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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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 Posted: Sun May 21, 2006 2:01 am Post subject:
JSH: Lying about the distributive property
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What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a
group you multiply the elements within that group:
a*(b + c) = a*b + a*c
and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
and here is where it gets really bizarre, as my position that the value
of the function has no impact on the distributive property is now key,
so I say that if at x=0, f(x)=0, then that is as valid a point as any
other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of
the distributive property.
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
where it's still just the distributive property, but with more stuff,
and now, like before, I'm going to put in functions, but then I have to
be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
because one of the functions just swallowed the 7 so that it is now
invisible.
How can a function do that? Easy. It's a function, so it can be
f(x) = 7*h(x)
so the function swallowing the visibility of the 7 is not a big deal,
but maybe though, it didn't so I add the rule that at x=0, f(x)=0 and
g(x) = 0, so that I can SEE what is going on at a particular value.
If you had some other convenient value, that would be ok as well.
7*(h(0) + b)*(c + d) = (0 + 7*b)*(0 + d)
and it's clear that the 7 multiplied through like with
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
while the functions just make things a little more complicated to
verify, but not impossible.
Now then, by the logical point that the value of the function does not
change the distributive property, I know what happened for ANY x, but
posters in fighting this argument have proclaimed x=0 to be a "special
case", defying the reality of how the distributive property operates.
I've explained, and explained, and explained so that the best
conclusion is that posters lie about this argument.
Otherwise they can't understand the basic principle that the value of
the function does not change the distributive property, which is a
major stretch.
Why is it such a big deal?
Because once the principle is established, I can get some complicated
functions that show a problem with the ring of algebraic integers.
Posters in defying what is mathematically correct are just slashing at
what they can, and in this case, that means questioning the
distributive property, and then claiming they are not doing so, while
they claim x=0 is a special case, but if I push them on the point that
the value of functions does not affect the distributive property, they
claim they don't disagree!
It's a case of where the lies just keep coming and it shows you how to
defy a mathematical proof.
Just claim it's wrong, keep claiming it's wrong, and get enough people
to claim it's wrong so that no one believes that it's correct.
And doing that you can block acceptance of mathematical proof.
These people are undermining the discipline of mathematics by showing
its true fragility.
It has few defenses against dedicated group lying about mathematical
arguments.
I mean, come on! The distributive property! How could people get away
with lying about that?
But they have now, for years.
James Harris |
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Mike Amling
science forum Guru
Joined: 05 May 2005
Posts: 525
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| Posted: Sun May 21, 2006 2:59 am Post subject:
Re: Lying about the distributive property
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<jstevh@msn.com> wrote in message
news:1148176879.874552.177500@j33g2000cwa.googlegroups.com...
| Quote: | What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a
group you multiply the elements within that group:
a*(b + c) = a*b + a*c
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you are misusing the term "group"
http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut8_property.htm
"Distributive Properties"
a(b + c) = ab + ac
or
(b + c)a = ba + ca
| Quote: |
and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
and here is where it gets really bizarre, as my position that the value
of the function has no impact on the distributive property is now key,
so I say that if at x=0, f(x)=0, then that is as valid a point as any
other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of
the distributive property.
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everybody already knows that, the ones that took highschool algebra.
| Quote: |
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
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no, it is 7*(ac+bd+bc+bd)
| Quote: |
where it's still just the distributive property, but with more stuff,
and now, like before, I'm going to put in functions, but then I have to
be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
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Wrong. you are leaving out information that is always add next to the
equation
"Where a = h(x), 7*a = f(x), and c = g(x)"
(Therefore 7* h(x) = f(x))
| Quote: |
because one of the functions just swallowed the 7 so that it is now
invisible.
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no swallowing, no invisable, just algebra !
| Quote: |
How can a function do that? Easy. It's a function, so it can be
f(x) = 7*h(x)
so the function swallowing the visibility of the 7 is not a big deal,
but maybe though, it didn't so I add the rule that at x=0, f(x)=0 and
g(x) = 0, so that I can SEE what is going on at a particular value.
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trivial.
| Quote: |
If you had some other convenient value, that would be ok as well.
7*(h(0) + b)*(c + d) = (0 + 7*b)*(0 + d)
and it's clear that the 7 multiplied through like with
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What does "multiplied through" mean ?
| Quote: |
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
while the functions just make things a little more complicated to
verify, but not impossible.
Now then, by the logical point that the value of the function does not
change the distributive property, I know what happened for ANY x, but
posters in fighting this argument have proclaimed x=0 to be a "special
case", defying the reality of how the distributive property operates.
I've explained, and explained, and explained so that the best
conclusion is that posters lie about this argument.
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bullpuppy, stir your own soup, nobody is going to lie about a trivial
discovery that you have on the distributive property.
It should be highly embarrising to post such an artical on such a trivial
matter on the internet.
| Quote: |
Otherwise they can't understand the basic principle that the value of
the function does not change the distributive property, which is a
major stretch.
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4 u it is.
go get a highschool book on maths and read it, or eat it, perhaps you will
absorb it better with the latter.
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Abstract Dissonance
science forum Guru Wannabe
Joined: 29 Dec 2005
Posts: 201
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| Posted: Sun May 21, 2006 5:09 am Post subject:
Re: Lying about the distributive property
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<jstevh@msn.com> wrote in message
news:1148176879.874552.177500@j33g2000cwa.googlegroups.com...
| Quote: | What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a
group you multiply the elements within that group:
a*(b + c) = a*b + a*c
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ok.
| Quote: | and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
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Sure, Real valued functions are just real numbers too... in fact, they are
just a mapping of something to the real numbers and hence you can do algebra
on them also.
| Quote: |
and here is where it gets really bizarre, as my position that the value
of the function has no impact on the distributive property is now key,
so I say that if at x=0, f(x)=0, then that is as valid a point as any
other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of
the distributive property.
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Well, technically the function has to have those properties defined on it...
if the function does not exact at those points then you can run into
trouble... If, say, f(x) exists for all x in some domain and you restrict
yourself to working in that domain then there is no problem because by
virtue of f(x) being a real number it will have all the properties of real
numbers.
| Quote: | Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
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and?
| Quote: | where it's still just the distributive property, but with more stuff,
and now, like before, I'm going to put in functions, but then I have to
be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
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ok?
| Quote: | because one of the functions just swallowed the 7 so that it is now
invisible.
How can a function do that? Easy. It's a function, so it can be
f(x) = 7*h(x)
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ok... not sure what the technical definition of swallowed is but I see what
you are doing.
| Quote: | so the function swallowing the visibility of the 7 is not a big deal,
but maybe though, it didn't so I add the rule that at x=0, f(x)=0 and
g(x) = 0, so that I can SEE what is going on at a particular value.
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hmm... You can't just arbitrarily place a condition on f(x) now... YOU HAVE
DEFINED f(x) TO BE 7*h(x)... f(x) cannot now be 0 too when x = 0...
because!!!!!!!!!!!!!
f(x) ======== 7*h(x)
what if h(x) = exp(x)+ cos(3x)*zeta(2/(x+1))
but h(0) != 0
but you have said f(0) = 0
so you are saying equivilently
0 = f(0) = 7*h(0) = 7*(exp(0) + cos(3*0)*zeta(2)) = 7*(1 + pi^2/6)
but surely 7 + 7*pi^2/6 != 0?
Why is this wrong?
because you have first said
f(x) = 7*h(x)
then said
f(0) = 0
and these two conditions are basicaly defining an object(f(x) here) in two
different ways... you can't do this in math.
now, it might work in some cases like when h(x) = x or h(x) = exp(x) - 1
but its only by coincidence.
Now if you continue to do math with f(x) you are actually violating basic
rules of logic because f(x) is not 7*h(x) but something potentially totally
different. You can end up with any result you please by doing "math" like
this.
| Quote: | If you had some other convenient value, that would be ok as well.
7*(h(0) + b)*(c + d) = (0 + 7*b)*(0 + d)
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nope. You are making a basic logical fallacy.
You first DEFINE f(x) := 7*h(x). (The := means defined as... not equal to
but in essence f(x) is just a symbolic reference to 7*h(x)).
Now you DEFINE f(0) := 0
then you say that since f(x) := 0 and f(x) := 7*h(x) THEN h(0) = 0. This is
not correct and you should figure out why before you start going on about
how all of mathematics is against you... its your fault for making a simple
logical fallacy. You really need to spend time to figure it out.
Your problem seems to stem from the fact that you don't understand what a
function is and how they are used. I'd suggest reading a simple book ok
pre-calculus and work through it very slowly.
It is true that the algebra of functions parallel the algebra of real
numbers... but this is only because there is a deeper structural similarity
between the two. But you problem has nothing to do with the algebra of
functions but has to do with the logical mechanics of mathematics.
Everyone knows that if you define an apple to be a red gorilla you can't
later say that an apple is now defined as a blue ape then say "hence a red
gorilla is a blue ape".
logically it is a fallacy to say
if A->B
and B
then A
Which is in essence what you are doing. A lot of people end up using this
fallacy and it even happens a lot in mathematics from beginners.
| Quote: | and it's clear that the 7 multiplied through like with
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
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ok.
| Quote: | while the functions just make things a little more complicated to
verify, but not impossible.
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not really. For all you know, a,b,c,d are functions. Real valued functions
are REAL VALUED... MEAING THEY BEHAVE JUST LIKE REAL NUMBERS... IN FACT,
THEY ARE REAL NUMBERS! THEY JUST HAPPENT TO BE DEFINED IN A LITTLE
DIFFERENT WAY.
Its similar to how all real valued functions are complex valued functions
to. We could define every real number as a constant valued real function.
hence
3 is really
f(x) := 3 for all x.
later on I might "redefine" f(x) such as
f(x) := 8 for all x
but this doesn't mean
3 = 8
but this is exactly the logic you have used above.
| Quote: | Now then, by the logical point that the value of the function does not
change the distributive property, I know what happened for ANY x, but
posters in fighting this argument have proclaimed x=0 to be a "special
case", defying the reality of how the distributive property operates.
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well, x=0 is not a special case and you have probably misinterpreted what
they were saying.
| Quote: | I've explained, and explained, and explained so that the best
conclusion is that posters lie about this argument.
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Yes... and you still insist everyone else is crazy.
| Quote: | Otherwise they can't understand the basic principle that the value of
the function does not change the distributive property, which is a
major stretch.
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It doesn't.
| Quote: | Why is it such a big deal?
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The big deal is that you are making basic logical fallacies but instead of
trying to figure out your mistake you just call everyone else stupid.
| Quote: | Because once the principle is established, I can get some complicated
functions that show a problem with the ring of algebraic integers.
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Not if your using the logic above. You can get any results you want by
making a mathematical error. You can even prove 0 = 1... which I suggest you
look at and make sure you understand the problem and the fallacy that
occurs.... you might then understand that you are making a mistake too.
(notice that one you prove 0 = 1 you can prove anything = anything else...
fish*0 = fish*1 ==> 0 = fish, i93jf0920*0 = i93jf0920*1 => 0 = i93jf0920
hence
fish = i93jf0920
and so on and so forth,
(James Harris)*0 = (James Harris)*1 ==> 0 = James Harris
Crazy*0 = Crazy*1 ==> 0 = Crazy
hence
James Harris = Crazy
Hence assuming a false statement allows you to prove things that are
false... it also allows you to prove things are true... the two examples
above demonstrate this respectively.
| Quote: | Posters in defying what is mathematically correct are just slashing at
what they can, and in this case, that means questioning the
distributive property, and then claiming they are not doing so, while
they claim x=0 is a special case, but if I push them on the point that
the value of functions does not affect the distributive property, they
claim they don't disagree!
It's a case of where the lies just keep coming and it shows you how to
defy a mathematical proof.
Just claim it's wrong, keep claiming it's wrong, and get enough people
to claim it's wrong so that no one believes that it's correct.
And doing that you can block acceptance of mathematical proof.
These people are undermining the discipline of mathematics by showing
its true fragility.
It has few defenses against dedicated group lying about mathematical
arguments.
I mean, come on! The distributive property! How could people get away
with lying about that?
But they have now, for years.
|
To bad you won't stop and listen to any one of them and try to learn your
mistakes and possibly actually contribute something to mathematics. If you
actually spent all this time learning about mathematics you might have a phd
by now... instead you insist that everyone else is stupid and your a genius
and they just can't understand your super human abilities.
One thing I learned long ago in life is that sometimes I'm wrong... infact I
was wrong a lot. What I started doing was always assuming I was wrong. By
doing that I ended up being right much more often. Ok, not exactly like that
but basicaly the point is that if you always believe you are right then you
will never look for other solutions.
if you believe x = 8 solves the equation x^2 - 4 = 0 then you will never
look for any other solutions and you will go along in your life being wrong.
Now maybe x = 8 does solve it but if you can't directly test it as in this
case its best to assume it just might be wrong and spend some time searching
for another way... maybe you will stumble on the quadratic formula and be
able to prove mathematically that it always gives the right answer instead
of having to guess. Maybe you will make some logical mistakes... but if you
keep on convincing yourself that you are always right you soon will ingrain
those logical mistakes into your head and never be able to get rid of
them... eventually those fallacies will not be fallacies for you but
logical...
The problem is that logic is not necessarily a universal property. Logic is
a democracy and if you are not in the majority then you are crazy...
regardless if you think you are logical or not. Now the thing about logic
and math is that for all pratical and theoretical purposes it is not
democracy but a natural law of nature. Maybe tomarrow everone will think
like you and think that 0 = 1 makes complete logical sense.
But for now you are in the minority and so you need to STFU because you are
the crazy one. If you actually have some desire to do real math then you
should listen to people here and assume you are not the genius you think you
are... if you really are a genius then chances are you wouldn't be having
all these problems. Chances are that if everyone is saying you are wrong
then you are wrong... instead of ignoring the problem of finding out why you
are wrong you might actually listen to people... and if its confusing then
take a break and some back later.
I really hope you will listen to some of the things I have to say. I am not
a mathematician(I do have a degree in math but that was a few years ago and
I've forgotten some stuff). I also have absolutely no reason to conspire
against you and it would in no way benefit me in any pratical way to do so.
Now maybe I have so psychological reason to do so but thats your choice to
believe. If you think every person has some psychological problem then
chances are you are the actual one with the problem.
Now, it might be true that there are some mathematicians out there who are
arrogant and some that thing they know everything... some might even be
"cooks" themselfs(not in the sense of a real cook but just that they are not
that good). But any sane person would look at the statistical nature and
realize that if EVERY SINGLE PERSON is saying your nuts then chances are you
are.
It would be nice if you would come back to reality and maybe do some real
math.
For example, I have spend many years working on a single problem and every
thing I tried did not work. Sometimes I might get a little farther down the
road but eventually I end up runinng into circular reasoning... something
like end up proving 1 = 1. Its not that mathematicians are conspiring
against me or that mathematics itself has problems but just that I haven't
found the right way.
One thing I don't do is make up mathematics to get results. But this is
what you are basicaly doing.
for example,
Sometimes I work on a problem and I end up making a logical mistake similar
to what you have done but I get some very good results.... I get all happy
that I might have broke through a barrier only to go back over my results to
realize I ended up writing a 1 for a 2 which caused everything to work
out... but when it was 2 it didn't lead me anywhere(maybe 1 = 1). Almost
always I end up checking my results in some numerical way and end up
realizing a mistake was made somewhere then I go and redo the proof then
realize what I did wrong.
One way to test your results and see if they hold any water is to put them
to use. All mathematical endeavors must have some way to test them(although
some might be extremly difficult or intractable). If, say, you find some
super fast factoring method, then just use it to factor several easy cases.
If it works then you might have something or you just got lucky. If it
works then you will go and reprove it from scratch and try to clear things
up and make a more elegant proof. If it doesn't work then YOU MADE A MISTAKE
SOMEWHERE. Even if it does work you might have made a mistake(like the
Wiles FLT proof). Maybe the mistake is not significant to the problem and
you can fix it easily and the proof will work... other times your results
directly depedended on that mistake and you gotta start over.
If you truely want to get somewhere and are not just trolling this NG for
attention then you must realize that you are making a simple logical mistake
and instead of going crazy about it you need to understand the mistake and
move on to hopefully bigger and better things.. Else you will be stuck in
this NG for the next 10 years doing the same s**t and never making any real
progress.
Anyways, I'm tired and my head hurts ;/
Jon |
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José Carlos Santos
science forum Guru
Joined: 25 Mar 2005
Posts: 1111
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| Posted: Sun May 21, 2006 6:24 am Post subject:
Re: Lying about the distributive property
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Juke wrote:
| Quote: | Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
no, it is 7*(ac+bd+bc+bd)
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What do you mean by that? First of all, you should have written
ac + bd + bc + ad instead. But are you denying that
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)?
It sure looks like.
Best regards,
Jose Carlos Santos |
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Gene Ward Smith
science forum Guru
Joined: 08 Jul 2005
Posts: 409
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| Posted: Sun May 21, 2006 6:27 am Post subject:
Re: Lying about the distributive property
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José Carlos Santos wrote:
| Quote: | Juke wrote:
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
no, it is 7*(ac+bd+bc+bd)
What do you mean by that? First of all, you should have written
ac + bd + bc + ad instead. But are you denying that
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)?
It sure looks like.
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I can't figure out what the "lie" is supposed to be, Harris just seems
to be saying trivial stuff like the above. |
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Hero
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 220
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| Posted: Sun May 21, 2006 8:50 am Post subject:
Re: Lying about the distributive property
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Abstract Dissonance schrieb:
| Quote: | ....
One thing I don't do is make up mathematics to get results. But this is
what you are basicaly doing.
for example,
Sometimes I work on a problem and I end up making a logical mistake similar
to what you have done but I get some very good results.... I get all happy
that I might have broke through a barrier.....
That's the point. I ask myself, why it makes one happy? |
Because one get's admiration of others, one earns respect and/or
money?- Look at math history: only very few of the real barrier-brakers
got their reward from others. And to get it depends on political,
personal and many other reasons and most of these guys weren't good in
promoting themselfes. In sci.math You'll get the response, when You
make mistakes, but with new results it seldom will be more than silent
attention.
Because one can see now, what is behind the barrier - something, which
was impossible to oneself before. It's like feeling the good vibes in
being in resonance with nature. But after this first feeling one needs
confirmation.
| Quote: | ....only to go back over my results to
realize I ended up writing a 1 for a 2 which caused everything to work
out... but when it was 2 it didn't lead me anywhere(maybe 1 = 1). Almost
always I end up checking my results in some numerical way and end up
realizing a mistake was made somewhere then I go and redo the proof then
realize what I did wrong.
And checking on Your own results is very difficult. One can do some |
training in checking on others (and it is comforting to see others
running into faults as well).
| Quote: |
One way to test your results and see if they hold any water is to put them
to use. All mathematical endeavors must have some way to test them(although
some might be extremly difficult or intractable). If, say, you find some
super fast factoring method, then just use it to factor several easy cases.
If it works then you might have something or you just got lucky. If it
works then you will go and reprove it from scratch and try to clear things
up and make a more elegant proof. If it doesn't work then YOU MADE A MISTAKE
SOMEWHERE. Even if it does work you might have made a mistake(like the
Wiles FLT proof). Maybe the mistake is not significant to the problem and
you can fix it easily and the proof will work... other times your results
directly depedended on that mistake and you gotta start over.
If you truely want to get somewhere and are not just trolling this NG for
attention then you must realize that you are making a simple logical mistake
and instead of going crazy about it you need to understand the mistake and
move on to hopefully bigger and better things.. Else you will be stuck in
this NG for the next 10 years doing the same s**t and never making any real
progress.
But others do their training on finding and rectifying these mistakes.
Anyways, I'm tired and my head hurts ;/
You wrote a nice letter full of sympathy, of feeling with someone, |
so we wish You a refreshing sleep.
Hero |
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fishfry
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 299
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| Posted: Sun May 21, 2006 11:35 am Post subject:
Re: JSH: Lying about the distributive property
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In article <1148176879.874552.177500@j33g2000cwa.googlegroups.com>,
jstevh@msn.com wrote:
| Quote: | What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a
group you multiply the elements within that group:
a*(b + c) = a*b + a*c
|
Weren't we all here a few weeks ago? Since then you've apologized for
your error, then ranted about the lying culture of mathematics, then
gone away for a while, and now you're back with your old stuff.
Didn't several people post specific counterexamples to your claims?
You are confused about something. It's true that in general,
a(f(x) + b) = af(x) + ab.
But it is NOT necessarily true that if the right side holds at one
single point, then it holds at all points. If I recall, you were
confused about that. |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Sun May 21, 2006 2:31 pm Post subject:
Re: JSH: Lying about the distributive property
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On 20 May 2006 19:01:19 -0700, jstevh@msn.com wrote:
| Quote: | What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
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Um, that's a lie. Nobody has disputed the distributive property
in all of this.
************************
David C. Ullrich |
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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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| Posted: Sun May 21, 2006 3:32 pm Post subject:
Re: Lying about the distributive property
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Gene Ward Smith wrote:
| Quote: | José Carlos Santos wrote:
Juke wrote:
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
no, it is 7*(ac+bd+bc+bd)
What do you mean by that? First of all, you should have written
ac + bd + bc + ad instead. But are you denying that
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)?
It sure looks like.
I can't figure out what the "lie" is supposed to be, Harris just seems
to be saying trivial stuff like the above.
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Well you need to get to the expressions with functions.
Like
7*(h(x) + b)*(g(x) + d) = (f(x) + 7*b)*(g(x) + d)
which is actually easy in a way as you can just divide g(x) + d from
both sides, so let's make it slightly harder and have
7*P(x) = (f(x) + 7*b)(g(x) + d)
with the rule that f(0) = 0, and g(0) = 0, and let's say that P(x) is a
polynomial.
Given the reality that the distributive property does not care about
the value of functions, what is true at x=0 is true at any other value
of x, and you have proof of a case like
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
showing how the 7 multiplied through.
The mathematics is so basic it can seem incredible that if you accept
that result with the functions, and agree that the value of the
functions allow me to use x=0, and don't accept that as a special case
then the argument in my paper on non-polynomial factorization looks
trivial--as it is--and you can realize that posters claiming it false
are wrong.
But then you have this unknown behavior where you are pushed out of the
ring of algebraic integers, and if you chase that--pulling on the
string metaphorically--you unravel ideal theory.
So, the start of it is to accept that with
7*P(x) = (f(x) + 7*b)(g(x) + d)
where the functions f(x) and g(x) equal 0 when x=0 that you have shown
a case like
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
as if you're in a ring like the ring of algebraic integers that would
seemingly mean that
(f(x) + 7*b)
has 7 as a factor.
Get it now?
The reason for lying about the basic result is to block use of that
result in other areas, specifically with the ring of algebraic
integers, where I can show this remarkable problem that happens to take
away ideal theory.
James Harris |
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Dik T. Winter
science forum Guru
Joined: 25 Mar 2005
Posts: 1359
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| Posted: Mon May 22, 2006 12:44 am Post subject:
Re: Lying about the distributive property
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In article <1148225560.756412.85040@j33g2000cwa.googlegroups.com> jstevh@msn.com writes:
....
| Quote: | Like
7*(h(x) + b)*(g(x) + d) = (f(x) + 7*b)*(g(x) + d)
which is actually easy in a way as you can just divide g(x) + d from
both sides, so let's make it slightly harder and have
7*P(x) = (f(x) + 7*b)(g(x) + d)
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7*(x - 7) = (sqrt(7x) + 7)(sqrt(7x) - 7).
| Quote: |
with the rule that f(0) = 0, and g(0) = 0, and let's say that P(x) is a
polynomial.
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Yes, correct.
| Quote: |
Given the reality that the distributive property does not care about
the value of functions, what is true at x=0 is true at any other value
of x,
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You have to prove that.
| Quote: | and you have proof of a case like
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
showing how the 7 multiplied through.
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That is an example, not a proof. It works in this simple case but does
not necessarily work in all cases. You have to show that it works in
*all* cases. But in my above example, which of sqrt(7x) and sqrt(7x)
is divisible (in the algebraic integers) by 7?
| Quote: | But then you have this unknown behavior where you are pushed out of the
ring of algebraic integers, and if you chase that--pulling on the
string metaphorically--you unravel ideal theory.
So, the start of it is to accept that with
7*P(x) = (f(x) + 7*b)(g(x) + d)
where the functions f(x) and g(x) equal 0 when x=0 that you have shown
a case like
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
as if you're in a ring like the ring of algebraic integers that would
seemingly mean that
(f(x) + 7*b)
has 7 as a factor.
Get it now?
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P(x) = x - 7, f(x) = g(x) = sqrt(7x). In this case *none* of the factors
is divisible by 7. Get it now?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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| Posted: Mon May 22, 2006 1:19 am Post subject:
Re: Lying about the distributive property
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Dik T. Winter wrote:
| Quote: | In article <1148225560.756412.85040@j33g2000cwa.googlegroups.com> jstevh@msn.com writes:
...
Like
7*(h(x) + b)*(g(x) + d) = (f(x) + 7*b)*(g(x) + d)
which is actually easy in a way as you can just divide g(x) + d from
both sides, so let's make it slightly harder and have
7*P(x) = (f(x) + 7*b)(g(x) + d)
7*(x - 7) = (sqrt(7x) + 7)(sqrt(7x) - 7).
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Yeah, and you picked d not coprime to 7, so that both have sqrt(7) as a
factor if you're in the ring of algebraic integers, but so?
That CHOICE is revealed at x=1.
Remember I said any value of convenience.
At x=0, you get an ambiguous answer.
The logical point is that the value of the functions does not matter to
the distributive property, and notice, your example is just a
demonstration of that reality.
x=0 is not a special case and neither is x=1
You just use whatever value gets you the answer.
| Quote: |
with the rule that f(0) = 0, and g(0) = 0, and let's say that P(x) is a
polynomial.
Yes, correct.
Given the reality that the distributive property does not care about
the value of functions, what is true at x=0 is true at any other value
of x,
You have to prove that.
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It's a logical point.
I don't know if you're finally serious and willing to accept the truth,
but I'll give you the benefit of the doubt.
The logical point is that the value of functions is irrelevant to the
distributive property, just like the value of the variables is, as that
a*(f(x) + b) = a*f(x) + a*b
without regard to the value of f(x), but functions can hide what is
going on, so you can have instances, like with my examples, where you
have to find a convenient value that shows you what is happening.
Once you find out what is happening then you know it happens for all x
BECAUSE THE VALUE of the function does NOT matter.
It is a crucial and simple logical point.
Do you disagree?
| Quote: | and you have proof of a case like
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
showing how the 7 multiplied through.
That is an example, not a proof. It works in this simple case but does
not necessarily work in all cases. You have to show that it works in
*all* cases. But in my above example, which of sqrt(7x) and sqrt(7x)
is divisible (in the algebraic integers) by 7?
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You use a different convenience value, like x=1, and it is revealed
what is happening.
The issue here is not making discussion points, impressing the crowd,
or seeing who you can get over, but what is mathematically correct.
Looking for what is mathematically correct, you just boil it down to
what has to be true, not what you can convince others is true.
The logical point that the distributive property does not care about
value is all that matters.
So you use whatever value of the function lets you see what's
happening.
| Quote: | But then you have this unknown behavior where you are pushed out of the
ring of algebraic integers, and if you chase that--pulling on the
string metaphorically--you unravel ideal theory.
So, the start of it is to accept that with
7*P(x) = (f(x) + 7*b)(g(x) + d)
where the functions f(x) and g(x) equal 0 when x=0 that you have shown
a case like
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
as if you're in a ring like the ring of algebraic integers that would
seemingly mean that
(f(x) + 7*b)
has 7 as a factor.
Get it now?
P(x) = x - 7, f(x) = g(x) = sqrt(7x). In this case *none* of the factors
is divisible by 7. Get it now?
--
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Yet, can't you use x=1, and see?
Do you get it?
This is not a contest. It's not a debate. It's not some social thing
where all that matters is if you are convincing or not.
Mathematically, the value of a function does NOT affect whether or not
it gets multiplied as the value does not change how multiplication
distributes.
It's such a basic point that it is strange to have to repeat it, and
repeat it and repeat it.
Here is your own example:
7*(x - 7) = (sqrt(7x) + 7)(sqrt(7x) - 7)
Trying to use x=0 gives no information, as you get
7*(-7) = (7)(-7)
and it's a wash, but if you use x=1, you get
7*(1 - 7) = (sqrt(7) + 7)(sqrt(7) - 7)
and it's clear that sqrt(7) was multiplied times both of the
expressions on the right.
It's not rocket science.
It's mathematics.
Follow the logic. There is nothing in mathematics that says the value
of a function can affect it being multiplied or not.
There is nothing in mathematics that says that
a*(f(x) + b) = a*f(x) + a*b
is only valid sometimes, depending on the value of the function.
Using convenience values does not create special cases.
I multiply 7 times a polynomial so that I have
7*P(x) = (f(x) + 7)(g(x) + 1)
and guess what? If at x=0, f(x) and g(x) both equal 0, there is no
ambiguity, no way there can be any other way for the 7 to have been
multiplied.
James Harris |
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Dik T. Winter
science forum Guru
Joined: 25 Mar 2005
Posts: 1359
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| Posted: Mon May 22, 2006 1:31 am Post subject:
Re: JSH: Lying about the distributive property
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In article <1148176879.874552.177500@j33g2000cwa.googlegroups.com> jstevh@msn.com writes:
| Quote: | What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
|
They do not.
| Quote: | I can state the argument simply enough.
The distributive property simply enough says that if you multiply a
group you multiply the elements within that group:
a*(b + c) = a*b + a*c
and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
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Indeed.
| Quote: | and here is where it gets really bizarre, as my position that the value
of the function has no impact on the distributive property is now key,
so I say that if at x=0, f(x)=0, then that is as valid a point as any
other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of
the distributive property.
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
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Yup.
| Quote: | where it's still just the distributive property, but with more stuff,
and now, like before, I'm going to put in functions, but then I have to
be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
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Please be much more careful, you mean 'g(x)' when you write 'c'. Or
do you assume 'g(x)' is constant?
I have problems with the remainder, considering this.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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Dik T. Winter
science forum Guru
Joined: 25 Mar 2005
Posts: 1359
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| Posted: Mon May 22, 2006 12:55 pm Post subject:
Re: Lying about the distributive property
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In article <1148260756.942524.191110@g10g2000cwb.googlegroups.com> jstevh@msn.com writes:
| Quote: | Dik T. Winter wrote:
In article <1148225560.756412.85040@j33g2000cwa.googlegroups.com> jstevh@msn.com writes:
...
Like
7*(h(x) + b)*(g(x) + d) = (f(x) + 7*b)*(g(x) + d)
which is actually easy in a way as you can just divide g(x) + d from
both sides, so let's make it slightly harder and have
7*P(x) = (f(x) + 7*b)(g(x) + d)
7*(x - 7) = (sqrt(7x) + 7)(sqrt(7x) - 7).
Yeah, and you picked d not coprime to 7, so that both have sqrt(7) as a
factor if you're in the ring of algebraic integers, but so?
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Ok, let me try something different (although you did *not* have that
restriction in your original formulation). Take P(x) = (x + 1)(x + 
and further:
f(x) = 0 if x = 0 else it is x + 1
g(x) = 0 if x = 0 else if is 7x - 1
we now have:
7*P(x) = (f(x) + 7)(g(x) + .
Is f(x) divisible by 7? All of f(x), g(x) and P(x) satisfy the requirements
you have set. But only for x = 0 the left factor is divisible by 7, for other
x it is the right factor that is divisible by 7.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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Rupert
science forum Guru
Joined: 18 May 2005
Posts: 372
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| Posted: Tue May 23, 2006 12:09 am Post subject:
Re: JSH: Lying about the distributive property
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jstevh@msn.com wrote:
| Quote: | What makes this story even more dramatic though, especially in terms of
the fraud involved, is that posters in fighting against my research are
willing to lie about the distributive property--and they get away with
it.
I can state the argument simply enough.
The distributive property simply enough says that if you multiply a
group you multiply the elements within that group:
a*(b + c) = a*b + a*c
and there is no reason not to use functions so you may have
a*(f(x) + b) = a*f(x) + a*b
and here is where it gets really bizarre, as my position that the value
of the function has no impact on the distributive property is now key,
so I say that if at x=0, f(x)=0, then that is as valid a point as any
other AS THE VALUE OF THE FUNCTION DOES NOT MATTER to the operation of
the distributive property.
Now let's make a more complicated example:
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
where it's still just the distributive property, but with more stuff,
and now, like before, I'm going to put in functions, but then I have to
be more careful than before:
7*(h(x) + b)*(c + d) = (f(x) + 7*b)*(g(x) + d)
because one of the functions just swallowed the 7 so that it is now
invisible.
How can a function do that? Easy. It's a function, so it can be
f(x) = 7*h(x)
so the function swallowing the visibility of the 7 is not a big deal,
but maybe though, it didn't so I add the rule that at x=0, f(x)=0 and
g(x) = 0, so that I can SEE what is going on at a particular value.
If you had some other convenient value, that would be ok as well.
7*(h(0) + b)*(c + d) = (0 + 7*b)*(0 + d)
and it's clear that the 7 multiplied through like with
7*(a + b)*(c + d) = (7*a + 7*b)*(c + d)
while the functions just make things a little more complicated to
verify, but not impossible.
Now then, by the logical point that the value of the function does not
change the distributive property, I know what happened for ANY x, but
posters in fighting this argument have proclaimed x=0 to be a "special
case", defying the reality of how the distributive property operates.
I've explained, and explained, and explained so that the best
conclusion is that posters lie about this argument.
Otherwise they can't understand the basic principle that the value of
the function does not change the distributive property, which is a
major stretch.
Why is it such a big deal?
Because once the principle is established, I can get some complicated
functions that show a problem with the ring of algebraic integers.
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What problem? Tell us the argument that there is a "problem" with the
ring of algebraic integers.
| Quote: | Posters in defying what is mathematically correct are just slashing at
what they can, and in this case, that means questioning the
distributive property, and then claiming they are not doing so, while
they claim x=0 is a special case, but if I push them on the point that
the value of functions does not affect the distributive property, they
claim they don't disagree!
It's a case of where the lies just keep coming and it shows you how to
defy a mathematical proof.
Just claim it's wrong, keep claiming it's wrong, and get enough people
to claim it's wrong so that no one believes that it's correct.
And doing that you can block acceptance of mathematical proof.
These people are undermining the discipline of mathematics by showing
its true fragility.
It has few defenses against dedicated group lying about mathematical
arguments.
I mean, come on! The distributive property! How could people get away
with lying about that?
But they have now, for years.
James Harris |
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Tim Peters
science forum Guru
Joined: 30 Apr 2005
Posts: 426
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| Posted: Tue May 23, 2006 6:50 am Post subject:
Re: JSH: Lying about the distributive property
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[jstevh@msn]
| Quote: | ...
Why is it such a big deal?
Because once the principle is established, I can get some complicated
functions that show a problem with the ring of algebraic integers.
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[Rupert]
| Quote: | What problem? Tell us the argument that there is a "problem" with the
ring of algebraic integers.
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Of course there isn't one. It's hard to know what his endlessly
repeated "7 multiplies through" argument intends to claim -- like this
version did, it typically starts with simple observations about
distributivity, and then magically leaps to a conclusion that's one of
{incoherent, wrong, trivial} depending on what you guess he might be
trying to say.
In any case, "building" on that non-result, he comes up with
divisibility results that are demonstrably false in the ring of
algebraic integers. His conclusion is that this ring must therefore
"have a problem", or "be incomplete". To everyone else, his underlying
"7 multiplies through" argument is just gibberish (or wrong, or trivial
and then misapplied in fancier settings).
For some reason, he can't understand that a result that _does_ follow
from distributivity is necessarily true in all rings, so that the
failure of his claim in the ring of algebraic integers proves he's
relying on something that's _not_ true in all rings. I can't make
coherent sense of his "7 multiplies through" argument, so I'm not clear
on what that may be.
| Quote: | I've explained, and explained, and explained so that the best
conclusion is that posters lie about this argument.
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Repeating false claims doesn't make them true. The reality is simply
that you're confused here (and seemingly determined to remain that way). |
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