). Can you understand
that?
> Rather than actually work to figure out the math, 
youve
instead
> engaged in chasing your tail, publicly through posts on
these forums.
What? Are you saying that the factorization Ive posted and
provided
all the necessary calculation to verify that, is somehow
inadequate?
Please explain. Show me the error.
That is, show how one of these is not correct:
1. 5 = q(a)*r(a)
2. a = r(a)*s(a)
3. r(a) is an algebraic integer
4. r(a) is NOT a unit.
>>This establishes the fact that a UNIT algebraic integer is
coprime to
>>every algebraic integer, in the ring of algebraic integers.
A very
>>slight rewording proves the result for a unit of any
commutative
>>ring.
> It is true that a unit algebraic integer is coprime to
every algebraic
> integer.
>>No wonder you think the ring of algebraic integers is
screwed up!
> Youre chasing your tail.
Sez you. You havent found an error in my posted
calculations, and
all you have as an explanation of why Im wrong is to claim
some
problem with the ring of algebraic integers. Ive displayed
verifiable algebraic integers that achieve a factorization 
that
you have claimed (and apparently are continuing to claim) is
impossible. For a reason, you appeal to pixies.
>>Lets get the language straight. I have been clear in this
that Ôa
>>refers to (-1) times any one of the coefficients ai in the
factorization
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>I claim that *each* of those numbers a has a non-unit
algebraic integer
>>factor in common with 5.
> But you do not prove it. Ill point out the gap in what 
you
have that
> follows.
>>PROOF: The above expressions q(a),r(a),s(a) all yield
algebraic integers
>>for any algebraic integer a. When, in addition, a is a root
of the
>>polynomial
>> x^3 - 12 x^2 + 65
>>then the products are as I mentioned above:
>> q(a) r(a) = 5
>> r(a) s(a) = a.
>>Your claim:
>> No it doesnt. What you can prove is that if 
Ôa is
>> coprime to 5 it cant be a unit in the ring of 
algebraic
>> integers, but as Ive said the ring of algebraic 
integers
>> is screwed up.
> Yup.
>>amounts to this:
>>PARAPHRASE:
>> It doesnt matter whether anyone provides algebraic
integers
>> q,r, and s for which
>> 5 = q*r
>> a = r*s
>> it doesnt prove that a and 5 are not coprime.
> That is correct, and in fact, for ONE of the three values
for Ôa they
> are in fact coprime.
No, each of the values of Ôa has a 
corresponding factor
r(a). That
factor is (1) an algebraic integer, (2) not a unit, and (3)
also a
factor of 5. This is what youre disputing, although you 
have
also
agreed that (1 - 3) is in fact correct.
>>However, it *does* prove that. Assume otherwise, that a is
coprime to 5.
>>Then there exist algebraic integers u and v for which
>> a u + 5 v = 1
>>and the above factorizations show this:
>> rs u + qr v = 1
>>so
>> r( su + qv ) = 1,
>>and we deduce that r is a unit.
> You do not deduce you claim, but you cannot prove.
Youre not following the argument, possibly because 
youve
forgotten
what it is Im proving in that statement.
Ill be more explicit in *what* the claim is:
CLAIM: if a is coprime to 5 in the ring of algebraic integers,
and if there is an algebraic integer r, for which the
following
two equations hold:
5 = qr
a = rs
for some algebraic integers q and s, then r must be a unit.
PROOF:
By the definition of coprime in the ring of algebraic
integers, and the assumption that a and 5 are coprime,
there are algebraic integers u and v such that:
a u + 5 v = 1.
Since a = rs, and 5 = qr, this equation can be written:
(rs) u + (qr) v = 1
factoring, we have:
r (su + qv) = 1.
Thus if w = su + qv, we have an algebraic integer w,
for which
r w = 1.
Therefore, r must be a unit in the ring of algebraic
integers.
> In fact, you *can* prove that it is not a unit.
Yes, I did that.
> If you assume that the ring of algebraic integers is ok,
then you
> think youre done.
At no place did I assume that the ring of algebraic integers
is OK.
I have no definition of OK for rings, and in fact only used 
the
following features of that ring:
1. Definition of coprime
2. Substitution of equals for equals (a = rs, 5 = qr)
3. Associativity of multiplication
4. Distributivity of multiplication over addition.
5. Definition of unit.
> But youre chasing your tail, as the ring of algebraic
integers is NOT
> ok, and in fact, in that situation where r *should* be a
unit, its
> NOT in the ring of algebraic integers, as there is a
fascinating
> problem with the ring.
No, it is YOU who claims that r should be a unit, for a and 5
to be
coprime. I did not claim that; in fact, the proof shows that
the
implication is:
(a and 5 coprime) & (r divides both a and r) ==> r is a unit.
I have stated that the proper conclusion of (r not a unit) is
this:
a and 5 are NOT coprime.
>>However, its not one of your fanciful
>>Unit with no inverse hallucinations. In fact, the above
equation
>>provides us with an inverse for r. Based on its
construction, it
>>*must be* an algebraic integer: u and v are specified to be
algebraic
>>integers, and q,r,s were already specified as algebraic
integers.
> <deleted Why dont you show that inverse.
I just did: if a and 5 are coprime, choose the u and v that
force
a u + 5 v = 1.
Then the inverse of r will be w = su + qv.
> Now thats bizarre as the poster has just attacked his own
argument.
> Now I dont claim that r is a unit, as its 
not *in the
ring of
> algebraic integers*.
Idiot. I have already shown that r(a) = 8 a^2 - 4 a - 45
Therefore, since the algebraic integers form a ring (you
still agree
to that, right?), r(a) must be an algebraic integer.
> It amazes me how often posters just throw out false
statements.
Like what?
> Here the posters claim is that a non-monic primitive
irreducible over
> Q, which is what gives you the inverse for r in context, is
an
> algebraic integer.
You are purposely misreading what I have written. Ill break
it down
again for you. Please try to pay attention:
5 = q(a) r(a)
a = r(a) s(a).
q,r, and s are given by these formulas:
q(a) = 8 a^2 - 76 a - 185
r(a) = 8 a^2 - 4 a - 45
s(a) = 4 a^2 - 37 a - 104
Whenever a is a root of the cubic x^3 - 12 x^2 + 65, the above
factorizations are correct, and consist of algebraic integers.
Further, I have shown that if, despite the above
factorization,
a and 5 are coprime (that is: IF YOUR CLAIM IS CORRECT), it
must
follow that r(a) is a unit algebraic integer.
I also show that the minimal polynomial for r(a) is this:
MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
That shows that r(a) CANNOT be a unit.
The contradiction, which youre trying very hard to twist to
suit your own purposes, is that your claim (a coprime to 5)
cannot be true.
> Dont see it?
> Well consider the statements:
>>Then there exist algebraic integers u and v for which
>> a u + 5 v = 1
>>and the above factorizations show this:
>> rs u + qr v = 1
>>so
>> r( su + qv ) = 1,
>>and we deduce that r is a unit.
> The poster wants to claim that r must share a non-unit
factor in
> common with 5 by showing a contradiction by assuming its 
a
unit,
> which actually only shows that it cant be a unit in the
ring of
> algebraic integers.
> So he should have just ended up chasing his tale.
> But he screwed up and managed to bite himself!
Yeah, right. If you could read and understand mathematical
proofs,
you would have just said, Oh, well, I was wrong there.
Instead,
youre fixated on the fantasy that 
youve gone where no
mathematician
has trod before. Fine, have your fantasy.
Its just not mathematics, nor is it supported by anything
other than
your massively-ßawed messing around with uber polynomials.
> James Harris
Dale
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
Nitpicking:
>What? Are you saying that the factorization Ive posted and
provided
>all the necessary calculation to verify that, is somehow
inadequate?
>Please explain. Show me the error.
>That is, show how one of these is not correct:
> 1. 5 = q(a)*r(a)
> 2. a = r(a)*s(a)
> 3. r(a) is an algebraic integer
as are q(a) and s(a). An important point.
> 4. r(a) is NOT a unit.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
lemma: coprime is dependent upon the definition of primality
in the natural numbers. thank you!
> Yes, I see that the rings Ive been working with have a
multiplicative
> identity. Your ring doesnt, does it?
what about m=0 ??... I havent followed any of this, because
no-one wants to put any sort of a hypothesis up front,
to justify it (I mean, I *might*
be able to follow it). I do have one hypothesis:
necessarily, monsieur Harris, iff he exists
in the real world (sik), will not reply to this proof
of his lack of a proof, but
taht is not sufficient to dysprove the results
of his 10-year Mission!
> Notes:
> * 1. The Galois theory argument has been presented before.
> That the polynomial in [3],
> -m*k*f^2 + 3*v*y^2 + y^3
> is irreducible in general is easily seen by letting m = 1
> and f = 5, where it becomes
> y^3 + 72*y^2 - 13825,
> which is easily shown to be irreducible.
> *2. This argument does not say anything about whether e1,
> e2, and e3 are coprime to f.
> *3. This is the key new step in the argument.
> *4. The same argument is implied in Harriss proof of
> FLT, except there he is using objects rather than
> algebraic integers. However it has not been established
> whether objects are different from algebraic integers,
> and a number of basic theorems for objects which are
> needed for his argument have not been proven.
--Dec.2000 ÔWAND Chairman Paul 
ONeill, reelected to Board.
Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: How I know, linchpin of my FLT proof
>
>
>
> <deleted >
>No it doesnt. What you can prove is that if 
Ôa is coprime
to 5 it
>cant be a unit in the ring of algebraic integers, but as
Ive said
>the ring of algebraic integers is screwed up.
>No, no one can prove that. After all, a unit *is* coprime to
every
>>number!
>
>
> Thats true, but it doesnt prove that in a 
given ring two
numbers
> that dont share unit factors *in the ring* are coprime,
given your
> definition of coprime.
For those who dont know W. Dale Hall distinguished himself
by being
a poster who has repeatedly made racist postings.
Interestingly enough, he also made a racist defense of his
racist
postings in at least one case.
I find him somewhat interesting for those reasons.
> So, why did you state that one *could* prove this, if it is
false?
I didnt say you could prove that two numbers that 
dont
share unit
factors are coprime by your definition of coprime.
> Actually, every element of a ring shares ALL unit factors
with EVERY
> OTHER element of the ring. Why would a person even talk
about numbers
> that DONT share unit factors? That is literally talking
about nothing!
It is true that every element of a ring shares all unit
factors with
every other element in the ring, and in fact, that has not
been
challenged.
>
> Its not complicated.
Still W. Dale Hall seeks to be emotional rather than logical.
My theory of racism is that racists are people who are afraid
of their
own inability to live up to the requirements of their world.
So they attack others in order to try and prove to themselves
that
they are worthy.
> Maybe it would help if you consider the ring of evens,
where 2 and 6
> dont share non-unit factors in the ring because 1 and 3
are not in
> the ring of evens, but by *your* definition they are NOT
coprime.
>
> Think about it.
>
> Yes, I see that the rings Ive been working with have a
multiplicative
> identity. Your ring doesnt, does it?
Well Arturo Magidin gave an example that might help you as he
talked
of Z[(sqrt(-5)] where 1+sqrt(-5) is NOT coprime to 2 by his
definition, while it does not share any non-unit factors in
the ring
with 2.
And that ring has a unit.
You can chase your tail, and stay emotional, or you can
follow the
math.
>
>>Let u be a unit algebraic integer, and let p be any
algebraic integer.
>>Then there is an algebraic integer v for which uv = 1, and
we get:
>> u v + 0 p = 1,
>>where 0 is zero.
>
>
> Yeah, where youre saying that it IS a unit.
>
> Im showing that a unit is coprime to everything. Duh.
That has not been challenged.
> Remember, youve tried to use examples where numbers have
been
> determined to NOT be units.
>
> Im showing that if you insist that the following setup:
> 5 = q(a)*r(a)
> a = r(a)*s(a)
> with Ôa and 5 being coprime, then it is *YOU* 
who is
insisting
> that r(a) is a unit.
Actually there are *three* as and for only one of them is
r(a) a
unit.
You claim otherwise but keep making circular arguments which
depend on
a false belief.
You are then, figuratively, chasing your tail.
> Going from there, youve jumped to concluding that they
then must
> share non-unit factors *in the ring of algebraic integers*
which you
> have not proven.
>
> I have shown the non-unit factor: r(a). Can you understand
that?
It is indeed not a unit for any of the as, but it IS 
coprime
for one
of them.
While you may think that simply asserting things makes
mathematical
truth, it does not.
Your emotional needs do not affect the math.
> Rather than actually work to figure out the math, 
youve
instead
> engaged in chasing your tail, publicly through posts on
these forums.
>
> What? Are you saying that the factorization Ive posted 
and
provided
> all the necessary calculation to verify that, is somehow
inadequate?
> Please explain. Show me the error.
> That is, show how one of these is not correct:
> 1. 5 = q(a)*r(a)
> 2. a = r(a)*s(a)
> 3. r(a) is an algebraic integer
> 4. r(a) is NOT a unit.
You have not given proof that r(a) cannot be coprime to 5.
>
>>This establishes the fact that a UNIT algebraic integer is
coprime to
>>every algebraic integer, in the ring of algebraic integers.
A very
>>slight rewording proves the result for a unit of any
commutative
>>ring.
>
>
> It is true that a unit algebraic integer is coprime to
every algebraic
> integer.
>
>
>>No wonder you think the ring of algebraic integers is
screwed up!
>
>
> Youre chasing your tail.
>
> Sez you. You havent found an error in my posted
calculations, and
> all you have as an explanation of why Im wrong is to 
claim
some
> problem with the ring of algebraic integers. Ive 
displayed
> verifiable algebraic integers that achieve a factorization
that
> you have claimed (and apparently are continuing to claim) is
> impossible. For a reason, you appeal to pixies.
You are chasing your tail. Emotion does not change
mathematical
truth.
>
>>Lets get the language straight. I have been clear in this
that Ôa
>>refers to (-1) times any one of the coefficients ai in the
factorization
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>I claim that *each* of those numbers a has a non-unit
algebraic integer
>>factor in common with 5.
>
>
> But you do not prove it. Ill point out the gap in what 
you
have that
> follows.
>
>
>>PROOF: The above expressions q(a),r(a),s(a) all yield
algebraic
integers
>>for any algebraic integer a. When, in addition, a is a root
of the
>>polynomial
>> x^3 - 12 x^2 + 65
>>then the products are as I mentioned above:
>> q(a) r(a) = 5
>> r(a) s(a) = a.
>>Your claim:
>>> No it doesnt. What you can prove is that if 
Ôa is
>> coprime to 5 it cant be a unit in the ring of 
algebraic
>> integers, but as Ive said the ring of algebraic 
integers
>> is screwed up.
>
>
> Yup.
>
>
>>amounts to this:
>>PARAPHRASE:
>> It doesnt matter whether anyone provides algebraic
integers
>> q,r, and s for which
>> 5 = q*r
>> a = r*s
>> it doesnt prove that a and 5 are not coprime.
>
>
> That is correct, and in fact, for ONE of the three values
for Ôa they
> are in fact coprime.
>
> No, each of the values of Ôa has a 
corresponding factor
r(a). That
> factor is (1) an algebraic integer, (2) not a unit, and (3)
also a
> factor of 5. This is what youre disputing, although you
have also
> agreed that (1 - 3) is in fact correct.
However, you have not proven that r cannot be coprime to 5,
as that
cant be proven, as in fact, for one of the three 
as, r IS
coprime to
5.
For the other two as, r does not share non-unit factors in
common
with 5 ***in the ring of algebraic integers***.
>
>>However, it *does* prove that. Assume otherwise, that a is
coprime to 5.
>>Then there exist algebraic integers u and v for which
>> a u + 5 v = 1
>>and the above factorizations show this:
>> rs u + qr v = 1
>>so
>> r( su + qv ) = 1,
>>and we deduce that r is a unit.
>
>
> You do not deduce you claim, but you cannot prove.
>
> Youre not following the argument, possibly because 
youve
forgotten
> what it is Im proving in that statement.
> Ill be more explicit in *what* the claim is:
> CLAIM: if a is coprime to 5 in the ring of algebraic
integers,
> and if there is an algebraic integer r, for which the
following
> two equations hold:
> 5 = qr
> a = rs
> for some algebraic integers q and s, then r must be a unit.
> PROOF:
> By the definition of coprime in the ring of algebraic
> integers, and the assumption that a and 5 are coprime,
> there are algebraic integers u and v such that:
> a u + 5 v = 1.
Youre chasing your tail.
By your definition in Z[(sqrt(-5)] 1+sqrt(-5) and 2 are not
coprime
though they dont share a non-unit factor in the ring.
But they *do* share a non-unit factor which is sqrt(2) in the
ring of
algebraic integers.
> Since a = rs, and 5 = qr, this equation can be written:
> (rs) u + (qr) v = 1
> factoring, we have:
> r (su + qv) = 1.
> Thus if w = su + qv, we have an algebraic integer w,
> for which
> r w = 1.
> Therefore, r must be a unit in the ring of algebraic
> integers.
You went in a circle.
> In fact, you *can* prove that it is not a unit.
>
> Yes, I did that.
Yes you did.
> If you assume that the ring of algebraic integers is ok,
then you
> think youre done.
>
> At no place did I assume that the ring of algebraic
integers is OK.
> I have no definition of OK for rings, and in fact only used
the
> following features of that ring:
> 1. Definition of coprime
> 2. Substitution of equals for equals (a = rs, 5 = qr)
> 3. Associativity of multiplication
> 4. Distributivity of multiplication over addition.
> 5. Definition of unit.
> But youre chasing your tail, as the ring of algebraic
integers is NOT
> ok, and in fact, in that situation where r *should* be a
unit, its
> NOT in the ring of algebraic integers, as there is a
fascinating
> problem with the ring.
>
> No, it is YOU who claims that r should be a unit, for a and
5 to be
> coprime. I did not claim that; in fact, the proof shows
that the
> implication is:
> (a and 5 coprime) & (r divides both a and r) ==> r is a
unit.
> I have stated that the proper conclusion of (r not a unit)
is this:
> a and 5 are NOT coprime.
You are chasing your tail.
>
>>However, its not one of your fanciful
>>Unit with no inverse hallucinations. In fact, the above
equation
>>provides us with an inverse for r. Based on its
construction, it
>>*must be* an algebraic integer: u and v are specified to be
algebraic
>>integers, and q,r,s were already specified as algebraic
integers.
>
>
> <deleted >
> Why dont you show that inverse.
>
> I just did: if a and 5 are coprime, choose the u and v that
force
> a u + 5 v = 1.
> Then the inverse of r will be w = su + qv.
Are you saying that you showed the inverse to a number 
youve
proven
is NOT a unit, and that it is an algebraic integer?
> Now thats bizarre as the poster has just attacked his own
argument.
>
> Now I dont claim that r is a unit, as its 
not *in the
ring of
> algebraic integers*.
>
> Idiot. I have already shown that r(a) = 8 a^2 - 4 a - 45
Showing your true colors, and chasing your tail.
> Therefore, since the algebraic integers form a ring (you
still agree
> to that, right?), r(a) must be an algebraic integer.
Yet you claim to have proven it cant be a unit, yet you 
made
a claim
about its inverse being a unit.
> It amazes me how often posters just throw out false
statements.
>
> Like what?
> Here the posters claim is that a non-monic primitive
irreducible over
> Q, which is what gives you the inverse for r in context, is
an
> algebraic integer.
>
> You are purposely misreading what I have written. Ill
break it down
> again for you. Please try to pay attention:
> 5 = q(a) r(a)
> a = r(a) s(a).
> q,r, and s are given by these formulas:
> q(a) = 8 a^2 - 76 a - 185
> r(a) = 8 a^2 - 4 a - 45
> s(a) = 4 a^2 - 37 a - 104
> Whenever a is a root of the cubic x^3 - 12 x^2 + 65, the
above
> factorizations are correct, and consist of algebraic
integers.
> Further, I have shown that if, despite the above
factorization,
> a and 5 are coprime (that is: IF YOUR CLAIM IS CORRECT), it
must
> follow that r(a) is a unit algebraic integer.
> I also show that the minimal polynomial for r(a) is this:
> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
> That shows that r(a) CANNOT be a unit.
> The contradiction, which youre trying very hard to twist 
to
> suit your own purposes, is that your claim (a coprime to 5)
> cannot be true.
Showing that r(a) cant be a unit does not prove that it 
must
have a
non-unit factor in common with 5.
Notice the following exchange.
> Dont see it?
>
> Well consider the statements:
>
>
>>Then there exist algebraic integers u and v for which
>> a u + 5 v = 1
>>and the above factorizations show this:
>> rs u + qr v = 1
>>so
>> r( su + qv ) = 1,
>>and we deduce that r is a unit.
>
>
> The poster wants to claim that r must share a non-unit
factor in
> common with 5 by showing a contradiction by assuming its 
a
unit,
> which actually only shows that it cant be a unit in the
ring of
> algebraic integers.
>
> So he should have just ended up chasing his tale.
>
> But he screwed up and managed to bite himself!
That is, he went on as if r were a unit, and claimed to have
given its
inverse, which is what followed but is now gone.
But luckily its still in the post, so copying from earlier.
>>However, its not one of your fanciful
>>Unit with no inverse hallucinations. In fact, the above
equation
>>provides us with an inverse for r. Based on its
construction, it
>>*must be* an algebraic integer: u and v are specified to be
algebraic
>>integers, and q,r,s were already specified as algebraic
integers.
There the poster claims that the inverse for r must be an
algebraic
integer, with an emphasis on the must be.
That is, in fact, false, and goes against the posters own
claims.
He quit chasing his tale for a while in order to bite himself,
figuratively speaking, of course.
> Yeah, right. If you could read and understand mathematical
proofs,
> you would have just said, Oh, well, I was wrong there.
Instead,
> youre fixated on the fantasy that 
youve gone where no
mathematician
> has trod before. Fine, have your fantasy.
> Its just not mathematics, nor is it supported by anything
other than
> your massively-ßawed messing around with uber polynomials.
Now then back to my theory. Is it true that racists feel a
need to
attack others based on their own fear of inferiority?
It makes sense.
Imagine that youre very successful, beloved in your
community, and
generally things go well for you in your life.
Why would you go after other people?
Wouldnt you have to think they were some kind of threat to
you?
James Harris
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
[.snip.]
>> Yes, I see that the rings Ive been working with have a
multiplicative
>> identity. Your ring doesnt, does it?
>Well Arturo Magidin gave an example that might help you as
he talked
>of Z[(sqrt(-5)] where 1+sqrt(-5) is NOT coprime to 2 by his
>definition, while it does not share any non-unit factors in
the ring
>with 2.
>And that ring has a unit.
You missed the point of that example, but that is neither
here nor
there (and you also failed to notice that I managed to prove
both
statements without having to appeal to a larger ring, as you
did
when you argued that they were not coprime because sqrt(2)
divides
both in a larger ring).
An important point for you to ponder, however, is that YOUR
definition
is insufficient for your eventual purposes. What you want to
do,
eventually, is to argue that you have a product
a*b*c = 0 (mod f)
and that two of a,b,c, say a and b, are coprime to f, and
therefore
conclude that c=0 (mod f).
But that step is invalid using only your definition. The
validity of
that step depends on the fact that the correct definition of a
and b
are coprime is equivalent, in rings with 1 (and assuming the
Axiom of
Choice), with the statement that there exist x and y in the
ring such
that ax+by = 1.
For, coming back to the example you are apparently agreeing
with, you
have that
(i) (1+sqrt(-5))*(1-sqrt(-5)) = 0 (mod 2) in Z[sqrt(-5)];
(ii) 2 and 1+sqrt(=5) have no non-unit common factors in
Z[sqrt(-5)], so
by
your definition are coprime;
(iii) but it is not the case that (1-sqrt(-5)) = 0 (mod 2) in
Z[sqrt(-5)].
Since the eventual purpose of your coprimality argument is to
terms in a congruence, if you rely on your definition, you
will not be
definition I have
been using, and more specifically the equivalent statement
mentioned.
THEOREM. Let R be a commutative ring with 1, and let a, b,
and c be
elements of R. Assume that a and c are coprime in R, in the
sense that
there exist elements x and y in R such that ax+cy = 1. If a*b
= 0 (mod
c), then b=0 (mod c) (congruences occurring in R as well).
Proof. Multiply ax+cy=1 by b to obtain abx + cby = b. Since
ab=0 (mod
c), there exists k in R such that ab=ck. Substituting, we have
b = abx + cby
= ckx + cby
= c(kx+by).
Since kx+by is in R, this means that b=0 (mod c). QED
This is just the general commutative ring theoretic version
of the
classical result for integers:
PROP. c|ab -> ( (c,a)=1 -> c|b).
So you really need to switch to the correct definition in any
case, or
else prove that in your Ring of Objects (assuming you ever
correct
the current problems with your definition, which has no actual
referent right now, so no such ring exists) a and b have no
common
factors
other than units is equivalent to there exists x and y such
that
ax+by=1 (the implication from the existence of x and y to the
no
common factors other than units is valid in every commutative
ring
with 1).
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
> Well its easy enough to show my ring 
definition removes all
> fractions that would make an integer other than -1 or 1 a
unit.
You implicitly reveal below that if your object ring J
contains an
algebraic integer w then J also contains all the conjugates
of w, i.e.
J contains all the roots of the monic min.poly of w over Q.
This then
implies that the only algebraic numbers in J are algebraic
integers.
The proof is very easy: Suppose that w is an algebraic number
in J and
let f(X) in Q[X] be the monic min.poly of w over Q. Every
coef of f(X)
is a (symmetric) polynomial in the roots (conjugates)
w,w,w...; e.g.
(X-w)(X-w)(X-w) = X^3 - (w+w+w) X^2 + 
(ww+ww+ww) X - www
Since J is _ring_ containing w,w,w... it must contain all
sums and
products of such; therefore J must contain every coef of the
min.poly.
However, by definition, the coefs of the min.poly are rational
numbers,
so if J contains only integral rationals then the min.poly
contains only
integral rational coefs. Thus w is an algebraic integer (by
definition).
Hence every algebraic number in J is in fact an algebraic
integer. QED
The above proof implies that the ring of algebraic integers
may be
characterized as the maximal subring I of the algebraic
numbers that
is closed under conjugates (as above) and furthermore
contains only
integral rational numbers: I / Q = Z. The latter constraint
is very
natural. It simply means that integer divisibility remains
the same
in I as in Z: m|n in I <=> m|n in Z -- a necessity if we wish
to
I to faithfully extend integrality and divisibility relations
on Z,
e.g. when solving polynomial Diophantine equations in Z by
factoring
them over a splitting field. Likewise, the former constraint,
that
if I contains an algebraic integer w then I also contains all
the
min.poly conjugates of w, is quite natural too since
conjugates
are _algebraically_ indistinguishable: number fields Q(w) and
Q(w)
are isomorphic, so it would very unnatural to define one of w
or w
to be integral without doing the same for the other. As an
example,
consider the prior-mentioned method of solving Diophantine
equations
over Z by factoring them over some number ring. The best
possible
simplification results in extension fields where 
the polynomial
splits completely into linear factors. For this one needs to
work
in a splitting field containing all of the conjugates. If your
object number rings are too weak to provide all the conjugates
then they wont provide a natural domain to support such
fundamental
problem solving techniques. Thus instead of reducing a
non-linear
equation to some linear equations in a higher number ring,
one would
instead be stuck working with non-linear equations in an
object ring.
Hence if you defined J to exclude certain conjugates then
youd be
unnecessarily limiting its power. On the other hand, if J
violates
the other constraint by containing noninteger rational
numbers then
it does not faithfully model the divisibility of integers and
again
this limits the power of deducing results about the integers.
For
example, if 2 becomes a unit in J then J loses any sense of
parity,
so all arguments involving parity are lost.
Thats not to say that rings violating these constraints
arent useful.
They do find use for particular purposes. For example, if one
is only
interested in determining information about a particular
prime p, such
as the power of p that divides a solution of a Diophantine
equation,
then one can often simplify matters greatly by working in a
ring where
all other primes are units -- a so-called localization at p.
It focuses
attention only on the prime p. This is the approach of
valuation theory,
which espouses a local (vs. global) approach to algebraic
number theory.
More generally one may focus in on a (finite) set of primes
p,q,r ...
To do such one simply works in the subring of the quotient
field where
all other primes are units. In other words one permits
denominators
containing only those primes one wishes to ignore. It turns
out every
subring of the algebraic numbers containing the algebraic
integers I
may be realized in this manner, i.e. by adjoining inverses of
certain
elements. Such rings are familiar when one studies algebraic
numbers
locally, via valuation theory.
Thus any attempt to generalize the algebraic integers is
severely
constrained by the innate mathematical structure as described
above.
If you ever succeed in rigorously specifying some useful
subring J
of the algebraic numbers then no doubt it will be one of the
very
well-known rings alluded to above. Theory doesnt allow
otherwise.
With that overview, lets move on to the 
specifics of your
post.
> Its kind of fun to take a convoluted path to check
something like
1/(1+i).
> Let u1 = 1/(1+i), then u2 = 1+i.
> Now (1+i)(1-i) = 2, so u2 (1-i) = 2, and multiplying both
sides by
> u1, gives
> 1-i = 2u1, and adding 2i to both sides gives
> 1+2i = 2u1 + 2, which is u2 = 2(u1 + 1),
> which would force 2 to be a factor of u2, which would make
it a unit,
> so 1/(1+i) is excluded.
Correcting arithmetical errors and removing extraneous
variables yields:
| Now (1+i)(1-i) = 2 and multiplying both sides by
| 1/(1+i), gives
|
| 1-i = 2/(1+i), and adding 2i to both sides gives
|
| 1+i = 2/(1+i) + 2i, which is
|
| 1+i = 2(1/(1+i) + i)
i.e. J has 1/(1+i) => J has (1+i)/2 = i + 1/(1+i)
=> J has (1+i)/2 * 1/(1+i) = 1/2
In fact this example does not depend on the ring R but only
on w,
since here w = 1/(1+i) has trace(w) = w+w = 1 in R; however
if trace(w) = w + w in R then w in R <=> w 
in R
Hence, as above, w and w in R => so are coefs of 
ws
min.poly,
so Z[w] / Q = Z <=> w is an algebraic integer, which its
not in this case since w = 1/(1+i) has norm ww = 1/2.
>Jim Propp asks:
>>
>> Does there exist an algebraic number w, NOT an algebraic
integer,
>> but nevertheless with the property that the only rational
numbers
>> in Z[w] are the rational integers?
> w = 1/(3 + sqrt(2)) [...]
> Here as you could have above, you just notice that
> (3+sqrt(2))(3-sqrt(2)) = 7,
> so if 1/(3+sqrt(2)) is a unit, so is 3+sqrt(3), which forces
> 3-sqrt(2) to have a factor that is 7.
> Now considering 7x = 3-sqrt(2), you get
> 7x^2 + 6x + 1 = 0, which gives you two solutions,
- 6
> one of which forces 3-sqrt(2) to be a unit, which means 7
is a unit.
The quoted post shows 7 isnt a unit in Z[w], w =
1/(3+sqrt(2)).
In any larger ring R containing w (and Z) it is easy to see,
(using ww = 1/7, and w in R => 1/w in R via 1/w in I)
1/7 in R <=> w in R
So your object ring J apparently satisfies the following
[cc] w in J => w in J
since that seems to be the only natural way to interpret your
(too sketchy) deduction that 1/7 in J. But the only natural
way for [cc] to hold in general is to hypothesize that your
ring J is closed under conjugates, as in the discussion which
opened this post. In that case your object ring J is subject
to all of the constraints mentioned there. On the other hand,
if your object ring isnt closed under conjugates then it is
not at all clear how you deduced that 1/7 in J. In any case
if J is not closed under conjugates then its not going to 
be
very useful for your intended purpose of solving (arbitrary)
Diophantine equations over the ring of integers. Try as you
may
you wont find another subring of algebraic 
numbers that has
more power for solving arbitrary Diophantine equations over Z.
Youd soon realize this if, instead of chasing wild pseudo-
mathematical gooses, you studied some algebraic number theory.
All the wrong paths towards the crystallization of the notion
of algebraic integer were well-explored over a century ago.
Why do you insist in wallowing in this mud when instead you
could stand on the shoulders of giants to view the majestic
panoramas of the queen of sciences?
-Bill Dubuque
===
Subject: Re: How I know, linchpin of my FLT proof
Visiting Assistant Professor at the University of Montana.
[.snip.]
>> That is, show how one of these is not correct:
>> 1. 5 = q(a)*r(a)
>> 2. a = r(a)*s(a)
>> 3. r(a) is an algebraic integer
>> 4. r(a) is NOT a unit.
>You have not given proof that r(a) cannot be coprime to 5.
r(a) is a divisor of 5 and not a unit. Therefore, r(a) is a
common
factor of r(a) and 5, and not a unit. Therefore, even
according to
your definition, there is a nonunit common factor of r(a) and
5, and
so they are not coprime in the ring of all algebraic integers.
[.snip.]
>By your definition in Z[(sqrt(-5)] 1+sqrt(-5) and 2 are not
coprime
>though they dont share a non-unit factor in the ring.
>But they *do* share a non-unit factor which is sqrt(2) in
the ring of
>algebraic integers.
So what? Coprime, like factor, divides, and unit, are not
absolute properties, but contextual ones. They depend on the
ambient
ring.
And yes, 1+sqrt(-5) and 2 are not coprime IN Z[sqrt(-5)],
even though
they do not have any non unit common factor IN Z[sqrt(-5)].
In Z[sqrt(-5),sqrt(2)], they are still not coprime (in that
ring), and
in this ring they do have a non unit common factor, namely
sqrt(2). And in Z[sqrt(-5), sqrt(2), 1/sqrt(2)], they are
coprime (in
that ring) and have no non unit common factor.
So what?
[.snip.]
>That is, he went on as if r were a unit, and claimed to have
given its
>inverse, which is what followed but is now gone.
>But luckily its still in the post, so copying from 
earlier.
>However, its not one of your fanciful
>Unit with no inverse hallucinations. In fact, the above
equation
>provides us with an inverse for r. Based on its
construction, it
>*must be* an algebraic integer: u and v are specified to be
algebraic
>integers, and q,r,s were already specified as algebraic
integers.
>There the poster claims that the inverse for r must be an
algebraic
>integer, with an emphasis on the must be.
Yes. By definition, r is a unit in the ring of algebraic
integers if
and only if it has a multiplicative inverse in the ring of
algebraic
integers.
Now, if you think of the ring of all algebraic integers as
sitting
inside the complex numbers, then we know that there is one
and only
one complex number which can be a multiplicative inverse for
r:
namely, the complex number 1/r. So r, an algebraic integer,
is a unit
in the ring of all algebraic integers, if and only if the
complex
number 1/r happens to also be an algebraic integer.
So we take the number 1/r, and show that, although it is an
algebraic
number, it is not an algebraic integer. And since this is the
ONLY
complex number which could possibly be a multiplicative
inverse for r,
it follows that r does NOT have an inverse in the ring of
algebraic
integers.
Clear now?
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: How I know, linchpin of my FLT proof
> Why do you insist in wallowing in this mud when instead you
> could stand on the shoulders of giants to view the majestic
> panoramas of the queen of sciences?
He seems to prefer slithering under the heels of midgets.
--
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: How I know, linchpin of my FLT proof
> Its just not mathematics, nor is it supported by anything
other than
> your massively-ßawed messing around with uber polynomials.
Now, now: in Harristotelian terms ersatz reasoning is
considered
perfectly valid, and uber polynomials are perfectly
respectable
objects - they even form a ring!
Kummer, Kummer, Uber factors.....
===
Subject: Re: How I know, linchpin of my FLT proof
><deleted>>No it doesnt. What you can prove is that if 
Ôa
is coprime to 5 it
>cant be a unit in the ring of algebraic integers, but as
Ive said
>the ring of algebraic integers is screwed up.
>No, no one can prove that. After all, a unit *is* coprime to
every
>>number!
>Thats true, but it doesnt prove that in a 
given ring two
numbers
>that dont share unit factors *in the ring* are coprime,
given your
>definition of coprime.
And now....
ITS OUT OF LEFT FIELD.......
Fresh from the mouth of lets keep with the math, 
its
lets lynch whitey:
> For those who dont know W. Dale Hall distinguished 
himself
by being
> a poster who has repeatedly made racist postings.
You can say what you like. I have made no racist postings.
Have I
reason. I do not accept abuse unchallenged, and I maintain
that it
is reasonable to taunt those who are abusive.
But as for the claim that I have *REPEATEDLY* made racist
postings, well Im sorry, but you are just plain wrong.
Correctness? Not at all. My position is that Political
Correctness
is what destroyed the vitality of the US left during the late
1960s
through the 1970s; Political Correctness is only
slightly-veiled
Stalinism, and I have no part of it.
I find it offensive that JSH finds it necessary to 
run to the
cry of
racism at this juncture. Find me one racist epithet that I
have
applied to you, Mister Smarty Pants.
> Interestingly enough, he also made a racist defense of his
racist
> postings in at least one case.
Interestingly enough, JSH finds racism wherever he cannot win
by
honest debate. Here is a pointer to the alleged racist
posting:
<3D372ABB.2000802@farir.com>
The alleged racist content is a 20-line passage from a
288-line
demean the African-American, or suggest in any way that
African-
Americans are less qualified, less hard-working, or less
deserving
of full participation in society? Not at all.
and here is a pointer to the racist defense.
<3E360594.6000908@farir.com>
concessions, but rather to outline my intentions behind the
consider the case closed. If you want to hold parades, start
new
newsgroups, and just have a big-old blowout on my account,
then
have at it. I dont consider it a topic worth any more
attention.
> I find him somewhat interesting for those reasons.
Yes, when you dont have a mathematical argument (which is
common)
you go for the slur.
>>So, why did you state that one *could* prove this, if it is
false?
> I didnt say you could prove that two numbers that 
dont
share unit
> factors are coprime by your definition of coprime.
Heres your direct quote, placed in context:
>>even>>: JSH
>odd>>: me
> In fact, the minimal polynomial of this number (r(a) for -a
=
> any of the above ais) is given as:
> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
> The above facts prove that this a has a non-unit algebraic
> integer as a factor.
>> No it doesnt. What you can prove is that if 
Ôa is
coprime to 5
>> it cant be a unit in the ring of algebraic integers, but
as Ive
>> said the ring of algebraic integers is screwed up.
> No, no one can prove that. After all, a unit *is* coprime
to every
> number!
>> Thats true, but it doesnt prove that in a 
given ring two
numbers
>> that dont share unit factors *in the ring* are coprime,
given your
>> definition of coprime.
> So, why did you state that one *could* prove this, if it is
false?
> Actually, every element of a ring shares ALL unit factors
with EVERY
> OTHER element of the ring. Why would a person even talk
about numbers
> that DONT share unit factors? That is literally talking
about
> nothing!
Note that JSH cant keep straight what it is that 
hes been
saying.
First, he says one could prove that if Ôa is 
coprime to 5
then
it cant be a unit. I pointed out that such a statement is
false,
since EVERY unit is coprime to EVERYTHING!
JSH backtracks and says (in effect) that I misunderstood him,
and
rephrases to talk about two numbers in a ring that dont
share unit
factors. Again, this is nonsense, since EVERY number shares
ALL unit
factors with EVERY OTHER number.
> It is true that every element of a ring shares all unit
factors with
> every other element in the ring, and in fact, that has not
been
> challenged.
So why did you bring it up? Ill quote again:
... but it diesnt prove that in a given ring two numbers
that dont share unit factors *in the ring* are coprime ...
Ill repeat myself here: you are talking about something 
that
doesnt
exist: two numbers that dont share unit factors. That
combination
of technical terms refers to something that fails to exist.
>Its not complicated.
> Still W. Dale Hall seeks to be emotional rather than
logical.
> My theory of racism is that racists are people who are
afraid of their
> own inability to live up to the requirements of their world.
> So they attack others in order to try and prove to
themselves that
> they are worthy.
responding?
My theory is that many would-be intellectuals cook up
conspiracies
to explain their personal failures. Racism is one of those,
as is
Anti-Semitism. It works on both sides.
As far as Im concerned, to accuse your critics of racism is
to
announce that you have just run out of real ideas.
>Maybe it would help if you consider the ring of evens, where
2 and 6
>dont share non-unit factors in the ring because 1 and 3 
are
not in
>the ring of evens, but by *your* definition they are NOT
coprime.
>Think about it.
>>Yes, I see that the rings Ive been working with have a
multiplicative
>>identity. Your ring doesnt, does it?
> Well Arturo Magidin gave an example that might help you as
he talked
> of Z[(sqrt(-5)] where 1+sqrt(-5) is NOT coprime to 2 by his
> definition, while it does not share any non-unit factors in
the ring
> with 2.
> And that ring has a unit.
Are you claiming that the ring of algebraic integers has
this peculiar property?
Are we discussing algebraic integers?
Are you pretending that you have studied ANY ring theory?
The example, however points this out:
If elements are coprime, then they may fail to share
any non-unit factors.
However, if they share non-unit factors, anyone can prove
that they are
NOT coprime. Well, maybe not ANYONE, but some people who are
familiar
with commutative algebra may be able to prove it.
> You can chase your tail, and stay emotional, or you can
follow the
> math.
You must imagine that I dont understand something, and that
Im
ßailing about. After all, you are rushing to claim Im 
being
hysterical, and racist, and that I am cruel to animals and
children.
>>Let u be a unit algebraic integer, and let p be any
algebraic integer.
>>Then there is an algebraic integer v for which uv = 1, and
we get:
>> u v + 0 p = 1,
>>where 0 is zero.
>Yeah, where youre saying that it IS a unit.
>>Im showing that a unit is coprime to everything. Duh.
> That has not been challenged.
where that argument was coming from.
>Remember, youve tried to use examples where numbers have
been
>determined to NOT be units.
>>Im showing that if you insist that the following setup:
>> 5 = q(a)*r(a)
>> a = r(a)*s(a)
>>with Ôa and 5 being coprime, then it is *YOU* 
who is
insisting
>>that r(a) is a unit.
> Actually there are *three* as and for only one of them is
r(a) a
> unit.
There are three as. Yes.
Each of these as will work in the above set of equations
(separately,
of course). In each case, the r(a) is a root of the cubic
MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
which shows it not to be a unit.
> You claim otherwise but keep making circular arguments
which depend on
> a false belief.
> You are then, figuratively, chasing your tail.
>Going from there, youve jumped to concluding that they 
then
must
>share non-unit factors *in the ring of algebraic integers*
which you
>have not proven.
>>I have shown the non-unit factor: r(a). Can you understand
that?
> It is indeed not a unit for any of the as, but it IS
coprime for one
> of them.
Coprime to what?
What does it mean for A number to be coprime (without any
reference to
another number)?
> While you may think that simply asserting things makes
mathematical
> truth, it does not.
This is the game of a five year-old. I have provided a proof,
and you
are simply refusing to acknowledge its validity.
> Your emotional needs do not affect the math.
That is yet another sign of your desperation. Get a grip.
>Rather than actually work to figure out the math, 
youve
instead
>engaged in chasing your tail, publicly through posts on
these forums.
>>What? Are you saying that the factorization Ive posted 
and
provided
>>all the necessary calculation to verify that, is somehow
inadequate?
>>Please explain. Show me the error.
>>That is, show how one of these is not correct:
>> 1. 5 = q(a)*r(a)
>> 2. a = r(a)*s(a)
>> 3. r(a) is an algebraic integer
>> 4. r(a) is NOT a unit.
> You have not given proof that r(a) cannot be coprime to 5.
I have proven
(1) r(a) is a factor of 5
(2) r(a) is not a unit.
This proves that r(a) cannot be coprime to 5.
Why? Well, heres a proof:
Suppose r(a) and 5 are coprime. Then there are algebraic
integers u and v for which:
u r + v 5 = 1
Now, 5 = q r, so we can rewrite the equation
u r + v q r = 1
and by factoring, we find this:
(u + v q) r = 1.
Thus, if 5 is coprime to r, then r must be a unit,
with inverse equal to u + vq.
Note that I started with statement 1, and deduced the
contradiction
to statement 2. The statement proved is that if (1) and (2)
hold,
then r(a) is NOT coprime to 5.
Satisfied?
Or, did you want your definition: x and y are coprime in R if
there is
no non-unit factor of both x and y in the ring R.
Assuming r(a) an algebraic integer
(1) r(a) is a factor of 5
(2) r(a) is not a unit.
Claim: There is a non-unit algebraic integer z which is
a factor of both r and 5, in the following sense: there
are algebraic integers v and w for which
r = wz
5 = vz
Proof: let z = r. Then
r = 1*z
5 = qr = qz
Therefore, there is a non-unit algebraic integer z that
is a factor of each of r and 5.
Therefore, r and 5 cannot be coprime.
>>This establishes the fact that a UNIT algebraic integer is
coprime to
>>every algebraic integer, in the ring of algebraic integers.
A very
>>slight rewording proves the result for a unit of any
commutative
>>ring.
>It is true that a unit algebraic integer is coprime to every
algebraic
>integer.
>>No wonder you think the ring of algebraic integers is
screwed up!
>Youre chasing your tail.
>>Sez you. You havent found an error in my posted
calculations, and
>>all you have as an explanation of why Im wrong is to 
claim
some
>>problem with the ring of algebraic integers. Ive 
displayed
>>verifiable algebraic integers that achieve a factorization
that
>>you have claimed (and apparently are continuing to claim) is
>>impossible. For a reason, you appeal to pixies.
> You are chasing your tail. Emotion does not change
mathematical
> truth.
Your only defense is the pixie defense mysterious, yet
fascinating
defect of the ring of algebraic integers. If you had a 
specific
example, you would have shown it by now.
>>Lets get the language straight. I have been clear in this
that Ôa
>>refers to (-1) times any one of the coefficients ai in the
factorization
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).
>>I claim that *each* of those numbers a has a non-unit
algebraic integer
>>factor in common with 5.
>But you do not prove it. Ill point out the gap in what you
have that
>follows.
>>PROOF: The above expressions q(a),r(a),s(a) all yield
algebraic
integers
>>for any algebraic integer a. When, in addition, a is a root
of the
>>polynomial
>> x^3 - 12 x^2 + 65
>>then the products are as I mentioned above:
>> q(a) r(a) = 5
>> r(a) s(a) = a.
>>Your claim:
>> No it doesnt. What you can prove is that if 
Ôa is
>> coprime to 5 it cant be a unit in the ring of 
algebraic
>> integers, but as Ive said the ring of algebraic 
integers
>> is screwed up.
>Yup.
>>amounts to this:
>>PARAPHRASE:
>> It doesnt matter whether anyone provides algebraic
integers
>> q,r, and s for which
>> 5 = q*r
>> a = r*s
>> it doesnt prove that a and 5 are not coprime.
>That is correct, and in fact, for ONE of the three values
for Ôa they
>are in fact coprime.
>>No, each of the values of Ôa has a 
corresponding factor
r(a). That
>>factor is (1) an algebraic integer, (2) not a unit, and (3)
also a
>>factor of 5. This is what youre disputing, although you
have also
>>agreed that (1 - 3) is in fact correct.
> However, you have not proven that r cannot be coprime to 5,
as that
> cant be proven, as in fact, for one of the three 
as, r IS
coprime to
> 5.
No,
for EVERY value of a (root of the appropriate cubic):
r(a) divides 5,
r(a) is not a unit.
Therefore r(a) is NOT coprime to 5. There are two proofs
above, if you
would care to scroll back to see them.
> For the other two as, r does not share non-unit factors 
in
common
> with 5 ***in the ring of algebraic integers***.
Still with this game, eh? r(a) is a divisor of 5. 5 is a
multiple of
r(a). Everything that divides r(a) ALSO divides 5. Every
non-unit
algebraic integer that divides r(a) is a non-unit algebraic
integer
that divides 5.
>>However, it *does* prove that. Assume otherwise, that a is
coprime to 5.
>>Then there exist algebraic integers u and v for which
>> a u + 5 v = 1
>>and the above factorizations show this:
>> rs u + qr v = 1
>>so
>> r( su + qv ) = 1,
>>and we deduce that r is a unit.
>You do not deduce you claim, but you cannot prove.
>>Youre not following the argument, possibly because 
youve
forgotten
>>what it is Im proving in that statement.
>>Ill be more explicit in *what* the claim is:
>> CLAIM: if a is coprime to 5 in the ring of algebraic
integers,
>> and if there is an algebraic integer r, for which the
following
>> two equations hold:
>> 5 = qr
>> a = rs
>> for some algebraic integers q and s, then r must be a unit.
>> PROOF:
>> By the definition of coprime in the ring of algebraic
>> integers, and the assumption that a and 5 are coprime,
>> there are algebraic integers u and v such that:
>> a u + 5 v = 1.
> Youre chasing your tail.
> By your definition in Z[(sqrt(-5)] 1+sqrt(-5) and 2 are not
coprime
> though they dont share a non-unit factor in the ring.
I suppose thats correct. They are not coprime, since they 
do
not
generate the unit ideal *in that ring*.
> But they *do* share a non-unit factor which is sqrt(2) in
the ring of
> algebraic integers.
In a larger ring, they acquire more factors. So what?
>> Since a = rs, and 5 = qr, this equation can be written:
>> (rs) u + (qr) v = 1
>> factoring, we have:
>> r (su + qv) = 1.
>> Thus if w = su + qv, we have an algebraic integer w,
>> for which
>> r w = 1.
>> Therefore, r must be a unit in the ring of algebraic
>> integers.
> You went in a circle.
A circle. I proved what I set out to prove: a divisor of 5
that is coprime to 5 is NECESSARILY a unit. That is not
circularity,
and if you had any notion of what a proof is about, you would
have recognized that.
Unfortunately, you have that ignorance is golden thing going
on.
>In fact, you *can* prove that it is not a unit.
>>Yes, I did that.
> Yes you did.
>If you assume that the ring of algebraic integers is ok,
then you
>think youre done.
>>At no place did I assume that the ring of algebraic
integers is OK.
>>I have no definition of OK for rings, and in fact only used
the
>>following features of that ring:
>> 1. Definition of coprime
>> 2. Substitution of equals for equals (a = rs, 5 = qr)
>> 3. Associativity of multiplication
>> 4. Distributivity of multiplication over addition.
>> 5. Definition of unit.
>But youre chasing your tail, as the ring of algebraic
integers is NOT
>ok, and in fact, in that situation where r *should* be a
unit, its
>NOT in the ring of algebraic integers, as there is a
fascinating
>problem with the ring.
Again! You continue to use the ring of algebraic integers is
NOT OK
as a factor in an argument dedicated to proving ring of
algebraic
integers is NOT OK . THAT is what is called circular
reasoning.
>>No, it is YOU who claims that r should be a unit, for a and
5 to be
>>coprime. I did not claim that; in fact, the proof shows
that the
>>implication is:
>> (a and 5 coprime) & (r divides both a and r) ==> r is a
unit.
>>I have stated that the proper conclusion of (r not a unit)
is this:
>> a and 5 are NOT coprime.
> You are chasing your tail.
You must really like that phrase. At last count, its ten
times in
>>However, its not one of your fanciful
>>Unit with no inverse hallucinations. In fact, the above
equation
>>provides us with an inverse for r. Based on its
construction, it
>>*must be* an algebraic integer: u and v are specified to be
algebraic
>>integers, and q,r,s were already specified as algebraic
integers.
><deleted>>Why dont you show that inverse.
>>I just did: if a and 5 are coprime, choose the u and v that
force
>> a u + 5 v = 1.
>>Then the inverse of r will be w = su + qv.
> Are you saying that you showed the inverse to a number
youve proven
> is NOT a unit, and that it is an algebraic integer?
Are you that stupid?
I showed that the only way for two numbers, that share a
given number
as a divisor, to be coprime, is for that divisor to have been
a unit.
Thats what I showed. No more and no less.
Your claim is that a and 5 are coprime. I showed a common
factor.
You claimed that such a demonstration was insufficient, and I
showed
that the ONLY way for the demonstration to have been
insufficient
would be that the common factor r(a) was a unit.
I then showed that r(a) is NOT a unit.
Yet you cant accept the facts.
>Now thats bizarre as the poster has just attacked his own
argument.
>Now I dont claim that r is a unit, as its 
not *in the ring
of
>algebraic integers*.
>>Idiot. I have already shown that r(a) = 8 a^2 - 4 a - 45
> Showing your true colors, and chasing your tail.
Oh, you can call people dishonest, frauds, lying bastards,
and worse,
yet when you persist in being purposely obtuse, no one can
show how
you are behaving? You just said that r was not in the ring of
algebraic integers, despite more than adequate demonstration
of that
fact. You are being idiotic.
>>Therefore, since the algebraic integers form a ring (you
still agree
>>to that, right?), r(a) must be an algebraic integer.
> Yet you claim to have proven it cant be a unit, yet you
made a claim
> about its inverse being a unit.
No, youre being obtuse again.
Get the argument straight:
I proved it cant be a unit.
I proved that if a and 5 were coprime, then it would have had
to be a
unit.
But its not a unit.
Therefore a and 5 are not coprime.
>It amazes me how often posters just throw out false
statements.
>>Like what?
>Here the posters claim is that a non-monic primitive
irreducible over
>Q, which is what gives you the inverse for r in context, is
an
>algebraic integer.
>>You are purposely misreading what I have written. Ill
break it down
>>again for you. Please try to pay attention:
>> 5 = q(a) r(a)
>> a = r(a) s(a).
>>q,r, and s are given by these formulas:
>> q(a) = 8 a^2 - 76 a - 185
>> r(a) = 8 a^2 - 4 a - 45
>> s(a) = 4 a^2 - 37 a - 104
>>Whenever a is a root of the cubic x^3 - 12 x^2 + 65, the
above
>>factorizations are correct, and consist of algebraic
integers.
>>Further, I have shown that if, despite the above
factorization,
>>a and 5 are coprime (that is: IF YOUR CLAIM IS CORRECT), it
must
>>follow that r(a) is a unit algebraic integer.
>>I also show that the minimal polynomial for r(a) is this:
>> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
>>That shows that r(a) CANNOT be a unit.
>>The contradiction, which youre trying very hard to twist 
to
>>suit your own purposes, is that your claim (a coprime to 5)
>>cannot be true.
> Showing that r(a) cant be a unit does not prove that it
must have a
> non-unit factor in common with 5.
> Notice the following exchange.
It certainly does, since Ive shown that r(a) is a divisor 
of
5.
>Dont see it?
>Well consider the statements:
>>Then there exist algebraic integers u and v for which
>> a u + 5 v = 1
>>and the above factorizations show this:
>> rs u + qr v = 1
>>so
>> r( su + qv ) = 1,
>>and we deduce that r is a unit.
>The poster wants to claim that r must share a non-unit
factor in
>common with 5 by showing a contradiction by assuming its a
unit,
>which actually only shows that it cant be a unit in the
ring of
>algebraic integers.
You misread all things that dont say what you want, is that
your
current game? Read correctly, and youll find 
out what I said.
Read incorrectly, and youll continue as always: no
credibility,
no results, no life.
>So he should have just ended up chasing his tale.
>But he screwed up and managed to bite himself!
> That is, he went on as if r were a unit, and claimed to
have given its
> inverse, which is what followed but is now gone.
> But luckily its still in the post, so copying from 
earlier.
>>However, its not one of your fanciful
>>Unit with no inverse hallucinations. In fact, the above
equation
>>provides us with an inverse for r. Based on its
construction, it
>>*must be* an algebraic integer: u and v are specified to be
algebraic
>>integers, and q,r,s were already specified as algebraic
integers.
> There the poster claims that the inverse for r must be an
algebraic
> integer, with an emphasis on the must be.
Right. I say
If a and 5 are coprime, then r must be a unit
My proof exhibits the inverse for r, UNDER THE ASSUMPTION that
a and 5 are coprime.
> That is, in fact, false, and goes against the posters own
claims.
Cant hack a proof by contradiction? Is that your problem?
> He quit chasing his tale for a while in order to bite
himself,
> figuratively speaking, of course.
>>Yeah, right. If you could read and understand mathematical
proofs,
>>you would have just said, Oh, well, I was wrong there.
Instead,
>>youre fixated on the fantasy that 
youve gone where no
mathematician
>>has trod before. Fine, have your fantasy.
>>Its just not mathematics, nor is it supported by anything
other than
>>your massively-ßawed messing around with uber polynomials.
> Now then back to my theory. Is it true that racists feel a
need to
> attack others based on their own fear of inferiority?
Did you ever notice that JSH cant keep an argument on 
track?
Ever notice how his critics are evil, whether by dint of
incompetence,
or fraud, or just by being infinitely powerful, and therefore
infinitely
corrupt?
Or racists?
> It makes sense.
I dont take kindly to being slurred in such a way. I do not
engage
in racist behavior, and dont condone it in those around me.
> Imagine that youre very successful, beloved in your
community, and
> generally things go well for you in your life.
> Why would you go after other people?
> Wouldnt you have to think they were some kind of threat 
to
you?
You are an unwelcome presence in sci.math, for reasons that I
have
posted on many occasions. There is some value, that of having
a buffoon
in the room, but for the most part, you represent the punk
who hangs
out behind the grocery store, razzing the older folks and
beating up
kids. A fart in an elevator.
> James Harris
Dale.
===
Subject: Re: Hyperbola for two arbitrary foci
solving for either x or y is very difficult for me. 
Ive
squared the
lot but there still remains some SQRT terms. Any ideas on
extracting a
lovely solitary x or y?
Jez
> Does anyone know of an expression for the hyperbola related
to two
> aribitrary foci at x1,y1 and x2,y2 for a constant distance
difference
> to x,y of d?
> sqrt[(x-x1)^2+(y-y1)^2]-sqrt[(x-x2)^2+(y-y2)^2}=d ;
Simplify. The
> coefficient of xy will be non-zero.
===
Subject: Re: Hyperbola for two arbitrary foci
>solving for either x or y is very difficult for me. 
Ive
squared the
>lot but there still remains some SQRT terms. Any ideas on
extracting a
>lovely solitary x or y?
>Jez
>Does anyone know of an expression for the hyperbola related
to two
>aribitrary foci at x1,y1 and x2,y2 for a constant distance
difference
>to x,y of d?
>
>>sqrt[(x-x1)^2+(y-y1)^2]-sqrt[(x-x2)^2+(y-y2)^2}=d ;
Simplify. The
>>coefficient of xy will be non-zero.
>>
sqrt(expression1) - d = sqrt(expression2).
Square. Move all non-square terms from left side to right
side. Square
again.
You have a 4th degree equation in x (also in y). Fortunately,
stuff
drops out.
The xy coefficient may or may not be non-zero. (Will be zero
if points
are vertically or horizontally aligned.)
Jon Miller
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
Okay, this is a fun game, only a game though, so lets name
famous
scientists who were inspired by, er, eastern religions
If there is any religion that would cope with modern
scientific needs
it would be Buddhism.
Albert Einstein
At dawn on July 16, 1945, Robert Oppenheimer (father of
Americas
nuclear programme) watched the first atomic bomb explode in
the New
Mexico desert. I remembered the line from the Hindu
scripture, the
Bhagavad Gita, he later said. I am become Death, the
destroyer of
worlds.
Oppenheimer and many of the Jewish/German physicists studied
sanskrit
literature in its original...
> Buddhism is the only religion that is compatible with
science and
> mathematics. It is a way of thinking which oroginated in
India in the
> 6the century BC. India chose to stay Hindu and destroyed
Buddhism and
> is paying the price for this till today. If India was
Buddhist, the
> Industrial revolution would have happened in India in 1000
AD.
>
> Excuse me, sir, but I beg to differ.
>
> The Industrial Revolution would not be allowed to happen
under *any*
form of
> Buddhism of which Im aware.
>
> tvp
> Buddhism is the most powerful way to understand ANY truth,
including
> science and mathematics. Japan is essentially a Buddhist
country and
> is very successful technologically, better than most western
> countries.
> There are many nobel-prize winning western scientists who
are
> Buddhists. Niels Bohr (who now has a Buddhist family crest)
and
> Heisenberg, both of whom created modern physics in the form
of Quantum
> Theory, both embraced Buddhism because they could not find
anyhting in
> another philosophy to explain modern physics.
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
> Okay, this is a fun game, only a game though, so lets 
name
famous
> scientists who were inspired by, er, eastern religions
> If there is any religion that would cope with modern
scientific needs
> it would be Buddhism.
> Albert Einstein
Here are a few quotes by this brilliant man. Enjoy. You might
find this
is a keeper!
The Religion of the future will be a cosmic religion. It will
transcend
a personal God and avoid dogma and theology. Covering both
the natural
and spiritual, it will based on a religious sense arising
from the
experience of all things natural and spiritual as a
meaningful unity.
Buddhism answers this description. If there is any religion
that can cope
with modern scientific needs, it is Buddhism.
ALBERT EINSTEIN, Ideas and Opinions
The intuitive mind is a sacred gift and the rational mind is
a faithful
servant. We have created a society that honors the servant
and has
forgotten
the gift. -Albert Einstein
Imagination is more important than knowledge. -- Albert
Einstein
A human being is part of the whole, called by us 
ÔUniverse;
a part
limited in time and space. He experiences himself, his
thoughts and
feelings as something separated from the rest--a kind of
optical
delusion of his consciousness. This delusion is a kind of
prison for
us, restricting us to our personal desires and affection for
a few
persons nearest us. Our task must be to free ourselves from
this prison
by widening our circle of compassion to embrace all living
creatures and
the whole nature in its beauty. Nobody is able to achieve this
completely but striving for such achievement is, in itself, a
part of
the liberation and a foundation for inner security.
--Albert Einstein >>
What does the man of confidence say about ethics? That we
should *believe*
in rewards in future lives or heavens? I dont think so.
Albert says:
A mans ethical behavior should be based effectually on
sympathy,
education, and social ties; no religious basis is necessary.
Man would
indeeded be in a poor way if he had to be restrained by fear
of punishment
and hope of reward after death. --Einstein, Albert
Does the man of confidence bow to religious authorities?
Albert says:
Great spirits have always found violent opposition from
mediocrities.
The
latter cannot understand it when a man does not thoughtlessly
submit to
hereditary prejudices but honestly and courageously uses his
ntelligence. --Albert Einstein
Can science ultimately explain everything that occurs in the
universe?
Gravity cannot be held responsible for people falling in love.
--Albert
Einstein
And are there any problems too complex for the human mind to
comprehend?
The hardest thing in the world to understand is the income
tax.
--Albert
Einstein
What about how to relate to the world spiritually?
There are only two ways to live your life. One is as though
nothing is a
miracle. The other is as though everything is a miracle. --A.
Einstein
And what does the man of confidence say about the ultimate or
the infinite?
Only two things are infinite, the universe and human
stupidity, and Im
not
sure about the former. --Albert Einstein
The further the spiritual evolution of mankind advances, the
more certain
it seems to me that the path to genuine religiosity does not
lie through
the
fear of life, and the fear of death, and blind faith, but
through striving
after rational knowledge.--Albert Einstein
It was, of course, a lie what you read about my religious
convictions, a
lie which is being systematically repeated. I do not believe
in a personal
God and I have never denied this but have expressed it
clearly. If
something is in me which can be called religious then it is
the unbounded
admiration for the structure of the world so far as our
science can reveal
it.--Albert Einstein, 1954, from _Albert Einstein: The Human
Side_,edited
by Helen Dukas and Banesh Hoffman, Princeton U Press
The finest emotion of which we are capable is the mystic
emotion. Herein
lies the germ of all art and all true science. Anyone to whom
this feeling
is alien, who is no longer capable of wonderment and lives in
a state of
fear is a dead man. To know that what is impenatrable for us
really exists
and manifests itself as the highest wisdom and the most
radiant beauty,
whose gross forms alone are intelligible to our poor
faculties - this
knowledge, this feeling ... that is the core of the true
religious
sentiment. In this sense, and in this sense alone, I rank
myself amoung
profoundly religious men.--Albert Einstein
Since you tend to share your enormous Buddhist sites list
from time to
time,
Ill give you my list of good sites for Einstein quotes:
http://rescomp.stanford.edu/~cheshire/EinsteinQuotes.html
http://www.foxberry.com/godiam/einstein/index.html-ssi
http://www.moosecom.com/~moosecom/einquotes.html
http://www.humboldt1.com/~gralsto/einstein/quotes.html
http://stripe.colorado.edu/~judy/einstein.html
http://www.aip.org/history/einstein/einlinks.htm
http://www.bluebirdweb.com/spirit/s-einstein.html
http://www.netunlimited.net/~spike/einstein_quotes.htm
http://users.cwnet.com/kwithee/einstein.html
http://members.aol.com/levatino/equotes.html
http://www.mindspring.com/~mmessall/quotes/einstein.txt
http://users.cwnet.com/kwithee/
And a good source of general Einstein links:
http://www.westegg.com/einstein/
Law never made men a whit more just; and by means of their
respect for
it,
even the well-disposed are daily made the agents of
injustice. -Henry
David Thoreau
The poets were not alone in sanctioning myths, for long
before the poets
the states and the lawmakers had sanctioned them as a useful
expedient.
They
needed to control the people by superstitious fears, and
these cannot be
aroused without myths and marvels. -Strabo (c. 58 B.C.--c. 24
A.D.),
Greek
geographer
ÔThe most beautiful thing we can experience is the
mysterious. It is the
source of all true art and science. - Albert Einstein
The mystical trend of our time, which shows itself
particularly in the
rampant growth of the so-called Theosophy and Spiritualism,
is for me no
more than a symptom of weakness and confusion. Since our
inner experiences
consist of reproductions, and combinations of sensory
impressions, the
concept of a soul without a body seems to me to be empty and
devoid of
meaning. -Albert Einstein
True education is to learn *how* to think, not *what* to
think. If you
know how to think, if you really have that capacity, then you
are a free
human being -- free of dogmas, superstitions, ceremonies --
and therefore,
you can find out what religion is. Then what *is* religion? If
you have
wiped the window clean --which means that you have actually
stopped
performing ceremonies, given up all beliefs, cease to follow
any leader or
guru --then your mind, like the window, is clean, polished,
and you can see
out of it very clearly. When the mind is swept clean of
image, or ritual,
of
belief, of symbol, of all words, mantras, and repetitions,
and of all fear,
then what you see will be the real, the timeless, the
everlasting, which
may
be called ÔGod by some; but this requires 
enormous insight,
understanding, patience; and it is only for those who really
*inquire* into
what is religion and pursue it day after day to the end. Only
such people
will know what is true religion. The rest are merely mouthing
words, and
all their ornaments and bodily decorations, their pujas and
ringing of
bells -- all that is just superstition without any
significance. It is only
when the mind is in revolt against all so-called religion
that it finds the
real.
Any intelligent fool can make things bigger, more complex,
and more
violent. It takes a touch of genius -- and a lot of courage
-- to move in
the opposite direction.
-Albert Einstein
Cosmic Religious Feeling Albert Einstein Quantum Questions
But there is a third stage of religious experience which
belongs to all
of
them, even though it is rarely found in a pure form: I shall
call it cosmic
religious feeling. It is very difficult to elucidate this
feeling to anyone
who is entirely without it, especially as there is no
anthropomorphic
conception of God corresponding to it
The individual feels the futility of human desires and aims
and the
sublimity and marvelous order which reveal themselves both in
nature and in
the world of thought. Individual existence impresses him as a
sort of
prison
and he wants to experience the universe as a single
significant whole. The
beginnings of cosmic religious feeling already appear at an
early stage of
development, e.g., in many of the Psalms of David and in some
of the
Prophets. Buddhism, as we have learned from the wonderful
writings of
Schopenhauer, contains a much stronger element of this.
The religious geniuses of all ages have been distinguished by
this kind of
religious feeling, which knows no dogma and no God conceived
in mans
image;
so that there can be no church whose central teachings are
based on it.
Hence it is precisely among the heretics of every age that we
find men who
were filled with this highest kind of religious feeling and
were, in many
cases, regarded by their contemporaries as atheists,
sometimes also as
saints. Looked at in this light, men like Democritus, Francis
of Assisi,
and
Spinoza are closely akin to one another.
How can cosmic religious feeling be communicated from one
person to another
if it can give rise to no definite notion of a God and no
theology? In my
view, it is the most important function of art and science to
awaken this
feeling and keep it alive in those who are receptive to it.
We thus arrive at a conception of the relation of science to
religion very
different from the usual one. When one views the matter
historically, one
is
inclined to look upon science and religion as irreconcilable
antagonists,
and for a very obvious reason. The man who is thoroughly
convinced of the
universal operation of the law of causation cannot for a
moment entertain
the idea of a being who interferes in the course of events --
provided, of
course, that he takes the hypothesis of causality really
seriously. He has
no use for the religion of fear and equally little for social
or moral
religion. A God who rewards and punishes is inconceivable to
him for the
simple reason that a mans actions are determined by
necessity, external
and
internal, so that in Gods eyes he cannot be responsible, 
any
more than an
inanimate object is responsible for the motions it undergoes.
Science has,
therefore, been charged with undermining morality, but the
charge is
unjust.
A mans ethical behavior should be based effectually on
sympathy,
education,
and social ties and needs; no religious basis is necessary.
Man would
indeed
be in a poor way if he had to be restrained by fear of
punishment and hope
of reward after death.
It is, therefore, easy to see why the churches have always
fought science
and persecuted its devotees. On the other hand, I maintain
that the cosmic
religious feeling is the strongest and noblest motive for
scientific
research. Only those who realize the immense effort and,
above all, the
devotion without which pioneer work in theoretical science
cannot be
achieved are able to grasp the strength of the emotion out of
which alone
such work, remote as it is from the immediate realities of
life, can
issue...
I have never imputed to Nature a purpose or a goal, or
anything that could
be understood as anthropomorphic. What I see in Nature is a
magnificent
structure that we can comprehend only very imperfectly, and
that must fill
a
thinking person with a feeling of humility. This is a
genuinely religious
feeling that has *nothing* to do with mysticism. - Albert
Einstein
It was, of course, a lie what you read about my religious
convictions, a lie which is being systematically repeated. I
do not
believe in a personal God and I have never denied this but
have
expressed it clearly. If something is in me which can be
called
religious then it is the unbounded admiration for the
structure of the
world so far as our science can reveal it. - Albert Einstein
--
Evelyn
Since everything is but an apparition, perfect in being what
it is,
having
nothing to do with good or bad, acceptance or rejection, one
may well burst
into laughter. -Longchenpa
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
> Here are a few quotes by this brilliant man. Enjoy.
Einstein was a brilliant physicist. So far as being
brilliant as meaning smart about everything, he wasnt. No
doubt his mother would have said about Einstein the same
thing Feynmans mother said about Feynman: If 
thats the
worlds smartest man, then god help us.
> The Religion of the future will be a cosmic religion.
It will transcend
> a personal God and avoid dogma and theology.
In other words, it wont have anything about it that lets us
recognize it AS a religion? Whats the point in using the
word at all, as it only tends to create confusion?
> Covering both the natural
> and spiritual, it will based on a religious sense arising
from the
> experience of all things natural and spiritual as a
meaningful unity.
> Buddhism answers this description. If there is any
religion that can cope
> with modern scientific needs, it is Buddhism.
And what is this religious sense Einstein talked about?
Why..
>If
> something is in me which can be called religious then it
is the unbounded
> admiration for the structure of the world so far as our
science can reveal
> it.--Albert Einstein, 1954, from _Albert Einstein: The
Human Side_,edited
> by Helen Dukas and Banesh Hoffman, Princeton U Press
If, then. But maybe there wasnt something in Einstein that
could be called religious. And that being the case, he
shouldnt have tried to.
> The finest emotion of which we are capable is the mystic
emotion. Herein
> lies the germ of all art and all true science. Anyone to
whom this feeling
> is alien, who is no longer capable of wonderment and lives
in a state of
> fear is a dead man. To know that what is impenatrable for
us really exists
> and manifests itself as the highest wisdom and the most
radiant beauty,
> whose gross forms alone are intelligible to our poor
faculties - this
> knowledge, this feeling ... that is the core of the true
religious
> sentiment.
Einstein is going to tell us the core of the true religious
sentiment?
>In this sense, and in this sense alone, I rank myself
amoung
> profoundly religious men.--Albert Einstein
By redefining the word to suit himself instead of the way
its been generally and historically used? I think of Lewis
Carrols Humpty Dumpty in Though the Looking Glass.
When I use a word, Humpty Dumpty said, in rather a
scornful tone, it means just what I choose it to
meanÖneither more nor less.
The question is, said Alice, whether you can make
words mean so many different things.
The question is, said Humpty Dumpty. which is to be
masterÖthats all.
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
> Buddhism is the only religion that is compatible with
science and
> mathematics. It is a way of thinking which oroginated in
India in the
> 6the century BC.
> India WAS Buddhist. Its not the most advanced country in
the world.
> Of course, if advanced means polluting the earth, taxing
water because
all
> drinking water is too dirty and needs processing, and
destroying the
earth
> and its resources India was really, really backward...
Ah yes, the bait and switch con. When an argument is proven
wrong,
redefine everything.
> if advanced means [sic] polluting the earth, [sic] taxing
> water [...]
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
> Organized Buddhism resembles the teachings of the Buddha
> about the same way Organized Xtianity resembles the
> teachings of Jesus of Nazareth. That is to say, not at all.
> Babtists want me to have a personal relationship with
> their Savior, then want me to get churched so I do it
> right. The Buddha never said a damn thing about chanting,
> vegetarianism or fasting (except in terms of moderation
> for helths sake), or total nonviolence, but suggested
> others _try_ his methodology for making life easier.
> Frinst, about the nonviolence myth:
> http://www.ßashbunny.org/content/gunnutsposter.html
> far right (no pun intended).
> Once a set of teachings is extended past three or four
> people, it usually devolves to crowd control (politics).
> Mark L. Fergerson
Im following the suggestions by the Buddha, not the 
Buddhist
establishment, and I can use the Buddhas teaching to 
justify
my
action:
1. In one of his famous parable, he compare truth to the moon,
and his guidance to a finger pointing to the moon. There is no
point
worshipping the finger, it is the moon that 
were after.
2. Knowing a lots of Sutra means nothing (just like memorizing
the Bible means nothing). Historical case in point: Ananda,
whose
remarkable memory helped preserve the Buddhas teaching, was
not
enlightened at the time of the Buddhas death (it was
remarkable that
he knew he wasnt enlightened despite the fact that he knew
more of
the Buddhas teaching by heart than anyone else.)
3. An interesting anecdote. The Buddha was dying and Ananda
was
crying. The Buddha said Congratulations! Why? asked Ananda.
The
Buddha gave a great reply Because you will not have me around
as an
authority figure, therefore you will find your own 
path.
Ever heard the zen saying: If you meet the Buddha on the road,
kill him! An exaggeration of the same exact meaning. Dont
rely on the
Buddha, you have to save yourself.
4. To me, the teaching of the Buddha is exactly the goal of
science. He taught the root of all problems is ignorance. The
goal,
then, is to overcome ignorance. This is also the goal of
science.
5. Most remarkable to me is the Buddhas prediction that his
teachings will be misinterpreted by later generations. In
fact, he
gave an approximate timeline. According to this timeline we
are now in
the time where his teachings are being abused for other
purposes
(instead of being understood and practiced.)
Thinh Tran
http://www.thinhtran.com
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
er, okay since you are talking about Einstein redefining
religion, can
you please tell us YOUR definition or what YOU believe the
historical original definition of religion is?
> Here are a few quotes by this brilliant man. Enjoy.
> Einstein was a brilliant physicist. So far as being
> brilliant as meaning smart about everything, he wasnt. No
> doubt his mother would have said about Einstein the same
> thing Feynmans mother said about Feynman: If 
thats the
> worlds smartest man, then god help us.
> The Religion of the future will be a cosmic religion.
> It will transcend
> a personal God and avoid dogma and theology.
> In other words, it wont have anything about it that lets 
us
> recognize it AS a religion? Whats the point in using the
> word at all, as it only tends to create confusion?
> Covering both the natural
> and spiritual, it will based on a religious sense arising
> from the
> experience of all things natural and spiritual as a
> meaningful unity.
> Buddhism answers this description. If there is any
> religion that can cope
> with modern scientific needs, it is Buddhism.
> And what is this religious sense Einstein talked about?
> Why..
>If
> something is in me which can be called religious then it
> is the unbounded
> admiration for the structure of the world so far as our
> science can reveal
> it.--Albert Einstein, 1954, from _Albert Einstein: The
> Human Side_,edited
> by Helen Dukas and Banesh Hoffman, Princeton U Press
> If, then. But maybe there wasnt something in Einstein 
that
> could be called religious. And that being the case, he
> shouldnt have tried to.
> The finest emotion of which we are capable is the mystic
> emotion. Herein
> lies the germ of all art and all true science. Anyone to
> whom this feeling
> is alien, who is no longer capable of wonderment and lives
> in a state of
> fear is a dead man. To know that what is impenatrable for
> us really exists
> and manifests itself as the highest wisdom and the most
> radiant beauty,
> whose gross forms alone are intelligible to our poor
> faculties - this
> knowledge, this feeling ... that is the core of the true
> religious
> sentiment.
> Einstein is going to tell us the core of the true religious
> sentiment?
>In this sense, and in this sense alone, I rank myself
> amoung
> profoundly religious men.--Albert Einstein
> By redefining the word to suit himself instead of the way
> its been generally and historically used? I think of 
Lewis
> Carrols Humpty Dumpty in Though the Looking Glass.
> When I use a word, Humpty Dumpty said, in rather a
> scornful tone, it means just what I choose it to
> mean?neither more nor less.
> The question is, said Alice, whether you can make
> words mean so many different things.
> The question is, said Humpty Dumpty. which is to be
> master?thats all.
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in
theWorld.
> A muslim? Ha, ha ! Most Muslims cant read forget about
> thinking...notice no Christian, Jew objected to any of the
facts...
> Read about muslims worshipping cows in Al Baqara...about
Allah being
> everywhere...there is a lot of Hinduism in there for you...
Ive tried reading the Al Baqara and tried to locate the 
holy
cow of the
Muslims, it is pretty much the same as the Hindu holy cow. It
is just that
due to many middle-easterner mlecchas from Afghanistan
claiming to be
Hindus
post 700AD, the holy cow in Hinduism has been confused with
the earthly cow
roaming the streets. Hindu texts clearly state that the
Divine Cow resides
with Indra in Heaven, thus the separation from the regular
cow.
===
Subject: Re: If India was Buddhist, It would be Most Advanced
Country in the
World.
The important thing to note is that the person who scolds
Sakra has
attained
to a special region above Sakra. This does not imply that
Sakra is not
worth attaining to, it implies that one should go beyond it.
I dont know
about the Kalama Sutra, but I have started reading quite a
few Buddhist
Sutras and I can tell their inner or how should I say
mystical meaning. I
have tried the Amitabha Sutra for example and I know why it
is the pure
land
in the west! Most people do not read the Sutras in the
correct fashion.
But rest assured I will be reading a lot of other sutras. All
such sutras
will all find a place in my collection under Vishnu, as Buddha
was Vishnus
incarnation and thus his tantra is associated with Vishnu.
Does anybody know of any book or internet site which has the
> When he died and Mahakasyapa called the gathering and
organized the
> various
> sutras he spoke of Sakra (Indra), he did not speak of a
non-Vedic
deity!
> But yes, Buddha had his own Ôsutras or in 
Hinduism called
Ômantras
> which
> Mahakasyapa was very interested in preserving. That makes
Buddha
> another
> King who introduced a new Ôtantra into 
Hinduism. Why
Hinduism ?
> Because
> the deities to which one attained to according to his
followers were
> VEDIC!
> Most deities spoken of by Buddhists from Dakinis to Yoginis
etc. are
> actually VEDIC/HINDU deities! Thus to claim that it is a new
religion
> is
> quite untenable when the religion itself verifies the
existence of
the
> deities of the Vedic religion!
> Buddhism was never about attaining to deities. Read the
sutta in the
> Majjhima where Moggallana goes to the Tavatimsa heaven and
scolds Sakka
> for wasting time in sense pursuits, addressing him somewhat
> disrespectfully as yakkha. The deities in Buddhism are just
other
> beings trapped in samsara.
> Bhante, you should ask him to read the Kalama Sutra too.
Never believe
> anything you read in the scripture too.
===
Subject: Inequality confusion
In Introduction to Analysis, Rosenlicht proves the following
lemma:
A real-valued function f on the interval [a,b] is integrable
on [a,b] if
and only if, given any epsilon > 0, there exists a number
delta > 0 such
that | S1 - S2 | < epsilon whenever S1 and S2 are Reimann
sums for f
corresponding
to partitions of [a,b] of width less than delta.
To prove the only if part of the lemma he continues:
First suppose f integrable on [a,b]. Then given any epsilon >
0 there is a
delta > 0 such that | S - Int(a..b)[f(x) dx] | < epsilon/2
whenever S is a
Riemann sum for f corresponding to a partition of [a,b] of
width less than
delta.
If S1 and S2 are two such Reimann sums then
(1) | S1 - S2 | =
(2) | (S1 - Int(a..b)[f(x) dx]) - (S2 - Int(a..b)[f(x) dx]) |
<=
(3) | S1 - Int(a..b)[f(x) dx] | + | S2 - Int(a..b)[f(x) dx] |
<
(4) epsilon/2 + epsilon/2 = epsilon.
I do not understand the justification for going from line (2)
to line (3),
unless the following assumption is made which I did not find
in the text.
Let a = S1, b = S2, c = Int(a..b)[f(x) dx].
First | (a - c) - (b - c) | = | (a - c) + (-b + c) | <= | a -
c | + | -b + c
|.
Assume (2) and (3) are correct: | (a - c) - (b - c) | <= | a
- c | + | b - c
|.
So assume: | -b + c | <= | b - c |,
in order to have | a - c | + | -b + c | <= | a - c | + | b -
c | which
would sustain the inequality (2) <= (3).
| c - b | <= | b - c |
+- (c - b) <= +- (b - c)
(c - b) <= (b - c)
2c <= 2b
c <= b
(c - b) <= -(b - c) = -b + c
0 <= 0
-(c - b) <= (b - c)
-c + b <= b + -c
0 <= 0
-(c - b) <= -(b - c)
-c + b <= -b + c
-2c <= -2b
c >= b
Then c <= b & c >= b implies that c = b.
Int(a..b)[f(x) dx] = S2 (this is assumed nowhere?).
If anyone can shed some light on this matter, I would
appreciate it.
===
Subject: Re: Inequality confusion
> (1) | S1 - S2 | =
> (2) | (S1 - Int(a..b)[f(x) dx]) - (S2 - Int(a..b)[f(x) dx])
| <=
> (3) | S1 - Int(a..b)[f(x) dx] | + | S2 - Int(a..b)[f(x) dx]
| <
> (4) epsilon/2 + epsilon/2 = epsilon.
Using the Triangle Inequality |z1-z2| <= |z1| + |z2|
( also |z1+z2| <= |z1|+|z2| would do ),
so you can obtain line(2) <= line(3).
Michael Leung
===
Subject: Re: Inequality confusion
Patrick M. Clot <patrickclot@hotmail.com> escreveu na mensagem
> First | (a - c) - (b - c) | = | (a - c) + (-b + c) | <= | a
- c | +
> | -b + c |.
> Assume (2) and (3) are correct: | (a - c) - (b - c) | <= |
a - c | +
> | b - c |.
> So assume: | -b + c | <= | b - c |,
> in order to have | a - c | + | -b + c | <= | a - c | + | b
- c | which
> would sustain the inequality (2) <= (3).
I think that your question is how one conclude that
|(a - c) - (b - c)| <= |a - c| + |b - c| (1)
You did reach the inequality
|(a - c) - (b - c)| <= |a - c| + |-b + c| (2)
Now we notice that
|-b + c| = |b - c| (3)
In fact, let x = -b + c. Then -x = b - c. Because we have |x|
= |-x| for
all
real x, then (3) is true. If in the right member of (2) we
replace |-b + c|
by |b - c| (which we can do because (3)), then we get (1).
-----
Another way of getting (1). Let z and w be any real numbers.
We know the
triangle inequality
|z + w| <= |z| + |w|
One can show that
|z - w| <= |z| + |w| (4)
In fact, one has
|z - w| =
|z + (-w)| <= (apply the triangle inequality)
|z| + |-w| = (notice that |-w| = |w|)
|z| + |w|
If you replace z by a - c and replace w by b - c in (4), you
get (1).
Jaime Gaspar
______________________________
Homepage: www.jaimegaspar.com
E-mail: e-mail@jaimegaspar.com
===
Subject: Re: Inequality confusion
> In Introduction to Analysis, Rosenlicht proves the
following lemma:
> A real-valued function f on the interval [a,b] is
integrable on [a,b] if
> and only if, given any epsilon > 0, there exists a number
delta > 0 such
> that | S1 - S2 | < epsilon whenever S1 and S2 are Reimann
sums for f
corresponding
> to partitions of [a,b] of width less than delta.
> To prove the only if part of the lemma he continues:
> First suppose f integrable on [a,b]. Then given any epsilon
> 0 there is
a
> delta > 0 such that | S - Int(a..b)[f(x) dx] | < epsilon/2
whenever S is
a
> Riemann sum for f corresponding to a partition of [a,b] of
width less
than
delta.
> If S1 and S2 are two such Reimann sums then
> (1) | S1 - S2 | =
> (2) | (S1 - Int(a..b)[f(x) dx]) - (S2 - Int(a..b)[f(x) dx])
| <=
> (3) | S1 - Int(a..b)[f(x) dx] | + | S2 - Int(a..b)[f(x) dx]
| <
> (4) epsilon/2 + epsilon/2 = epsilon.
> I do not understand the justification for going from line
(2) to line
(3),
> unless the following assumption is made which I did not find
in the text.
> Let a = S1, b = S2, c = Int(a..b)[f(x) dx].
> First | (a - c) - (b - c) | = | (a - c) + (-b + c) | <= | a
- c | + | -b
+
c |.
> Assume (2) and (3) are correct: | (a - c) - (b - c) | <= |
a - c | + | b
-
c |.
> So assume: | -b + c | <= | b - c |,
> in order to have | a - c | + | -b + c | <= | a - c | + | b
- c | which
> would sustain the inequality (2) <= (3).
> | c - b | <= | b - c |
> +- (c - b) <= +- (b - c)
> (c - b) <= (b - c)
> 2c <= 2b
> c <= b
> (c - b) <= -(b - c) = -b + c
> 0 <= 0
> -(c - b) <= (b - c)
> -c + b <= b + -c
> 0 <= 0
> -(c - b) <= -(b - c)
> -c + b <= -b + c
> -2c <= -2b
> c >= b
> Then c <= b & c >= b implies that c = b.
> Int(a..b)[f(x) dx] = S2 (this is assumed nowhere?).
> If anyone can shed some light on this matter, I would
appreciate it.
-|x| <= x <= |x|
-|y| <= y <= |y|
Adding yields
-( |x| + |y| ) <= x + y <= ( |x| + |y| )
then
|x + y| <= |x| + |y|.
Steve
===
Subject: interval logic on ordinals
Please consider the following construction,
For ordinals x and y let
x pp y
indicate that x and y are initial segments of Ord and x is an
initial
segment of y. The relation pp reßects the term proper part
from
mereology and is an irreßexive transitive order.
Let
AaAb( [a, b] <-> Ec(a pp c / c pp b) )
where a, b, and c are ordinals define an interval relation on
Ord. We
can define a logic on pairs of ordinals satisfying this
relation.
First, we need an irreßexive order between the intervals,
A[a, b] A[c, d] ( [a, b] < [c, d] <-> Ex(b pp x / x pp c))
In addition, we need an inclusion relation. With respect to
the natural
ordering on Ord, we might write
A[a, b] A[c, d] ( [a, b] nat-include [c, d] <-> (Ax(x pp a ->
x ppc) /
Ax(x pp d -> x pp b)))
so that monotonicity
(([a, b] nat-include [c, d]) / ([a, b] < [x, y])) -> [c, d] <
[x, y]
((([x, y] < [a, b]) / ([a, b] nat-include [c, d]) -> [x, y] <
[c, d]
and convexity
( ([a, b] < [c, d])
/
([c, d] < [e, f])
/
([x, y] nat-include [a, b])
/
([x, y] nat-include [e, f]) )
([x, y] nat-include [c, d])
are preserved.
However, taking lead from Cohen forcing and the construction
of Suslin
trees from Suslin lines, we invert the sense of the natural
inclusion
for intervals. That is,
A[a, b] A[c, d] ( [a, b] include [c, d] <-> [c, d]
nat-include [a, b])
So, for example, we would have
A[a, b]([a, b] include [emptyset, Ord])
Finally, we wish to eliminate any connectivity between
intervals under
this inclusion. That is, the endpoints for an interval are
simply
interpreted as language delimiters so that the order-type of
an interval
under this inclusion is decreased by 2 from what it would
have under the
natural inclusion.
I have no idea how to interpret this structure.
The only thing I have ever found that remotely seems to refer
to it is
at the end of Aczels paper on replacement systems,
Note: Instead of using the category of all the
algebras we must use the subcategory of full
algebras and homomorphisms between them.
In fact, the final algebras are the trivial
one-element algebras, which are in general
not full, while the tree algebra is full
Does this inclusion tree reßect the category of all algebras
in any
way?
:-)
mitch
===
Subject: inverse of the gamma function
if the gamma function is restricted to the real interval
[2,infinity] than
the inverse function exists. Does someone know a closed
expression of this
inverse gammafunction or any other usable form, like an
integral function
or
power series,...?
Dirk Osterkamp
===
Subject: Re: inverse of the gamma function
> if the gamma function is restricted to the real interval
[2,infinity]
> than the inverse function exists. Does someone know a
closed expression
> of this inverse gammafunction or any other usable form,
like an integral
> function or power series,...?
dated 2001 Oct 25, in the thread Inverse gamma function at
<http://mathforum.org/discuss/sci.math/a/t/367123>.
David Cantrell
===
Subject: Is there someone to help me PLEASE...
This is the third time that I send these questions..
please help me to solve the questions below:
1-The first qustion is to find the largest set
of natural numbers that if we choose two members a and b then
ab+1 is
square
of a natural number.(is there any limit for the number of
members of the
set?)
2-And the second is To find all the (a,b,c,d)s that the
inequality
a|x| + b|y| + c|x+y| + d|x-y| >=0
is true for all the (x,y)s. and generalizing the qustion for
(x,y,z,...)
(there we will have the phrases like |x+y+z| , |x-y+z| , |z|
, .....)
http://newsone.net/ -- Free reading and anonymous posting to
60,000+
groups
other posts
made through NewsOne.Net violate posting guidelines, email
abuse@newsone.net
===
Subject: Re: Is there someone to help me PLEASE...
> This is the third time that I send these questions..
> please help me to solve the questions below:
> 1-The first qustion is to find the largest set
> of natural numbers that if we choose two members a and b
then ab+1 is
square
> of a natural number.(is there any limit for the number of
members of the
> set?)
suppose you had some a in such a set, then this would force
also
a^2 + 1 to be a square, lets say a^2 + 1 = c^2 for some
natural number
c.
Now what can you derive from the equation 1 = c^2 - a^2 =
(c+a)(c-a) ?
Marc
===
Subject: Re: Is there someone to help me PLEASE...
> This is the third time that I send these questions..
> please help me to solve the questions below:
> 1-The first qustion is to find the largest set
> of natural numbers that if we choose two members a and b
then ab+1 is
square
> of a natural number.(is there any limit for the number of
members of the
> set?)
I suspect that the described set has an infinite number of
members.
Can you study the behavior of sets having three members or
supply an
example?
Clearly { 3,4,5}
does not qualify for such a set , but {9,6} would, because, I
think you
mean
that for ANY two Different numbers, your rule holds, and
{3,4} is obviously
disqualified.
I dont know how to prove this, ( in Grad School, I stayed
away from Number
Theory, because the teacher was a Chinese lady who screamed
at the
blackboard with a heavy accent and never talked to people )
... but
wouldnt
the No largest prime type of approach be productive??
I.E. Suppose ( or show) there is a number that is larger than
any in the
set..then find some member in the set that WOULD follow the
rule with it,
thus increasing the number of members of the set..etc..
Bob Pease
===
Subject: Re: Is there someone to help me PLEASE...
> This is the third time that I send these questions..
> please help me to solve the questions below:
> 1-The first qustion is to find the largest set
> of natural numbers that if we choose two members a and b
then ab+1 is
square
> of a natural number.(is there any limit for the number of
members of the
> set?)
I conjecture four (counting only the positive numbers).
Martin Gardner, in one of his puzzle books, mentions the
following
set
{1, 3, 8, 120}
as having the interesting property that any two members of
the set,
multiplied together, are always one less than a perfect
square.
He then asks for a fifth integer which can be added to the set
without
destroying its unique property. Solution left as exercise for
the
reader. ;-)
Using a computer program, I found sets like
{1, 3, 1680, 23408}
{1, 8, 15, 528}
{1, 15, 24, 1520}
that also seem to be closed in the sense that no other
positive
integer can be added - but I have not proved this.
> 2-And the second is To find all the (a,b,c,d)s that the
inequality
> a|x| + b|y| + c|x+y| + d|x-y| >=0
> is true for all the (x,y)s. and generalizing the qustion
for (x,y,z,...)
> (there we will have the phrases like |x+y+z| , |x-y+z| ,
|z| , .....)
Start with a=b=c=d=0. That is obviously a solution.
Try induction on the individual a,b,c,d, and see where
that takes you. ;-)
[I assume |x| means absolute value of x; if it doesnt, then
my apologies for the winky-face.]
-Arthur
===
Subject: Re: Is there someone to help me PLEASE...
> 1-The first qustion is to find the largest set
> of natural numbers that if we choose two members a and b
then ab+1 is
square
> of a natural number.(is there any limit for the number of
members of the
> set?)
4 or 5. See http://www.math.hr/~duje/dtuples.html - summary
is, there
are lots of ways of getting 4, its impossible to get 6, no
one has
ever gotten 5, and if there are any 5s there are only finitely
many.
The reference given is A. Dujella, There are only finitely 
many
Diophantine quintuples, J. Reine Angew. Math., to appear.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Is there someone to help me PLEASE...
>2-And the second is To find all the (a,b,c,d)s that the
inequality
>a|x| + b|y| + c|x+y| + d|x-y| >=0
>is true for all the (x,y)s. and generalizing the qustion for
(x,y,z,...)
>(there we will have the phrases like |x+y+z| , |x-y+z| , |z|
, .....)
Lets stick with the non-generalized form.
If x and y both have the same sign, |x+y| = |x|+|y| and
|x-y|=||x|-|y||
If they have different signs, |x+y|=||x|-|y|| and
|x-y|=|x|+|y|.
So, taking u = min(|x|,|y|) and v = max(|x|,|y|) - u, we need
a u + b (v+u) + c(2u+v) + dv = (a+b+2c)u + (b+c+d)v >= 0
a (v+u) + b u + c(2u+v) + dv = (a+b+2c)u + (a+c+d)v >= 0
a u + b (v+u) + cv + d(2u+v) = (a+b+2d)u + (b+c+d)v >= 0
a (v+u) + b u + cv + d(2u+v) = (a+b+2d)u + (a+c+d)v >= 0
whenever u >= 0 and v >= 0.
This works if and only if
a+b+2c >= 0
a+b+2d >= 0
b+c+d >= 0
a+c+d >= 0
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: James Harris Double
In sci.math, G Frege
<f25gjvs6l20t1o1uf2c33pbffbuhkajh46@4ax.com>:
>> Just trying to be a silly asshole?
> Im sorry, I think I mis.took your intention... :-(
> [...] where if f is coprime to 3, one of the as is
> coprime to f, and here Ive given it the indice 3.
> (etc. etc.)
> I see.
> F.
Sorry; my humor is a little off. :-) But my math is usually
spot on, which is more than I can say for Mr. Harriss.
(Not that his algebraic manipulation is all that bad, but
the conclusions he jumps to!)
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: James Harris Double
> Sorry; my humor is a little off. :-) But my math is usually
> spot on, which is more than I can say for Mr. Harriss.
> (Not that his algebraic manipulation is all that bad, but
> the conclusions he jumps to!)
To be honest, I really dont understand why you guys STILL
try to argue
with this ehrrr... you know.
F.
===
Subject: JSH: Resolution?
Note: The argument below seems to settle claims of James
Harris
in some generality. I would be grateful for comments on
it from anyone (incl. James Harris of course). I cannot
see anything wrong with it, but there is something about it
that makes me slightly nervous.
-------------------------------------------------------------
----
James Harris in his paper Advanced Polynomial Factorization
(see also Note *4) considers polynomials of the form
[1] P(x) = (v^3 + 1)*x^3 - 3*v*(u*f)^2*x + (u*f)^3,
where v = -1 + m*f^2, f is a prime number > 3, m is an
integer coprime to f, and u is an integer coprime to f.
He wants to show that if P(x)/f^2 is factored in the form
[2] P(x)/f^2 = (a1*x + c1)*(a2*x + c2)*(a3*x + c3),
where a1, a2, a3, c1, c2, and c3 are algebraic integers, then
at least one of a1, a2, or a3 must be coprime to f.
-------------------------------------------------------------
------
It is easily shown that v^3 + 1 is divisible by m*f^2. Let
us say
v^3 + 1 = m*f^2*k,
where k is an integer. Thus
P(x) = m*f^2*k*x^3 - 3*v*u^2*f^2*x + u^3*f^3.
Make the substitution x = -u*f/y. Then P(x) = 0 becomes
-m*f^2*k*u^3*f^3/y^3 + 3*v*u^3*f^3/y + u^3*f^3 = 0,
or, factoring out u^3*f^3,
[3] -m*k*f^2 + 3*v*y^2 + y^3 = 0.
Let y1, y2, and y3 be the roots of this equation. Then y1,
y2, and y3 are algebraic integers, and each is *not* coprime
to f [by an elementary Galois theory argument, provided the
polynomial in [3] is irreducible. In general it is (see
Note *2 below)].
Note that [3] implies
y1^2*(y1 + 3*v) = m*k*f^2
where (y1 + 3*v) *IS* coprime to f. [Note: f is a
prime > 3, and v = -1 + m*f^2 is necessarily coprime to f.]
Therefore y1^2 must be a multiple of f^2. [Note *3]. Therefore
y1 = e1*f where e1 is an algebraic integer. Similarly for y2
and y3.
Therefore a factorization of P(x) is
P(x) = (e1*f*x + u*f)*(e2*f*x + u*f)*(e3*f*x + u*f).
This implies
P(x)/f^3 = (e1*x + u)*(e2*x + u)*(e3*x + u),
where e1, e2, and e3 are all algebraic integers.
But this implies
P(x)/f^2 = f*(e1*x + u)*(e2*x + u)*(e3*x + u).
Now let f1, f2, and f3 be any three nonunit algebraic integers
such that
f1*f2*f3 = f.
Then
P(x)/f^2 = (f1*e1*x + u*f1)*(f2*e2*x + u*f2)*(f3*e3*x + u*f3),
which implies a factorization in the form of Equation [2],
with
a1 = f1*e1, a2 = f2*e2, a3 = f3*e3. Note that EACH of a1, a2
and
a3 is *not* coprime to f. This shows that Harriss claim 
that
one
of a1, a2, or a3 must be coprime to f is incorrect in general
(*2).
Nora B.
Notes:
* 1. The Galois theory argument has been presented before.
That the polynomial in [3],
-m*k*f^2 + 3*v*y^2 + y^3
is irreducible in general is easily seen by letting m = 1
and f = 5, where it becomes
y^3 + 72*y^2 - 13825,
which is easily shown to be irreducible.
*2. This argument does not say anything about whether e1,
e2, and e3 are coprime to f.
*3. This is the key new step in the argument.
*4. The same argument is implied in Harriss proof of
FLT, except there he is using objects rather than
algebraic integers. However it has not been established
whether objects are different from algebraic integers,
and a number of basic theorems for objects which are
needed for his argument have not been proven.
===
Subject: Re: JSH: Resolution?
Visiting Assistant Professor at the University of Montana.
>Note: The argument below seems to settle claims of James
Harris
>in some generality. I would be grateful for comments on
>it from anyone (incl. James Harris of course). I cannot
>see anything wrong with it, but there is something about it
>that makes me slightly nervous.
[.snip.]
>[3] -m*k*f^2 + 3*v*y^2 + y^3 = 0.
>Let y1, y2, and y3 be the roots of this equation. Then y1,
>y2, and y3 are algebraic integers, and each is *not* coprime
>to f [by an elementary Galois theory argument, provided the
>polynomial in [3] is irreducible. In general it is (see
>Note *2 below)].
>Note that [3] implies
> y1^2*(y1 + 3*v) = m*k*f^2
>where (y1 + 3*v) *IS* coprime to f. [Note: f is a
>prime > 3, and v = -1 + m*f^2 is necessarily coprime to f.]
>Therefore y1^2 must be a multiple of f^2. [Note *3].
Yeah, this is definitely wrong. That would give that y1 is a
multiple
of f, and the same for y2, y3. That would require that
m*k*f^2 =
y1*y2*y3 be a multiple of f^3, and so that mk be a multiple
of f, but
there is certainly no reason to assume this is the case, and
in fact
can be seen not to be the case for specific values. Since k is
defined
to be the integer such that v^3+1 = mf^2*k, with v=mf^2-1,
set m=1,
f=5, and we have v=24, v^3+1 = 13825 = 25*553, so k=553, m=1,
and mk
is not a multiple of f.
I think the error is what I said: you have not shown that
y1+3v is
coprime to f, merely that any common factor is coprime to
(y1,f).
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: JSH: Resolution?
what about m=0 ??... I havent followed any of this, because
no-one wants to put any sort of a hypothesis up front,
to justify it (I mean, I *might*
be able to follow it). I do have one hypothesis:
necessarily, monsieur Harris, iff he exists
in the real world (sik), will not reply to this proof
of his lack of a proof, but
taht is not sufficient to dysprove the results
of his 10-year Mission!
> Notes:
> * 1. The Galois theory argument has been presented before.
> That the polynomial in [3],
> -m*k*f^2 + 3*v*y^2 + y^3
> is irreducible in general is easily seen by letting m = 1
> and f = 5, where it becomes
> y^3 + 72*y^2 - 13825,
> which is easily shown to be irreducible.
> *2. This argument does not say anything about whether e1,
> e2, and e3 are coprime to f.
> *3. This is the key new step in the argument.
> *4. The same argument is implied in Harriss proof of
> FLT, except there he is using objects rather than
> algebraic integers. However it has not been established
> whether objects are different from algebraic integers,
> and a number of basic theorems for objects which are
> needed for his argument have not been proven.
--Dec.2000 ÔWAND Chairman Paul 
ONeill, reelected to Board.
Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
===
Subject: Re: JSH: Resolution?
> Note: The argument below seems to settle claims of James
Harris
> in some generality. I would be grateful for comments on
> it from anyone (incl. James Harris of course). I cannot
> see anything wrong with it, but there is something about it
> that makes me slightly nervous.
>
-------------------------------------------------------------
----
> James Harris in his paper Advanced Polynomial Factorization
> (see also Note *4) considers polynomials of the form
> [1] P(x) = (v^3 + 1)*x^3 - 3*v*(u*f)^2*x + (u*f)^3,
> where v = -1 + m*f^2, f is a prime number > 3, m is an
> integer coprime to f, and u is an integer coprime to f.
> He wants to show that if P(x)/f^2 is factored in the form
> [2] P(x)/f^2 = (a1*x + c1)*(a2*x + c2)*(a3*x + c3),
> where a1, a2, a3, c1, c2, and c3 are algebraic integers,
then
> at least one of a1, a2, or a3 must be coprime to f.
>
-------------------------------------------------------------
------
> It is easily shown that v^3 + 1 is divisible by m*f^2. Let
> us say
> v^3 + 1 = m*f^2*k,
> where k is an integer. Thus
> P(x) = m*f^2*k*x^3 - 3*v*u^2*f^2*x + u^3*f^3.
> Make the substitution x = -u*f/y. Then P(x) = 0 becomes
> -m*f^2*k*u^3*f^3/y^3 + 3*v*u^3*f^3/y + u^3*f^3 = 0,
> or, factoring out u^3*f^3,
> [3] -m*k*f^2 + 3*v*y^2 + y^3 = 0.
> Let y1, y2, and y3 be the roots of this equation. Then y1,
> y2, and y3 are algebraic integers, and each is *not* coprime
> to f [by an elementary Galois theory argument, provided the
> polynomial in [3] is irreducible. In general it is (see
> Note *2 below)].
> Note that [3] implies
> y1^2*(y1 + 3*v) = m*k*f^2
> where (y1 + 3*v) *IS* coprime to f. [Note: f is a
> prime > 3, and v = -1 + m*f^2 is necessarily coprime to f.]
Arturo Magidin has pointed out that this step is not
correct. y1 may not be coprime to f, but that is not enough
to imply that y1 + 3*v is coprime to f. I dont see a way
to fix the problem. As I noted at the beginning of the post,
I felt like there was something fish about this - it seemed
to prove too much - but I overlooked what Arturo caught.
Harriss claim in Advanced Polynomial Factorization is
actually somewhat stronger than what I noted above. He
says that if P(x) is factored in the form
[*] P(x) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f),
where a1, a2, and a3 are algebraic integers, then
... two of the as have a factor that is f. [See p.3
of APF.]
To see if this is true, let f = 5, u = 1, m = 1.
Then one readily sees that
P(x) = 25*(553*x^3 - 72*x + 5).
If this is factored in the form [*] above, clearly
r1 = -u*f/a1 = -5/a1 is a root. If a1 has a factor that is
f, then a1 = 5*g1, where g1 is an algebraic integer.
Therefore r1 = -1/g1. This implies
553*(-1/g1^3) - 72*(-1/g1) + 5 = 0, or
5*g1^3 + 72*g1 - 553 = 0.
The left side of this equation is a non-monic,
irreducible, primitive polynomial in g1. Therefore
g1 cannot be an algebraic integer, showing that
Harriss claim in APF is false.
I dont have doubts about this one.
Nora B.
===
Subject: Lagrangian & Differential Form
Lagrangian on the standard-model is written so as to be
invariable
under gauge transformation. However it is indistinct why
Lagrangian
should satisfies gauge invariability. I thought that reason
might be
revealed explicitly by rewriting Lagrangian with inner
products and
differential forms.
For details:
http://139.134.5.123/tiddler2/gauge4/gauge.htm
===
Subject: Lines Affecting Themselves (Within Polygons)
This is more a fun project than heavy math. But there are
some good
questions to ask about this, but others will have to ask them
...
Start with any arbitrary convex polygon.
Draw a path consisting of connected straight line-segments
such that:
The path starts in any location within the polygon, and heads
in any
direction.
The path does not change direction, except...
When the path hits the edge of the polygon, it reßects as if
the
polygon has a mirrored inner-surface, or by some other rule,
remaining
inside the polygon.
And when the path crosses itself, as it has already been
drawn to that
point, the path changes direction by some rule. For example,
it
crosses a previously-drawn segment, but acts as if it was
reßected by
a mirror which crosses the previously-drawn segment
perpendicularly.
Or, another example, it crosses unimpeded every 3rd time it
encounters
a previously-drawn portion of the path, and is reßected the
other
times.
(But beware of ambiguous cases, such as when the path hits a
previously-drawn intersection of 2 or more segments, or when
the path
hits a vertex of the polygon. Such cases can be dealt with as
you
wish.)
End the path after drawing a finite number of segments, or for
some
other reason (such as the path approaching a limit-point or
loop or
being reßected exactly back along itself).
So, I have drawn a few of these paths by-hand, and I have
noticed that
they can look interesting, even beautiful. And they seem to
act often
in chaotic ways: such as settling into a partially-repeating
cycle
(strange attractor?), or simply spiraling into a point, etc.
I wonder if these paths are as really as interesting-looking
as I
assume, since I am currently unable to plot these with a
computer.
But the question occurs to me, given such a path, is there a
closed-form (ie non-recursive) way to determine where the
path is
drawn?
I highly doubt this.
Leroy Quet
===
Subject: m/phi(m) average
Let phi(m) be
the Euler totient function, the number of positive integers
<= m and coprime
to m.
Now, I am guessing that:
limit{n -> oo}
(1/n) *sum{m=1 to n} m/phi(m)
= zeta(2) zeta(3) /zeta(6),
where zeta(j) = sum{k=1 to oo} 1/k^j.
Is this well-known or even true?
(I think it is already well-known that:
limit{n -> oo}
(1/n) *sum{m=1 to n} phi(m)/m
= 1/zeta(2) = 6/pi^2.)
Leroy
Quet
===
Subject: Masters, doctorate, post-doc opportunity
NNTP-Posting-User: mckay
I have an opportunity for a graduate student (Masters or
doctorate)
or a post-doctoral student - to study automorphic functions
and
finite group characters as exemplified by 
Monstrous Moonshine.
URL: http://cicma.concordia.ca/faculty/cummins/moonshine.html
John McKay (Professor of Mathematics and of Computer Science)
mac@mathstat.concordia.ca
--
But leave the wise to wrangle, and with me
the quarrel of the universe let be;
and, in some corner of the hubbub couched,
make game of that which makes as much of thee.
--
But leave the wise to wrangle, and with me
the quarrel of the universe let be;
and, in some corner of the hubbub couched,
make game of that which makes as much of thee.
===
Subject: more easy questions
http://mygate.mailgate.org/mynews/sci/sci.math/
7d1e13fa81061b34f63795e7792b38
53.35661%40mygate.mailgate.org
Im drawing on several resources as I try to get my little
brain round
measure theory and Lebesgue integration. Please can someone
help answer
the following questions:
1). Is it fair to say that sigma rings are a subset of rings
- so that
the number of sigma rings is less than the number of rings?
2). When applying Ômeasure why does the set you 
are applying
it to have
to belong to some bigger space called the measure space? Why
cant you
measure any given set without making sure it is a subset of
some larger
set?
3). If the set E is the collection of all finite unions of
disjoint
intervals (on R or R^n, I think), why is E a ring but not a
sigma ring?
Surely if we add say two intervals together we get another
interval
which is still part of E, dont we?
4). In the definition of outer measure we have something along
the lines
of m*(A) = inf SUM (k=1 to inf.) m(I_k), where I_k is some
interval. Now
I thought that outer measure was of the form a - b i.e. the
length of
the interval, but according to this definition, it is the
infimum
(minimum) of the interval that is important - so Im not 
sure
whats
going on here.
David.
--
===
Subject: Re: more easy questions
> 4). In the definition of outer measure we have something
along the lines
> of m*(A) = inf SUM (k=1 to inf.) m(I_k), where I_k is some
interval. Now
> I thought that outer measure was of the form a - b i.e. the
length of
> the interval, but according to this definition, it is the
infimum
AAAAAAAAARRGGGGGGGGGGGHHHHHHHHHHHHHHHH!
The point is that we dont want just to measure lengths of
INTERVALS,
we want to measure sizes of all sorts of subsets of R (or
R^n).
They can be pretty wacky you know. Just think of the Cantor
set
or the Mandelbrot set and just think ..... there are subsets
of R^n
which are much wackier than those .....
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: more easy questions
http://mygate.mailgate.org/mynews/sci/sci.math/
aa550f0649fda4048de594ffcfb7f6
b2.35661%40mygate.mailgate.org
> The point is that we dont want just to measure lengths of
INTERVALS,
> we want to measure sizes of all sorts of subsets of R (or
R^n).
Lets try an example - if my interval is (2,3), then the
measure I get
using b - a is 1. But if I apply the infimum 
rules/definition
what
answer do I get then?
David.
--
===
Subject: Re: more easy questions
>> The point is that we dont want just to measure lengths 
of
INTERVALS,
>> we want to measure sizes of all sorts of subsets of R (or
R^n).
> Lets try an example - if my interval is (2,3), then the
measure I get
> using b - a is 1. But if I apply the infimum 
rules/definition
what
> answer do I get then?
1, of course.
You get this as the infinimum of m(U) over all open sets
containing (2,3).
If U contains (2,3) then m(U) >= 1, so the infimum is >= 1.
But it is eaxctly one as there is a U containing (2,3) with
m(U) = 1, viz.,
U = (2,3) itself. So the infimum is 1.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: more easy questions
http://mygate.mailgate.org/mynews/sci/sci.math/
400b60ca1c460e81f3f9df9373f9b9
9d.35661%40mygate.mailgate.org
>
>
> You get this as the infinimum of m(U) over all open sets
containing
(2,3).
> If U contains (2,3) then m(U) >= 1, so the infimum is >= 1.
> But it is eaxctly one as there is a U containing (2,3) with
m(U) = 1,
viz.,
> U = (2,3) itself. So the infimum is 1.
Hah! Got it (at last)!
David.
--
===
Subject: Re: more easy questions
> Im drawing on several resources as I try to get my little
brain round
> measure theory and Lebesgue integration. Please can someone
help answer
> the following questions:
> 1). Is it fair to say that sigma rings are a subset of
rings - so that
> the number of sigma rings is less than the number of rings?
Yes. A sigma ring is a ring, but the converse need not be
true.
> 2). When applying Ômeasure why does the set 
you are
applying it to have
> to belong to some bigger space called the measure space?
Why cant you
> measure any given set without making sure it is a subset of
some larger
> set?
Consider defining a measure on the subsets of R^3. If this
measure is to
agree
with the notion of volume we are trying to capture, then (1)
the measure of
a
finite or countable union of disjoint sets should equal the
sum of the
measures
of those sets; (2) moving a set E via translations,
rotations, or
reßections
should not change its measure; and (3) the measure of the
unit cube should
be 1.
It turns out that no measure can satisfy all three of these
requirements for
all
subsets of R^3. (You need the axiom of choice to find one that
does not.)
So
instead one chooses a collection of subsets of R^3 on which
such a measure
can
be defined. This gives rise to Lebesgue measure. (See, e.g.,
Folland,
Real
Analysis, chapter 1.)
> 3). If the set E is the collection of all finite unions of
disjoint
> intervals (on R or R^n, I think), why is E a ring but not a
sigma ring?
> Surely if we add say two intervals together we get another
interval
> which is still part of E, dont we?
If E is a sigma ring, it must be closed under countable
unions, not just
finite
ones. Clearly every interval of the form (n - 1/2, n + 1/2),
where n is an
integer, lies in E, but the union of these intervals does
not, so E is not
a
sigma ring.
> 4). In the definition of outer measure we have something
along the lines
> of m*(A) = inf SUM (k=1 to inf.) m(I_k), where I_k is some
interval. Now
> I thought that outer measure was of the form a - b i.e. the
length of
> the interval, but according to this definition, it is the
infimum
> (minimum) of the interval that is important - so Im not
sure whats
> going on here.
It is not clear what you are asking. The goal here is to
*define* the
length
of a set that is not an interval, by covering it with
intervals (whose
length we
have already defined), and choosing the smallest such 
covering.
===
Subject: Re: My FLT proof, *last* warning to critics
|> Please stop replying to the imposter at yahoo as if
|> he/she/they were jstev@msn.com.
|
|Why? Replying to the Yahoo poster and to the MSN poster
|both seem about equally productive.
Unproductive, you mean.
Replying to jstev@yahoo wont give you any idea of how
anyone manages to hold the kind of beliefs that jstev@msn
has. Moreover, I have no problem with people addressing
jstev@yahoo as long as they dont pretend he/she/they
are jstev@msn, which is mean.
Keith Ramsay
===
Subject: Re: non-Euclidean geometry
Bhu Joshipura
> I am not a mathematician but I love mathematics. I love the
warmth
> with which this group responds to my questions. Here is one
more:
> Every web-page I read on non-Euclidean geometry refered to
problem
> with Euclids fifth postulate.
> Why does none talk about geometries with any other postulate
> re-examined?
> Is it not possible or is it not interesting?
Quite possible and quite interesting. Various axioms can be
discarded or
modified. E.g. look around for affine geometry and 
projective
geometry.
Hilberts _Foundations of Geometry_ studied various axiom
systems. That
book
went through seven editions from around 1899 to 1930. Its
not a long book,
but had a huge inßuence, inaugurating the axiomatic method as
a tool of
research, and not just a technical necessity. Some version of
the book is
still in print, I believe. Another good one is Coxeters
_Introduction to
Geometry_.
Larry
Vancouver
===
Subject: Re: non-Euclidean geometry
> I am not a mathematician but I love mathematics. I love the
warmth
> with which this group responds to my questions. Here is one
more:
> Every web-page I read on non-Euclidean geometry refered to
problem
> with Euclids fifth postulate.
> Why does none talk about geometries with any other postulate
> re-examined?
> Is it not possible or is it not interesting?
> -Bhu
The postualtes are listed at
http://mathworld.wolfram.com/EuclidsPostulates.html
Notice the nature of the first four postulates: You can do
_____
The fifth one, when phrased differently, is the only one that
asserts
the *unique* *existence* of an object. As a result, it is
easier to ask
what if when viewing this than the others.
Also, if you look at the surface of a sphere and think of a
line as a
great circle, the fifth postulate is the one that is most
obviously
violated. This observation leads naturally to investigating
the nature
of parallel lines.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: non-Euclidean geometry
> I am not a mathematician but I love mathematics. I love the
warmth
> with which this group responds to my questions. Here is one
more:
> Every web-page I read on non-Euclidean geometry refered to
problem
> with Euclids fifth postulate.
> Why does none talk about geometries with any other postulate
> re-examined?
> Is it not possible or is it not interesting?
> -Bhu
> The postualtes are listed at
> http://mathworld.wolfram.com/EuclidsPostulates.html
> Notice the nature of the first four postulates: You can do
_____
> The fifth one, when phrased differently, is the only one
that asserts
> the *unique* *existence* of an object. As a result, it is
easier to ask
> what if when viewing this than the others.
> Also, if you look at the surface of a sphere and think of a
line as a
> great circle, the fifth postulate is the one that is most
obviously
> violated.
And another of Euclids axioms is violated in this case too.
In the
Hyperbolic case _only_ the fifth postulate is violated.
> This observation leads naturally to investigating the nature
> of parallel lines.
> --
> Will Twentyman
> email: wtwentyman at copper dot net
--
G.C.
===
Subject: Re: online/self-study 4 year math degrees?
I am not aware of any such program. You may be able to pick
up an Analysis
or Abstract algebra course, though. I would have loved to
finish a math
program when I was in the Army, but, alas, I wasnt so 
lucky.
If you find
such a program, please post back here with the details. But,
as far as I
know, there is no college or university with a complete math
degree program
online.
Lurch
> Can anyone provide me a list of some universities which
offer such a
> program? I am in the military and so attending classes like
normal
> people is pretty much nixed. But I very much desire to take
classes
> in math. Im already self teaching myself alot of maths
(heck, of all
> the maths i HAVE managed to take here and there at various
colleges,
> the only one i hadnt already taught myself- thus the only
one that
> wasnt just a matter of warming a chair to get a credit-
was linear
> algebra) so i am well disciplined and prepared to survive
such a
> program.
> PS please post answers to sci.math since this email account
doesnt
> exist
===
Subject: Re: online/self-study 4 year math degrees?
> Can anyone provide me a list of some universities which
offer such a
> program? I am in the military and so attending classes like
normal
> people is pretty much nixed. But I very much desire to take
classes
> in math. Im already self teaching myself alot of maths
(heck, of all
> the maths i HAVE managed to take here and there at various
colleges,
> the only one i hadnt already taught myself- thus the only
one that
> wasnt just a matter of warming a chair to get a credit-
was linear
> algebra) so i am well disciplined and prepared to survive
such a
> program.
> PS please post answers to sci.math since this email account
doesnt
> exist
I do not know any online-programs.
However, if it is only a matter of attendance, tradidional
programs of
distance learning (i.e. via correspondence) might help too.
Look at
<http://www.dmoz.org/Reference/Education/Distance_Learning/>
Marc
===
Subject: outer morphism
hi all,
i am looking for concrete examples about outer morphism for
simple
finite Lie algebra.
TF
===
Subject: Re: probabilistic way of looking at life
> With a large enough sample, any outrageous thing is likely
to happen
> (Diaconis and Mosteller 1989). Littlewood (1953) considered
an event
> which occurs one in a million times to be surprising.
Taking this
> definition, close to 100,000 surprising events are expected
each
> year in the United States alone and, in the world at large,
we can be
> absolutely sure that we will see incredibly remarkable
events
> (Diaconis and Mosteller 1989).
> -- from
> http://mathworld.wolfram.com/LawofTrulyLargeNumbers.html
> Ah, that means that 10^11 events happen in the USA, per
year.
> The area of the United States is about 10^7 km^2.
> So: 10^4 events per km^2, per year.
> That is, one event, in every 100 square meters, per year.
> I think they took the statistic for Des Moines and
extrapolated it over
the
> whole country.
When it is infinitely unlikely that something will ever
happen, it
will happen almost immediately -- Douglas Adams.
Herman
http://www.AerospaceSoftware.com
===
Subject: Probability (permutations)
I came across a problem of calculating a probability that in
two randomly
chosen permutations there is a given number of equal elements.
E.g The two 6-element permutations:
P=(1 2 4 3 0 5) and
Q=(1 5 4 0 3 2) have two equal elements P(0)=Q(0)=1 and
P(2)=Q(2)=4.
I need this probability for up to 256-element permutations
having
0,1,2,3,4 and 5 equal elements.
Can somebody suggest a way to calculate it? At the first
glance I tried to
ignore the fact that P and Q are permutations and assume they
are sets of
numbers, But it showe to be not corect...
Bartosz Zoltak
qpbzoltak@vmpcfunction.com without qp
===
Subject: Re: Probability (permutations)
> I came across a problem of calculating a probability that
in two randomly
> chosen permutations there is a given number of equal
elements.
> E.g The two 6-element permutations:
> P=(1 2 4 3 0 5) and
> Q=(1 5 4 0 3 2) have two equal elements P(0)=Q(0)=1 and
P(2)=Q(2)=4.
> I need this probability for up to 256-element permutations
having
> 0,1,2,3,4 and 5 equal elements.
> Can somebody suggest a way to calculate it? At the first
glance I tried
to
> ignore the fact that P and Q are permutations and assume
they are sets of
> numbers, But it showe to be not corect...
I think you can assume that one of the chosen permutations is
the
identity permutation without loss of generality.
Now, you want to come up with a count of the second chosen
permutations
that match the condition of having a given number of equal
elements.
Let K(n, m) be that count where m <= n.
I have the following diagram (for n = 5):
0
(1 2 3 4 5) ---> ?
1
(1 2 3 4 5) ---> 5 * K(4,0) = 5 * 9
2
(1 2 3 4 5) ---> (5*4)/(2*1) * K(3,0) = 10 * 2
3
(1 2 3 4 5) ---> (5*4*3)/(3*2*1) * K(2,0) = 10 * 1
4
(1 2 3 4 5) ---> (5*4*3*2)/(4*3*2*1) * K(1,0) = 5 * 0
5
(1 2 3 4 5) ---> (5*4*3*2*1)/(454*3*2*1) * K(0,0) = 1 * 1
But the sum of all the counts on the right hand side is n!,
so the first question mark can be evaluated.
My analysis may be wrong. You should check it. Even if it
is wrong, it may give you an approach to the problem.
Note that I dont have a closed form for K(n,m). If you
can program it on a computer, then the above should be
sufficient for the computer to handle it.
-- Bill Hale
===
Subject: Re: Probability (permutations)
>I came across a problem of calculating a probability that in
two randomly
>chosen permutations there is a given number of equal
elements.
>E.g The two 6-element permutations:
>P=(1 2 4 3 0 5) and
>Q=(1 5 4 0 3 2) have two equal elements P(0)=Q(0)=1 and
P(2)=Q(2)=4.
>I need this probability for up to 256-element permutations
having
>0,1,2,3,4 and 5 equal elements.
I assume you mean that all (n!)^2 ordered pairs of
permutations are
equally likely to be chosen.
The first simplification is that P(k) = Q(k) if 
and only if
P^(-1) Q(k) = k, and P^(-1) Q is equally likely to be any of
the n!
permutations. So we can restate the problem as: what is the
probability
that a randomly chosen permutation on n letters has exactly r
fixed
points.
A permutation on n letters with no fixed points is called a
derangement.
The number of derangements on n letters is D_n = n!
sum_{k=0}^n (-1)^k/k!.
A permutation on n letters with a given r letters as fixed
points
corresponds to a derangement of the other n-r letters. There
are
(n choose r) = n!/(r! (n-r)!) ways to choose r fixed points,
so the number of permutations of n letters with exactly r
fixed points
is (n choose r) D_{n-r} = n!/r! sum_{k=0}^{n-r} (-1)^k/k!
and the probability of having exactly r fixed points is
1/r! sum_{k=0}^{n-r} (-1)^k/k!
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Problem Solving Strategies by Arthur Engel
> I hope you are not taking part in the german competition
Bundeswettb.
> I would ask the readers of this newsgroup not to to
publicize further
So is some major competition actually setting a problem from
a well-known problem book (albeit one whose solution is not
included in the book)?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
Visiting Assistant Professor at the University of Montana.
><deleted> What I showed in my post is that w1*w2 can be ANY
two algebraic
integer
>> factors of f = 3. It is NOT necessarily the case that
w1*w2 =
3^{2/3}
>> in that case. The case f = 3 does not give you any
information
about
>> the ws.
>>
>> Well it turns out I was in error
>> ... not for the first time, or, I suspect, the last.
>Oh, so now you want to attack me for admitting a mistake.
When people point out your mistake, you almost invariable
engage in
personal attacks. It is hard not to at least point out, on
the rare
occassion when the truth finally sinks in, that you were quite
strident in your defense of what you now admit to be an error.
Particularly when you are so wont to hark back on your record
in
support of your cause.
>That tells a lot about you Nora Baron.
>Why dont you tell the newsgroups who you actually are and
quit hiding
>behind the pseudonym?
Sigh. If people post with their true names, you accuse them
of trying
to use their position to inßuence people. If people post with
(apparent) pseudonyms, you accuse them of trying to hide
something. When people simply point out your mistakes, you
accuse them
of not being clear. When people go to great lengths to
explain their
mistakes, you accuse them of trying to snow you.
When people post rude messages, you accuse them of attacking
you. When
people post calm, reasoned messages, you accuse them (as you
did me a
couple of days ago) of trying to persuade people.
When people post that they are mathematicians, you accuse
them of
trying to use their education to inßuence. When people post
that they
are NOT mathematicians, you accuse them of trying to put
their noses
where they dont belong.
Never mind that you have used your true name, tried to use
your
history to inßuence people, made great deals of even the most
trivial
mistakes of others, been routinely rude and engaged in
outright libel
against individuals, when not harassment, tried to claim your
undergraduate degree in physics is somehow relevant to the
discussion, etc.
In short, you accuse anyone who doesnt agree with you.
Doesnt matter
why, youll find something. Its 
not that there is really an
issue
with what you choose to attack. Its just that you are so
scared that
you need to grab onto something to attack them.
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to
answer
on like occasions - A mans capacity is no measure of his
power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does
more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A
great
many people are staggered to this extend, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
> <deleted > What I showed in my post is that w1*w2 can be
ANY two algebraic
integer
> factors of f = 3. It is NOT necessarily the case that w1*w2
=
3^{2/3}
> in that case. The case f = 3 does not give you any
information
about
> the ws.
>
> Well it turns out I was in error
>
>
> ... not for the first time, or, I suspect, the last.
> Oh, so now you want to attack me for admitting a mistake.
> That tells a lot about you Nora Baron.
> Why dont you tell the newsgroups who you actually are and
quit hiding
> behind the pseudonym?
> as the answer is given by considering
> the factorization
>
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
>
> 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> as with f=3, its clear that *each*
>
>
> each what?
> Each of the as has 3 as a factor when f=3, and that is
without the
> dependency on m that you seem so dedicated on claiming
exists when f
> is coprime to 3.
> has 3 as a factor, which gives you
>
> P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 -
>
> (-1+m 3^2 )x u^2 + u^3 =
>
> (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u).
>
>
>
> What I showed in my previous post implies that if
> a1, a2, and a3 are the negatives of the roots of
> the polynomial
>
> u^3 - 8*u^2 - 19,
>
> then a perfectly legit factorization is:
>
> [****] (a1*x + 1)*(a2*x + 1)*(27*a3*x + 27)
>
> which happens to equal
>
> 27*(a1*x + 1)*(a2*x + 1)*(a3*x + 1).
>
> I think you can work out the reasoning
> for yourself on that. If not, let me know.
> But Nora Baron the as are NOT the negatives of the roots 
of
> u^3 - 8*u^2 - 19
> as in fact,
> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f) =
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> so the as are the roots of
> a^3 + 3(-1+9m)a^2 - ((-1+9m)^3+1).
{see below for correction of this}
P(1) = 3^2*((3^4 - 3*3^2 + 3)*x^3 - 3(8)*x + 3)
= 3^3*(19*x^3 - 8*x + 1).
It can be readily shown that
(19*x^3 - 8*x + 1) = (c1*x + 1)*(c2*x + 1)*(c3*x + 1),
where c1, c2, and c3 are the negative of the roots of
u^3 - 8*u^2 - 19 = 0.
Each of the roots of the latter equation is:
(1) an algebraic integer, and
(2) coprime to 3 [because 19 is coprime to 3].
Now consider
P(1)/3^2 = 3*(c1*x + 1)*(c2*x + 1)*(c3*x + 1).
So the whole question is: how can I distribute the 3
just to the right of the equals sign, among the three
linear factors?
Suppose t1, t2, and t3 are ANY THREE algebraic integers
whose product, t1*t2*t3 = 3. Then the following is
a factorization that meets your requirements:
(t1*c1 + t1)*(t2*c2 + t2)*(t3*c3 + t3).
Note that t1*c1, t2*c2, and t3*c3 are algebraic
integers, as of course are t1, t2, and t3.
Now: how many factorizations like this are there?
Answer: infinitely many: one for each way of representing
3 as a product of 3 integers.
Conclusion: no UNIQUE way of doing it. Any of an
infinite number of ways of factoring will work.
Further conclusion: this in no way confirms your
claim that the ws are constant with respect to m.
This adds no information about the ws at all.
> Now using the bs and ws gives
>
> P(m)/3^2 = 3((m^3 3^3 - 3^2 m^2 + m) x^3 -
>
> (-1+m 3^2 )x u^2 + u^3) =
>
> (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
> but at m=0, I have
>
> P(0)/3^2 = u^2(3x - 3u) = u^2(b_3 w_1 w_2 - 3u), so
>
> and dividing off the last 3 gives
>
> P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 -
>
> (-1+m 3^2 )x u^2 + u^3 =
>
> (b_1 x + u)(b_2 x + u)(b_3/3 x + u)
>
>
>
> Nope, this is NOT OK. Why? Because you are assuming
> what you want to prove. You want to show that the
> factorization when m <> 0 is similar to that when m = 0.
> Here m = 3. In case you have not noticed, 3 is not equal
> to 0.
> But m does not equal 3, as readers can verify by looking
above at the
> expressions where they see m all over the place, as well as
3,
and
> there is no claim that m=3.
My mistake here. Your intention was that m is any value, not
necessarily 3. But my *point* was still correct.
My point was, you assumed a factorization for m = 0, and then
you assumed the same form for the factorization when m <> 0.
This
occurs in your line above where you have P(m)/3^3 on the
right.
Note that the previous line has P(0)/3^2. In other words, you
jumped from P(0) to P(m) with no justification. Right there 
you
were assuming what you want to prove: that the form of the
factorization is the same for m = 0 as it is for m <> 0.
> Plus, if you look at what I said above at [****}, you
> will see a counterexample.
> <deleted remainder to focus on strange statements>
????? You deleted a lot.
> Im curious about how youd like you to 
explain the weird
statements
> youve already made,
> and Ill finish out by focusing on the last 
one
> Ive included.
> Heres what you had above:
>
> What I showed in my previous post implies that if
> a1, a2, and a3 are the negatives of the roots of
> the polynomial
>
> u^3 - 8*u^2 - 19,
>
> then a perfectly legit factorization is:
>
> [****] (a1*x + 1)*(a2*x + 1)*(27*a3*x + 27)
> But in fact, the as are the roots of
> a^3 + 3(-1+9m)a^2 - ((-1+9m)^3+1).
No - your original polynomial was
((-1 + 9m)^3 + 1)*x*3 - 3*(-1 + 9m)*x*3^2 + 3^3.
Letting m = 1, this is
513*x^3 - 3^3*8*x + 3^3
= 3^3*(19*x^3 - 8*x + 1).
[Please feel free to check this]
This means that the coefficients I called c1, c2, and c3
are the negatives of the roots of
u^3 + 8*u^2 - 19 = 0
which is (almost) what I claimed. (I had -8 instead of +8.
As it happens that makes no difference in the conclusions.)
You too made an algebra error again somewhere.
> Yet after you said if in one place, further down in your
post you
> said:
> Plus, if you look at what I said above at [****}, you
> will see a counterexample.
> Why dont you try and explain yourself to the newsgroups?
See above. I think it is more clearly explained there.
The point of all this, again, is that when f = 3, the ws
are not uniquely determined. In other words, f = 3 adds
no information regarding the ws. It does not 
confirm your
claim that the ws are constant with respect to m.
Moreover:
Even if you had obtained the result you wanted for f = 3,
you still have not shown anything for f = 5, 7, 11, 13, ....
That is, if you had your desired result for f = 3, you would
STILL have to prove that it holds also for f = primes > 3.
You have been nibbling on the underbrush at the edge of a
very large forest. To generalize about whats in the deep
interior, you would still need to provide a proof. Knowing
what happens on the edge (which you dont have anyway) would
not be enough.
Bottom line:
You are right back where you started: no proof whatsoever
that the pattern of factorization for m = 0 generalizes to
m <> 0.
Nora B.
> James Harris
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
===
>Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
>Message-id: <3F379EE0.1C6EB7F1@ix.netcom.com> You, and some
other posters, have displayed signs of codependancy. As
>> if James sickness draws you to him, like ßies on .
>Say, arent you the same guy who posts all the James Harris
parodies?
Flatrings was a pseudonym once used by the real James Harris,
so I think
we
can safely say yes to that question.
>(Talk about codependancy!)
>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
--
Mensanator
2 of Clubs
http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm
===
Subject: Profinite Galois cohomology as a sheaf cohomology?
Hi everyone.
I was just pondering...
I know how to get Galois cohomology classically and through
.8etale
cohomology of a point (field spectrum), but...
Can you get profinite Galois cohomology more directly through
some kind
of sheaf of G-modules over the underlying topological space
of G?
That way you would have a topological group which acts on its
sheaves -
does this type of structure have a name?
Any inputs?
Martin
--
Tout ce quil y a de b.8eb.90te. -- Grothendieck
===
Subject: Projections of bivariate gaussian distributions
I am looking for a formula to calculate the projection of a
bivariate
gaussian function under an arbitrary angle theta in function
of
sigma^2_x, sigma^2_y and sigma_xy. Is there anyone who knows
such a
===
Subject: p vs. np problem
Im looking for knowledgeable readers to give feedback on a
paper that
I dont want to claim that it is correct unless others 
agree,
as its
possible that I made an error. Its a very 
difficult problem
that
many have tried and if I am wrong, then Im not the 
first and
wont be
the last. It is listed at http://arxiv.org/abs/cs.CC/0305035
Understanding the paper requires a knowledge of posets and
Dilworths
theorem, as this is the main insight.
Craig
===
Subject: Re: p vs. np problem
> Im looking for knowledgeable readers to give feedback on 
a
paper that
> I dont want to claim that it is correct unless others
agree, as its
> possible that I made an error. Its a very 
difficult problem
that
> many have tried and if I am wrong, then Im not the 
first
and wont be
> the last. It is listed at http://arxiv.org/abs/cs.CC/0305035
> Understanding the paper requires a knowledge of posets and
Dilworths
> theorem, as this is the main insight.
> Craig
It seems to me the difficulty of P vs. NP lies precisely in
trying to
corral the notion of what all possible computations might be
(for
solving, in your case, the SUBSET-SUM problem). As you know,
it is not enough to prove that various methods of solution
require
a more than polynomial number of operations (in the size of
the
input); one must somehow demonstrate this for _all_ possible
methods of solution.
Your paper seems to be quite terse in its treatment of why
methods
of solution must fall into one of the three categories
described,
and these descriptions themselves seem not to have the
clarity of
definition that would be required to convince me of a proof.
Id be
happy to read a revised version with more attention to that
aspect.
===
Subject: Re: Question about consecutive prime numbers
> Question: Given a positive integer K, is it always possible
to find
> three consecutive primes p,q,r so that q > p+K and r > q+K?
I dont know whether there is an unconditional proof, but 
you
can
easily prove that its true on the prime k-tuples 
conjecture.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Question about consecutive prime numbers
>Question: Given a positive integer K, is it always possible
to find
>three consecutive primes p,q,r so that q > p+K and r > q+K?
Yes.
Suppose q is a prime congruent to 1 mod p_1, p_2, ..., p_m
and congruent
to 2 mod p_{m+1}, where p_j is the jth prime, m large 
enough
that
p_m > K + 1.
Infinitely many of these q exist by Dirichlets 
Theorem on
primes in
arithmetic progressions. Then q + k = 1 + k mod p_j for 1 <=
j <= m.
If K >= |k| >= 2, 1 + k is divisible by some p_j with j <= m,
so
(if q is large enough) q + k is composite. This leaves only q
- 2,
which is composite because its divisible by p_{m+1}. So
there are no
other primes within K of q, and we must have p < q-K and r >
q+K.
> (If so,
>theres an obvious generalization to more than three
consecutive
>primes.)
That looks much harder to prove.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: question about dual space
Is it possible to find form of space dual to C_0
where C_0 contains all of continous f:[0,inf)->R such that
lim f(x) = 0 when x->inf and C_0 is endowned with sup norm ?
Is C_0 reßexive?
olej
===
Subject: Re: question about dual space
>Is it possible to find form of space dual to C_0
>where C_0 contains all of continous f:[0,inf)->R such that
>lim f(x) = 0 when x->inf and C_0 is endowned with sup norm ?
Yes. Note that [0,infinity) is homeomorphic to [0,1). The
members
of C_0 correspond to continuous functions on [0,1] with f(1)
= 0.
Use the Riesz-Markov theorem.
>Is C_0 reßexive?
No. Note e.g. that its unit ball has no extreme points, and
therefore
(by Krein-Milman) cant be weakly compact.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Question regarding Collatz Conjecture (Optimization?)
I have a question regarding the Collatz Conjure, first for
people who are
unfamiliar...
The Collatz conjecture, also known as the 3n+1 conjecture,
the Ulam
conjecture or the Hailstone sequence, was first stated around
1950 and
concerns the following process:
1. Pick any positive integer n.
2. If n is even, divide it by two; if it is odd, multiply it
by three and
add one.
3. If n = 1, stop; else go back to step 2.
For instance, starting with n = 6, we get the sequence 6, 3,
10, 5, 16, 8,
4, 2, 1.
6, 3, 10, 5, 16, 8, 4, 2, 1.
So My Question is:
Would it not be faster to add +1 to an odd number rather then
multiply by
three and add 1? If I add +1 to an odd number instead, I get
faster
results...
For instance, starting with n = 6, we get the sequence 6, 3,
4, 2, 1.
Much quicker... is there a problem with me doing it this way?
Mike Curry
===
Subject: Re: Question regarding Collatz Conjecture
(Optimization?)
> I have a question regarding the Collatz Conjure, first for
people who are
> unfamiliar...
> The Collatz conjecture, also known as the 3n+1 conjecture,
the Ulam
> conjecture or the Hailstone sequence, was first stated
around 1950 and
> concerns the following process:
> 1. Pick any positive integer n.
> 2. If n is even, divide it by two; if it is odd, multiply
it by three and
> add one.
> 3. If n = 1, stop; else go back to step 2.
> For instance, starting with n = 6, we get the sequence 6,
3, 10, 5, 16,
8,
> 4, 2, 1.
> 6, 3, 10, 5, 16, 8, 4, 2, 1.
> So My Question is:
> Would it not be faster to add +1 to an odd number rather
then multiply by
> three and add 1? If I add +1 to an odd number instead, I
get faster
> results...
> For instance, starting with n = 6, we get the sequence 6,
3, 4, 2, 1.
> Much quicker... is there a problem with me doing it this
way?
Wouldnt it be even quicker just to map all integers to 1?
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Question regarding Collatz Conjecture
(Optimization?)
> Would it not be faster to add +1 to an odd number rather
then multiply by
> three and add 1? If I add +1 to an odd number instead, I
get faster
> results...
If you sere sitting an exam and were asked to multiply 148263
by 237893,
what would you expect to happen if instead of answering the
question
as set, you argued that it would be faster to multiply 100000
by 200000
and did that instead? :-)
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: Question regarding Collatz Conjecture
(Optimization?)
> If you sere sitting an exam and were asked to multiply
148263 by 237893,
> what would you expect to happen if instead of answering the
question
> as set, you argued that it would be faster to multiply
100000 by 200000
> and did that instead? :-)
You have a good point! ;)
Mike
===
Subject: Re: Question regarding Collatz Conjecture
(Optimization?)
Mike Curry <mcurry8173@rogers.com> escreveu na mensagem
> Would it not be faster to add +1 to an odd number rather
then multiply
> by three and add 1? If I add +1 to an odd number instead, I
get
> faster results...
If the current number hits a even number, then the current
number
decreases by a factor of 2. If it hits a even number, then
the current
number increases by a factor of 3 (more or less, because we
add a unit
after
multiply by 3). So, if the number increases more when he hits
a even
number than he decreases when he hits a odd number, and it
seems that
he
have the same probability of hit an odd number that he have
of hit
a
even number, then one would expect that, in average terms,
the current
number increases more that decreases, so that he would go to
infinite.
Collatz Conjecture says the inverse: that the current number
goes to 1. If
the Collatz Conjecture is true, then this behavior is a
surprise.
If you change the 3n + 1 rule for odd numbers by the n + 1
rule, then you
have a factor of increase of 1 (more or less) for odd numbers
and a factor
of decrease of 2 for even numbers. Because the factor of
decrease is larger
that the factor of increase, one would expect that the
current number goes
to 1. If that was true, then that would not be a surprise.
So, I think that your rule would lose the interest of the
problem, because
if the current number really goes to 1, that would not be
surprising. But
it
would be surprising with the 3n + 1 rule.
Sorry my English.
Jaime Gaspar
______________________________
Homepage: www.jaimegaspar.com
E-mail: e-mail@jaimegaspar.com
===
Subject: Questions on the properties of inter-mingled samples
All messages from thread
Message 1 in thread
===
Subject: Questions on the properties of inter-mingled samples
suppose that you have 2 different sets, S1 and S2. there are
many,
many sample points in each one. S1 has a mean of x1 and a
standard
deviation, sd, of sd1. moreover, S2 has a mean and sd of x2
and sd2.
moreover, S1 and S2 are slightly correllated with each other
with R =
-0.4.
suppose that i created a new set called S* with 45% of its
sample
points from S1 and 55% from S2, what would the new standard
deviation
and mean be? is there an algorithm/calculation for this?
finally, suppose that you have 3 different sets, S1, S2, and
S3.
there are many, many sample points in each one. S1 has a mean
of x1
and a standard deviation, sd, of sd1. moreover, S2 has a mean
and sd
of x2 and sd2, and s3 has a mean of x3 and sd of sd3. S1, S2,
and S3
are all correllated with each other like this:
S1 S2 S3
S1 1.00 -0.55 0.22
S2 -0.55 1.00 0.02
S3 0.22 0.02 1.00
and suppose that i create a new set with a mixture of k1% of
S1, k2%
of S2, and k3% of S3 (k1 + k2 + k3 = 1.00). what would be the
new
mean and sd of this? does a math exist for these kinds of
problems?
finally, what measurement of correllation do i use: R^2, or R?
===
Subject: Re: Registration fee ??
> I am anxious to university registration fee for a term at
your nation.
> in my case(korea)
> if national university, usually 850~1300 dollar
> if private university, 2000~2500 dollar
> your nation ?? USA, Britain, Switzerland, Germany....etc...
> In Britain there is only one private university, the
University of
> Buckingham. Here are their fees
> http://www.buckingham.ac.uk/study/fees/undergrad.html.
> Im trying to find the fees for the University 
of Birmingham
but Im
> having Internet problems...
Now Ive found it:
http://www.undergraduate.bham.ac.uk/cost.htm.
--
G.C.
===
Subject: Re: Registration fee ??
Well, in the USA it can vary greatly!
Harvard, Yale, i.e. Ivy league --> one year $30, 000+ ; one
semester $
15,000+
State Universities --> one semester $12, 000- $15, 000
These figure include room & board, tuition, books, etc...
Lurch
> I am anxious to university registration fee for a term at
your nation.
> in my case(korea)
> if national university, usually 850~1300 dollar
> if private university, 2000~2500 dollar
> your nation ?? USA, Britain, Switzerland, Germany....etc...
> please, tell me what price~
===
Subject: Re: Registration fee ??
> I am anxious to university registration fee for a term at
your nation.
> in my case(korea)
> if national university, usually 850~1300 dollar
> if private university, 2000~2500 dollar
> your nation ?? USA, Britain, Switzerland, Germany....etc...
> please, tell me what price~
In Helsinki.fi you pay about 150 USD / year.
It is actually the membership fee for the student association
including health care.
===
Subject: Re: Registration fee ??
> I am anxious to university registration fee for a term at
your nation.
> in my case(korea)
> if national university, usually 850~1300 dollar
> if private university, 2000~2500 dollar
> your nation ?? USA, Britain, Switzerland, Germany....etc...
> please, tell me what price~
Germany, public universities, aprox. 40 EUR per term.
There are no further tuiton fees.
Some universities have contracts with regional transportation
services,
resulting in a higher fee (approx. 120 EUR) including free
transport.
You can find some additional information at
<http://www.campus-germany.de/> or via Google.
If you consider distance learning, there is the
FernUniversitaet Hagen:
<http://www.fernuni-hagen.de>. Their fees are higher, since
you have to
pay for the material they send.
Marc
===
Subject: Re: Registration fee ??
>Well, in the USA it can vary greatly!
>Harvard, Yale, i.e. Ivy league --> one year $30, 000+ ; one
semester $
>15,000+
I think this is low. Unless its tuition only (contrary to
your figure
for State Universities). (Cite? We dont need no 
steenken
cites!) I
recall the figure being closer to $38,000 (including
transportation to
Nashville several times a year). No academic (or athletic)
scholarships. Once upon a time, all need-based assistance was
loans
(according to stories), but this is now being eroded (also
according to
stories).
>State Universities --> one semester $12, 000- $15, 000
Well, some. This is the high end of out-of-state tuition.
In-state
tuition is much lower, often in the range of $2-5,000 per
year. Room
and board is not appreciably different from the cost of
living, about
$8000 a year. (By my reckoning.) Also, many state
universities offer
scholarships (including academic scholarships), which bring
the price
down. A standard option (but not at the University of
Michigan) is to
have a fairly large class of students (top 1, 3 or even 5% of
graduating
class, + national merit scholars, + several other categories)
automatically qualify for in-state tuition and maybe even
additional
scholarships. Texas Tech seems to be pretty aggressive in
using
financial incentives to attract students. Academic
scholarships seem to
seldom include room and board. (Michigan States Alumni
Distinguished
Scholarship being one notable exception.) I havent the
foggiest notion
of what is available to foreign students.
>These figure include room & board, tuition, books, etc...
Theyre also based on an academic year, rather than assuming
year-round
studies. I dont know how much of a difference that makes 
when
comparing universities across national borders.
Also, for graduate school, things are much different, because
graduate
assistantships basically mean that youre slave labor. But 
at
least it
doesnt cost you anything to be a slave. I 
dont know if its
different
for the Ivies. (I was not offered any money, but that 
doesnt
mean no
one is. I also dont know if undergrads get graduate 
students
at the
Ivies. Im pretty ignorant of them, really, simply because I
dont
believe that a Mercedes is enough better a car than the
standard makes
to merit the price difference. Although Ive never owned a
Mercedes, so
I could be wrong. Same with the Ivies.)
Jon Ive covered professors, in ivy covered halls Miller
===
Subject: Re: Registration fee ??
>>I am anxious to university registration fee for a term at
your nation.
>>in my case(korea)
>>if national university, usually 850~1300 dollar
>>if private university, 2000~2500 dollar
>>your nation ?? USA, Britain, Switzerland, Germany....etc...
>>please, tell me what price~
>>
>Germany, public universities, aprox. 40 EUR per term.
>There are no further tuiton fees.
>Some universities have contracts with regional
transportation services,
>resulting in a higher fee (approx. 120 EUR) including free
transport
>You can find some additional information at
><http://www.campus-germany.de/> or via Google.
Which indicates (but doesnt actually *say*) that, if you 
are
accepted,
the price is the same for foreign students. Its almost 
worth
it to do
your undergrad work in Germany just to save the tuition.
Hey! Maybe thats why the International Baccalaureate 
program
is
gaining in popularity.
It is interesting that the semester abroad in Europe (Germany
included) costs about the same as US tuition. Is there a
double
standard for pricing? The courses seem comparable. Well,
theyre real
courses, but those are available in the US, too.
Jon Miller
===
Subject: Re: Registration fee ??
> I havent the foggiest notion of what is available to
foreign students.
At Harvard, Princeton and Yale, for undergrads foreign
students, the
financial
aid is in form of loans, on-campus jobs and if thats not
enough,
reconsideration of the fee. (in that order IIRC)
Sam
--
Dont be afraid, Im gonna give you the choice 
I never had...
- Lestat in Interview with the Vampire (Ann Rice, 1976)
===
Subject: Re: Registration fee ??
> I am anxious to university registration fee for a term at
your nation.
In France, the registration fee is around 150 EUR for a year
in regular
non-selective universities (the ones you can attend as long
as you have
graduated from high school).
For selective universities (2-year colleges, which lead to
competitive
exams
after which you enter the higher-level universities depending
on your
ranking) its usually free.
Sam
--
[..] but the delight and pride of Aule is in the deed of
making, and in
the
thing made, and neither in possession nor in his own mastery;
wherefore he
gives and hoards not, and is free from care, passing ever on
to some new
work.
- J.R.R. Tolkien, Ainulindale (Silmarillion)
===
Subject: Re: Registration fee ??
> I am anxious to university registration fee for a term at
your nation.
> in my case(korea)
> if national university, usually 850~1300 dollar
> if private university, 2000~2500 dollar
> your nation ?? USA, Britain, Switzerland, Germany....etc...
Come to Britain and you will be a hot girl according to the
news
headlines.
> please, tell me what price~
--
G.C.
===
Subject: Relationship between Undecidability, Incompleteness,
and NP
Completeness??
I am a student who is about to start studying Undecidability
and
Incompleteness. However, before I began studying it, it
struck my
curiosity as to how it related to something else I studied, NP
Completeness. I am very curious to know if these are related
and if my
understanding of NP Completeness will help me in this new
area.
Blake Manner.
===
Subject: Re: Relationship between Undecidability,
Incompleteness, and NP
Completeness??
> I am a student who is about to start studying
Undecidability and
> Incompleteness. However, before I began studying it, it
struck my
> curiosity as to how it related to something else I studied,
NP
> Completeness. I am very curious to know if these are
related and if my
> understanding of NP Completeness will help me in this new
area.
You can partition the space of all problems into those that
can be
solved and those that cant. Then you can partition the 
space
of those
that can be solved into those that are easy and those that
are hard.
NP Completeness has to do with the second partition,
Undecidability
and Incompleteness with the first.
I dont think youll see much relation between 
them, but
maybe youre
destined to be the person who sees what the rest of us have
missed.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Relationship between Undecidability,
Incompleteness, and NP
Completeness??
> I am a student who is about to start studying
Undecidability and
> Incompleteness. However, before I began studying it, it
struck my
> curiosity as to how it related to something else I studied,
NP
> Completeness. I am very curious to know if these are
related and if my
> understanding of NP Completeness will help me in this new
area.
> You can partition the space of all problems into those that
can be
> solved and those that cant. Then you can partition the
space of those
> that can be solved into those that are easy and those that
are hard.
> NP Completeness has to do with the second partition,
Undecidability
> and Incompleteness with the first.
> I dont think youll see much relation 
between them, but
maybe youre
> destined to be the person who sees what the rest of us have
missed.
> --
> Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
Model theory assumes consistent global labeling.
:-)
mitch
===
Subject: Re: Relationship between Undecidability,
Incompleteness, and NP
Completeness??
>I am a student who is about to start studying Undecidability
and
>Incompleteness. However, before I began studying it, it
struck my
>curiosity as to how it related to something else I studied,
NP
>Completeness. I am very curious to know if these are related
and if my
>understanding of NP Completeness will help me in this new
area.
Most combinatorial problems, of the sort that might be in NP,
are
trivial from the point of view of decidability, because there
are
(in any particular instance) only finitely many possible cases
to
check, and the number of cases to be checked can be estimated
in
advance. The only trouble is it can be a very large number.
Decidability is only interesting for statements that would
involve
infinitely many possible cases, or a finite number 
that cant be
estimated in advance.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Rule 90 and left-to-right reversal
> Definitely related to whats on
> <http://www.richholmes.net/games/cellaut/r90.html>, then.
>
> Your site has moved: http:...
>
> Its a redirect. The first URL should always 
work; the
second might
> not.
> I get a 404 from the URL above.
Works on my system. But Ive changed domain name registrars
and DNS;
maybe thats taking a while to propagate.
===
Subject: School hours (was: anxiety of high school??)
>To what country are you refering? USA?
>High school in the USA is generally between the hours of
8:00am - 2 or
>3:00pm.
>Lurch
This is not really a sci.math topic, but Im unsure where to
post it.
During my college days, 8:00 am classes were unpopular.
Now that Im in the workplace, I often sleep until 8 am,
working
into the evening. I see more colleagues in the office at 8 pm
than when I arrive 8 am. [But the cafeteris is open at 8 am
and
close at 8 pm.]
When I attend local transportation meetings, I repeatedly
hear that
21% of morning commute-hour traffic is school-related.
I ask, why not push school openings later into the day?
Schedule the high school 9-4 rather than 8-3, or 9:30-3:30
rather than 8-2. Are there historical reasons, such as
a need to tend the farm in the afternoon, for the early
school starts?
The only good reason Ive heard for moving the clocks 
backward
in the Fall (i.e., ending daylight savings time) is to have
more sunlight as youngsters go to school -- delaying school
openings would solve this problem a different way.
--
Wanted: Experts at choosing the best of 100+ applicants for a
position.
Register as a California voter by September 22, and vote on
October 7.
Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
===
Subject: sir~differ topology problem
let X = {a,b,c,d,e}
For topology T of X that generated by A={{a},{a,b,c},{c,d}}
find local basis Bc for point c in X
-----------------------------------------------
i think that local basis seems to come out variously.
but in the solution paper, the one local basis written down
it.
i am wrong thinking??
i ask advice of that.
===
Subject: Re: sir~differ topology problem
>let X = {a,b,c,d,e}
>For topology T of X that generated by A={{a},{a,b,c},{c,d}}
>find local basis Bc for point c in X
>-----------------------------------------------
>i think that local basis seems to come out variously.
>but in the solution paper, the one local basis written down
it.
>i am wrong thinking??
No, there are many different local bases for a topology
at a point - you can have the right answer even though
your answer is not the same as the one in the solution paper.
What did you get for the local basis?
>i ask advice of that.
************************
David C. Ullrich
===
Subject: sir~~topology problem
let X = {a,b,c,d,e}
find topology T of X that generated by A={{a},{a,b,c},{c,d}}
----------------------------------
it is possible??
i think that A is not subbasis.
thus, not find.
sir~ check up problem. please.
===
Subject: Re: sir~~topology problem
> let X = {a,b,c,d,e}
> find topology T of X that generated by A={{a},{a,b,c},{c,d}}
> ----------------------------------
> it is possible??
> i think that A is not subbasis.
> thus, not find.
> sir~ check up problem. please.
Are you sure you understand the difference between a basis
and a
subbasis?
Every collection of subsets of a set X is a subbasis for some
topology on
X.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: sir~~topology problem
>let X = {a,b,c,d,e}
>find topology T of X that generated by A={{a},{a,b,c},{c,d}}
>----------------------------------
>it is possible??
>i think that A is not subbasis.
>thus, not find.
>sir~ check up problem. please.
Do you know the _definition_ of the topology generated by?
************************
David C. Ullrich
===
Subject: Re: sir~~topology problem
> let X = {a,b,c,d,e}
> find topology T of X that generated by A={{a},{a,b,c},{c,d}}
> ----------------------------------
> it is possible??
> i think that A is not subbasis.
It isnt a basis because there is a point in X which is not
in an
element of A. But that doesnt mean that A 
cant _generate_ a
topology.
Let the elements of A be open subsets of X. A collection W of
all open
subsets of X must be closed under unions and finite
intersections, so
0 (the empty set) must be in W since 0 = {a} u {c,d}
X must be in W since X = the intersection of the empty subset
of A
{c} is in W because {c} = {a,b,c} u {c,d}
... and so on...
If W so generated _is_ a topology on X then youre done, 
take
T to be
W. Otherwise add in as little as you need to make it a
topology.
Formally that means make W into a topology in all possible
ways and take
T to be the intersection of all those topologies.
> thus, not find.
> sir~ check up problem. please.
--
G.C.
Distribution: world
===
Subject: Solving Sums of Exponential and Linear Terms?
Does anybody know how one would approach solving something
like this
for n?
a^n + n = 1
Starling
Who has only had up to Linear Algebra, so be kind. :/
===
Subject: Re: Solving Sums of Exponential and Linear Terms?
Distribution: world
> Does anybody know how one would approach solving something
like this
> for n?
> a^n + n = 1
Numerically. See Newtons Method.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Solving Sums of Exponential and Linear Terms?
> Does anybody know how one would approach solving something
like this
> for n?
> a^n + n = 1
> Starling
> Who has only had up to Linear Algebra, so be kind. :/
Assuming that the unknown is n: if a > 0 then an obvious real
solution is n=0. If a>1, this solution is unique (something is
strictly increasing). The case 0<a<1 may lead to two, or one
double, or no, real solution. Make a sketch.
(There has been a lot of anguish here about what to do with
a=0, and I wont touch that.)
Whats being kind? Guessing what you meant by something like
Extend the problems to read
a^n + n = b
and then specialize for b=1. The answer uses a non-elementary
function called LambertW, namely:
n = (-lambertW(log(a)*exp(b*log(a)))+b*log(a))/log(a)
Conduct your own search about lambertW; its fun.
===
Subject: Some questions on methematical analysis
Hello everyone,
I am a newbie to methematical analysis and I am reading the
book
ADVANCED CALCULUS by Patrick M.Fitzpatrick. I have two
questions about
the exercises in this book, but the full answer is not given
at the
1) Let the function f:R->R be continous and suppose that its
image
f(R) is bounded. Prove that there is a solution of the
equation
f(x)=x, x in R.(done)
Now choose a number a with f(a)>a and define the sequence {an}
recursively by defining a_1=a and a_n+1=f(a_n) if n is a
natural
number for which a_n is defined. If f:R->R is strictly
increasing,
show that {a_n} converges to a solution of the above equation.
I can prove it by limits. But the hint said Use the density
of the
irrationals and the Intermediate Value Theorem. So, could any
one
tell me how to prove the assersion by that?
2) Suppose that the functions f:(a, b)->R is uniformly
continuous.
Prove that
f:(a, b)->R is bounded.
The hint said Express (a, b) as the union of finitely many 
such
subintervals. Could any one tell me how to express this idea?
It the first time I post here. I dont konw 
wether my question
is too
naive...If true, is there any proper place to post such newbie
questions?
# -----------------
# Gong Wuming
# Department of Biology
# Wuhan University, China
# -----------------
===
Subject: Re: Some questions on methematical analysis
===
Subject: Some questions on methematical analysis
>1) Let the function f:R->R be continous and suppose that its
image
>f(R) is bounded. Prove that there is a solution of the
equation
>f(x)=x, x in R.(done)
>Now choose a number a with f(a)>a and define the sequence 
{an}
>recursively by defining a_1=a and a_n+1=f(a_n) if n is a
natural
>number for which a_n is defined. If f:R->R is strictly
increasing,
>show that {a_n} converges to a solution of the above
equation.
lim a_n <= f(lim a_n) = lim f(a_n) = lim a_(n+1) = lim a_n
>I can prove it by limits. But the hint said Use the density
of the
>irrationals and the Intermediate Value Theorem. So, could
any one
>tell me how to prove the assersion by that?
Case 1: some x,y in R with f(x)-x <= 0 <= f(y)-y,
then by IVT, some z with f(z) - z = 0
Case 2: for all x,y in R, f(x) > x or y < f(y)
thus f is monotone and you can use above sequence method.
----
===
Subject: Re: Some questions on methematical analysis
> 2) Suppose that the functions f:(a, b)->R is uniformly
continuous.
> Prove that
> f:(a, b)->R is bounded.
> The hint said Express (a, b) as the union of finitely many
such
> subintervals. Could any one tell me how to express this
idea?
Take epsilon>0 with an arbitrary value. Then there is delta>0
such
that for x,y in (a,b)
|x-y|<delta => |f(x)-f(y)|<epsilon
Now consider (a,b) as the union of finitely many intervals
with length
<delta. Then f is bounded on each of these intervals by the
above
relation. And because two of these (open) intervals which are
adjacent
to each other have a non-empty intersection, f is bounded on
the whole
interval (a,b).
===
Subject: Re: Some questions on methematical analysis
> 2) Suppose that the functions f:(a, b)->R is uniformly
continuous.
> Prove that
> f:(a, b)->R is bounded.
> The hint said Express (a, b) as the union of finitely many
such
> subintervals. Could any one tell me how to express this
idea?
Since f is uniformly continuous, there is some r > 0 such that
|x - y| < r => |f(x) - f(y)| < 1. Since you can write (a, b)
as
a reunion of a finite set of intervals with length < r, then
the
image of f has length at most n, where n is the number of
intervals.
Jose Carlos Santos
===
Subject: Re: Standard Topology
>Bishop and Goldberg said that the standard topology on R^n is
> d_p(x,y) = [sum |u^i x|^p + |u^i y|^p]^(1/p)
>for p >= 1.
>> Thats nonsense - how does the formula actually read in
the book?
>Its a category mistake too. Thats a metric 
not a topology.
That too - in fact that was the first thing I said, before
looking
at the formula carefully; then when I noticed the formula had
been garbled the original correction got lost.
>The book says something like
>[sum_i|u^i x - u^i y|^p]^(1/p),
>u^i having been previously defined as the i-th projection 
from
>R^n to R (ugh!).
Well, actually I knew thats what it said even though I
havent
seen the book - was curious whether the OP could read
it properly.
>They sure dont have a lot of pictures in their book.
>What is the use of a book without pictures or conversations?
:-)
************************
David C. Ullrich
===
Subject: Re: Standard Topology
>>Bishop and Goldberg said that the standard topology on R^n
is
>> d_p(x,y) = [sum |u^i x|^p + |u^i y|^p]^(1/p)
>>for p >= 1.
>Thats nonsense - how does the formula actually read in the
book?
Eh. I threw a few extra bars up there, including a bar across
the minus
sign. Dunno how I did that.
>>I dont think they defined the u^i, are they 
just some
operator that
>>picks a component from a representation of a point?
>Probably.
>>I dont think they explained why p>=1, although it may 
have
been in a
>>problem that I solved incorrectly. Looks to me like theyd
all be
>>strongly equivalent if p!=0, for instance. And whats 
wrong
with p=1/2?
>If 0 < p < 1 then the correct version of the formula above
does
>not define a metric. (The fact that d(x,y) <= d(x,z) + d(z,y)
needs
>to be proved; for p < 1 it cant be proved because 
its
false.)
Oh, the triangle inequality.
I suppose that makes perfect sense since a metric space is
ßat in a small
enough region. But is there some reason we cant relax that
restriction?
--
A good plan executed right now is far better than a perfect
plan
executed next week.
-Gen. George S. Patton
===
Subject: Re: Standard Topology
>> Bishop and Goldberg said that the standard topology on R^n
is
>> d_p(x,y) = [sum |u^i x|^p + |u^i y|^p]^(1/p)
>> for p >= 1.
>They dont.
Now that we have that cleared up, lets have a beer.
--
A good plan executed right now is far better than a perfect
plan
executed next week.
-Gen. George S. Patton
===
Subject: Re: Standard Topology
>Bishop and Goldberg said that the standard topology on R^n is
> d_p(x,y) = [sum |u^i x|^p + |u^i y|^p]^(1/p)
>for p >= 1.
>> Thats nonsense - how does the formula actually read in
the book?
>Its a category mistake too. Thats a metric 
not a topology.
>The book says something like
>[sum_i|u^i x - u^i y|^p]^(1/p),
>u^i having been previously defined as the i-th projection 
from
>R^n to R (ugh!).
Yeah, I was a little generous with the bars there.
The book also says The standard topology on R is that of the
metric
defined by absolute value of differences. They seem to be
identifying
the metric with the topology.
>They sure dont have a lot of pictures in their book.
>What is the use of a book without pictures or conversations?
:-)
Well, Im a physics guy, and this reads like a book written
by a
mathematician. It takes a lot of time before I turn a page.
Im liking
Flanders a bit better.
--
A good plan executed right now is far better than a perfect
plan
executed next week.
-Gen. George S. Patton
===
Subject: Re: Standard Topology
...
>Oh, the triangle inequality.
>I suppose that makes perfect sense since a metric space is
ßat in a small
>enough region.
Thats a really ridiculous statement, which I assume you 
have
made because you have conßated the notion of a Riemannian
metric
on a manifold with the abstract notion of a metric space.
Though if (for instance) John Mitchell were to weigh in with
some argument that a metric space is ßat in a small enough
region, I would listen respecfully and no doubt learn
something,
in the absence of some such informed comment I will throw out
my opinion that its very rare indeed for a metric space--
even a metric space on R^2 which induces the standard
topology,
such as you seem to be trying to study--to be ßat in a small
enough region for any reasonable sense of ßat. I dont
think that the various non-Euclidean Minkowski p-norms on
R^2 are ßat, for example.
In particular, I dont see any reasonable sense in which
the triangle inequality is closely related to ßatness.
If you care to suggest such a sense, go for it.
Lee Rudolph
===
Subject: Re: Standard Topology
> The book also says The standard topology on R is that of
the metric
> defined by absolute value of differences. They seem to be
identifying
> the metric with the topology.
They arent. They explained earlier how a metric on a set
defines
a topology on a set. But thats not the same as identifying
metrics
with topologies. For a start, there are some topologies which
cannot
be defined by metrics. Also a topology defined by 
one matric
is also defined by uncountably many other matrics.
In brief the map
{metrics} --> {topologies}
is neither injective nor surjective.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
His mind has been corrupted by colours, sounds and shapes.
The League of Gentlemen
===
Subject: Re: Standard Topology
>Bishop and Goldberg said that the standard topology on R^n is
> d_p(x,y) = [sum |u^i x|^p + |u^i y|^p]^(1/p)
>for p >= 1.
>>Thats nonsense - how does the formula actually read in 
the
book?
>Eh. I threw a few extra bars up there, including a bar
across the minus
>sign. Dunno how I did that.
>I dont think they defined the u^i, are they 
just some
operator that
>picks a component from a representation of a point?
>>Probably.
>I dont think they explained why p>=1, although it may have
been in a
>problem that I solved incorrectly. Looks to me like theyd
all be
>strongly equivalent if p!=0, for instance. And whats wrong
with p=1/2?
>>If 0 < p < 1 then the correct version of the formula above
does
>>not define a metric. (The fact that d(x,y) <= d(x,z) +
d(z,y) needs
>>to be proved; for p < 1 it cant be proved because 
its
false.)
>Oh, the triangle inequality.
>I suppose that makes perfect sense since a metric space is
ßat in a small
>enough region.
Huh??? (See Rudolphs post for a more detailed huh?...)
> But is there some reason we cant relax that restriction?
Not sure what restriction you have in mind. But if you want
to relax it go ahead and relax it. The question was why the
(correct version of the garbled) formula does not define a
metric when p < 1. This has nothing to do with whether
one can modify the definition of metric...
************************
David C. Ullrich
===
Subject: Synergetics coordinates
What Eric W. Weisstein calls the x,y,z axes of the triangle
are rotated
one position to the right from mine in his description, but,
it looks
like at least some version of Synergetics coordinates might
become
respectable in academia now that they are mentioned at:
http://mathworld.wolfram.com/SynergeticsCoordinates.html
Cliff Nelson
===
Subject: Re: Tenser said the tensor, The Demolished Man by
Alfred
Bester
>I dont know much about what tensor analysis is all about 
or
what a
>tensor is, but Im wonder if a simple 
laymans explaination
of what
>one is would help to better appreciate this song.
No. Bester did not have a mathetical or scientific background,
and
just picked the word for the sound.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action.
I reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
reply
===
Subject: Re: Tenser said the tensor, The Demolished Man by
Alfred
Bester
>So in order to describe stress we wind up using 9 components,
But not nine independent components.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Any unsolicited bulk E-mail will be subject to legal action.
I reserve the
right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do not
reply
===
Subject: Re: The basic idea behind my great forthcoming proof
[stuff snipped]
> You got a point. (I smell a sense of irritation on your
side; i apologize
> :-) )
[...]
> And, of course, i dont know how you talk about your
subject to students.
You have a good sense of smell, but Im only slightly
irritated ;-).
I try to talk about mathematics in as concrete a fashion as
possible, but
there are interesting discrepancies in how people in my field
write and
talk about their research. Some papers, even though written
by very
intuitive people who speak very vividly, drawing very nice
pictures, and
so forth, are the model of terseness and esoteric combination
of set
theoretic symbols. Sometimes this is necessary, but I often
think that if
you are trying to say something like, Arc alpha intersects
these discs
transversely, writing a bunch of set theoretic symbols 
doesnt
illuminate. Of course, theres nothing wrong with doing
*both*, but too
frequently I think people just write the symbols and dont
draw the simple
picture.
>> The point here is that Im working in a 
specific context.
Im fine with
>> other foundational theories, but not all of them will be
useful to me.
>> Im not espousing that ZFC (or just ZF or whatever) is
*the* foundation
of
>> mathematics or anything of that sort. For me, thats
irrelevant.
> That is very good, and very interesting.
> People doing foundations should be interested in what you
do, and in
> explicitizing
> the assumptions you actually use / need in this kind of
geometry.
> But are they? And are geometers themselves interested in
making these
> assumptions
> more explicit?
The only cases Im familiar with this kind of axiomatization
happening are
few. For example, R.L. Moore axiomatized planar topology; one
of his
axioms was the Jordan curve theorem, but some of them were
less intuitive.
I suppose Hilbert was a geometer, but he was also a logician,
and I think
he was primarily motivated by reasons of logic rather than
geometry in
formalizing Euclidean geometry.
In Moores case, I think he believed that in the technical
situations he
worked in, axiomatization would make things a lot
easier/clearer. I think
that was true to a degree, but on the other hand, how many
topologists
that work with the plane, know these axioms? (Im not very
knowledgable
about Moores axioms, so I may be mistaken about some of 
what
I say here).
I dont believe geometers/topologists would be very
interested in
axiomatization, or even making assumptions even more explicit
(than they
currently are), *unless* some pressing need for them arose.
After all,
they didnt come to their current field of 
expertise because
they wanted
to be logicians.
I think this kind of axiomatization of the assumptions that
geometric
topologists use would be very interesting to people in
foundations. Im
not aware of anyone that has done a significant body of work
on this.
There are people that for example, have tried to axiomatize
proofs by
picture in Euclidean planar geometry, but I dont think a 
lot
has been
done on it. In any case, I think thats probably easier than
trying to
axiomatize the proofs by picture that topologists use.
> I know most mathematicians are not; and when they teach
their subject to
> students,
> they simply continue the tradition: ZFC or an informal
version of it,
> without thinking further.
> In that sense, the math culture has stopped questioning the
basics.
> Doesnt that worry you? It does worry me. Because i think
it is this that
> distracts
> young people from a study or a career in math. And it is
also not a good
> sign,
> in general, for the future of a science.
It used to worry me. Or perhaps its better to say it
irritated me.
After all, when a lecturer says the proof is by axiom of
choice, and
someone (like me) doesnt accept the axiom of choice, then
the situation
is equivalent to listening to some crank speaking about
solving the
Riemann hypothesis by using space-time discontinuities.
On the other hand, Im no longer irritated by it (although I
remain
somewhat concerned) because I discovered people are not
purposely trying
to discourage alternative ways of thinking but rather are
simply trying to
avoid trouble. To illustrate, I once challenged a speaker on
his use
of
the axiom of choice in a particular theorem. He responded by
saying that
he didnt believe in it, but thought of it as just a
convenient
short-hand for I can prove it constructively for the examples
and
situations Im really interested in; he found it was just
easier to
invoke AC.
I agree its not a good sign, anymore than 
its never a good
sign when
people stop thinking about fundamental ideas. But are things
really that
bad? I think not. Ive found its relatively 
easy to talk
about these
kinds of issues during coffee break or whatever. In any
event, real
paradigm shifts occur because of the young accept what the
old cannot. I
think young people in mathematics are far more interested in
things like
Goedels theorems and such than the older generations. Maybe
the fact
that we are in the computer age has something to do with it.
===
Subject: Re: The basic idea behind my great forthcoming proof
> After all, when a lecturer says the proof is by axiom of
choice, and
> someone (like me) doesnt accept the axiom of choice, then
the situation
> is equivalent to listening to some crank speaking about
solving the
> Riemann hypothesis by using space-time discontinuities.
Im impressed!
:-)
===
Subject: Re: The Definition of a Manifold
> Dirac Delta Function. Renormalization. Formal power series
that may
> not converge.
After rereading some of the posts and considering the above
example I
guess ill interpret cargo cult mathematics to be that
intuitive
application of an mathematical idea without setting up (or
checking
for) its formal foundations. Diracs delta function,
heavisides
function, the imaginary unit, analysis (for a few centuries
following
newton) and such. Is this what you mean?
Kevin
===
Subject: Re: The paternity of Christ (was: something
disrespectful if not
blasphemous)
> So are you saying that the Y-chromosome that God made for
Jesus would
> somehow be different than the one he made for Adam?
> Why would you ask such a silly question? There isnt 
nearly
enough
> information in scripture to draw that sort of conclusion.
Odds favor
> the conjecture that they were different, of course, unless
you propose
> that Jesus and Adam were identical twins.
> I made the fairly simple statement that scripture calls
Christ the
> only _begotten_ son of God, while distinctly calling others
than
> Christ sons of God. This clearly implies a distinction.
> They dont have to be identical twins to have the same
Y-chromosome.
> All they need to have is the same father.
Not even that much is required. Consider your statement, then
how the sons
of two such men could inherit this identical Y, etc.
===
Subject: Re: The two envelope paradox
> I must respectfully disagree with the seeming consensus of
the
> mathematicians whose analyses I scanned. The seeming
paradox has
little
> to
> do with the specific distributions assumed for the source of
the
pairs.
> It
> is resolved in all cases by inclusion of the boundary cases
in the
> calculation of expected gain over any finite series.
> I havent thought it through completely, but it would
seem that a
> winning
> strategy would be to switch in every case in which you
observe a
value
> less
> than the maximum value you have previously observed,
otherwise to
stick
> with
> your first choice.
> Of course if you are told a maximum bound ahead of time
there is no
> problem.
> Always switch unless you see that value.
> knowing the bound IS having knowledge of the distribution!
> Herc
> Of course. But the resolution of the paradox does not
depend on knowing
a
> bound, or on any reasoning about any particular kinds of
distribution.
The
> paradox arises because of the conßict between the provable
impossibility
of
> overall gain from unconditional switching, and a seemingly
valid
calculation
> which says one will indeed gain from unconditional
switching. To defeat
the
> paradox, one need only recognize that the estimation of
benefit from
> unconditional switching must also include the guaranteed
loss at the
> (typically unknown) upper bound of the values encountered
in any finite
> series of trials.
this seems like youre saying its not a paradox because
eventually you can
stop switching, but switching twice is still paradoxical.
> In an infinite series of trials with no finite 
upper bound,
switching
might
> be said to improve the rate of gain, but your expectation
in any case is
> either infinite or undefined depending on which 
aspects of
the picture
you
> want to quibble about, so there is no paradox to be worried
about in the
> first place...
> I mentioned knowing the bound only in context with
explicating the
> conditions under which one could actually gain by a
conditional switching
> strategy.
OK, I have trouble with this problem, when I do simulations
on computer you
get the same average return switching or not.
I imagine the problem as giving instructions to someone on a
show. You
tell
them, hes going to give you an envelope, you look at the
amount, then
ask
to switch to the second envelope.
Which would be equivalent to saying remember to ask for the
second
envelope.
I.e. as you put it, unconditional switching wins.
Now the trick to the problem is when you see the dollar
amount, it tells you
2 things.
One is the amount in that envelope, the other is the expected
range for
both
envelopes. e.g. $20 tells you the average outcome of the
second envelope is
$25.
So does giving the option to swap make the game company lose
more than the
average envelope content over time?
I think there are similar problems. When you snap a domain
name from being
recently
expired, if more than one person snapped it you have a
private auction.
Say the name might go for $1000 under public auction, but
typically would
get $100.
The private auction will go for higher than the typical
public auction,
several hundred dollars.
A more related one. Ive tried to figure a 
blackjack system.
Apparently
with most
casino rules the odds are 48.x % towards the player. i.e.
only slightly
favored
towards the casino if you play optimally. So a few %
advantage over the
best
known system would break 50%.
So I had the theory, if you sit at the spot just before the
dealer, then the
card you
buy affects his result. Say a 10 is the next card and it will
give you 21
but it
will also bust the banker, and several of your accomplices at
the table are
sitting
on low scores. Then it benefits the group overall for the
banker to get the
10.
After a decade I could never figure out the logic if an
advantagous
situation arose
because it relies on using a deck of cards in place. And
computerised
systems
just generate a random card on the spot so the 
Ôanchor
player would have
no
effect on what card the dealer got. But all the logic for
reasoning about
expected
scores are the same for shufßed cards as they are for random
cards. If you
program
it you randomise the cards into an array (shufße) or you
select a random
card when
required, they have the same effect but a random card will
never allow the
system
to have benefit. Maybe thats a proof the system cannot work?
anyway.... Ill work it out sometime.
Herc
this is an IOU for 5c. You can trade for another IOU for
either 1000 times
that
amount or 1000 times less!
===
Subject: Re: The two envelope paradox
>> I must respectfully disagree with the seeming consensus of
the
>> mathematicians whose analyses I scanned. The seeming
paradox has
little
>> do with the specific distributions assumed for the source
of the
pairs.
>> is resolved in all cases by inclusion of the boundary
cases in the
>> calculation of expected gain over any finite series.
>> I havent thought it through completely, but it would 
seem
that a
>winning
>> strategy would be to switch in every case in which you
observe a value
>less
>> than the maximum value you have previously observed,
otherwise to
stick
>with
>> your first choice.
>> Of course if you are told a maximum bound ahead of time
there is no
>problem.
>> Always switch unless you see that value.
>> knowing the bound IS having knowledge of the distribution!
>> Herc
>Of course. But the resolution of the paradox does not depend
on knowing a
>bound, or on any reasoning about any particular kinds of
distribution.
The
>paradox arises because of the conßict between the provable
impossibility
of
>overall gain from unconditional switching, and a seemingly
valid
calculation
>which says one will indeed gain from unconditional
switching. To defeat
the
>paradox, one need only recognize that the estimation of
benefit from
>unconditional switching must also include the guaranteed
loss at the
>(typically unknown) upper bound of the values encountered in
any finite
>series of trials.
I do not think that you have resolved the paradox. To do
that, you would
have to find an error in the mathematical argument which
proves that you
always have a positive expected gain from switching. The fact
that any
finite
series of trials will in practice have an upper bound is
obvious, and does
not represent an error in the argument.
Of course if you knew a specific upper bound in advance, then
that would
resolve the problem. One possible resolution is to say that
since the
envelope in question has finite dimensions, there must in fact
be such an
upper bound on the number that you could write on a piece of
paper and
put in the envelope - some function of the total number of
atoms that could
occupy the space of the envelope would probably work.
I have never found that resolution very satsifactory, because
I would
instinctively prefer a mathemtical rather than a physical
resolution.
Derek Holt.
>In an infinite series of trials with no finite 
upper bound,
switching
might
>be said to improve the rate of gain, but your expectation in
any case is
>either infinite or undefined depending on which 
aspects of the
picture you
>want to quibble about, so there is no paradox to be worried
about in the
>first place...
>I mentioned knowing the bound only in context with
explicating the
>conditions under which one could actually gain by a
conditional switching
>strategy.
>Bill
===
Subject: Re: The two envelope paradox
>> I must respectfully disagree with the seeming consensus of
the
>> mathematicians whose analyses I scanned. The seeming
paradox has
little
>to
>> do with the specific distributions assumed for the source
of the
pairs.
>It
>> is resolved in all cases by inclusion of the boundary
cases in the
>> calculation of expected gain over any finite series.
>>> I havent thought it through completely, but it would
seem that a
>winning
>> strategy would be to switch in every case in which you
observe a
value
>less
>> than the maximum value you have previously observed,
otherwise to
stick
>with
>> your first choice.
>>> Of course if you are told a maximum bound ahead of time
there is no
>problem.
>> Always switch unless you see that value.
>>> knowing the bound IS having knowledge of the
distribution!
>> Herc
>Of course. But the resolution of the paradox does not depend
on knowing
a
>bound, or on any reasoning about any particular kinds of
distribution.
The
>paradox arises because of the conßict between the provable
impossibility
of
>overall gain from unconditional switching, and a seemingly
valid
calculation
>which says one will indeed gain from unconditional
switching. To defeat
the
>paradox, one need only recognize that the estimation of
benefit from
>unconditional switching must also include the guaranteed
loss at the
>(typically unknown) upper bound of the values encountered in
any finite
>series of trials.
> I do not think that you have resolved the paradox. To do
that, you would
> have to find an error in the mathematical argument which
proves that you
> always have a positive expected gain from switching. The
fact that any
finite
> series of trials will in practice have an upper bound is
obvious, and
does
> not represent an error in the argument.
Perhaps I am missing the point, but so far I dont see it...
I still think
this does resolve the paradox by showing that the calculation
of a positive
expected gain from switching is in error.
Im a lousy mathematician, but Ill try to 
explain what I
mean.
There is some maximum value M in any finite series of pairs
drawn from any
distribution.
There is some non-zero probability P(M) that the first
envelope chosen in a
trial
will contain that maximal value M.
In the naive calculation of estimated gain we implicitly
assume the
gain calculation applies to all trials. But with probability
P(M) it
does not apply to a given trial, and the correct calculation
in that case
is that we will switch from the value M to the next lower
possible value,
a negative gain.
We dont know M. But we know that some M exists in any 
finite
series,
and I think it is an error to ignore that knowledge,
specifically it is
the error that leads to paradoxical results.
The expected gain calculation is based on an assumption known
to be false:
that the probability of winning or losing by switching is
independent of
the
value first chosen. The two values are not independent, they
are coupled.
Depending on the value chosen, you may be guaranteed a higher
value by
switching, or guaranteed a lower.
How is it not an error to assume that two random variables
are independent
when they are known to be interdependent?
A correct calculation of expected gain must include an
estimate of the
probabilities of the bounding cases and a weighted assessment
of the
overall
expectation over those and the non-bounding cases, and the
naive
calculation
is simply wrong.
I know that in any particular case a numeric analysis will
reveal that all
this works out that the expected loss on the high valued
boundary cases
exactly balances the expected gain from non-boundary cases. I
imagine that
this could be shown in the general case by a proper depiction
of the
probabilities as the result of sums over the finite
combinations possible,
but my head spins when I attempt it. For me it seems that
pointing out
the error suffices to defuse the paradox.
> Of course if you knew a specific upper bound in advance,
then that would
> resolve the problem. One possible resolution is to say that
since the
> envelope in question has finite dimensions, there must in
fact be such an
> upper bound on the number that you could write on a piece
of paper and
> put in the envelope - some function of the total number of
atoms that
could
> occupy the space of the envelope would probably work.
> I have never found that resolution very satisfactory,
because I would
> instinctively prefer a mathematical rather than a physical
resolution.
> Derek Holt.
I agree... the physical resolution seems to miss the point
that we
dont
have to know a specific upper bound to know that one exists
for any finite
set of numbers. I thought that was one of those things that
mathematicians
frequently call on in the course of other proofs. Im sure 
it
is readily
derivable in any plausible system of arithmetic, why can we
not assume it
here?
>In an infinite series of trials with no finite 
upper bound,
switching
might
>be said to improve the rate of gain, but your expectation in
any case is
>either infinite or undefined depending on which 
aspects of the
picture
you
>want to quibble about, so there is no paradox to be worried
about in the
>first place...
>I mentioned knowing the bound only in context with
explicating the
>conditions under which one could actually gain by a
conditional
switching
>strategy.
>Bill
===
Subject: Re: Total derivative
> Can anybody give me a more intuitive description of the
total derivative?
> Not the definition, please.
Suppose f maps a neighborhood of a in R into R. The graph of
f is then a
curve C in R^2. Forget about the definition of a derivative
for the moment
and think about what it should mean for C to have a tangent
line L at
(a,f(a)). The condition is that as x -> a, the corresponding
points on C
and L should be o(|x-a|) apart from each other. In other
words, the
absolute difference in y-values on C and L is ever smaller in
comparsion
with |x-a| as x -> a.
Same intution if f goes from R^2 to R. The graph of f is then
a surface S
in R^3. S has a tangent plane P at (a,f(a)) iff, as (x,y) ->
(a,b), the
corresponding points on S and P have z-values on that are
o(|(x,y) -
(a,b)|).
Now L and P above are given by affine linear functions, and
once you
rephrase the above in terms of these functions, youll have
exactly the
definition of the total derivative. The extension to higher
dimensions
is
then immediate.
===
Subject: Re: Total derivative
> Hi all,
> Can anybody give me a more intuitive description of the
total derivative?
> Not the definition, please.
> TIA,
> Lurch
Try this site for some help:
http://www.ajnpx.com/html/SD.html
What I say on this page is some what controversial. I have
posted this
stuff here a couple times with mixed reviews, though no
mathematician
has found fault with the basic reformulation. They tend to
think that
the reformulation is inferior to the standard approach,
though,
naturally, I disagree with this view.
The whole point of SD is to clarify what we mean by total and
partial
derivatives as they are used in math and physics.
Mathematicians and
physicists dont always agree on notations between them. I
provided
lots of examples to back up my claims.
Patrick
===
Subject: Re: Total derivative
> Hi all,
> Can anybody give me a more intuitive description of the
total derivative?
> Not the definition, please.
> TIA,
> Lurch
===
Subject: Re: Ullrich says probability 1 does not imply
certainty
>> Suppose that you ßip a coin infinitely many times. What 
is
the
>> probability that you get all heads? Zero. Does that mean
that its
> ^^^^
>> impossible to get all heads? No,...
>
> Yes.
>
>> you get _some_sequence, and any given sequence is just as
likely as
any
>> other.
>
> Although, a priori, any_finite_sequence of a given length is
as likely
as
> any other, I claim: When ßipping a fair coin, an 
infinite
sequence of
> heads is impossible, theoretically.
If an infinite sequnce of all heads is impossible, then so is
any
other predetermined infinite sequence equally impossible for
the
same reason. Then your conclusion must be that an infinite
sequence
of tosses is also impossible, so something like the coin must
wear
out before an infinite number of tosses.
===
Subject: Re: Well-ordering R
>* Stephen J. Herschkorn
> Under AC, of course, does there exist an algortihm which
develops a
> well-ordering of the reals (not necessarily in finite time)
and which
> has the property that, given any two real numbers, it
determines in
> finite time which of the two is no greater than the other
under this
> ordering?
Im not sure whether existence of a well-ordering of R is
known to be
independent of ZF without AC. Existence of well-orderings for
all sets
is equivalent to AC.
One could imagine, though, that you could have an algorithm
that does
the well-ordering, but you need some form of AC to prove that
it does.
> An algorithm is a finitely-specified list of 
steps. Such an
algorithm
> would define a well-ordering S on the reals. The algorithm
would take
> as input two reals x and y (specified as you wish) and, in
finite
> time, return as an output either (x,y) in S or (y,x) in S.
> I see that the specification of x and y may present
problems. For
> example, if we wish to specify a real by a decimal
expansion, we may not
> even be able to specify it in finite time.
Well, we can say theres an oracle that gives the digits as
required.
The first problem is that if x and y happen to be equal you
cant
determine in finite time that they are equal. But 
thats not a
problem
if we require that the inputs to the algorithm are distinct
numbers. If
x and y are different, you can tell that by looking at enough
digits.
However, it still wont work.
Lets suppose that in the case of a number with two 
different
decimal
expansions, the algorithm will work with either of the two
(and of
course give the same result). Then not only cant you get a
well-ordering from an algorithm, you cant get any linear
ordering
except the usual one or its opposite.
Suppose you had such an algorithm F, computing a linear
ordering <<.
Consider any real number y.
Given any x not equal to y, your algorithm will tell in a
finite
number of steps whether x << y or y << x. Lets say F(x,y) =
1 if
x << y and 0 if y << x. In computing this, it will look at
only
finitely many of the digits of x. So there will be an open
interval
V around x such that if w is in V, F(w,y) = F(x,y). But then
x -> F(x,y) is a continuous function from R {y} to {0,1}.
Since
(-infinity, y) and (y,infinity) are connected, the 
conclusion is
that the function is constant on those sets.
There are four cases:
1) x << y iff x < y
2) x << y iff x > y
3) x << y for all x <> y (so y is maximal wrt <<)
4) y << x for all x <> y (so y is minimal wrt <<)
Consider y_1 < y_2. If y_1 << y_2, then y_1 is in case 1) or
4) and
y_2 is in case 1) or 3), while if y_1 >> y_2 then y_1 is in
case 2)
or 3) and y_2 is in case 1) or 4). So except for the minimal
and
maximal elements (if any), all real numbers are in the same
case 1)
or 2). Ill leave it as an easy exercise to show that there
actually
cant be a minimal or maximal element.
If you only require the algorithm to work with one of the two
decimal
expansions in the cases where there are two (lets say only
with the
one that doesnt end in all 9s), then other 
orderings are
possible,
but its still true that the algorithm must give a 
continuous
function
on the Cartesian product of the irrationals with itself. And
then
its easy to see that you cant get a 
well-ordering: there
would have
to be a least irrational number y_1 and a second-least
irrational
number y_2 in the ordering, but some open set V around y_1 in
which
x << y_2, and this must contain other irrationals,
contradicting the
assumption on y_2.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Well-orders, again
>>Forget about algorithms. Here is a simpler question:
>>Does there exist an explicit well-ordering of 2^N (the set
of all
>>binary sequences)?
>No.
>>Again, I suspect not, for
>>a) it probably would be well-known;
>>b) it would yield a well-ordering of R
>Yes.
>>and there wouldnt have been so
>>much controversy about the axiom of choice, and
>>c) it seems to be somehow equivalent to exhibiting an
explicit Hamel
>>basis, which I gather is also impossible.
>Hmm, never noticed that. Its clear that a well-ordering of
R allows
>one to construct a Hamel basis for R over Q; come to think of
>it a Hamel basis also allows one to construct a
well-ordering.
A well-ordered Hamel basis allows this to be done; if a
Hamel basis of the reals over the rationals cannot be
well-ordered, neither can the reals.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Well-orders, again
>>[...]
>>Hmm, never noticed that. Its clear that a well-ordering 
of
R allows
>>one to construct a Hamel basis for R over Q; come to think
of
>>it a Hamel basis also allows one to construct a
well-ordering.
>A well-ordered Hamel basis allows this to be done;
>if a
>Hamel basis of the reals over the rationals cannot be
>well-ordered, neither can the reals.
************************
David C. Ullrich
===
Subject: What is counting?
This question came up in a discussion I had with colleague
about
counting and permutations and combinations.
Its not easy to get any definitions at all 
regarding
counting. One I
found is this: counting is a technique for assessing the
sizes of
sets.
In the pdf paper at
http://www.maths.qmw.ac.uk/~pjc/notes/counting.pdf
we find some interesting characterizations of counting, but no
clear
characterization of exactly what kinds of things can be
counted.
My own definition is this: Counting is any procedure that
correctly
determines the cardinality of a finite set.
Any comments on the topic?
Patrick
===
Subject: Re: What is counting?
>This question came up in a discussion I had with colleague
about
>counting and permutations and combinations.
>Its not easy to get any definitions at all 
regarding
counting. One I
>found is this: counting is a technique for assessing the
sizes of
>sets.
>In the pdf paper at
>http://www.maths.qmw.ac.uk/~pjc/notes/counting.pdf
>we find some interesting characterizations of counting, but
no clear
>characterization of exactly what kinds of things can be
counted.
>My own definition is this: Counting is any procedure that
correctly
>determines the cardinality of a finite set.
>Any comments on the topic?
I definitely disagree. Counting is a procedure which
assigns a distinct ordinal to every member of a finite
set. Placing the elements of a set in a rectangle and
counting the number in each coordinate is not counting
the elements of the set, although it correctly determines
the cardinality.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: What is counting?
>I definitely disagree. Counting is a procedure which
>assigns a distinct ordinal to every member of a finite
>set. Placing the elements of a set in a rectangle and
>counting the number in each coordinate is not counting
>the elements of the set, although it correctly determines
>the cardinality.
I can count money even when it consists of different
denominations.
What am I assigning ordinals to then?
-- Richard
--
Spam filter: to mail me from a .com/.net site, put my surname
in the
headers.
FreeBSD rules!
===
Subject: Re: What is counting?
>>I definitely disagree. Counting is a procedure which
>>assigns a distinct ordinal to every member of a finite
>>set. Placing the elements of a set in a rectangle and
>>counting the number in each coordinate is not counting
>>the elements of the set, although it correctly determines
>>the cardinality.
>I can count money even when it consists of different
denominations.
>What am I assigning ordinals to then?
You are not counting; you are computing a measure.
It would have worked when there were half-cents,
or tenth-cents.
In fact, my characterization is ßawed; the ordinals
assigned must be consecutive, and starting from 1.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: What is counting?
> This question came up in a discussion I had with colleague
about
> counting and permutations and combinations.
> Its not easy to get any definitions at all 
regarding
counting. One I
> found is this: counting is a technique for assessing the
sizes of
> sets.
> In the pdf paper at
> http://www.maths.qmw.ac.uk/~pjc/notes/counting.pdf
> we find some interesting characterizations of counting, but
no clear
> characterization of exactly what kinds of things can be
counted.
> My own definition is this: Counting is any procedure that
correctly
> determines the cardinality of a finite set.
> Any comments on the topic?
> Patrick
Im not sure I agree with your definition, and 
also it does
not seem
to answer your own question of what kind of things can be
counted.
What does determine the cardinality exactly mean? A procedure
to
determine the number of, say, the possible games of chess is
certainly
available, but brute force enumeration is rarely what
counting, in
the combinatorial sense, is about.
In another direction, your definition misses the subtle and
crucial
art of approximations, which is related to another ingredient
youre
ignoring - the fact that, typically, one doesnt determine
the size of
one given set, but rather looks for a formula to capture (or
approximate) the size of sets in some naturally defined 
infinite
family.
All of this is exaplained much more clearly in the first
chapter (or
the introduction, I cant remember) of 
Stanleys superb
Enumerative
Combinatoris, volume 1.
Hope that helps,
- EM
===
Subject: Re: What is counting?
>My own definition is this: Counting is any procedure that
correctly
>determines the cardinality of a finite set.
(That sounds right to me ...)
> I definitely disagree. Counting is a procedure which
> assigns a distinct ordinal to every member of a finite
> set.
I think the procedure you describe would be called numbering,
indexing,
or enumerating. As theres already at least three words that
refer
precisely to this procedure, it would seem abusive to take
the word
counting as another synonym, especially since counting
ordinarily
refers
to merely determining how many of something you have (i.e.,
the size of
a
set), as the original poster suggested.
- Brent
===
Subject: Re: What is counting?
>>I can count money even when it consists of different
denominations.
>>What am I assigning ordinals to then?
>You are not counting; you are computing a measure.
In that case, you are making up your own definition of
counting.
The king was in his counting house,
measuring his money.
-- Richard
--
Spam filter: to mail me from a .com/.net site, put my surname
in the
headers.
FreeBSD rules!
===
Subject: Re: What is counting?
>>I definitely disagree. Counting is a procedure which
>>assigns a distinct ordinal to every member of a finite
>>set. Placing the elements of a set in a rectangle and
>>counting the number in each coordinate is not counting
>>the elements of the set, although it correctly determines
>>the cardinality.
>I can count money even when it consists of different
denominations.
>What am I assigning ordinals to then?
> You are not counting; you are computing a measure.
> It would have worked when there were half-cents,
> or tenth-cents.
> In fact, my characterization is ßawed; the ordinals
> assigned must be consecutive, and starting from 1.
But you only allow finite sets to be counted. What then
becomes of the
definition of countABLE?
Accept the axiom of choice so that every set can be well
ordered, and
then every set can be counted with a long enough sequence of
ordinals.
--
G.C.
===
Subject: Re: What is counting?
> This question came up in a discussion I had with colleague
about
> counting and permutations and combinations.
>
> Its not easy to get any definitions at all 
regarding
counting. One I
> found is this: counting is a technique for assessing the
sizes of
> sets.
>
> In the pdf paper at
>
> http://www.maths.qmw.ac.uk/~pjc/notes/counting.pdf
>
> we find some interesting characterizations of counting, but
no clear
> characterization of exactly what kinds of things can be
counted.
>
> My own definition is this: Counting is any procedure that
correctly
> determines the cardinality of a finite set.
>
> Any comments on the topic?
>
> Patrick
> Im not sure I agree with your definition, and 
also it does
not seem
> to answer your own question of what kind of things can be
counted.
> What does determine the cardinality exactly mean?
I know my definition doesnt seem to present an 
in your face
meaning
of what kinds of things can be counted, but I believe that it
is
accurate anyway: By implication ONLY, all finite sets are
considered
to be countable. How they can be counted is really a whole
other
matter. My definition treats the means of arriving at a 
finite
sets
cardinality as primitive, i.e., as left unspecified. But to be
precise, my definition is only a definition, not 
an existence
axiom.
All it says is that if you find a procedure that is able to
arrive at
the cardinality of some finite set that procedure amounts to
Ôcounting the elements of that set.
The simplest question to ask regarding my definition is: What
things
are countable that it leaves out, and what uncountable things
does it
include? If you can come up with specific examples, then we
have
something definite to go on.
> A procedure to
> determine the number of, say, the possible games of chess
is certainly
> available, but brute force enumeration is rarely what
counting,
in
> the combinatorial sense, is about.
> In another direction, your definition misses the subtle and
crucial
> art of approximations, which is related to another
ingredient youre
> ignoring - the fact that, typically, one doesnt determine
the size of
> one given set, but rather looks for a formula to capture (or
> approximate) the size of sets in some naturally defined
infinite
> family.
True, but I believe that counting is meant to produce an
exact number,
not an approximation, though our common language often allows
for this
abuse of meaning. Anyway, this is what I wanted for my
definition of
counting. For example, the US census never really counts the
number of its citizens, though some may claim that it does.
This is
one of those abuses of terminology I mentioned.
Mathematicians should
do better, though.
> All of this is exaplained much more clearly in the first
chapter (or
> the introduction, I cant remember) of 
Stanleys superb
Enumerative
> Combinatoris, volume 1.
> Hope that helps,
> - EM
OK, then, what did Stanley offer as his definition of 
counting?
Patrick