Simple principle in core error proof

Discussion in 'Undergraduate Math' started by James Harris, Oct 20, 2003.

  1. James Harris

    James Harris Guest

    What makes my situation especially frustrating is how simple the
    argument is that proves there's this problem with algebraic integers,
    yet still I have to keep explaining.

    The basic principle is like with

    P(m) = 2(m^2 + 2m + 1) = (a_1 m + 2)(a_2 m + 1)

    where most of you can probably guess what factor a_1 must have!!!

    Now the polynomial I use is more complicated such that I need to set
    m=0 to figure things out, but notice here what happens:

    P(0) = 2, and dividing off 2 from P(m) gives

    P(m)/2 = m^2 + 2m + 1

    and notice that P(0)/2 = 1, which tell you that the independent term
    changed.

    Given

    (a_1 m + 2)(a_2 m + 1)

    it's clear that a_1 has a factor that is 2.

    It's a simple idea that I use with a more complicated polynomial, and
    I think that the reason so many math people go out of their way to
    make it seem like it's wrong is that they're embarrassed by the over
    hundred year old error that I found.

    I've been surprised at how dedicated they can be at trying to hide the
    truth.

    After all, I've communicated with top mathematicians like Barry Mazur,
    Andrew Granville, and Ralph McKenzie, who may be people many of you
    haven't heard of, but in certain math circles they're well-known.

    In McKenzie's case I explained it all to him ***in-person*** and he
    basically blew me off.

    These mathematicians are hellbent on trying to hide that their
    discipline could actually have a flaw like this for as long as they
    can get away with it.

    It's wrong, and it can't help world society.

    Meanwhile it's not doing me a bit of good either, and I think that
    part of their motivation is making me miserable, knowing that what I
    found *should* get me accolades.

    Yup, call me crazy, but I think the bastards are out to get me!!!


    James Harris
     
    James Harris, Oct 20, 2003
    #1
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  2. James Harris

    Virgil Guest

    There is a meaning of "simple" which applies to JSH's explanation,
    meaning authored by a simpleton, which JSH is mathematically at
    least.
     
    Virgil, Oct 20, 2003
    #2
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  3. This is a red herring. Here, the polynomial has a coefficients which
    are constant with respect to the variable of factorization, and what
    you are varying is the polynomial variable m. In your actual
    situation, you are factoring with respect to a DIFFERENT variable, X,
    and the coefficients are not constant, but are instead functions of
    m.

    This example is thus worthless as a guide.
    What happens is you engage in a red herring. In your actual
    application, your factorization is not with respect to m, but with
    respect to x. And you do not set x equal to 0, you set m equal to
    0. What you are doing here is in no way parallel to what you do in
    your actual application, and thus conclusions you draw from this
    example are not necessarily applicable to your actual application.

    [.red herring left in place so James cannot complain that I
    'removed the math'.]
    Because a1 and a2 are constant. They do not vary. So whatever happens
    at one point is necessarily what happens at all points. This is not what
    happens in your actual application; in your application, the values of
    the coefficients change as you change m, and it does not follow that
    what happens at one value of m is what happens at all values of m. In
    fact, it does not, has been amply demonstrated by explicit
    calculation.
    Then you got confused. here, you have a polynomial in m being factored
    as linear terms in m, where the coefficients with respect to m are
    constant. In your application, you are factoring a polynomial which, as
    m varies, is reducible or irreducible over Q depending on the value of
    m; here, you have a polynomial which is always reducible over Q
    regardless of the value of the (non-existent) non-polynomial variable.

    In your application, you have a polynomial in m being factored into
    linear terms WITH RESPECT TO A DIFFERENT VARIABLE x, and the
    coefficients with respect to m are ->functions of x<-, and the
    coefficients of x are functions of m. Here you set the polynomial
    variable equal to 0, but in your application you set m, not the
    polynomial variable at issue, equal to 0. Here you KNOW the
    coefficients are constant and therefore whatever happens at one point
    will happen at all points, but in your application the coefficients
    are NOT constant and therefore it does not follow that what happens at
    one point will happen at all points. The situations are NOT similar,
    and therefore the situations are not comparable.

    [.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 man's 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 criticize. A great
    many people are staggered to this extent, 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
     
    Arturo Magidin, Oct 20, 2003
    #3
  4. James Harris

    fishfry Guest

    Ask yourself this. How can a definition lead to a contradiction?
    Suppose I define a set B = {all objects in the universe that are
    left-handed giraffes} and then I look at every object in the universe
    and ask, is it a left-handed giraffe? If it is, it's in B. If not,
    it's not in B. Perhaps there are no left-handed giraffes, in which case
    B is still well-defined, it just happens to be empty.

    If I define the algebraic integers as real numbers that are the root of
    a polynomial with monic leading coefficient, then I can look at each
    real number and say it's either in this set or isn't.

    Can you explain what you mean when you say you are getting a
    contradiction from the definition? For example, have you found a real
    number that both is and isn't an algebraic integer? What contradiction,
    exactly, have you found?
     
    fishfry, Oct 20, 2003
    #4
  5. James Harris

    James Harris Guest

    Yet readers can see the reality here:

    P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7)

    The form of the polynomial allows me to factor P(m) into
    non-polynomial factors, and the factorization with those factors is

    P(m) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)

    where ONLY m is a variable.

    And yes I used more symbols before but found that posters like Arturo
    Magidin could too easily confuse people about what was going on, so
    I've put in values where once there were symbols.

    Unfortunately, Arturo Magidin is a rather evil person who lies about
    the math as if the core error should stay hidden.

    He's bad, he's evil, and I'm sick of his crap.


    James Harris
     
    James Harris, Oct 20, 2003
    #5
  6. And can you PROVE that a_1, a_2, and a_3 will satisfy the properties
    you want under these conditions? What are those properties?

    You can ALWAYS get such a factorization, if you allow a_1, a_2, and
    a_3 to be algebraic numbers; but then your congruence arguments bite
    the dust.
    Unfortunately, James Harris is a rather dishonest person who lies about me,
    having been unable to find a single instance of me "lying about the
    math", and who, each time he accuses me of such behavior, ends up
    retracting his claim but not his accusation.

    ======================================================================
    "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 man's 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 criticize. A great
    many people are staggered to this extent, 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
     
    Arturo Magidin, Oct 21, 2003
    #6
  7. Yet, when I offered to stop posting if you would only tell me to do
    so, you ->refused<-. Were you too scared to tell someone to stop
    posting, or are you just lying here about being "sick of [my] crap"?
    (No, I won't count as a lie the fact that you have given no evidence
    whatsoever of anything I have posted being false, let along 'crap';
    that's just your usual lying to yourself).

    Why, if as you claim I am "bad", "evil", "rather evil", someone who
    "lies about the math", and if you are "sick of [my] crap", did you
    refuse my offer to 'step out of your way', as it were, by no longer
    posting to anything you might say?

    ======================================================================
    "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 man's 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 criticize. A great
    many people are staggered to this extent, 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
     
    Arturo Magidin, Oct 21, 2003
    #7
  8. James Harris

    flip Guest

    .... Worthless dribble deleted! ***

    I'll go for calling you crazy.

    Does anyone second the motion.

    You are like a broken record, replaying the same old nonsense and you have
    lied to yourself for so long, you believe your own incompetence.

    Please go get some psychotropic drugs or jump off a bridge, you raving
    lunatic.

    How pathetic you have proven with your feeble attempts for fame, fortune and
    recognition. You are so sad and an example of schooling gone horribly wrong
    in the US (as much as it hurts to admit that).

    Perhaps you should renounce your degree and go back to grade school and see
    if people still think you are "smart".

    Get a life!
     
    flip, Oct 21, 2003
    #8
  9. James Harris

    Wayne Brown Guest

     
    Wayne Brown, Oct 21, 2003
    #9
  10. isn't a "term," just a monomial, 6abm^2 e.g.?... so,
    what do you mean by "independent term?..." so,
    i followed that example, but what is it supposed to prove?... I mean,
    it's a good, artificial construction to prove ... what?
    you could clear this up, by using an even simpler equation, and
    applying your alleged method to it. as far as I can see,
    you simply refuse to assert your definitions --
    whether or not they are from "over-100-year-old math."
    what is it in "substitution of variables,"
    that exposes the wrongness of the definition
    of algebraic integers (and, of course,
    the evil of Arturo) ??

    <deletives impleted>

    --les ducs d'Enron!
    http://larouchepub.com/other/2003/3041shultz_arnie.html
    http://larouchein2004.net/pages/speeches/2003/031022webcast.htm
     
    Brian Quincy Hutchings, Oct 22, 2003
    #10
  11. Tell me what happens when

    P(m) = 5*(m^2 + 7*m + 5)

    and you assume

    P(m) = (a_1*m + 5)*(a_2*m + 5).

    Andrzej
     
    Andrzej Kolowski, Oct 22, 2003
    #11
  12. James Harris

    James Harris Guest

    Readers should note the challenge to the *reality* that setting m=0
    gives the independent, or constant term.

    But trivially with the poster's example the constant term is 25.

    The independent term is 25.
    There's nothing to assume as what I do is focus on constant terms.

    Again, trivially, the constant term is 25.

    The independent term is 25.

    I want readers to pay *careful* attention as in fact several posters
    have gotten away with misrepresenting what I do when in fact I focus
    on constant or independent terms.

    The math is basic. These posters, however, need to get you to
    question basic algebra which many of you learned while still children.

    And they've apparently been VERY successful up to now.

    I focus on the independent terms.

    That's how the argument works.

    To win the war of confidence, posters like Andrzej Kolowski have to
    get you to question your own mathematical knowledge, to get you to
    lack confidence in your own mathematical understanding, so that you
    rely on them as the experts, and then they lie to you.

    It's not complicated people, and it's an old story in human history as
    I'm challenging a powerful group that'd rather not let the truth get
    in its way.


    James Harris
     
    James Harris, Oct 22, 2003
    #12
  13. James Harris

    C. Bond Guest

    James Harris wrote:

    [snip]
    Your use of the qualifier 'independent' is incorrect. The function, P(m), is a polynomial in 'm'. Presumably *all*
    coefficients of the individual terms involving products with powers of 'm' are 'independent' of 'm'. Each of the
    coefficients in the polynomial is a constant, including the coefficient of the term corresponding to m^0 ( the standard
    constant term). If you *evaluate* the polynomial at m=0, only the constant term remains, but not because it is
    'independent' of 'm' -- is is neither more nor less dependent on 'm' than any other coefficient -- rather that you have
    simply replaced 'm' with 0 to zero out all other coefficients and evaluated P(0).
     
    C. Bond, Oct 22, 2003
    #13
  14. I appreciate all the comments, but I don't think you
    answered my question at all, though admittedly it was a
    little vague.

    What I wanted to know was about the divisibility of a_1 and
    a_2 by 5 for various values of m. Is one or the other of
    them always divisible by 5 regardless of m, or what? Your
    answer on this could greatly strengthen your credibility as
    far as I am concerned.

    Andrzej
     
    Andrzej Kolowski, Oct 22, 2003
    #14
  15. James Harris

    James Harris Guest

    Here's what you have.
    The constant term is 25. The most you can divide through by with m an
    algebraic integer m, is 5, which gives you a constant term of 5 for
    your P(m)/5.

    So there's no relevance to my argument as you still have 5 as a factor
    of the constant term.

    The simple and basic algebra argument that proves my case, and shows
    that there IS an error in core because of the definition of algebraic
    integers, depends on coprimeness of the constant term, which is how I
    filled what had been called a "gap".

    Remember that Andrzej Kolowski? It was almost two years ago, when I
    admitted to having a gap, as several posters had claimed, and then
    when I filled it, you all just lied, so I know about credibility, and
    you have none.

    You've been lying for almost two years now, over an issue that I
    handled back in November of 2001.

    People like you've been torturing me, in an inhumane behavior that's
    hard to explain, as I see you as evil incarnate, dedicated to doing
    evil until someone stops you.

    And I will stop you.


    James Harris
     
    James Harris, Oct 22, 2003
    #15
  16. Which is somewhat parallel to what happens in your
    argument when you divide through by f^2: note that
    when you factor f^2 out of the first *two* terms of

    (a1*x + uf)*(a2*x + uf)*(a3*x + uf),

    you have

    f^2*((a1/f)*x + u)*((a2/f)*x + u)*(a3*x + uf),

    and you still have "uf" in that last term.

    I note that you didn't really answer my question, whether
    one, both, or neither of a1 and a2 are divisible
    by 5 when m <> 0.

    But I admit, I do see your point. You are talking about
    the constant term with respect to m, not the constant
    term with respect to x. My example, unlike your P(m),
    doesn't have any x's in it.

    Let me modify my example a little bit to get closer
    to your P(m): let

    Q(m) = f*(m^2*x^2 + 3*x + f).

    Then Q(0) = f*3*x + f^2 = f*(3*x + f).

    This is the constant term, "independent of m".

    Note that Q(0)/f = 3*x + f. This is similar to
    your polynomial P(m), where

    P(0)/f^2 = u^2*(3*x + u*f), or, when u = 1,

    P(0)/f^2 = 3*x + f.

    Now assume that my polynomial Q(m) is factored in the form

    Q(m) = (b1*x + f)*(b2*x + f).

    As in your P(m), b1 and b2 are going to be functions
    of m: b1(m) and b2(m).

    The fact that Q(0) = f*(3*x + f) is consistent with
    b1(0) = 0 and b2(0) = 3. Agree?

    So b1(0) is divisible by f and b2(0) is relatively
    prime to f (if f is a prime <> 3).

    This too is similar to your polynomial P(m), where
    a1(0) = a2(0) = 0 and a3(0) = 3, the latter being
    coprime to f (if f is a prime <> 3).

    Anyway, for my new polynomial Q, I would like to see
    your answer to the question: for my polynomial, for m
    in general, is b1(m) or b2(m) divisible by f? Both,
    only one, or neither ?

    Yes, well, of course thanks for the reply.

    Andrzej
     
    Andrzej Kolowski, Oct 23, 2003
    #16
  17. ho hum; same ****, different day.

    --les ducs d'Enron!
    http://larouchepub.com/other/2003/3041cody_on_recall.html
     
    Brian Quincy Hutchings, Oct 23, 2003
    #17
  18. James Harris

    James Harris Guest


    I figured out that posters were having a field day with all the
    variables, and that it made it *easy* for them to lie and confuse
    people.

    But

    P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078

    takes that away.

    But notice how they try to go back to more variables, like the twisted
    and despicable people they are.

    I'm telling you, these people are EVIL.

    Now if they just admit that constant terms are constant, then the rest
    of the argument follows.

    That's because the constant terms CANNOT change dependent on the value
    of m.

    That means that focusing exclusively on the constant terms and how
    they do change when 49 divides off of

    P(m) = 14706125 m^3 - 900375 m^2 + 17640 m + 1078

    tells the tale.

    Now I'm informing all of you that the people arguing against me are
    EVIL, yes they are real, live EVIL people as mathematics is that
    important, so it's important enough for Evil itself to send minions
    like them.


    James Harris
     
    James Harris, Oct 23, 2003
    #18
  19. so, just who *else* do you think is reading your stuff,
    other than the Axis of Ones Who Argue with JSH?
    Arturo gave a "tutorial on indepeence of terms"
    in another item, in reply to you, but
    i have a feeling taht that's the last you'll "go"
    to it.
    I'm not arguing *against* you, just with you!

    --les ducs d'Enron!
    http://larouchepub.com/other/2003/3041cody_on_recall.html
     
    Brian Quincy Hutchings, Oct 23, 2003
    #19
  20. Q(m) = 6125 m^3 + 6125 m^2 - 6370 m + 1078

    tells another tale.
     
    Dik T. Winter, Oct 24, 2003
    #20
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