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| JSH... |
 Posted: Sat Jul 04, 2009 7:51 pm |
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Guest
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On Jul 4, 7:56 pm, "Achava Nakhash, the Loving Snake"
<ach... at (no spam) hotmail.com> wrote:
[quote:9b3d5451f3]James,
I realize that this is an exercise in futility, but I am going to try
to demonstrate to you why the mathematical community does not take
anything you say seriously. Many of us do, however, find you
[/quote:9b3d5451f3]
I've read your reply and it fits more with the psychological aspects
of this issue.
Readers who are puzzled by his reply need to understand how
devastating the core error is.
He may actually believe he has cogent rebuttals, but he also may be
faking in case later he believes he needs to say his mind snapped in
his own defense.
[quote:9b3d5451f3]marvelously entertaining. As I remarked several times already, you
write very well, and you spin an absorbing, if silly, science fiction
conspiracy involving mathematicians and, in particular, number
theorists. Never before have we been considered so powerful. We
certainly don't feel that way.
I will intersperse my specific comments withing your text, where
appropriate, to try to show you how mathematicians view the
mathematics of what you do. I and others have done this before to
little effect, but I happen to be in the mood. Beethoven's fifth is
playing in my speakers, so I guess I just feel empowered or something.
On Jul 4, 6:41 pm, JSH <jst... at (no spam) gmail.com> wrote:
Turns out that if you follow rigorous mathematics it is trivial to
show a problem with the use of the ring of algebraic integers with a
quadratic factorization that I've often given before:
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.
This appears to be a monic polynomial of degree 2 with coefficients
in Z[x]. As such, its roots would be in some sort of integral
extension (this is a technical term you may look up) of Z[x] living
within a field that is an extension of degree 2 over Z(x). It is not
an algebraic integer. I am unfamiliar with the factorization
[/quote:9b3d5451f3]
That is false. Given an algebraic integer x the roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
are always algebraic integers.
So readers keeping count have one obvious error from the poster to
start.
[quote:9b3d5451f3]properties of this particular extension of Z[x], but I have no reason
to suppose that it is a unique factorization domain, nor do I have any
[/quote:9b3d5451f3]
That is irrelevant. There is no reason to wonder or care if it is a
unique factorization domain.
Readers have a second error.
[quote:9b3d5451f3]idea what the structure of its unit group might be. These are
interesting questions, but I don't have the time and energy right no,
if ever, to try and figure this out.
[/quote:9b3d5451f3]
There is nothing to figure out at this point. Algebraically I've
given a factorization. At best you might determine if it is correct--
it is--but it's not something that should involve a serious amount of
time and energy.
Third error.
[quote:9b3d5451f3]The primary problem here is that if you let the ring be the ring of
algebraic integers, you get something never before seen in human
mathematics which is a constraint on a constant factor revealed on the
right hand side of
7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
which is obviously not there on the left hand side.
I simply don't know what you mean by this. As previously mentioned,
[/quote:9b3d5451f3]
Fourth error. Clearly the 7 is just some constant multiplied by the
polynomial. It should be trivially removable, and that should be
evident on both sides.
(Um, I made a mistake here in the original post! Reverse what I said:
the 7 is clearly unconstrained on the left hand side, but not so
clearly free on the right hand side.)
[quote:9b3d5451f3]a_1 and a_2 are NOT algebraic integers. If you do out the solution of
[/quote:9b3d5451f3]
Repeat of the original mistake as they are algebraic integers when x
is an algebraic integer, trivially. I don't know if that should count
as a fifth error or just a repeat of an earlier one, but then yes,
such a repeat is an error.
So 5 mathematical errors so far from the poster.
[quote:9b3d5451f3]the quadratic equation you presented, getting a bunch of square roots
of polynomial expressions involving x, you will probably, with perhaps
some effort, be able to see that the two sides are obviously equal,
[/quote:9b3d5451f3]
There is no effort--you just "plug and chug". Possibly the poster
wishes to now cast doubt on the original factorization?
Sixth error.
[quote:9b3d5451f3]assuming you have done the calculation correctly. My real point is
that, since you are not dealing in the ring of algebraic integers when
you use this expression, you can't pull out anything very direct that
[/quote:9b3d5451f3]
Seventh error. Repeat of earlier mistake that the ring cannot be the
ring of algebraic integers.
[quote:9b3d5451f3]you can say about the algebraic integers. The fact that you are
getting confused here and have done so now for a very long time does
not increase the respect you get from the mathematical community.
[/quote:9b3d5451f3]
Non sequitur. And the poster after making 7 mathematical errors in
his reply digresses to make an insult which is in context, ironic.
[quote:9b3d5451f3]To understand what I mean--I'm sure some posters would reply that they
find it incomprehensible--consider a simpler example in the ring of
integers:
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
where the 7 is still unconstrained, and you can in fact move it around
at will, so
I have no idea what an unconstrained 7 might be. I don't even know
[/quote:9b3d5451f3]
Eighth error.
I always find it remarkable when my attempts at simple explanation are
simply denied as easily as someone claiming to be confused by 2+2 = 4,
but it has happened so many times over the years that I'm no longer
very surprised at seeing it.
You see, if no explanation is ever simple enough then the grad student
or math professor whose professional research is torn into shreds by
the devastating core error can pretend to himself that it still is
intact.
You see, they may rationalize that all they have to do is keep
claiming that it doesn't make sense, no matter how often it's
explained or how simply.
I still think that undergrads who have NOT built a body of supposed
success on the core error are the ones who will end the charade if
only when they find themselves unable to generate the energy necessary
to learn bogus math--no matter how much their professors and grad
students beg and plead.
And they may. On the darker side of this human drama some of you may
see behavior from people you respected that troubles you for the rest
of your lives, as they may beg, and plead, and then threaten, or even
attack.
Their LIVES are what they are defending in their own minds. To some
of them, forcing them to see the truth is like taking something away
from them, and they may behave as if they were physically threatened.
James Harris |
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| fishfry... |
| Posted: Sat Jul 04, 2009 9:12 pm |
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Guest
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In article
<390f8145-82ca-45f6-8bcb-a79711bb8bd2 at (no spam) g1g2000pra.googlegroups.com>,
JSH <jstevh at (no spam) gmail.com> wrote:
[quote:07459a9eef]Turns out that if you follow rigorous mathematics it is trivial to
show a problem with the use of the ring of algebraic integers with a
quadratic factorization that I've often given before:
[/quote:07459a9eef]
Trivial though it may be, it's now been a couple of years and you still
have not said what the problem is. You have never said, "This shows that
the algebraic integers have property X, where X is impossible," or some
such. You've never done it.
Can you state clearly what the problem is? |
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| JSH... |
| Posted: Sun Jul 05, 2009 8:52 am |
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On Jul 5, 12:13 am, Bacle <ba... at (no spam) yahoo.com> wrote:
[quote:346bd981c1]On Jul 4, 7:56 pm, "Achava Nakhash, the Loving Snake"
ach... at (no spam) hotmail.com> wrote:
James,
I realize that this is an exercise in futility, but
I am going to try
to demonstrate to you why the mathematical
community does not take
anything you say seriously. Many of us do,
however, find you
I've read your reply and it fits more with the
psychological aspects
of this issue.
Readers who are puzzled by his reply need to
understand how
devastating the core error is.
You mean that you decided to spend 15 years of your
life in something without getting any useful return?.
I would be devastated too.
[/quote:346bd981c1]
Sounds like you think a "useful return" is acceptance and accolades?
Maybe money? As well?
Other readers, see where I'm going here?
These people are NOT mathematicians. They want money and power.
To them, without those things there is no success!
[quote:346bd981c1]He may actually believe he has cogent rebuttals, but
he also may be
faking in case later he believes he needs to say his
mind snapped in
his own defense.
You, OTOH can just say that you're stupid in your
own defense. Everyone will believe you.
[/quote:346bd981c1]
For someone who makes a living doing mathematics if later there is a
question about whether or not they were perpetuating a fraud the
answer can mean their livelihood.
For you this may be just a game, but for some of these people it could
mean losing their jobs, and having to start their careers over
completely, possibly with a fraud conviction hanging over them.
Certain entities like the US Government do take academic fraud
seriously. The repercussions could be quite significant.
An insanity defense in the face of prosecution might seem like an
option to some.
[quote:346bd981c1]marvelously entertaining. As I remarked several
times already, you
write very well, and you spin an absorbing, if
silly, science fiction
conspiracy involving mathematicians and, in
particular, number
theorists. Never before have we been considered so
powerful. We
certainly don't feel that way.
I will intersperse my specific comments withing
your text, where
appropriate, to try to show you how mathematicians
view the
mathematics of what you do. I and others have done
this before to
little effect, but I happen to be in the mood.
Beethoven's fifth is
playing in my speakers, so I guess I just feel
empowered or something.
On Jul 4, 6:41 pm, JSH <jst... at (no spam) gmail.com> wrote:
Turns out that if you follow rigorous mathematics
it is trivial to
show a problem with the use of the ring of
algebraic integers with a
quadratic factorization that I've often given
before:
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.
This appears to be a monic polynomial of degree 2
with coefficients
in Z[x]. As such, its roots would be in some sort
of integral
extension (this is a technical term you may look
up) of Z[x] living
within a field that is an extension of degree 2
over Z(x). It is not
an algebraic integer. I am unfamiliar with the
factorization
That is false. Given an algebraic integer x the
roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
are always algebraic integers.
Convince me. It is not even clear if you intend
[/quote:346bd981c1]
It's math. It's not about what you personally believe!!!
Those who wonder how YEARS can go by with stupid arguments about basic
math need only look at that request to understand the Hell these
people have put me through, as they keep asking to be convinced of
basic mathematics, over and over and over and over and over and over
and over again.
It is a deliberate frustration strategy in my opinion as they are con
artists. Remember this guys lust for money and power revealed in his
notions of success above to which I replied.
To such a person, truth is just a word. Success is getting something
from people and his demand is for me to convince HIM, as if
mathematical correctness is just such trash.
For such a person, public opinion is all that matters as THAT is how
you get money and power.
[quote:346bd981c1] 7x-1 to be taken as a polynomial, or if it will
assume numerical values. I think Achava was assuming
[/quote:346bd981c1]
That's not a mathematical statement. Seems to be something you just
made up.
a^2 - (7x-1)a + (49x^2 - 14x) = 0
IS in the ring of algebraic integers if that is the declared ring.
The error by the earlier poster in claiming it is not is not fixed by
your support!!!
People like you seem insistent in a democratic process against
mathematics.
[quote:346bd981c1] the first case. Then he was right. So you should
[/quote:346bd981c1]
He was wrong. And you are wrong in reply.
It doesn't matter how many people like you reply claiming he was
right. He was wrong.
[quote:346bd981c1] specify which one you are working with.
This shows you lack basic understanding. I am
not by any means a number theorist and I can see
how this could be confusing. This shows you are
[/quote:346bd981c1]
a^2 - (7x-1)a + (49x^2 - 14x) = 0
If you say ring of algebraic integers there is NO PROBLEM with that
expression.
And there is NO PROBLEM with the rest either:
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.
[quote:346bd981c1] not too sharp nor knowledgeable.
deleted[/quote:346bd981c1]
That insistence on insults is so telling, and it is amazing how
solidly people who rely on insults sit on the sci.math newsgroup.
Year after year after year I've refuted people on basic mathematical
points, and they will just repeat the error year after year after year
and rely on the group opinion.
But you're doing fake math!!!
You're STILL wasting your lives and energy. So what? You can squeak
out a few years with the world unaware?
YOUR LIFE WAS WASTED ON FAKE WORK regardless.
Your life was wasted on fake work.
Your people lie about mathematics. When error is revealed you simply
repeat the error and gang up against the mathematics that shows your
error--and then you lie about what you are doing.
I think some of you have some weird fantasy that you can break the
entire human race, get over some hump, and have the mathematical
errors so entrenched in human belief that NO one can ever break it
out--and then you think you win?
By betraying the human race.
Your win is the betrayal of the human race, the human species.
You are anti-people.
James Harris |
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| Fishcake... |
| Posted: Sun Jul 05, 2009 9:59 am |
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[quote:9eb7e41c51]
 7x-1 to be taken as a polynomial, or if it will
 assume numerical values. I think Achava was
assuming
That's not a mathematical statement. Seems to be
something you just
made up.
[/quote:9eb7e41c51]
You do not have the ability nor the authority to decide what a "mathematical" statement is.
[quote:9eb7e41c51]
a^2 - (7x-1)a + (49x^2 - 14x) = 0
IS in the ring of algebraic integers if that is the
declared ring.
[/quote:9eb7e41c51]
What?
[quote:9eb7e41c51]
The error by the earlier poster in claiming it is not
is not fixed by
your support!!!
[/quote:9eb7e41c51]
Are you treating x as an element of Z[X]? Or are you treating x as an arbitrary algebraic integer?
Bacle clearly asks you of this, but instead, you start ranting.
[quote:9eb7e41c51]
People like you seem insistent in a democratic
process against
mathematics.
 the first case. Then he was right. So you should
He was wrong. And you are wrong in reply.
It doesn't matter how many people like you reply
claiming he was
right. He was wrong.
[/quote:9eb7e41c51]
No. You aren't being precise.
[quote:9eb7e41c51]
 specify which one you are working with.
  This shows you lack basic understanding. I am
 not by any means a number theorist and I can see
 how this could be confusing. This shows you are
a^2 - (7x-1)a + (49x^2 - 14x) = 0
If you say ring of algebraic integers there is NO
PROBLEM with that
expression.
And there is NO PROBLEM with the rest either:
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.
[/quote:9eb7e41c51]
This looks like a typo.
[quote:9eb7e41c51]
 not too sharp nor knowledgeable.
deleted
That insistence on insults is so telling, and it is
amazing how
solidly people who rely on insults sit on the
sci.math newsgroup.
[/quote:9eb7e41c51]
Yes, namely YOU.
[quote:9eb7e41c51]
Year after year after year I've refuted people on
basic mathematical
points, and they will just repeat the error year
after year after year
and rely on the group opinion.
[/quote:9eb7e41c51]
I only recently joined this forum, but I can see from a few of your posts that you don't have a clear understanding of what you are talking about. So by resorting to abusive anguage you try to win arguments.
[quote:9eb7e41c51]
But you're doing fake math!!!
You're STILL wasting your lives and energy. So what?
You can squeak
out a few years with the world unaware?
YOUR LIFE WAS WASTED ON FAKE WORK regardless.
Your life was wasted on fake work.
Your people lie about mathematics. When error is
revealed you simply
repeat the error and gang up against the mathematics
that shows your
error--and then you lie about what you are doing.
I think some of you have some weird fantasy that you
can break the
entire human race, get over some hump, and have the
mathematical
errors so entrenched in human belief that NO one can
ever break it
out--and then you think you win?
By betraying the human race.
Your win is the betrayal of the human race, the human
species.
You are anti-people.
James Harris
[/quote:9eb7e41c51]
If you have nothing better to do, then read a book. |
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| JSH... |
| Posted: Sun Jul 05, 2009 10:18 am |
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Guest
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On Jul 5, 12:33 pm, Bacle <ba... at (no spam) yahoo.com> wrote:
[quote:b87215e882]On Jul 5, 12:13 am, Bacle <ba... at (no spam) yahoo.com> wrote:
On Jul 4, 7:56 pm, "Achava Nakhash, the Loving
Snake"
ach... at (no spam) hotmail.com> wrote:
James,
I realize that this is an exercise in futility,
but
I am going to try
to demonstrate to you why the mathematical
community does not take
anything you say seriously. Many of us do,
however, find you
I've read your reply and it fits more with the
psychological aspects
of this issue.
Readers who are puzzled by his reply need to
understand how
devastating the core error is.
You mean that you decided to spend 15 years of
your
life in something without getting any useful
return?.
I would be devastated too.
Sounds like you think a "useful return" is acceptance
and accolades?
Maybe money? As well?
Other readers, see where I'm going here?
Not really. No one I know has sided with you. But
you will surely use your circular logic to make this
a sign you are right.
There are hundreds of journals to submit to, yet
you don't do it.
Other readers see where I am going?
These people are NOT mathematicians. They want money
and power.
If they did and thought you were right, they would
have stolen your work. Then there are two choices:
i)They don't think you are right, and did not take
and therefore submit your work.
ii)They are power-hungry, and submitted your work as theirs.
But no one has then accepted their (and therefore your)
work as valid once it was submitted.
To them, without those things there is no success!
Here go the exclamation points again:
Top of the Muffin to You!!
He may actually believe he has cogent rebuttals,
but
he also may be
faking in case later he believes he needs to say
his
mind snapped in
his own defense.
Ever heard of crying wolf. I saw some of your old
posts and you have been predicting this for very long.
It is tiresome. And a lot of posts really convince me
that you are not in good mental health: prophecies
of doom, for once, talk about cannibalism, belief
that you are of some superior species, etc.
Seriously, I am not mocking you, get some therapy.
You, OTOH can just say that you're stupid in your
own defense. Everyone will believe you.
For someone who makes a living doing mathematics if
later there is a
question about whether or not they were perpetuating
a fraud the
answer can mean their livelihood.
Cry wolf. Submit a paper or stop whining, please.
For you this may be just a game, but for some of
these people it could
mean losing their jobs, and having to start their
careers over
completely, possibly with a fraud conviction hanging
over them.
And it could mean you being a nothing, which you
seem unable to accept.
Certain entities like the US Government do take
academic fraud
seriously. The repercussions could be quite
significant.
Same goes for the threats you have issued to people.
An insanity defense in the face of prosecution might
seem like an
option to some.
Same for your insanity.
marvelously entertaining. As I remarked
several
times already, you
write very well, and you spin an absorbing, if
silly, science fiction
conspiracy involving mathematicians and, in
particular, number
theorists. Never before have we been
considered so
powerful. We
certainly don't feel that way.
I will intersperse my specific comments withing
your text, where
appropriate, to try to show you how
mathematicians
view the
mathematics of what you do. I and others have
done
this before to
little effect, but I happen to be in the mood.
Beethoven's fifth is
playing in my speakers, so I guess I just feel
empowered or something.
On Jul 4, 6:41 pm, JSH <jst... at (no spam) gmail.com
wrote:
Turns out that if you follow rigorous
mathematics
it is trivial to
show a problem with the use of the ring of
algebraic integers with a
quadratic factorization that I've often given
before:
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.
This appears to be a monic polynomial of degree
2
with coefficients
in Z[x]. As such, its roots would be in some
sort
of integral
extension (this is a technical term you may
look
up) of Z[x] living
within a field that is an extension of degree 2
over Z(x). It is not
an algebraic integer. I am unfamiliar with the
factorization
That is false. Given an algebraic integer x the
roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
are always algebraic integers.
Convince me. It is not even clear if you intend
It's math. It's not about what you personally
believe!!!
Those who wonder how YEARS can go by with stupid
arguments about basic
math need only look at that request to understand the
Hell these
people have put me through, as they keep asking to be
convinced of
basic mathematics, over and over and over and over
and over and over
and over again.
It is a deliberate frustration strategy in my opinion
as they are con
artists. Remember this guys lust for money and power
revealed in his
notions of success above to which I replied.
To such a person, truth is just a word. Success is
getting something
from people and his demand is for me to convince HIM,
as if
mathematical correctness is just such trash.
For such a person, public opinion is all that matters
as THAT is how
you get money and power.
7x-1 to be taken as a polynomial, or if it will
assume numerical values. I think Achava was
assuming
That's not a mathematical statement. Seems to be
something you just
made up.
a^2 - (7x-1)a + (49x^2 - 14x) = 0
IS in the ring of algebraic integers if that is the
declared ring.
The error by the earlier poster in claiming it is not
is not fixed by
your support!!!
People like you seem insistent in a democratic
process against
mathematics.
No, you imbecile, 7x-1 is a _polynomial_ , an
algebraic integer is a complex solution to a poly
[/quote:b87215e882]
An algebraic integer is, of course, in the field of complex numbers
but it can be an integer.
It's not part of the definition that it be a complex solution.
You made that up.
[quote:b87215e882] with _integer coefficients_. Do you get that?.
Do you not get that 7x-1 is not an integer unless
you specify an assignment to x?.
[/quote:b87215e882]
If the declared ring is the ring of algebraic integers then x is an
algebraic integer, and yes, for reasons that escape me, some
mathematicians insist on then calling it an "integer", but the earlier
poster is still incorrect as with the ring declared it must be true
that
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
exists in the ring of algebraic integers without error.
What's of interest to other readers in such arguments is that in this
case at least other posters long ago agreed on this point.
It allows you to see how many of you live in your own little world
when it comes to what is "mathematically correct" where a lot of that
seems to be about opinion and perception of group agreement.
Like, I doubt the poster would still be debating here if he knew how
stupid it was to debate a point already generally agreed as correct on
the sci.math newsgroup so he DOES NOT KNOW THAT so he is instrumental
in highlighting the kind of deficient conversations I end up facing
over and over and over and over and over and over and over and over
again.
You people make up rules, and then break them, but mostly you just
disagree. And you rely on group opinion so mathematics is a moving
target with you.
And THAT is sad.
James Harris |
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| amzoti... |
| Posted: Sun Jul 05, 2009 11:38 am |
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On Jul 5, 1:18 pm, JSH <jst... at (no spam) gmail.com> wrote:
[quote:99bd199f07]And THAT is sad.
James Harris
[/quote:99bd199f07]
Your life is sad.
The sad truth about the welfare state and what it creates.
You are a cheat, a liar and a charlatan.
Making up your own truth to justify your lack in ability.
Then, blaming the rest of the world for your sad life.
Look at how hard you project your life's over and over again.
Look at how I puff and I puff and I'll blow your establishment down.
Yawn!
You have been saying that for over 15 years now and we still continue
getting fat with all of the tax dollars, research grants and corporate
sponsorship.
We will continue to silence you so the world of mathematicians can
continue the massive charade.
Doesn't that all sound logical?
You are a nothing more than a spoiled and misguided delusional
narcissist!
Have a nice day! |
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| amzoti... |
| Posted: Sun Jul 05, 2009 11:38 am |
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Guest
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On Jul 5, 1:18 pm, JSH <jst... at (no spam) gmail.com> wrote:
[quote:020e8caf7d]And THAT is sad.
James Harris
[/quote:020e8caf7d]
Your life is sad.
The sad truth about the welfare state and what it creates.
You are a cheat, a liar and a charlatan.
Making up your own truth to justify your lack in ability.
Then, blaming the rest of the world for your sad life.
Look at how hard you project your life's over and over again.
Look at how I puff and I puff and I'll blow your establishment down.
Yawn!
You have been saying that for over 15 years now and we still continue
getting fat with all of the tax dollars, research grants and corporate
sponsorship.
We will continue to silence you so the world of mathematicians can
continue the massive charade.
Doesn't that all sound logical?
You are a nothing more than a spoiled and misguided delusional
narcissist!
Have a nice day! |
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| Junoexpress... |
| Posted: Sun Jul 05, 2009 5:53 pm |
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Guest
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On Jul 4, 9:41 pm, JSH <jst... at (no spam) gmail.com> wrote:
[quote:93580ae3a7]
With errors in mathematics you can get this odd ability to appear to
prove just about anything.
James Harris
[/quote:93580ae3a7]
Produce just ONE example. If the results are *so* catastrophic, as you
claim they are, this should be simple to do.
OW, who cares one way or the other?
M |
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