Understanding the technicality

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

  1. James Harris

    James Harris Guest

    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

    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

    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

    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.

    Instead, leading "mathematicians" try to ignore my work, despite
    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

    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

    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

    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

    James Harris
    James Harris, Oct 17, 2004
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  2. 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
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  3. James Harris

    Oppie Guest

    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. 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
    suppose your paper was supposed to be about? Read the introductory
    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
    David Kastrup, Oct 17, 2004
  5. 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. James Harris

    fishfry Guest

    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. 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
  8. James Harris

    Saint Cad Guest

    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".
    Saint Cad, Oct 17, 2004
  9. 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. 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. James Harris

    Tim Smith Guest

    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
    follow an interesting pattern.

    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. 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.

    Read a bit about abstract algebra...
    Dik T. Winter, Oct 18, 2004
  13. 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. James Harris

    W. Dale Hall Guest

    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
    your question?
    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

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

    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...

    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!
    HARD WORK!!!
    Yes, the Voice of Wisdom: poop or get off the pooper.
    Lemme hear an AMEN!

    Amen, brother!
    Lemme hear an AAAAAMEEENNNN!!



    Praise be to MATH!!!


    W. Dale Hall, Oct 18, 2004
  15. 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. James Harris

    Mike Terry Guest

    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...

    Mike Terry, Oct 18, 2004
  17. James Harris

    Tim Smith Guest

    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. Probably every other mathematician would get fired.
    David Kastrup, Oct 18, 2004
  19. 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. 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.

    --A HYDROGEN (sic; cracked methane) ECONOMY?...
    The Three Phases of Exploitation of the Protocols
    of the Elders of Kyoto:
    (FOSSILISATION [McCainanites?] (TM/sic))/
    BORE/GUSH/NADIR "@" http://www.tarpley.net.
    Http://www.tarpley.net/bushb.htm (partial contents, below):
    17 -- THE ATTEMPTED COUP D'ETAT, 3/30/81 (87K)
    18 -- IRAN-CONTRA (140K)
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    Brian Quincy Hutchings, Oct 18, 2004
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