JSH: To parents
JSH
with curiosity about what your child is reading, or you have been
directed here by that child to read this message, which is somewhat of
a difficult one for me, as I'm here to ask you to do something which
may sound strange.
You need to ask a mathematician to do something which may reveal to
you that that person is knowingly living a fraud, which is important
to you as they may be TEACHING that fraud to your child.
Your child may have directed you here as she or he needs your help
with this extraordinary situation.
The request is, ask a mathematician to divide off the 7:
7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0 in the ring of algebraic integers.
The mathematics is as simple as I can get it. I've worked for years
to boil things down to a level where it is very easy to explain, but
still there are I'm sure elements which will not make sense to you.
That is ok. You are to use your knowledge of human nature here, and
everything in that piece above is needed, including the part about
algebraic integers.
If you're wondering about the principle here it is like how with 7x =
7x, you can get to x = x. Trivial, I know. But over a hundred years
ago a bizarre but somewhat subtle error entered into the mathematical
field, where that trivial thing is violated in the example I wish you
to use in this thing called the ring of algebraic integers.
I discovered this some years ago. I even got a paper on it published,
but some math people got it retracted by spooking the editors, who
yanked it against my wishes: http://www.emis.de/journals/SWJPAM/vol2-03.html
That link is to the archives of the journals maintained by a European
organization called EMIS, which keeps up with electronic mathematical
journals from around the world: http://www.emis.de/journals/index.html
The mathematical journal itself shut down not too long after pulling
my paper. Its hosting university scrubbed all mention of its
existence from its website. EMIS actually saved that journal from
vanishing completely.
If your child has brought you to this post, consider what he or she
faces: a massive error in the field of number theory, for which a
journal may have been destroyed, where people in the field may be
knowingly perpetuating a fraud. For that child to progress in the
mathematical field she or he may have to knowingly learn false
mathematics, just to get by, take tests, do homework, knowing it is
bogus.
Why would mathematics professors do such a thing?
You as an adult may be aware that people can do very wrong things when
faced with extreme loss. The error may invalidate the entire careers
of people who may have spent decades, working hard, thinking they were
brilliant, believing in their work.
With so much of their lives invested, it may seem easier to them to
live a lie, maybe even to deny in their own minds the truth.
But to that they are sacrificing the future of your child in
mathematics.
You do not have to believe all of that to act. If you know a
mathematician simply ask them to divide the 7 off. If they satisfy
you in answering that request, and ask why, you can say some strange
nutty person on Usenet or the Internet was babbling nonsense.
The point here is not your mathematical knowledge. It is your human
knowledge. If they are under the stress of the knowledge of a fraud
as immense as I say, you'll pick that up.
Your situation is shared by parents all over the world.
This problem is an issue for parents in China, as well as the United
States, in Russia, as well as Japan, in Singapore, Australia, New
Zealand, Mozambique, Mexico, Czechoslovakia, Britain, Ireland,
Venezuela, Malta, Vietnam, Iran, South Korea, North Korea and hundreds
more.
If you have read to here, then your judgement is what's necessary for
the next decision. I'm trying to enlist the aid of parents around the
world to stop this nonsense of fraudulent information being taught to
their children as all else has failed.
It is up to you now. And all you have to do is ask some
mathematician, to divide off a 7.
James Harris
Rotwang
> I you are reading this message maybe you're looking at the newsgroup
> with curiosity about what your child is reading, or you have been
> directed here by that child to read this message, which is somewhat of
> a difficult one for me, as I'm here to ask you to do something which
> may sound strange.
>
> You need to ask a mathematician to do something which may reveal to
> you that that person is knowingly living a fraud, which is important
> to you as they may be TEACHING that fraud to your child.
>
> Your child may have directed you here as she or he needs your help
> with this extraordinary situation.
>
> The request is, ask a mathematician to divide off the 7:
>
> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
> where the a's are roots of
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0 in the ring of algebraic integers.
First, however, you should ask James Harris to divide off the 7:
7*(25x^2 + 5x + 2) = (5a_1(x) + 7)(5a_2(x) + 7)
where the a's are the roots of
a^2 - (x - 1)a + 7x^2 = 0 in the object ring.
Why do you keep ignoring this challenge, James?
Ki Song
> On 4 Dec, 01:45, JSH <jst...@gmail.com> wrote:
>
>
>
> > I you are reading this message maybe you're looking at the newsgroup
> > with curiosity about what your child is reading, or you have been
> > directed here by that child to read this message, which is somewhat of
> > a difficult one for me, as I'm here to ask you to do something which
> > may sound strange.
>
> > You need to ask a mathematician to do something which may reveal to
> > you that that person is knowingly living a fraud, which is important
> > to you as they may be TEACHING that fraud to your child.
>
> > Your child may have directed you here as she or he needs your help
> > with this extraordinary situation.
>
> > The request is, ask a mathematician to divide off the 7:
>
> > 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
> > where the a's are roots of
>
> > a^2 - (7x-1)a + (49x^2 - 14x) = 0 in the ring of algebraic integers.
>
> First, however, you should ask James Harris to divide off the 7:
>
> 7*(25x^2 + 5x + 2) = (5a_1(x) + 7)(5a_2(x) + 7)
>
> where the a's are the roots of
>
> a^2 - (x - 1)a + 7x^2 = 0 in the object ring.
>
> Why do you keep ignoring this challenge, James?
>
What's the object ring?
Rotwang
> On Dec 3, 10:42 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
> > First, however, you should ask James Harris to divide off the 7:
>
> > 7*(25x^2 + 5x + 2) = (5a_1(x) + 7)(5a_2(x) + 7)
>
> > where the a's are the roots of
>
> > a^2 - (x - 1)a + 7x^2 = 0 in the object ring.
>
> > Why do you keep ignoring this challenge, James?
>
> What's the object ring?
The object ring is some crap that James has made up. See
http://mymath.blogspot.com/2005/03/object-ring.html
for his (nonsensical) definition. If you wish to know the relevance of
my question (and also the reason why James is pretending he can't hear
me whenever I ask it), see my reply to MichaelW from Wednesday:
http://groups.google.co.uk/group/alt.math.undergrad/msg/9590b3a7962440b2
BTW, regarding your other post in this thread: James is familiar with
the term UFD, and aware that there are rings which aren't UFDs, though
I don't think he knows the definition and he certainly doesn't
understand the implications. But I don't believe the fact that the AIs
fail to be a UFD is actually relevant here. (Warning: I may be about
to tell you a load of stuff you already know, in which case I
apologise) That's because any two AIs have a greatest common divisor,
and if gcd(a,b) = f then there exist AIs x and y such that
ax + by = f.
Applying this to an identity of the form
7P = ab (1)
and letting
f = gcd(a,7)
g = 7/f
h = a/f
ax + 7y = f (2)
we see that (1) is equivalent to (note that f is never 0)
gP = hb
while (2) becomes
hx + gy = 1
and therefore
b = (hx + gy)b = gyb + xgP
so that g divides b. This means that there exist AIs f and g such that
fg = 7, f divides a and g divides b. So for any given value of x in
James's original factorisation we can divide the 7 off from the RHS by
splitting it into two. The "flaw" is not that one can't divide the 7
from the individual factors (5a_1(x) + 7) and (5a_2(x) + 7), it's that
one can't do it in a constant way (i.e. by dividing each factor by a
constant factor of 7). James thinks this represents a problem with the
algebraic integers, rather than simply a fact which runs counter to
his intuition. My example shows that, if the object ring exists, it
must have the exact same problem.
Ki Song
> On 4 Dec, 07:50, Ki Song <kiwisqu...@gmail.com> wrote:
>
> > On Dec 3, 10:42 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > [...]
>
> > > First, however, you should ask James Harris to divide off the 7:
>
> > > 7*(25x^2 + 5x + 2) = (5a_1(x) + 7)(5a_2(x) + 7)
>
> > > where the a's are the roots of
>
> > > a^2 - (x - 1)a + 7x^2 = 0 in the object ring.
>
> > > Why do you keep ignoring this challenge, James?
>
> > What's the object ring?
>
> The object ring is some crap that James has made up. See
>
> http://mymath.blogspot.com/2005/03/object-ring.html
Oh.
>
> for his (nonsensical) definition. If you wish to know the relevance of
> my question (and also the reason why James is pretending he can't hear
> me whenever I ask it), see my reply to MichaelW from Wednesday:
>
> http://groups.google.co.uk/group/alt.math.undergrad/msg/9590b3a7962440b2
>
> BTW, regarding your other post in this thread: James is familiar with
> the term UFD, and aware that there are rings which aren't UFDs, though
> I don't think he knows the definition and he certainly doesn't
> understand the implications. But I don't believe the fact that the AIs
> fail to be a UFD is actually relevant here.
Perhaps not. I think I'm more concerned about the logic OP is using,
which is:
"You can't divide things off in a ring" ----> "The ring is broken" ----
> "Modern Mathematics is broken."
I'm just trying to provide him with examples where just because an
element in a ring admits two factorizations, it doesn't necessarily
mean that "dividing off" one of the factors gives a new factorization
of the other side.
It was my impression from all of his previous claims about how "you
should be able to divide off what you multiply" that OP believes all
rings "should" be a UFD.
> (Warning: I may be about
> to tell you a load of stuff you already know, in which case I
> apologise) That's because any two AIs have a greatest common divisor,
Hmm... is that true? I didn't know. Wait, so what would be the gcd
(2,3) in the algebraic integers?
> and if gcd(a,b) = f then there exist AIs x and y such that
>
> ax + by = f.
>
> Applying this to an identity of the form
>
> 7P = ab (1)
>
> and letting
>
> f = gcd(a,7)
Why is gcd (a,7) = gcd (a,b)?
Okay, so I'll accept this (though I don't really get it) and continue.
> g = 7/f
> h = a/f
> ax + 7y = f (2)
Are these the same x and y for a and b? I guess that doesn't matter.
Rotwang
> On Dec 4, 10:20 am, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
> > BTW, regarding your other post in this thread: James is familiar with
> > the term UFD, and aware that there are rings which aren't UFDs, though
> > I don't think he knows the definition and he certainly doesn't
> > understand the implications. But I don't believe the fact that the AIs
> > fail to be a UFD is actually relevant here.
>
> Perhaps not. I think I'm more concerned about the logic OP is using,
> which is:
>
> "You can't divide things off in a ring" ----> "The ring is broken" ----
>
> > "Modern Mathematics is broken."
>
> I'm just trying to provide him with examples where just because an
> element in a ring admits two factorizations, it doesn't necessarily
> mean that "dividing off" one of the factors gives a new factorization
> of the other side.
Sure, that makes sense.
> It was my impression from all of his previous claims about how "you
> should be able to divide off what you multiply" that OP believes all
> rings "should" be a UFD.
Quite possibly - he has strenuously denied believing that all rings
are UFDs, but since he doesn't understand what it means that doesn't
count for much. Really, he thinks that all rings should behave exactly
like the integers in every respect he wants to make use of in his
"proof" of FLT, and that rings which fail to do so aren't really rings
(even though they satisfy the definition - that just means the
definition is wrong). He has explicitly stated in the past that there
is only one ring.
> > (Warning: I may be about
> > to tell you a load of stuff you already know, in which case I
> > apologise) That's because any two AIs have a greatest common divisor,
>
> Hmm... is that true?
Apparently. I haven't seen a complete proof myself, but I have seen a
proof that relies on the finiteness of the class number (a result of
Dedekind's, I think). The latter may be seen here, courtesy of Bill
Hale; my comments in reply to his post fill in some of the gaps:
http://groups.google.co.uk/group/alt.math.undergrad/msg/f56abf2d2e44ee44
> I didn't know. Wait, so what would be the gcd
> (2,3) in the algebraic integers?
1. Note that if there exist x and y s.t. ax + by = f then any number
that divides both a and b also divides f, so the gcd of a and b
divides f. In the case at hand, note that -1*2 + 1*3 = 1, so the gcd
(2,3) divides 1, and since 1 also divides gcd(2,3) we may take gcd
(2,3) = 1 (note that gcds are only unique up to multiplication by
units, that is, if u is a unit and f is a gcd for a and b then uf is
also a gcd for a and b).
> > and if gcd(a,b) = f then there exist AIs x and y such that
>
> > ax + by = f.
>
> > Applying this to an identity of the form
>
> > 7P = ab (1)
>
> > and letting
>
> > f = gcd(a,7)
>
> Why is gcd (a,7) = gcd (a,b)?
It isn't. Sorry, bad choice of notation on my part - when I wrote gcd
(a,b) = f before I was just using the letters a and b to denote
arbitrary AIs. Ignore every equation before (1) in what follows, they
referred to a different a, b and f.
> Okay, so I'll accept this (though I don't really get it) and continue.
>
> > g = 7/f
> > h = a/f
> > ax + 7y = f (2)
>
> Are these the same x and y for a and b? I guess that doesn't matter.
As above, they are a new x and y.
Arturo Magidin
> On 4 Dec, 07:50, Ki Song <kiwisqu...@gmail.com> wrote:
>
> > On Dec 3, 10:42 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > [...]
>
> > > First, however, you should ask James Harris to divide off the 7:
>
> > > 7*(25x^2 + 5x + 2) = (5a_1(x) + 7)(5a_2(x) + 7)
>
> > > where the a's are the roots of
>
> > > a^2 - (x - 1)a + 7x^2 = 0 in the object ring.
>
> fail to be a UFD is actually relevant here. (Warning: I may be about
> to tell you a load of stuff you already know, in which case I
> apologise) That's because any two AIs have a greatest common divisor,
> and if gcd(a,b) = f then there exist AIs x and y such that
>
> ax + by = f.
Note that gcds in domains are generally only determined up to units;
in the integers, there is a clear choice between the (up to) two gcds,
so we take the nonnegative one and call it "the" gcd. In the algebraic
integers, there is no such clear choice; of course, everything you do
below has to do mainly with divisibility, so this non-uniqueness is
immaterial. Just thought I would mention it.
--
Arturo Magidin
Ki Song
> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
....
> If you're wondering about the principle here it is like how with 7x =
> 7x, you can get to x = x. Trivial, I know.
If we were to follow this analogy, it would be
7(175x^2 - 15x + 2) = 7(175x^2 - 15x + 2)
(175x^2 - 15x + 2) = (175x^2 - 15x + 2)
> But over a hundred years
> ago a bizarre but somewhat subtle error entered into the mathematical
> field, where that trivial thing is violated in the example I wish you
> to use in this thing called the ring of algebraic integers.
Do you understand what a ring is?
Do you understand why the sum of two algebraic integers are algebraic
integers?
Do you understand why the product of two algebraic integers are
algebraic integers?
Do you understand what a unique factorization domain is?
Do you understand that there are rings that are not unique
factorization domains?
I've shown you examples of rings where dividing off doesn't make
sense, but it seems like the point never got across.
Take the following example.
Let Z[\sqrt[5]] be the ring of all complex numbers of the following
form: a + i b \sqrt[5].
In this ring, 6 can be factored in two different ways:
6 = 2 * 3 = (1 + i \sqrt[5] ) ( 1 - i \sqrt[5])
6 has two different factorizations in the ring Z[\sqrt[5]]. And no,
you can't "divide off the 2" from the right hand side.
Does that mean it's also broken?
Note: I edited this post, because I said the "sum of algebraic
integers is an integer" rather than "sum of algebraic integers is an
algebraic integer."
Rotwang
>
> Your situation is shared by parents all over the world.
>
> This problem is an issue for parents in China, as well as the United
> States, in Russia, as well as Japan, in Singapore, Australia, New
> Zealand, Mozambique, Mexico, Czechoslovakia,
Yes, that's right, Czechoslovakia. Right now parents in the nation of
Czechoslovakia, which definitely exists, are having to deal with
undergraduates (who, being typically between the ages of 18 and 22,
just /love/ being referred to as children) tearfully explaining how
their mean Czechoslovakian maths professors are polluting their minds
with lies about the algebraic integers.
Truly, James, your knowledge of world affairs is second to none.
fishfry
<9cf84f79-8815-434f...@f20g2000prn.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> The request is, ask a mathematician to divide off the 7:
>
> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
JSH, you can't divide through by 7 in general, because 7 may well not
divide the expression for all values of x. It may divide the evaluation
of the expression for SOME values of x, but that does not justify
dividing the expression by 7.
This has been pointed out to you hundreds of times over the past couple
of years.
Rotwang
> In article
> <9cf84f79-8815-434f-adb8-0b5bce767...@f20g2000prn.googlegroups.com>,
Huh? If by "the expression" you mean (5a_1(x) + 7)(5a_2(x)+ 7) then 7
*does* divide the expression for all algebraic integer values of x, as
is clear from the fact that the LHS is equal to the RHS.
Tim Peters
>> The request is, ask a mathematician to divide off the 7:
>>
>> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
[fishfry]
> JSH, you can't divide through by 7 in general, because 7 may well not
> divide the expression for all values of x.
Sure it does. Whenever all of
7(175x^2 - 15x + 2)
5a_1(x) + 7
and
5a_2(x)+ 7
are algebraic integers, then the LHS is quite obviously 7 times an
algebraic integer, so it must be possible to split 7 into a product of
two algebaic integers that divide the RHS factors. Indeed, Rotwang
routinely ;-) shows JSH exactly how to do that.
The minor subtelty James still can't wrap his mind around is that the
precise way 7 so splits depends on the value of x.
> It may divide the evaluation of the expression for SOME values of x,
> but that does not justify dividing the expression by 7.
James's idiocy here is different: he picks a value of x such that /one/
of the two RHS factors is divisible by 7, and then insists (without
benefit of logic -- it's pure wishful thinking "supported" by nothing
but endlessly repetitive temper tantrums and crazy ranting) that the
same factor must be divisible by 7 for /all/ values of x.
This is exactly as baseless as noting that, in the ring of integers,
x*(x+1)
is always divisible by 2, then noting that at x=1 it's the "x+1" part
that's divisible by 2, and then insisting that the "x+1" part must be
divisible by 2 regardless of the value of x.
> This has been pointed out to you hundreds of times over the past
> couple of years.
Well, the x*(x+1) example has (in various forms) been pointed out to him
at least dozens of times.
Alas, his FLT "proof" depends on him continuing to misunderstand this
stuff -- and giving up that "proof" would be a huge blow to his ego.
It's not like James has a productive working relationship with reality
in any visible area ;-)
Mark Murray
> The request is, ask a mathematician to divide off the 7:
>
> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
> where the a's are roots of
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0 in the ring of algebraic integers.
How's this: http://people.freebsd.org/~markm/Algebraics.pdf ?
(I did it with Maple to [a] make DAMN sure I had no self-induced
algebra errors and [b] when I type this level of stuff out it looks
lousy in ascii)
M
Tim Peters
>> The request is, ask a mathematician to divide off the 7:
>>
>> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>>
>> where the a's are roots of
>>
>> a^2 - (7x-1)a + (49x^2 - 14x) = 0 in the ring of algebraic integers.
[Mark Murray]
> How's this: http://people.freebsd.org/~markm/Algebraics.pdf ?
>
> (I did it with Maple to [a] make DAMN sure I had no self-induced
> algebra errors and [b] when I type this level of stuff out it looks
> lousy in ascii)
If it were that easy, it's possible even James would understand it after
all these years ;-)
The fundamental problem is at the end:
> Now if each of fac_1(x) and fac_2(x) are algebraics, then my thinking
> is that they still are when divided by 7.
7 is an algebraic integer (proof: 7 is a root of the monic polynomial x-7),
but the AIs are not closed under division. For example, it turns out the
only rational numbers that are algebraic integers are exact integers: 1 is
an AI, and 7 is an AI, but 1/7 is not (1/7 is not the root of any monic
polynomial with integer coefficients). IOW, given an AI y, there's no
reason at all to suppose that y/7 is also an AI.
James's "challenge" has been met several times already, at least by Rotwang
and Marcus Bruckner. A correct answer is constructed via computing the
gcds of various AIs, but that's a computationally difficult procedure --
and it's highly non-trivial just to prove that a sensible notion of
"greatest common divisor" /exists/ for AIs. There's no chance James will
ever understand it -- and in his world, when something is beyond his
understanding, it's "a lie".
Mark Murray
> [Mark Murray]
>> How's this: http://people.freebsd.org/~markm/Algebraics.pdf ?
>>
>> (I did it with Maple to [a] make DAMN sure I had no self-induced
>> algebra errors and [b] when I type this level of stuff out it looks
>> lousy in ascii)
>
> If it were that easy, it's possible even James would understand it after
> all these years ;-)
:-)
> The fundamental problem is at the end:
>
>> Now if each of fac_1(x) and fac_2(x) are algebraics, then my thinking
>> is that they still are when divided by 7.
>
> 7 is an algebraic integer (proof: 7 is a root of the monic polynomial x-7),
> but the AIs are not closed under division. For example, it turns out the
> only rational numbers that are algebraic integers are exact integers: 1 is
> an AI, and 7 is an AI, but 1/7 is not (1/7 is not the root of any monic
> polynomial with integer coefficients). IOW, given an AI y, there's no
> reason at all to suppose that y/7 is also an AI.
Aaaah, OK.
If y is /irrational/, is y/n an algebraic (for arbitrary integer values
of n <> 0)?
> James's "challenge" has been met several times already, at least by Rotwang
> and Marcus Bruckner. A correct answer is constructed via computing the
> gcds of various AIs, but that's a computationally difficult procedure --
> and it's highly non-trivial just to prove that a sensible notion of
> "greatest common divisor" /exists/ for AIs. There's no chance James will
> ever understand it -- and in his world, when something is beyond his
> understanding, it's "a lie".
Thanks. More stuff to look up. :-) :-)
I've seen JSH's notion of a lie. I didn't quote the page number of a
reference, thus making my book references lies.
Arturo Magidin
> Tim Peters wrote:
> > 7 is an algebraic integer (proof: 7 is a root of the monic polynomial x-7),
> > but the AIs are not closed under division. For example, it turns out the
> > only rational numbers that are algebraic integers are exact integers: 1 is
> > an AI, and 7 is an AI, but 1/7 is not (1/7 is not the root of any monic
> > polynomial with integer coefficients). IOW, given an AI y, there's no
> > reason at all to suppose that y/7 is also an AI.
>
> Aaaah, OK.
>
> If y is /irrational/, is y/n an algebraic (for arbitrary integer values
> of n <> 0)?
Do you mean an algebraic integer? In general, no; it will be for only
finitely many integers (the number could be just two, to wit 1 and
-1). This is true for *any* algebraic integer (whether it is a
rational integer or an irrational algebraic integer).
--
Arturo Magidin
MichaelW
>
> The fundamental problem is at the end:
>
> > Now if each of fac_1(x) and fac_2(x) are algebraics, then my thinking
> > is that they still are when divided by 7.
>
> 7 is an algebraic integer (proof: 7 is a root of the monic polynomial x-7),
> but the AIs are not closed under division. For example, it turns out the
> only rational numbers that are algebraic integers are exact integers: 1 is
> an AI, and 7 is an AI, but 1/7 is not (1/7 is not the root of any monic
> polynomial with integer coefficients). IOW, given an AI y, there's no
> reason at all to suppose that y/7 is also an AI.
>
> James's "challenge" has been met several times already, at least by Rotwang
> and Marcus Bruckner. A correct answer is constructed via computing the
> gcds of various AIs, but that's a computationally difficult procedure --
> and it's highly non-trivial just to prove that a sensible notion of
> "greatest common divisor" /exists/ for AIs. There's no chance James will
> ever understand it -- and in his world, when something is beyond his
> understanding, it's "a lie".
I am hoping to generalize the approach. Here's my shot at using the
same logic in the ring of integers:
[begin]
Consider the equation in the ring of integers where x is an integer.
a^2 - a*(2-2x) + (x^2-2x) = 0
The roots of this equation are designated a_1(x) and a_2(x) where a_1
(x) is less than a_2(x) and both are in the ring of integers. In this
case we have
6(x^2 + 3x + 2) = (2*a_1(x) + 6)(3*a_2(x) + 6)
This can be easily verified. We arbitrarily set a_1(x) = 3*b_1(x).
Substituting we have
6(x^2 + 3x + 2) = 6(b_1(x) + 1)(3*a_2(x) + 6)
Dividing through by 6 we get
x^2 + 3x + 2 = (b_1(x) + 1)(3*a_2(x) + 6)
But note what happens when x=0. By substitution
2 = (b_1(0)+1)*3*(a_2(0)+2)
This says that in the ring of integers the number 2 has as a factor
the number 3! This is crazy! Obviously the ring of integers is broken.
[end]
Basically as I understand it you take a ring and slip in a stealth
illegal division (a_1(x)=3*b_1(x) in this case). From there it is a
simple matter to "break" the ring. Have I got this right?
Regards, Michael W.
Rotwang
> On Dec 5, 1:21 pm, Mark Murray <w.h.o...@example.com> wrote:
>> Tim Peters wrote:
>
>>> 7 is an algebraic integer (proof: 7 is a root of the monic polynomial x-7),
>>> but the AIs are not closed under division. For example, it turns out the
>>> only rational numbers that are algebraic integers are exact integers: 1 is
>>> an AI, and 7 is an AI, but 1/7 is not (1/7 is not the root of any monic
>>> polynomial with integer coefficients). IOW, given an AI y, there's no
>>> reason at all to suppose that y/7 is also an AI.
>> Aaaah, OK.
>>
>> If y is /irrational/, is y/n an algebraic (for arbitrary integer values
>> of n <> 0)?
Point of pedantry (sorry)...
> Do you mean an algebraic integer? In general, no; it will be for only
> finitely many integers (the number could be just two, to wit 1 and
> -1). This is true for *any* algebraic integer (whether it is a
^ non-zero
> rational integer or an irrational algebraic integer).
@Mark: to prove this, suppose that a_1 is a non-zero algebraic integer,
and let P be its minimal polynomial, which is a monic polynomial with
integer coefficients; if its other roots are a_2, ... a_m, then
P(x) = (x - a_1)(x - a_2)...(x - a_m)
so that the constant term (let's call it a) of P is a_1*a_2*...a_m. If
an integer n divides a_1 in the algebraic integers, then n therefore
divides a, so a/n is an algebraic integer; but the only rational numbers
which are algebraic integers are integers, so that n must be a factor
(in the integers) of the constant term of P. Clearly there are only a
finite number of such n. No idea if there's an easier way to prove it.
Mark Murray
> @Mark: to prove this, suppose that a_1 is a non-zero algebraic integer,
> and let P be its minimal polynomial, which is a monic polynomial with
> integer coefficients; if its other roots are a_2, ... a_m, then
>
> P(x) = (x - a_1)(x - a_2)...(x - a_m)
>
> so that the constant term (let's call it a) of P is a_1*a_2*...a_m. If
> an integer n divides a_1 in the algebraic integers, then n therefore
> divides a, so a/n is an algebraic integer; but the only rational numbers
> which are algebraic integers are integers, so that n must be a factor
> (in the integers) of the constant term of P. Clearly there are only a
> finite number of such n. No idea if there's an easier way to prove it.
Nope, that's clear enough. Thanks!
M
Tim Peters
>> ...
>> IOW, given an AI [algebraic integer] y, there's no reason
>> at all to suppose that y/7 is also an AI.
[Mark Murray]
> Aaaah, OK.
>
> If y is /irrational/, is y/n an algebraic (for arbitrary integer
> values of n <> 0)?
No -- and I see Rotwang already gave you a clear proof.
> ...
> Thanks. More stuff to look up. :-) :-)
First time I saw JSH beating this dead horse (it was stillborn, of
course), I vaguely recalled the algebraic integers from an abstract
algebra course I took over 30 years ago. They didn't come up all that
often in my subsequent career ;-)
To have real fun in a JSH thread (beyond just poking at him), you have
to do what he won't: learn something about the topic. The theory of
AIs is elegant, and while there are subtleties, they're not all /that/
mind-paralyzing.
I think the best intro to AI theory is still the one Dedekind wrote at
the start, over 120 years ago. There's a terrific English translation
due to John Stillwell, which you can find easily by searching for
"Dedekind algebraic integers". I bought a copy from amazon.com.
"Modern" treatments are more succinct, due to greater abstraction and
generalization, but that also makes them more "mysterious" at first.
Dedekind had specific questions in mind, and that keeps his presentation
grounded -- easier to follow. I'm not really qualified to judge this,
but I think his writings on the topic are works of true genius. Which
doesn't mean they're incomprehensible -- to the contrary, the genius I
see was in developing non-obvious abstractions that made the answers to
seemingly intractable questions suddenly /clear/ -- the opposite of what
you get from JSH's flavor of self-proclaimed "genius", where the more
effort you put into trying to make sense of what he says, the less sense
you find ;-)
Mark Murray
> First time I saw JSH beating this dead horse (it was stillborn, of
> course), I vaguely recalled the algebraic integers from an abstract
> algebra course I took over 30 years ago. They didn't come up all that
> often in my subsequent career ;-)
Mine neither. I'm a Physics/Applied maths sort of person (That's in my
spare time; for bread I do software engineering).
> To have real fun in a JSH thread (beyond just poking at him), you have
> to do what he won't: learn something about the topic. The theory of
> AIs is elegant, and while there are subtleties, they're not all /that/
> mind-paralyzing.
AI's are an interesting side-track to what I'm really trying to pick up;
enough about groups/rings/etc to make better sense of Hestenes' "New
Foundations of Classical Mechanics".
> I think the best intro to AI theory is still the one Dedekind wrote at
> the start, over 120 years ago. There's a terrific English translation
> due to John Stillwell, which you can find easily by searching for
> "Dedekind algebraic integers". I bought a copy from amazon.com.
On my Christmas list. Thanks!
> "Modern" treatments are more succinct, due to greater abstraction and
> generalization, but that also makes them more "mysterious" at first.
> Dedekind had specific questions in mind, and that keeps his presentation
> grounded -- easier to follow. I'm not really qualified to judge this,
> but I think his writings on the topic are works of true genius. Which
> doesn't mean they're incomprehensible -- to the contrary, the genius I
> see was in developing non-obvious abstractions that made the answers to
> seemingly intractable questions suddenly /clear/ -- the opposite of what
> you get from JSH's flavor of self-proclaimed "genius", where the more
> effort you put into trying to make sense of what he says, the less sense
> you find ;-)
Hehehehe! That's partly why I don't throw too much effort into JSH's
maths. It's sometimes fun to find his errors with a bit of idle algebra
software doodling.
Thanks for the recommendation.
M
Arturo Magidin
> I think the best intro to AI theory is still the one Dedekind wrote at
> the start, over 120 years ago. There's a terrific English translation
> due to John Stillwell, which you can find easily by searching for
> "Dedekind algebraic integers". I bought a copy from amazon.com.
A couple of notes: that was actually his third (or fourth) exposition
of his theory of ideals. (In his first exposition, for example, he
defined what it meant for an ideal to divide another ideal before
defining multiplication of ideals, and the "unique factorization
result" was defined in terms of intersections of certain kinds of
ideals). So it wasn't quite "at the start." Also, the description
assumes that the reader will be familiar with a lot of analysis (as
people were then) which is not necessarily the case here; though in
that respect Stillwell's introduction helps to overcome those gaps.
--
Arturo Magidin
Tim Peters
> I am hoping to generalize the approach. Here's my shot at using the
> same logic in the ring of integers:
>
> [begin]
>
> Consider the equation in the ring of integers where x is an integer.
>
> a^2 - a*(2-2x) + (x^2-2x) = 0
A fancy way of writing
(x+a-2)*(x+a) = 0
That's very JSH-like: take something simple and complicate it in
unmotivated ways until you can't understand it anymore :-( Then project
hopes of greatness into your resulting ignorance ;-)
> The roots of this equation are designated a_1(x) and a_2(x) where a_1
> (x) is less than a_2(x) and both are in the ring of integers. In this
> case we have
>
> 6(x^2 + 3x + 2) = (2*a_1(x) + 6)(3*a_2(x) + 6)
>
> This can be easily verified.
If you say so. That's very JSH-like too (i.e., claim something is
easily verified when it's actually wrong). That is, since the roots of
the original equation are a_1(x) = -x and a_2(x) = 2-x (trivial to see
from the de-obfuscated spelling), that new claim reduces to
6*(x+1)*(x+2) = (2*-x + 6) * (3*(2-x) + 6) =
2*(3-x) * 3*(4-x) =
6*(3-x)*(4-x)
At least that's true at x=1. Sometimes James's easily verified truths
are always false.
> We arbitrarily set a_1(x) = 3*b_1(x).
/That's/ the spirit! :-)
> Substituting we have
>
> 6(x^2 + 3x + 2) = 6(b_1(x) + 1)(3*a_2(x) + 6)
>
> Dividing through by 6 we get
>
> x^2 + 3x + 2 = (b_1(x) + 1)(3*a_2(x) + 6)
>
> But note what happens when x=0. By substitution
>
> 2 = (b_1(0)+1)*3*(a_2(0)+2)
>
> This says that in the ring of integers the number 2 has as a factor
> the number 3! This is crazy! Obviously the ring of integers is broken.
>
> [end]
>
> Basically as I understand it you take a ring and slip in a stealth
> illegal division (a_1(x)=3*b_1(x) in this case). From there it is a
> simple matter to "break" the ring. Have I got this right?
James is the only judge of that. And he's extremely literal-minded when
it suits his purposes. People have given him dozens of analogous
arguments over the years, but if there's /any/ way in which they don't
literally match his spelling of his argument, he claims the poster is
lying.
For example, when he's responded to the dead simple demonstration that,
in the ring of integers, the even factor in x*(x+1) depends on the value
of x, he's outraged because there isn't a left-hand side to that with a
literal "2*" portion. So spell it
2*f(x) = x*(x+1)
and then he's outraged because "2" isn't his "7". Or something like
that -- at heart, at some point you just have to accept that the lad
doesn't make any sense ;-)
Tim Peters
>> ...
>> I think the best intro to AI [albegraic integer] theory is still
>> the one Dedekind wrote at the start, over 120 years ago. �There's
>> a terrific English translation due to John Stillwell, which you
>> can find easily by searching for "Dedekind algebraic integers".
>>�I bought a copy from amazon.com.
[Arturo Magidin]
> A couple of notes: that was actually his third (or fourth) exposition
> of his theory of ideals. (In his first exposition, for example, he
> defined what it meant for an ideal to divide another ideal before
> defining multiplication of ideals, and the "unique factorization
> result" was defined in terms of intersections of certain kinds of
> ideals). So it wasn't quite "at the start."
After 120+ years, a few years plus or minus is in the noise to me -- I'm an
engineering type ;-)
> Also, the description assumes that the reader will be familiar
> with a lot of analysis (as people were then) which is not
> necessarily the case here; though in that respect Stillwell's
> introduction helps to overcome those gaps.
Yes, Stillwell's introduction is extremely valuable on several counts,
filling in gaps and providing historical perspective. It makes the book a
nearly self-contained intellectual treat for non-professionals with some
non-trivial math background. I enjoyed both the subject matter and the
history lessons, but - of course - they may not be everyone's cup of tea.
For example, I wouldn't recommend the book to James ;-)
MichaelW
> [MichaelW]
>
> > I am hoping to generalize the approach. Here's my shot at using the
> > same logic in the ring of integers:
>
> > [begin]
>
> > Consider the equation in the ring of integers where x is an integer.
>
> > a^2 - a*(2-2x) + (x^2-2x) = 0
>
> A fancy way of writing
>
> (x+a-2)*(x+a) = 0
>
> That's very JSH-like: take something simple and complicate it in
> unmotivated ways until you can't understand it anymore :-( Then project
> hopes of greatness into your resulting ignorance ;-)
Thanks! The style is remarkably difficult to emulate. One's natural
inclination for clarity and simplification are hard to suppress. For
example in the JSH version I see no reason whatever for the
multiplication of a_1 and a_2 by 5; the whole thing works just as well
using (a_1(x)+7) and (a_2(x)+7). If it were me I would automatically
drop the multiplication.
>
> > The roots of this equation are designated a_1(x) and a_2(x) where a_1
> > (x) is less than a_2(x) and both are in the ring of integers. In this
> > case we have
>
> > 6(x^2 + 3x + 2) = (2*a_1(x) + 6)(3*a_2(x) + 6)
>
> > This can be easily verified.
>
> If you say so. That's very JSH-like too (i.e., claim something is
> easily verified when it's actually wrong). That is, since the roots of
> the original equation are a_1(x) = -x and a_2(x) = 2-x (trivial to see
> from the de-obfuscated spelling), that new claim reduces to
>
> 6*(x+1)*(x+2) = (2*-x + 6) * (3*(2-x) + 6) =
> 2*(3-x) * 3*(4-x) =
> 6*(3-x)*(4-x)
>
> At least that's true at x=1. Sometimes James's easily verified truths
> are always false.
>
> > We arbitrarily set a_1(x) = 3*b_1(x).
>
> /That's/ the spirit! :-)
I got the phrase from one of his posts. Of course this is in fact the
trick used to break the ring since there is no b_1(x) = a_1(x)/3 for
all x in the integers. The analogy is those algebra tricks to prove
0=1 by sneaking in division by zero.
>
> > Substituting we have
>
> > 6(x^2 + 3x + 2) = 6(b_1(x) + 1)(3*a_2(x) + 6)
>
> > Dividing through by 6 we get
>
> > x^2 + 3x + 2 = (b_1(x) + 1)(3*a_2(x) + 6)
>
> > But note what happens when x=0. By substitution
>
> > 2 = (b_1(0)+1)*3*(a_2(0)+2)
>
> > This says that in the ring of integers the number 2 has as a factor
> > the number 3! This is crazy! Obviously the ring of integers is broken.
>
> > [end]
>
> > Basically as I understand it you take a ring and slip in a stealth
> > illegal division (a_1(x)=3*b_1(x) in this case). From there it is a
> > simple matter to "break" the ring. Have I got this right?
>
> James is the only judge of that. And he's extremely literal-minded when
> it suits his purposes. People have given him dozens of analogous
> arguments over the years, but if there's /any/ way in which they don't
> literally match his spelling of his argument, he claims the poster is
> lying.
>
> For example, when he's responded to the dead simple demonstration that,
> in the ring of integers, the even factor in x*(x+1) depends on the value
> of x, he's outraged because there isn't a left-hand side to that with a
> literal "2*" portion. So spell it
>
> 2*f(x) = x*(x+1)
>
> and then he's outraged because "2" isn't his "7". Or something like
> that -- at heart, at some point you just have to accept that the lad
> doesn't make any sense ;-)- Hide quoted text -
>
> - Show quoted text -
Well, I don't hold out much hope of convincing JSH; I am simply
treating the exercise as one in recreational maths and a chance to
learn more about rings. If I get time I might try and find the
smallest possible ring that can be "broken"; Z-9 looks doable.
Regards, Michael W.
amzoti
Is this the book you are referring to?
In other words - google books.
amzoti
> If you have read to here, then your judgement is what's necessary for
> the next decision. I'm trying to enlist the aid of parents around the
> world to stop this nonsense of fraudulent information being taught to
> their children as all else has failed.
>
> It is up to you now. And all you have to do is ask some
> mathematician, to divide off a 7.
>
> James Harris
James,
what an amazing gift to parents!
You are protecting the world's children from the filthy mathematicians
and their many lies, deceit and continued false funding.
Those lying and cheating bastards - down with them all!
Merry XMAS James - you are not a scrooge after all!
Arturo Magidin
> On Dec 6, 8:56 am, Tim Peters <tim....@comcast.net> wrote:
> > I think the best intro to AI theory is still the one Dedekind wrote at
> > the start, over 120 years ago. There's a terrific English translation
> > due to John Stillwell, which you can find easily by searching for
> > "Dedekind algebraic integers". I bought a copy from amazon.com.
> Is this the book you are referring to?
>
> http://books.google.com/books?id=F5zI9KYieTQC&dq=Dedekind+algebraic+i...
>
> In other words - google books.
Yes, that would be the book.
--
Arturo Magidin
Tim Peters
> Is this the book you are referring to?
Yes -- and the following URL is one of the top hits if you perform the
Google search I suggested.
> http://books.google.com/books?id=F5zI9KYieTQC&dq=Dedekind+algebraic+int
> egers&printsec=frontcover&source=bl&ots=StobeZYWgF&sig=ACMlm_9qDq3Jzkg_
> LQXoKM7KXvM&hl=en&ei=cDAdS4zJJ9CTkAWB-YjdAw&sa=X&oi=book_result&ct=resu
> lt&resnum=3&ved=0CBYQ6AEwAg#v=onepage&q=&f=false
>
> In other words - google books.
It's hard to know what question "Google Books" answers here ;-) Did you
look at the preview? Stillwell's translation is still covered by active
copyright, and very few of the pages are available. For example, all of
pages 2 through 101 are missing in the Google Books version. You can do
just as well looking at amazon.com's preview.