# Understanding the technicality

Discussion in 'Recreational Math' started by James Harris, Oct 17, 2004.

1. ### James HarrisGuest

You may have noticed that I repeatedly talk of a technicality, and I'd
like to explain simply and directly in a single post.

Consider a world where some math people decide to have a special ring
where 3 cannot be a factor.

Now they have the ring of integers, but they also have the ring of the
excluded 3, where 3 is not a factor.

Now in this special ring, you have that 9 is prime.

But in the ring of integers 3(3) equals 9, but not in the special ring
of the excluded 3.

In that ring 12 has 2 as a factor but not 4, but 35 has both 5 and 7
as a factor.

It's arbitrary, and 9 is prime on a technicality that 3 is excluded on
that strange world where the math people have decided to exclude it by
a definition.

Now in our world math people decided to have a special ring that only
included roots of monic polynomials with integer coefficients.

In that special ring numbers that cannot be roots of monic polynomials
with integer coefficients are excluded, just like before in the
special ring of the excluded 3, but here, on our world, the rule is
that numbers not roots of monic polynomials with integer coefficients
are excluded.

So you can have some number xy where xy is an algebraic integer, x is
an algebraic integer but y is not, so the number, in the ring of
algebraic integers, cannot have x as a factor. Nver.

It can never have x as a factor by the exclusion rule for any number
xy, where x is an algebraic integer, but y is not the root of a monic
polynomial with integer coefficients!

It's an arbitrary rule.

It's harder in this world of ours with the special ring of algebraic
integers than in a world with the special ring of the excluded 3 as in
that world, they know of the ring of integers, but in this world, math
people act as if all that matters is that a number does or does not
have a factor in the ring of algebraic integers.

Put that way it should sound silly.

Why in this world is there a special ring with an arbitrary exclusion?

Convenience and mathematical naivette are the answers.

Unlike arbitrarily excluding 3, arbitrarily excluding numbers that
cannot be roots of monic polynomials with integer coefficients can
*appear* to be ok.

For over a hundred years math people used it believing the arbitrary
was not only ok, but indicated significant and important properties of
numbers!!!

They built an edifice on that belief which they call Galois Theory,
where they made various leaps from the work of Evariste Galois,
building on the error that depends on the arbitrary exclusion--a
technicality.

The mistake is so basic that proving it's a mistake requires rather
basic algebra, in a proof that is so short it could be a homework
problem, but people have argued with me about it for over two years
now.

Imagine aliens considering this world who have correct mathematics.
To them the emphasis on algebraic integers would be as silly as that
world where they have the special ring of the excluded 3.

If math society played by its own rules then I could simply highlight
the error, and prove it exists using mathematical proof.

But math society does not play by its own rules.

having been made aware of it, while sci.math'ers go out of their way
to attack it, and use bogus arguments that I repeatedly shoot down.

Their arguments are so bizarre that a telling one that has recently
played out here which has played out before has to do with
constants!!!

I argue that constants are constant.

Sci.math'ers argue against the algebra and you let them as if algebra
itself, might be wrong.

And many of you clearly believe the sci.math'ers based on replies I
see on Usenet, and from the inaction on this issue.

You're no better than a world where 3 is arbitrarily excluded as a
factor.

This issue mathematically is trivial. Socially it's huge.

The math is not very complicated, and is easily shown such that I
could explain it to a kid.

The social issues are so huge that they will probably wreck
departments and destroy careers.

Do you wish me to feel sorry for you? If that's what it takes then
maybe I could say I feel sorry for you.

Or maybe I can tell you that there are other great achievements in
mathematics in other fields outside number theory, so it's not so bad.

Or I could try to think of other ways to comfort you, or salve the
emotional impact, or otherwise make you feel good about yourself
despite the situation.

But it won't change the math.

The lesson is that mathematics is a harsh and difficult discipline and
you must tell the truth, as well as not teach more than follows from
the math.

So mathematics is a difficult discipline requiring your best.

Is that not fair to you now? Suddenly too hard for you?

Either deal with it, or get out of the field.

If the rest of the lesson is that even "mathematicians" no matter how
high their social standing can be tossed into prison for believing
they own mathematical truth, then so be it.

If the rest of the lesson is that academia can get a black mark that
will shine forever as a dubious distinction, where diverse fields far
from mathematics will take the hit along with you, then so be it.

I assure you there will be a lesson taught, and if that lesson is that
math society could take the hit, and prove its own love of mathematics
in one of the greatest tests ever, then so be it.

Now you can shine, and you will shine, either brightly or darkly.

Now you continue to make your choice as you've been making choices all
along.

Either some of you are mathematicians, for real, or you will fight
this, and the math is what you will try to fight.

And that's like running headfirst into a brick wall, but worse, as you
might find something to break a brick wall, but the math will never
yield.

James Harris

James Harris, Oct 17, 2004

2. ### Marcel MartinGuest

James Harris a écrit :
A very special ring, indeed. It is no more a ring. The remaining
integers and the addition are no more a group. What is the value of
2 + 1? And don't suppress 2, I will ask you what is the value of
1 + 1 + 1.

Marcel Martin, Oct 17, 2004

3. ### OppieGuest

Oh gee whiz, Marcel! You're confusing the issue with FACTS!

Let poor Jimbo rant. Maybe if he listens to the echoes of his ramblings,
he'll shut up.

Oppie, Oct 17, 2004
4. ### David KastrupGuest

That's not something you can decide in a whim. If you have a ring
with identity, this is not possible, since 1+1+1 exists, and so
x+x+x = (1+1+1)x = 3x exists. What can happen is that 3 is a unit, so
that 3 is coprime to everything. But it will still divide everything.
Since you are talking about a mythical beast, there is not much sense
in getting excited over the size of its horns.
Do you have a fscking clue what the algebraic integers are about in
the first place? Of _course_ things like 1/2 are excluded _by_
_definition_. That's what makes the algebraic integers interesting.
In contrast to excluding 3 by definition, we still have a workable
ring. It's not a technicality. It's the whole _purpose_ of the
algebraic integers. If it annoys you, then fscking take the algebraic
numbers instead. There everything except 0 has an inverse because we
have a fscking field. There things are simple. The algebraic
integers are interesting _precisely_ because things are not so simple
there. If you don't want the complications (and those complications
are there with any case of nontrivial factoring), then don't take the
algebraic integers.
Wrong way round. Math people saw that the roots of monic polynomials
with integer coefficients form a ring! That the roots of monic
polynomials with integer coefficients are closed under addition and
multiplication! That's fscking amazing! That's what makes it a ring
in the first place.
There is no ring with excluded 3, unless you are talking about a ring
without identity (like the vector product).
Well, like if you have 3*(1/3) where 3 is an integer but (1/3) is not,
so 1, in the integers, cannot have 3 as a factor. Fscking trivial,
right? If you don't like it, take fractions.
It's the definition of algebraic integers. I mean, you preposterous
idiot try publishing a mathematical paper about algebraic integers and
about a year later are annoyed to find out that they are defined to be
algebraic integers? How stupid can you get?
If you are trying to publish a paper about factorization in algebraic
integers, it does matter whether a number does or does not have a
factor in the ring of algebraic integers.

I mean, you are a loon. You are completely off the wall. What do you
paragraph. Do you claim that you were just blindly dropping your head
on the keyboard without understanding what words came out?
One? There are millions of them. Take a different one if you don't
like it.
If you are talking about a ring defined as consisting of the roots of
monic polynomials with integer coefficients, it is not exactly
arbitrary to exclude numbers that aren't.
Oh, algebraic integers are a weird lot, certainly in some way more
arbitrary and less obvious than normal integers. But if you don't
want to deal with them, then don't write papers about them.
At the moment your arguments are so stupid that not even a kid would
swallow them.

Get sober before posting again, for heaven's sake.
Right. And the math concerned with the roots of monic polynomials
with integer coefficients will be exactly that: concerned with the
roots of monic polynomials with integer coefficients.
It requires

a) that you know the definitions of the stuff you work with,
b) that you are reasonably sober.

Actually, neither of the two are required all the time for all the
work, but you better get them catered for before judging and
presenting your ideas if you don't want to appear a gibbering idiot.
You fail on both accounts.
Yes, get out. You are the person that complains.
You are raving again. Get sober.
You are raving. Get sober.
Get sober. And then read the whole kaboodle you have written and
realize that you have written it to yourself. You are yelling at
yourself.

David Kastrup, Oct 17, 2004
5. ### David KastrupGuest

Threedimensional vectors with normal addition and vector product are a
ring without identity, and without 3.

Once you have identity, however, 3=1+1+1 is hard to avoid to have. It
can be a unit, though.

David Kastrup, Oct 17, 2004
6. ### fishfryGuest

Factor of what?

By the way, nobody's asked you this for quite some time, but you've
never actually answered. Do you happen to know what a ring is?

fishfry, Oct 17, 2004
7. ### Will TwentymanGuest

You mean something like Z_2? To exclude 3 as a factor, you would have
to somehow exclude 3. How about the ring of 2x2 matrices? I don't
think 3 is in that ring.

Hint: there are a lot of rings out there. Some of them are not subsets
of the complex numbers.

If you exclude 3 from the ring, 3 is necessarily not a factor of
anything *in that ring*. Just be careful to make sure you've defined
your ring in such a way that 3 is not required by additive closure.
Now you are talking nonsense. If your ring includes 2, it *must*
include 4. 2+2 = 4. To exclude 4 is to also exclude 2.
When you exclude 3, you also exclude 1. 9 may or may not be in the
ring, depending on how it's defined. The ring 2Z excludes both 3 and 9.
The ring 9Z excludes 3 but not 9.

Will Twentyman, Oct 17, 2004

It's a round thing. You know, they made that movies with that guy about a
ring. And I've looked very closely when he put the ring in the fire and the
writing showed. Sure enough, there was no "3" in the writing.

I guess James was right, there is a ring without a "3".

9. ### Marcel MartinGuest

David Kastrup a écrit :
But JSH didn't ask whether or not rings without {3} can exist. He
wrote that Z - {3} is a ring. This is wrong.

Now, of course, there are rings without {3}. The null ring, Z/2Z or
even Z/3Z, to quote the most obvious (while considering rings of
which the elements are integers).

Marcel Martin, Oct 17, 2004
10. ### David KastrupGuest

Uh, 3 is 1+1+1 and is in Z_2. It is the same as 1.
It's pretty much the same as [3,0;0,3] when we are talking about
normal matrix multiplication.
And if it includes 1, it must include 1+1+1. Game over. Unless we
don't even have 1 (multiplicative identity), like in the ring over
vector sum and product.
How does it do that? Are you sure you are not talking about 3Z?
Anyway, might be easier to understand than vector products.

David Kastrup, Oct 17, 2004
11. ### Tim SmithGuest

So? You don't need to go to algebraic integers to see this sort of thing.
You can see it in the regular integers. Let x = 2, xy = 5. Then xy is an
integer, and x is an integer, but y is not.

It looks like you are moving to the next phase of your argument now. You

1. Make claims that are false. Argue about it forever, despite clear
demonstrations by many people that they are false. For example, the
counterexamples to the main "theorem" of your APF paper.

2. Change the emphasis of what you are talking about (helped by the fact
that you have a poor grasp of mathematical terminology, so your writing in
#1 is very muddled) so that your results change from false results about
interesting things to trivially true results about uninteresting things.

Tim Smith, Oct 18, 2004
12. ### Dik T. WinterGuest

James, showing an extreme lack of comprehension of abstract algebra, not
even understanding (still, after many years), what a ring is:

I assume you are referring to subrings of the integers...
If 3 is not in the ring, 1 cannot be in the ring either.
Because if 1 is in the ring, so is 1+1+1 = 3.
If you have 9 in the ring, but neither 1 or 3, the ring is 9Z (i.e.
all multiples of 9).
Neither 12, nor 2, 4, 35, 5 and 7 can be in such a ring. Only multiples
of 9. When 7 and 2 are in a ring, so is 7 - 2 - 2 - 2 = 1. So 3 is in
that ring.
In 9Z, 9 can be called a prime because no element of the ring divides 9.

Dik T. Winter, Oct 18, 2004
13. ### Arturo MagidinGuest

In 9Z, 9 can be called a prime because no element of the ring divides 9.[/QUOTE]

Hmmm... Depends on your definition of "prime". What you describe is
closer to "irreducible". Prime usually means:

Let R be a ring; the element a in R is a "prime" if and only if
a|bc --> a|b or a|c.

In 9Z, 9 is not a prime under this definition, however: 9 divides 9*9
= 81 in 9Z, but it does not divide 9.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin

Arturo Magidin, Oct 18, 2004
14. ### W. Dale HallGuest

Not too surprising. The ring consists of all integral multiples of 9.
The principal ideal generated by 9 is the set of all integral multiples
of 81. All lesser multiples of 9 are irreducible, despite being obtained
as sums of various quantities of nines: 9+9+...+9. Similarly, if you
have a multiple of 9 that is not a multiple of 81, it is irreducible.
Suppose there exists such a ring. I'll imagine it's as much like the
integers as possible, only it omits 3 (in particular, I'll suppose
it is a subring of the ring of integers). First we note that the
ring cannot contain 1, since 1+1+1 = 3.

Let's suppose it contains the number 9. Then, the number 6 cannot be
contained in that ring, because 9 - 6 = 3. Also, 2 cannot exist, since
2 + 2 + 2 = 6. If 2 cannot be in the ring, then 9 - 2 = 7 cannot be in
the ring.

Perhaps the ring contains 4. But if it contains 4, then it must contain
4 + 4 = 8, and thus 9 - 8 = 1. So neither 4 nor 5 can be in the ring.

So far, we have the following numbers *not* in the ring:

1, 2, 3, 4, 5, 6, 7, 8.

Further arithmetic will show that the only elements of your ring that
includes 9 but excludes 3 will be integral multiples of 9.

As a result, your ring is simply the integral multiples of 9, considered
as a subring of the ordinary integers.
So, tell me. Exactly what problem is more easily understood by looking
at this ring?
Yes, and it has been proven to be a ring, with identity, and to have
properties that give us insight into the nature of the ring of integers.
They're only excluded from the ring of algebraic integers. If we cared
to look at roots of other equations, those can certainly be identified
in larger rings.
Is this even desirable? For instance, 30 is an ordinary integer, as is
14. However, in the ring of integers, 30 cannot have 14 as a factor.
Never. Look in the ring of rationals Z[1/7], with denominators given
by powers of 7. There, 30 *does* have 14 as a factor:

30 = 14*15/7
Same with 30 and 14 in my example.
Say silly stuff and you'll sound silly. My mama told me that, I'm
surprised yours didn't rear you that way. However, much of the recent
discussion has had to do with whether certain things can be done
*within that ring*! I wasn't the one to make the claim that two
particular numbers were coprime in that ring, *you* were! All I
did was to say that if you want this to be true in the ring of
algebraic integers, then certain manipulations had to apply in
that ring.

I showed that the numbers in question had a common factor within
the ring of algebraic integers, and proved that this common factor
was not a unit (and it is not a unit by virtue of the property of
not having a multiplicative inverse in that ring). Further, I showed
that if the numbers were coprime, then any common factor would
necessarily be a unit.
Why can I define something and prove what its properties are? Is that
Much of science is put in place the way it is, for the sake
of convenience. There is no reason to make already difficult
work even more difficult.

As far as the other, it's spelled naivete, and frequently
(when one has the appropriate font) spelled naiveté.

The idea that one is exhibiting naiveté by producing a definition
of some concept, proving what properties the resulting structure
has, and proceeding to do mathematics based on use of that structure
is patently absurd. Naiveté is the quality (according to the online
Webster's Dictionary) of being naive, which has the following
definition:

(1) Marked by unaffected simplicity : ARTLESS, INGENUOUS
(2) Deficient in worldly wisdom or informed judgment;
especially : CREDULOUS
(3) SELF-TAUGHT, PRIMITIVE

If the definition is any guide, it ain't sci.math or the world of
mathematicians who have been naive.
So how come all those proofs actually work?
Yes, and every so-called example that you've come up with has been
shown not to be an example at all? Every time you've said something
couldn't be factored over the algebraic integers, every time you've
said numbers were coprime over the algebraic integers, when it went
contrary to what the alleged error using Galois Theory, the facts
showed you to be in error.

Why does that make you correct?
You have tried, you have been shown to be incorrect, and what do you
do? You say that the people who show you the truth are liars, you
say they are cheating. You act as though the truth is something that
you and you alone possess.
You mean the bogus argument that uses only arithmetic to prove that
your alleged coprime elements of the ring of algebraic integers are
in fact *not* coprime? Are you talking about that argument?
Here's a news flash: it isn't algebra that's wrong, it's your bungling
of basic algebra. Live it, or live with it.
Huge, you say!
Wreck departments! Destroy careers!!
Rivers and seas boiling! 40 years of darkness, earthquakes, volcanos.
The dead rising from the grave!
Human sacrifice, dogs and cats, living together...

MASS HYSTERIA!!!
Did you ever give thought to the fact (yes, it's a fact) that Galois
Theory has contributed to mathematics outside of Number Theory?
Ah, yes, there's the Math!
It's...
HARD WORK!!!
Yes, the Voice of Wisdom: poop or get off the pooper.
Lemme hear an AMEN!

Amen, brother!
Lemme hear an AAAAAMEEENNNN!!

AAAAAAMMMEEEENNNNN, BROTHER!!!!

GLORY BE!

Praise be to MATH!!!

AMEN!!!
Dope.

Dale.

W. Dale Hall, Oct 18, 2004
15. ### Dik T. WinterGuest

Yup, I know. See my "can be called". However, when I went to school (upto
University), there was not yet a distinction between prime and irreducible.
The reason is that on that level you did not encounter rings where the
distinction was important (and it is at approximately that level where
James resides with his mathematics). Before University the only definition
of prime I had seen was: "is only divisible by 1 and itself".

Note that, because of such a definition, until about 80 years ago, 1 was
considered prime by many mathematicians.

Dik T. Winter, Oct 18, 2004
16. ### Mike TerryGuest

Funny thing, that ring 2Z... We can prove that there exist numbers like 17,
which are excluded from it purely on the technicality of not being of the
form 2*z for some integer z. The maths required to prove this is not
difficult, and is short enough to be treated as a homework assignment.

Looks like mathematicians have a lot of reworking to do on the body of
published mathematics over the centuries. Not only must they throw away any
results based on the flawed ring of algebraic integers, they're also going
to have to throw away anything based on 2Z. I forsee many universities
being bankrupted by the huge overtime claims if the 2Z flaw ever comes to
light, so there will be a huge financial incentive to suppress such flaws.
Also professional mathematicians will no doubt try to suppress the 2Z flaw,
as their positions with their universities will become untenable once
students come to realise they are being taught algebra using flawed rings...

Regards,
Mike.

Mike Terry, Oct 18, 2004
17. ### Tim SmithGuest

I don't think he's trying to *actually* talk about a ring that excluded 3,
but rather just using that as an example of the *kind* of thing he's saying
mathematicians do.

Tim Smith, Oct 18, 2004
18. ### David KastrupGuest

Probably every other mathematician would get fired.

David Kastrup, Oct 18, 2004
19. ### David KastrupGuest

Wouldn't it be easier for the layman to understand if he used child
molestation as an example for that purpose?

David Kastrup, Oct 18, 2004
20. ### Brian Quincy HutchingsGuest

that's the second mention in this thread
of the little fact that not allowing 3 in the ring,
doesn't allow one in it, either;
such a simple thing, that nine becomes the unit!
so much for actually doing teh math.

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