Trashing Galois Theory
JSH
with a simple mathematical example:
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.
The 7 on the left of the equals with the first expression cannot in
general be removed on the right hand side IN THE RING OF ALGEBRAIC
INTEGERS.
The a's are the roots of the second expression, so you can use the
quadratic formula to solve for them:
a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2, a_2(x) =
((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
I've brought this issue up time and time again, and one objection I've
heard repeatedly is that you CAN divide the 7 off for INDIVIDUAL
CASES, like for x=1, you can do some convoluted crap using Galois
Theory to find the "factors" of 7 in the ring of algebraic integers
that would divide off FOR THAT SPECIAL CASE.
But hey, isn't this algebra? Why don't you have special case for:
7(x^2 + 3x + 2) = (7x + 7)(x + 2)?
Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x +
2)?
The answer is: the ring of algebraic integers is CRAP. It was crap
years ago when I found this problem and it's still crap now though
YOUR STUPID PROFESSORS STILL TEACH YOU CRAP.
Yes, they are stupid for teaching you crap and you are stupid if you
keep learning it after this example.
Or are you such a moron that you don't know that if you multiply
something times say, 7, you should be able to divide the freaking 7
back off?
I don't care if your professor is at Harvard. If he's teaching you
stupid crap it's still stupid crap.
It doesn't matter if she is at Oxford. If she is teaching you stupid
crap it's still stupid crap.
And you're still stupid for learning it.
James Harris
David Moran
"JSH" <jst...@gmail.com> wrote in message
news:48f9f0bd-7f11-465c...@k13g2000prh.googlegroups.com...
What a stupid post.
Dave
fishfry
<48f9f0bd-7f11-465c...@k13g2000prh.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> Turns out you can destroy the underpinnings of modern number theory
> with a simple mathematical example:
>
> 7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
I feel ... I feel as if I've been here before ...
JSH, just to refresh my memory ... can you please state, without looking
it up, the Fundamental Theorem of Galois Theory?
Even better ... go ahead and look it up, then paraphrase it for us.
Tim Peters
> Turns out you can destroy the underpinnings of modern number theory
> with a simple mathematical example:
Well, yes, perhaps I can. However, you cannot. You've been pushing this
baseless argument for many years now, and it's still nonsense (for reasons
that have been explained to you countless times already):
> 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.
> ...
> It was crap years ago ... and it's still crap now
Indeed! Progress.
JSH
> [JSH <jst...@gmail.com>]
>
> > Turns out you can destroy the underpinnings of modern number theory
> > with a simple mathematical example:
>
> Well, yes, perhaps I can. However, you cannot. You've been pushing this
> baseless argument for many years now, and it's still nonsense (for reasons
> that have been explained to you countless times already):
Including the explanation for why publication in this area supposedly
was meaningless?
Or how destroying a math journal is an everyday thing?
http://www.emis.de/journals/SWJPAM/
Students should note sci.math people mounted an email assault against
an earlier argument in this area that was published, and ultimately
the journal itself was destroyed. They are anarchists who are immune
to the truth. They despise mathematical proof!
They are in actions--anti-mathematicians.
> > 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.
> > ...
> > It was crap years ago ... and it's still crap now
>
> Indeed! Progress.
Obviously people who would kill a math journal and celebrate it are
not going to talk reason now, but for those of you who actually enjoy
mathematics and believe in it, the demonstration is stunningly
irrefutable.
Multiplication is NOT controlled by what is multiplied, so when 7
multiplies times 175x^2 - 15x + 2, the result should be as trivial as
what you see with, say:
7(x^2 + 3x + 2) = (7x + 7)(x+2),
and in fact if you try a_1(x) = 7b_1(x), a_2(x) = b_2(x), it looks
almost as trivial:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x)+ 7)
but THAT possibility is blocked by the ring of algebraic integers if
b_1(x) is non-rational. Yeah, it's that specific, as trivially that
case IS allowed with x=0, when you have a rational case. Yup,
completely bonkers.
The ring of algebraic integers is functionally insane. It has stupid
arbitrary rules because it is a screwed up ring.
What I did was find an error using a non-standard construction but it
is a HUGE error.
To disbelieve the error you must throw out what you were taught that a
constant multiplied can simply be divided off as trivial, which it CAN
be with:
7(x^2 + 3x + 2) = (7x + 7)(x+2),
as trivially:
(x^2 + 3x + 2) = (x + 1)(x+2)
and the 7 is GONE.
YOU CANNOT DO THAT WITH THE EXAMPLE GIVEN IN THE RING OF ALGEBRAIC
INTEGERS.
That ring holds on to the 7, entangles it, and you can read a detailed
paper I wrote on that subject which you should find with a Google
search (has to be Google): algebraic integer entanglement
But the error allows a person to appear to prove things that are NOT
true, so it is seductive to academics who have careers built upon it,
or too scary to face, as a challenge to their egos.
It is a way for FAKES to get away with bogus math as they appear to
prove things not actually proven!!!
So they love it!
It's easy math. Many of your professors are turds. Incompetent on a
massive scale, so they rely on bogus math to hold their positions when
a trivial demonstration shows how decrepit it is.
They are weak. Intellectually they are inferior. Which is why they
maintain the error and have done so for years now.
The weak will not change. They will hold on to the error for as long
as you let them. They have no choice.
They have done so here for YEARS already. Just search the history
through the web.
They live in error, and TEACH error, as anything else would be
admitting to themselves their intellectual inferiority.
That they are not really mathematicians.
James Harris
Tim Peters
>>> Turns out you can destroy the underpinnings of modern number theory
>>> with a simple mathematical example:
[Tim Peters]
>> Well, yes, perhaps I can. �However, you cannot. �You've been pushing
>> this baseless argument for many years now, and it's still nonsense
>> (for reasons that have been explained to you countless times
>> already):
[JSH]
> Including the explanation for why publication in this area supposedly
> was meaningless?
Yes, of course. Countless times indeed. You have nothing new to say here.
> ...
> but for those of you who actually enjoy mathematics and believe
> in it, the demonstration is stunningly irrefutable.
Nevertheless, it's been refuted countless times already. Repeating your
errors a thousand times doesn't make them correct. Neither does refusing
to recognize your errors when they're pointed out.
> [... pretty good goofy rant elided, because you've done better
at /that/ before -- "pretty good" ain't good enough ;-) ...]
Rotwang
> Turns out you can destroy the underpinnings of modern number theory
> with a simple mathematical example:
>
> 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.
>
> The 7 on the left of the equals with the first expression cannot in
> general be removed on the right hand side IN THE RING OF ALGEBRAIC
> INTEGERS.
>
> The a's are the roots of the second expression, so you can use the
> quadratic formula to solve for them:
>
> a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2, a_2(x) =
> ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
>
> I've brought this issue up time and time again, and one objection I've
> heard repeatedly is that you CAN divide the 7 off for INDIVIDUAL
> CASES, like for x=1, you can do some convoluted crap using Galois
> Theory to find the "factors" of 7 in the ring of algebraic integers
> that would divide off FOR THAT SPECIAL CASE.
>
> But hey, isn't this algebra? Why don't you have special case for:
>
> 7(x^2 + 3x + 2) = (7x + 7)(x + 2)?
>
> Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x +
> 2)?
Because that is a factorisation in the ring of polynomials with
integer coefficients, and 7 is prime in that ring. That means that if
7 divides x*y then 7 divides at least one of x or y. On the other
hand, the factorisation with which you open this post is not a
factorisation in the ring of polynomials with integer coefficients.
It's a factorisation in the ring of functions with the algebraic
integers as domain and codomain. 7 is not prime in this ring, which is
why there is no reason to think that if 7*P = x*y then either x or y
must be divisible by 7. This fact is well understood by any
mathematician, and I do not believe that there is anything written in
any book or paper on modern number theory which contradicts it. If you
know of somewhere that does, by all means cite.
Do you know how to prove that 7 is prime in the ring of polynomials
with integer coefficients, btw? Since the answer to that is "no",
would you like to see how to prove it?
JSH
Because? Mathematically that has nothing to do with the inability to
simply divide the 7 off.
Or do you believe that whether or not 7 is prime or not is why you can
divide it off with:
7(x^2 + 3x + 2) = (7x + 7)(x+2)?
x^2 + 3x + 2 = (x+1)(x+2)
regardless.
The mathematics is trivial here. The denial though is a sign of rank
stupidity.
> 7 divides x*y then 7 divides at least one of x or y. On the other
> hand, the factorisation with which you open this post is not a
> factorisation in the ring of polynomials with integer coefficients.
Divide the 7 off with that factorization.
A simple request.
James Harris
Rotwang
> On Nov 30, 6:37 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
> > > 7(x^2 + 3x + 2) = (7x + 7)(x + 2)?
>
> > > Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x +
> > > 2)?
>
> > Because that is a factorisation in the ring of polynomials with
> > integer coefficients, and 7 is prime in that ring. That means that if
>
> Because? Mathematically that has nothing to do with the inability to
> simply divide the 7 off.
>
> Or do you believe that whether or not 7 is prime or not is why you can
> divide it off with:
>
> 7(x^2 + 3x + 2) = (7x + 7)(x+2)?
>
> x^2 + 3x + 2 = (x+1)(x+2)
>
> regardless.
Yes, the fact that 7 is prime is why you can divide it off. Look, 6 is
not prime in the ring of polynomials with integer coefficients, and
that's why it's easy to come up with examples like this:
6(x^2 + 3x + 2) = (3x + 3)(2x + 4)
Note that you can't divide 6 from either factor on the RHS. You can
divide each factor by a ring element such that the product of those
two ring elements is 6, and this is exactly what happens in your non-
polynomial factorisation; the only difference is that the
aforementioned ring elements in the non-polynomial factorisation case
are functions, which is completely unsurprising since your
factorisation takes place in a ring of functions.
> The mathematics is trivial here. The denial though is a sign of rank
> stupidity.
Agreed.
> > 7 divides x*y then 7 divides at least one of x or y. On the other
> > hand, the factorisation with which you open this post is not a
> > factorisation in the ring of polynomials with integer coefficients.
>
> Divide the 7 off with that factorization.
With what factorisation? You mean the factorisation x*y? Surely you
realise how silly this request is - it depends what x and y are, since
they denote arbitrary elements in an arbitrary ring.
You didn't respond to my offer to show you the proof that 7 is prime
in the ring of polynomials with integer coefficients. I'll make it
again, and add that it's an almost immediate corollary of a famous
result by Gauss. Would you like to see it?
JSH
> On 1 Dec, 03:14, JSH <jst...@gmail.com> wrote:
>
>
>
>
>
> > On Nov 30, 6:37 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > [...]
>
> > > > 7(x^2 + 3x + 2) = (7x + 7)(x + 2)?
>
> > > > Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x +
> > > > 2)?
>
> > > Because that is a factorisation in the ring of polynomials with
> > > integer coefficients, and 7 is prime in that ring. That means that if
>
> > Because? Mathematically that has nothing to do with the inability to
> > simply divide the 7 off.
>
> > Or do you believe that whether or not 7 is prime or not is why you can
> > divide it off with:
>
> > 7(x^2 + 3x + 2) = (7x + 7)(x+2)?
>
> > x^2 + 3x + 2 = (x+1)(x+2)
>
> > regardless.
>
> Yes, the fact that 7 is prime is why you can divide it off. Look, 6 is
> not prime in the ring of polynomials with integer coefficients, and
> that's why it's easy to come up with examples like this:
>
> 6(x^2 + 3x + 2) = (3x + 3)(2x + 4)
>
> Note that you can't divide 6 from either factor on the RHS. You can
But you can divide the 6 off.
x^2 + 3x + 2 = (x+1)(x+2)
Now then, does whether or not 6 is prime have anything to do with
that?
Answer is, no.
<deleted>
> > The mathematics is trivial here. The denial though is a sign of rank
> > stupidity.
>
> Agreed.
I'm being blunt for a purpose here, and you seem to not quite
understand how stupid your position is yet.
I made a simple request below...
> > > 7 divides x*y then 7 divides at least one of x or y. On the other
> > > hand, the factorisation with which you open this post is not a
> > > factorisation in the ring of polynomials with integer coefficients.
>
> > Divide the 7 off with that factorization.
>
> With what factorisation? You mean the factorisation x*y? Surely you
> realise how silly this request is - it depends what x and y are, since
> they denote arbitrary elements in an arbitrary ring.
That's stupid.
Now then, you are being stupid. When 7 is multiplied times something
then it can be divided back off, correct?
Answer is, yes in ANY RING except the ring of algebraic integers!!!
ANY RING EXCEPT THE RING OF ALGEBRAIC INTEGERS.
The ring is fatally flawed. You are stupidly trying to dodge around
real issues and brought up an idiot objection with the 6 example as if
the primeness of 7 means it matters what you multiply it times.
Here is the example again--the crushing example which simply blows
apart the delusions for people who still can think--if not, then
DIVIDE THE 7 OFF:
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.
Mathematicians continuing to teach the bogus math are feeding garbage
to YOUNG MINDS. They are training young people how to be in ERROR.
Fighting for that system makes you more than just stupid, though yes,
from your replies so far, clearly stupid you are.
It also makes you the enemy of any kid with a math dream.
James Harris
Rotwang
> On Nov 30, 7:27 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
>
> > On 1 Dec, 03:14, JSH <jst...@gmail.com> wrote:
>
> > > On Nov 30, 6:37 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > > [...]
>
> > > > > 7(x^2 + 3x + 2) = (7x + 7)(x + 2)?
>
> > > > > Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x +
> > > > > 2)?
>
> > > > Because that is a factorisation in the ring of polynomials with
> > > > integer coefficients, and 7 is prime in that ring. That means that if
>
> > > Because? Mathematically that has nothing to do with the inability to
> > > simply divide the 7 off.
>
> > > Or do you believe that whether or not 7 is prime or not is why you can
> > > divide it off with:
>
> > > 7(x^2 + 3x + 2) = (7x + 7)(x+2)?
>
> > > x^2 + 3x + 2 = (x+1)(x+2)
>
> > > regardless.
>
> > Yes, the fact that 7 is prime is why you can divide it off. Look, 6 is
> > not prime in the ring of polynomials with integer coefficients, and
> > that's why it's easy to come up with examples like this:
>
> > 6(x^2 + 3x + 2) = (3x + 3)(2x + 4)
>
> > Note that you can't divide 6 from either factor on the RHS. You can
>
> But you can divide the 6 off.
>
> x^2 + 3x + 2 = (x+1)(x+2)
Yes, and you can divide off the 7 in your non-polynomial
factorisation. The fact that there's no simple expression for the
resulting factors doesn't change the fact that it can be done.
> Now then, does whether or not 6 is prime have anything to do with
> that?
>
> Answer is, no.
No, but whether or not 6 is prime has everything to do with the fact
that it can't be divided out of the product in the simple manner you
think should apply to your non-polynomial factorisation.
> <deleted>
>
> > > The mathematics is trivial here. The denial though is a sign of rank
> > > stupidity.
>
> > Agreed.
>
> I'm being blunt for a purpose here, and you seem to not quite
> understand how stupid your position is yet.
>
> I made a simple request below...
>
> > > > 7 divides x*y then 7 divides at least one of x or y. On the other
> > > > hand, the factorisation with which you open this post is not a
> > > > factorisation in the ring of polynomials with integer coefficients.
>
> > > Divide the 7 off with that factorization.
>
> > With what factorisation? You mean the factorisation x*y? Surely you
> > realise how silly this request is - it depends what x and y are, since
> > they denote arbitrary elements in an arbitrary ring.
>
> That's stupid.
An integer x is even iff it is divisible by 2 in the ring of integers.
Let x be an integer. Is x even?
That's how stupid your request is, James.
> Now then, you are being stupid. When 7 is multiplied times something
> then it can be divided back off, correct?
>
> Answer is, yes in ANY RING except the ring of algebraic integers!!!
>
> ANY RING EXCEPT THE RING OF ALGEBRAIC INTEGERS.
Try to think abstractly about what you're saying here. "When 7 is
multiplied by something" means that there is something, and something
else, such that
something else = 7*something.
What you're saying is that, if something and something else are
algebraic integers, then you can't divide the 7 back off something
else. That is, you can't do this:
something = something else/7
You are wrong. As it happens, there are rings where you can't "divide
off" a 7 from something which has 7 as a factor, namely rings where 7
is a divisor of 0. The ring of algebraic integers is not one of them.
> The ring is fatally flawed. You are stupidly trying to dodge around
> real issues and brought up an idiot objection with the 6 example as if
> the primeness of 7 means it matters what you multiply it times.
>
> Here is the example again--the crushing example which simply blows
> apart the delusions for people who still can think--if not, then
> DIVIDE THE 7 OFF:
>
> 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.
The 7 may be divided off as follows:
(175x^2 - 15x + 2)
= ((5a_1(x) + 7)/GCD(5a_1(x) + 7,7))((5a_2(x) + 7)*GCD(5a_1(x) + 7,7)/
7)
Note that GCD(5a_1(x) + 7,7) and 7/GCD(5a_1(x) + 7,7) are both
elements of the ring in which your factorisation takes place, namely
the ring of functions with domain and codomain the algebraic integers,
and that their product is 7. This is just like the manner in which you
divided 6 from the polynomial example I gave above, except that the
factors of 7 are not constant functions. But they're still elements of
the appropriate ring.
> Mathematicians continuing to teach the bogus math are feeding garbage
> to YOUNG MINDS. They are training young people how to be in ERROR.
>
> Fighting for that system makes you more than just stupid, though yes,
> from your replies so far, clearly stupid you are.
No, you're stupid. That's why you've just spent a year believing you
had proved P = NP, when I was able to show that you hadn't by spending
an hour with Python. The reason you've managed to continue believing
this crap about the "flaw" in Galois theory for a period of years is
that it's not possible for you or anyone else to test such vague,
meaningless claims against reality by writing a script, which means
that the only way anybody could make you realise that you're mistaken
is to make you understand the actual mathematics involved. But nobody
can, because you're too stupid.
You're also a coward, with no desire to learn the truth. That's why,
when I tried to get you to look at another one of the many
demonstrations of your errors I've provided the other day, you called
me "greedy" - as if taking the time to write proofs and test your
algorithms is something I do purely for /my/ benefit. It isn't. I'd be
happy for you to spend the rest of your life making laughably false
claims to newsgroups, but /you/ shouldn't be.
Now, would you like to see that proof of Gauss that I mentioned?
Wouldn't it be nice to actually learn something?
Mark Murray
> Now, would you like to see that proof of Gauss that I mentioned?
> Wouldn't it be nice to actually learn something?
No, he wouldn't.
a) That would get in the way of his search for the truth.
b) He knows al the important stuff anyway. If he doesn't know
something, he'll soon discover it and inform YOU.
Sheesh.
M
JSH
It can't in general be done.
So it can't be done, like with
7(x^2 + 3x + 2) = (7x + 7)(x+2)
where dividing off the 7 just gives:
x^2 + 3x + 2 = (x+1)(x+2)
Notice the 7 is GONE! Isn't that a miracle? It's like, it just
completely GOES AWAY!
Amazing!!!
Readers who think I'm being silly need to look below to see the
example the poster gave for dividing the 7 off.
He left ghosts of 7.
> > Now then, does whether or not 6 is prime have anything to do with
> > that?
>
> > Answer is, no.
>
> No, but whether or not 6 is prime has everything to do with the fact
> that it can't be divided out of the product in the simple manner you
> think should apply to your non-polynomial factorisation.
That's stupid. It has nothing to do with anything.
The request I'm making is to divide the 7 off, with:
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
You're a freaking idiot primitive if you believe that whether or not 7
is prime or not has anything to do with that request.
Defending your stupidity here does not remove it.
7 being prime is mathematically completely irrelevant.
It was irrelevant when you first brought up the issue, is still
irrelevant now, and will be irrelevant to infinity.
Even basic math is hard when trying to communicate with people who
turn to idiocy when confronted with mathematical truth.
> > <deleted>
>
> > > > The mathematics is trivial here. The denial though is a sign of rank
> > > > stupidity.
>
> > > Agreed.
>
> > I'm being blunt for a purpose here, and you seem to not quite
> > understand how stupid your position is yet.
>
> > I made a simple request below...
>
> > > > > 7 divides x*y then 7 divides at least one of x or y. On the other
> > > > > hand, the factorisation with which you open this post is not a
> > > > > factorisation in the ring of polynomials with integer coefficients.
>
> > > > Divide the 7 off with that factorization.
>
> > > With what factorisation? You mean the factorisation x*y? Surely you
> > > realise how silly this request is - it depends what x and y are, since
> > > they denote arbitrary elements in an arbitrary ring.
>
> > That's stupid.
>
> An integer x is even iff it is divisible by 2 in the ring of integers.
> Let x be an integer. Is x even?
>
> That's how stupid your request is, James.
The ring is the ring of algebraic integers. The request is to divide
the 7 off with:
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.
That is highly specific. Very rigorous. And does not require
stupidity on your part in answering.
Either it can be done, and if so, do it (I know you have an example
below which I'll blow apart), or admit it cannot be done.
How hard can it be? The MATH WILL NOT CHANGE!!!
Why can't you understand that mathematics is not about your idiot
opinion?
Are you not at all educated? How stupid can one person be?
Circular.
Now, divide off the 7 with 7(x^2 + 3x + 2) = (7x + 7)(x+2).
You get x^2 + 3x + 2 = (x+1)(x+2), and guess what? NO mention of 7.
Why should 7 be there when you divide it off?
But look at YOUR:
(175x^2 - 15x + 2)
= ((5a_1(x) + 7)/GCD(5a_1(x) + 7,7))((5a_2(x) + 7)*GCD(5a_1(x) +
7,7)/
7)
Why does your result still mention 7 in it when you've supposedly
divided 7 off?
Does 7 now have some familial affiliation with 175x^2 - 15x + 2?
Maybe a blood connection that cannot be broken by mere mathematics?
Is 7 possibly MYSTICALLY connected? Like from the dawn of time 7 and
175x^2 + 15x + 2 have been bonded but only now has this been
revealed!!!
The 7 WILL NOT JUST GO AWAY!!! It's a miracle! It's a ghost.
Scary.
You are showing idiocy. When you divide off a factor what's left does
not have to know what was multiplied times it.
In your denial of mathematical truth, you have humiliated yoursel.
I can massacre your ego indefinitely.
And you can reply and reply and reply showing how great a fool you are
as you stepped into the one arena where your stupidity can always be
carefully described and pointed out: mathematics.
You are a being a fool.
Your education failed to inform you that in mathematics, the person
who is right can SHRED you.
You are not educated then.
James Harris
MichaelW
>
> Try to think abstractly about what you're saying here. "When 7 is
> multiplied by something" means that there is something, and something
> else, such that
>
> something else = 7*something.
>
> What you're saying is that, if something and something else are
> algebraic integers, then you can't divide the 7 back off something
> else. That is, you can't do this:
>
> something = something else/7
>
> You are wrong. As it happens, there are rings where you can't "divide
> off" a 7 from something which has 7 as a factor, namely rings where 7
> is a divisor of 0. The ring of algebraic integers is not one of them.
>
>
>
> (175x^2 - 15x + 2)
> = ((5a_1(x) + 7)/GCD(5a_1(x) + 7,7))((5a_2(x) + 7)*GCD(5a_1(x) + 7,7)/
> 7)
>
> Note that GCD(5a_1(x) + 7,7) and 7/GCD(5a_1(x) + 7,7) are both
> elements of the ring in which your factorisation takes place, namely
> the ring of functions with domain and codomain the algebraic integers,
> and that their product is 7. This is just like the manner in which you
> divided 6 from the polynomial example I gave above, except that the
> factors of 7 are not constant functions. But they're still elements of
> the appropriate ring.
>
follow this and related discussions but I am getting stuck on one
point. I get how a_1(x) is in the ring of algebraic integers but if b_1
(x) =a_1(x)/7 then I get
7b^2-(7x-1)b+7x^2-2x=0
which means b_1(x) is not in the ring. I can divide through by 7 but
then there are non-integer co-efficients which I gather (possibly
wrongly) implies the same. Either my understanding is wrong or the
status of b_1(x) does not matter to the argument...?
Thanks, Michael W.
Rotwang
> On Dec 1, 3:31 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
> Rotwang (or anyone with the time for a dumb question), I am trying to
> follow this and related discussions but I am getting stuck on one
> point. I get how a_1(x) is in the ring of algebraic integers but if b_1
> (x) =a_1(x)/7 then I get
>
> 7b^2-(7x-1)b+7x^2-2x=0
>
> which means b_1(x) is not in the ring.
Yes, this is the whole point. James knows that b_1(x) is not an
algebraic integer for all x even though a_1(x) is, and thinks that
this fact means that the ring is "flawed". What it actually means is
that the ring is not closed under division, which is an utterly
trivial observation. But James thinks that the fact that he defined
b_1 by
7*b_1(x) = a_1(x) (1)
means that he's been "pushed out of the ring" by "valid operations".
You may notice that whenever he's talking about this stuff, he always
defines b_1 by writing an equation like (1), and never by writing
b_1(x) = a_1(x)/7;
I suppose that giving the actual definition for b_1, rather than
writing down an equation which he wants it to solve and pretending
that said equation ought to have a solution, would make it obvious to
him that the operation that actually pushes him out of the ring is
division, so he doesn't.
By the way, his alleged solution to this "flaw" is to introduce the
object ring, and then state that b_1(x) is an element of the object
ring for all x, even when it isn't an algebraic integer. The object
ring is a ring whose elements may only be determined by asking James,
but whose wish-list of properties includes the facts that it contains
the algebraic integers, and also contains no non-integer rational. But
it's very easy to show that any such ring must have the same "flaw" as
the algebraic integers: consider, for example, the polynomial
Q(x) = 25x^2 + 5x + 2
and note that
7*Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) (2)
where the a's are now the roots of
a^2 - (x - 1)a + 7x^2 = 0.
James's argument applies to this polynomial just like it applies to
the other one: at x = 0, (2) gives (making an arbitrary choice about
which a_i is which root)
7*Q(0) = 7*2.
Note that 2 is not divisible (in the AIs) by any non-unit factor of 7,
so the only way up to associates to divide off the 7 in the manner
James wishes is to let b_1(x) = a_1(x)/7 at x = 0. But at x = 1, the
a's are +/- sqrt(7), so b_1(1) cannot be an object (for if it were
then b_1(1)^2 = 1/7 would be an object, contradicting his desire that
the object ring should contain no non-integer rationals). He has been
shown examples like this one many times, and responds to them with his
usual evasiveness.
Rotwang
> On Nov 30, 8:31 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
> > Yes, and you can divide off the 7 in your non-polynomial
> > factorisation. The fact that there's no simple expression for the
> > resulting factors doesn't change the fact that it can be done.
>
> It can't in general be done.
Yes it can, just not in the manner you think it should, namely by
dividing each factor by a constant factor of 7. So what? Nothing in
modern number theory contradicts this elementary fact.
> So it can't be done, like with
>
> 7(x^2 + 3x + 2) = (7x + 7)(x+2)
>
> where dividing off the 7 just gives:
>
> x^2 + 3x + 2 = (x+1)(x+2)
>
> Notice the 7 is GONE! Isn't that a miracle? It's like, it just
> completely GOES AWAY!
Yes. That's because you started out with a factorisation of x^2 + 3x +
2 in which one of the two factors was divisible by 7 for all x. In
your non-polynomial factorisation that is no longer the case. So what?
> > > Now then, does whether or not 6 is prime have anything to do with
> > > that?
>
> > > Answer is, no.
>
> > No, but whether or not 6 is prime has everything to do with the fact
> > that it can't be divided out of the product in the simple manner you
> > think should apply to your non-polynomial factorisation.
>
> That's stupid. It has nothing to do with anything.
>
> The request I'm making is to divide the 7 off, with:
>
> 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
>
> You're a freaking idiot primitive if you believe that whether or not 7
> is prime or not has anything to do with that request.
The fact that 7 is prime has nothing to do with the ability to divide
7 from the product 7*P(x). It does have something to do with the
ability to divide it off in the manner you're asking for, though,
namely by dividing it off from one of the factors on the RHS as a
constant. Since 7 is not prime in the ring of functions with domain
and codomain the AIs, there is no reason to suppose that, just because
7 divides the product for all x, it should divide one of the factors
for all x.
You keep saying that this utterly trivial fact shows that there's an
error in taught mathematics. But you can't present a single example of
a mathematician claiming that it /is/ possible to divide off the 7 in
the manner you think should be possible. If taught mathematics doesn't
actually say something that isn't true, then there isn't an error in
taught mathematics. The fact that you think one of the factors
"should" be divisible by 7 for all x, even though it isn't, doesn't
mean that mathematicians have made an error. It simply means that your
intuition about what should happen is wrong. This is nobody's problem
but yours.
> Even basic math is hard when trying to communicate with people who
> turn to idiocy when confronted with mathematical truth.
No, this is all very easy maths. Your inability to understand it
doesn't change that fact.
> The ring is the ring of algebraic integers.
For fixed x your factorisation takes place in the ring of algebraic
integers, yes. And for each such x it's possible to divide the two
factors by a pair of algebraic integers whose product is 7, just like
you divided the 6 from my earlier example by dividing one factor by 3
and the other by 2. On the other hand, if x variable then your
factorisation does not take place in the algebraic integers. It takes
place in the ring of functions with domain and codomain the algebraic
integers. In this context it's still possible to divide the two
factors by a pair of ring elements whose product is 7, but the ring
elements in question are now non-constant functions. This is not a
"flaw", it's just a fact, and nothing in modern number theory
contradicts it.
> The request is to divide
> the 7 off with:
>
> 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.
>
> That is highly specific. Very rigorous. And does not require
> stupidity on your part in answering.
>
> Either it can be done, and if so, do it (I know you have an example
> below which I'll blow apart), or admit it cannot be done.
>
> How hard can it be? The MATH WILL NOT CHANGE!!!
>
> Why can't you understand that mathematics is not about your idiot
> opinion?
>
> Are you not at all educated? How stupid can one person be?
>
> [...]
>
> > The 7 may be divided off as follows:
>
> > (175x^2 - 15x + 2)
> > = ((5a_1(x) + 7)/GCD(5a_1(x) + 7,7))((5a_2(x) + 7)*GCD(5a_1(x) + 7,7)/
> > 7)
>
> > Note that GCD(5a_1(x) + 7,7) and 7/GCD(5a_1(x) + 7,7) are both
>
> Circular.
>
> Now, divide off the 7 with 7(x^2 + 3x + 2) = (7x + 7)(x+2).
>
> You get x^2 + 3x + 2 = (x+1)(x+2), and guess what? NO mention of 7.
>
> Why should 7 be there when you divide it off?
>
> But look at YOUR:
>
> (175x^2 - 15x + 2)
> = ((5a_1(x) + 7)/GCD(5a_1(x) + 7,7))((5a_2(x) + 7)*GCD(5a_1(x) +
> 7,7)/
> 7)
>
> Why does your result still mention 7 in it when you've supposedly
> divided 7 off?
Because the definition of the two factors in your product involved 7.
Remember? The reason you were able to write down the original
factorisation is because you found a monic polynomial, parameterised
by x, whose roots a_1(x) and a_2(x) satisfied
7*P(x) = (5a_1(x) + 7)(5a_2(x) + 7)
See that 7 on the LHS? And those two 7's on the RHS? They mean that
the definition of the a's involved the number 7. Had you instead found
functions c_1(x) and c_2(x) such that
5*P(x) = (3c_1(x) + 5)(3c_2(x) + 5)
then the factors of P(x) you found by dividing the RHS by 5 would
still involve 5. They would be different factors. Since the definition
of your factors of 7*P involves the number 7, it is completely
unsurprising that the definition of any factors of P derived from them
should also involve the number 7.
> Does 7 now have some familial affiliation with 175x^2 - 15x + 2?
> Maybe a blood connection that cannot be broken by mere mathematics?
>
> Is 7 possibly MYSTICALLY connected?
Have you ever noticed that the people with whom you argue never
actually need to invoke mysticism or anthropomorphism or metaphors
about dogs to make their point, but that you do? What do you make of
this?
> You are showing idiocy. When you divide off a factor what's left does
> not have to know what was multiplied times it.
>
> In your denial of mathematical truth, you have humiliated yoursel.
Huh? No, you haven't. Perhaps you are confused about what
"humiliation" means. See, humiliation is the feeling you'll feel when
you wake up sober and look at what you wrote last night on your
twitter page. Nothing like that is happening here.
> I can massacre your ego indefinitely.
No you can't. You can try, but you continue to make a fool of yourself
in the process, whether you realise it or not.
> And you can reply and reply and reply showing how great a fool you are
> as you stepped into the one arena where your stupidity can always be
> carefully described and pointed out: mathematics.
>
> You are a being a fool.
>
> Your education failed to inform you that in mathematics, the person
> who is right can SHRED you.
You know, this talk about my education is slightly sillier than your
usual rants (and that's saying something). You realise that you have
had no actual mathematics education since leaving school, right? And
that many of the people you argue with have degrees and Ph.D.'s in
mathematics, and have therefore received much more of the relevant
education than you? If you think that somebody's lack of education is
preventing them from seeing the truth here, why on Earth do you
imagine that somebody isn't you?
MichaelW
helpful.
Regards, Michael W.
JSH
No, that IS the point, b_1(x) isn't allowed to be in the ring of
algebraic integers.
The ring of algebraic integers BLOCKS you from just dividing off the
7, which is a wacky thing which can't be seen with any other ring
known to humanity. No other ring known can grab hold of a number like
7 and not let go.
If it didn't have such a devastating impact it'd just be kind of cool:
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.
So you see the 7 on the left, and from what students are taught as
basic mathematics you SHOULD be able to just divide that 7 off! But
you cannot in the ring of algebraic integers.
You can divide off the 7 though. Doing so is as simple as letting a_1
(x) or a_2(x) have 7 as a factor, but provably NEITHER can have 7 as a
factor in the ring of algebraic integers as that ring is bonkers. It
is insane.
It actually leads you to a mathematical contradiction if you do not
resolve its insanity as it is stopping algebra itself--not allowing
you in general to divide that 7 off.
The fix is not to believe the ring of algebraic integers is ok. If
it's ok, then mathematics collapses.
If it's not ok, you can figure out mathematically what it's doing.
There are no other choices.
For VALID mathematics you must be able to divide back off what you
multiply.
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
is trivially: x^2 + 3x + 2 = (x+1)(x+1), with no trace of the 7.
Basic mathematics.
Problem though is that if you accept mathematical truth then oh, about
a hundred or so years of number theory goes out the window, so you
have this social b.s. of people fighting to hold on. To hold on
desperately.
But I can take them apart at will on this issue.
I can confront anyone on it. Could confront Kenneth Ribet, or Taylor,
or, yup, even Andrew Wiles with this thing.
Can take any of them apart on this issue.
James Harris
Rotwang
> On Dec 2, 12:25 am, MichaelW <ms...@tpg.com.au> wrote:
>
>
> No, that IS the point, b_1(x) isn't allowed to be in the ring of
> algebraic integers.
>
> The ring of algebraic integers BLOCKS you from just dividing off the
> 7, which is a wacky thing which can't be seen with any other ring
> known to humanity. No other ring known can grab hold of a number like
> 7 and not let go.
>
> If it didn't have such a devastating impact it'd just be kind of cool:
>
> 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.
>
> So you see the 7 on the left, and from what students are taught as
> basic mathematics you SHOULD be able to just divide that 7 off! But
> you cannot in the ring of algebraic integers.
>
> You can divide off the 7 though. Doing so is as simple as letting a_1
> (x) or a_2(x) have 7 as a factor, but provably NEITHER can have 7 as a
> factor in the ring of algebraic integers as that ring is bonkers. It
> is insane.
>
> It actually leads you to a mathematical contradiction if you do not
> resolve its insanity as it is stopping algebra itself--not allowing
> you in general to divide that 7 off.
Since you rejected the answer I gave when you asked me to divide off
the 7 from your non-polynomial factorisation, neither I nor,
apparently, anybody except you knows what you mean when you talk about
dividing off the 7. So perhaps you should illustrate with an example:
in the object ring,
7*(25x^2 + 5x + 2) = (5a_1(x) + 7)(5a_2(x) + 7)
where the a's are the roots of
a^2 - (x - 1)a + 7x^2 = 0.
Since the object ring isn't "flawed", it must be possible to divide
off the 7. Please do so, to show us how it's done.
> [...]
>
> I can confront anyone on it. Could confront Kenneth Ribet, or Taylor,
> or, yup, even Andrew Wiles with this thing.
>
> Can take any of them apart on this issue.
Unless they just ignore you, like you've repeatedly ignored the above
factorisation. Why not lead by example?
MichaelW
>
> No, that IS the point, b_1(x) isn't allowed to be in the ring of
> algebraic integers.
>
> The ring of algebraic integers BLOCKS you from just dividing off the
> 7, which is a wacky thing which can't be seen with any other ring
> known to humanity. No other ring known can grab hold of a number like
> 7 and not let go.
>
Ring of integers.
Ki Song
> For VALID mathematics you must be able to divide back off what you
> multiply.
>
This is a false statement.
JSH
Nope. If you disagree, demonstrate.
James Harris
JSH
It's about equality, and is equivalent to a statement of identity.
i.e. 7x = 7x, therefore, x = x.
That is true in all valid rings. Only the ring of algebraic integers
can violate that principle
James Harris
Ki Song
This is another false statement.
Consider Z_4, integers modulo 4.
2 * 2 = 0 mod 4, but this doesn't mean that 2 = 0 mod 4.
2 x = 2 y does not imply x = y.
Consider matrices (of a fixed dimension) with real value entries.
There are many non-zero matrices whose produce is 0.
Consider ring of real valued functions on the real line.
The list goes on and on.
JSH
> On Dec 2, 9:47 pm, JSH <jst...@gmail.com> wrote:
>
> > On Dec 2, 6:33 pm, Ki Song <kiwisqu...@gmail.com> wrote:
>
> > > > For VALID mathematics you must be able to divide back off what you
> > > > multiply.
>
> > > This is a false statement.
>
> > It's about equality, and is equivalent to a statement of identity.
>
> > i.e. 7x = 7x, therefore, x = x.
>
> > That is true in all valid rings. Only the ring of algebraic integers
> > can violate that principle
>
> > James Harris
>
> This is another false statement.
>
> Consider Z_4, integers modulo 4.
>
> 2 * 2 = 0 mod 4, but this doesn't mean that 2 = 0 mod 4.
That's not an identity!!!!!!!!!!!!!
I'm using identities.
Understand yet?
James Harris
MichaelW
5*7/7 = 5*7/7
Divide off the 7
5/7 = 5/7
No longer in the ring of integers.
Regards, Michael W.
JSH
> On Dec 3, 1:46 pm, JSH <jst...@gmail.com> wrote:
>
>
>
>
>
> > On Dec 2, 6:28 pm, MichaelW <ms...@tpg.com.au> wrote:
>
> > > On Dec 3, 12:38 pm, JSH <jst...@gmail.com> wrote:
>
> > > > No, that IS the point, b_1(x) isn't allowed to be in the ring of
> > > > algebraic integers.
>
> > > > The ring of algebraic integers BLOCKS you from just dividing off the
> > > > 7, which is a wacky thing which can't be seen with any other ring
> > > > known to humanity. No other ring known can grab hold of a number like
> > > > 7 and not let go.
>
> > > Ring of integers.
>
> > Nope. If you disagree, demonstrate.
>
> > James Harris
>
> 5*7/7 = 5*7/7
7/7 = 1 in the ring of integers.
> Divide off the 7
So you're dividing 5 by 7. Invalid in that ring as 7 is not a factor
of 5.
> 5/7 = 5/7
>
> No longer in the ring of integers.
Your fallacy is in believing that division is a ring operation. It's
not.
> Regards, Michael W.
Review ring operations.
James Harris
Ki Song
>
> James Harris
What I understand is that you made a ridiculous claim
> > > > > For VALID mathematics you must be able to divide back off what you
> > > > > multiply.
And I've shown you very simple examples in which you can't divide back
off what you multiply.
> > > i.e. 7x = 7x, therefore, x = x.
Did you mean to say 7x = 7y, therefore, x = y?
> > 2 * 2 = 0 mod 4, but this doesn't mean that 2 = 0 mod 4.
>
> That's not an identity!!!!!!!!!!!!!
Do you know what an "identity" is?
Are you telling me that
4 = 0 mod 4
is not an identity?
(2*2) = (0*1) (mod 4)
is not an identity?
Good luck dividing off the 2 up there.
> > > i.e. 7x = 7x, therefore, x = x.
>
> > > That is true in all valid rings. Only the ring of algebraic integers
> > > can violate that principle
I'm very sorry, but you are making very little sense to me. You have
two different factorizations of an algebraic integer. Say:
x = 7*b = c*d, where x, b, c, and d are algebraic integers.
Are you saying that algebraic integers are broken because there is no
factorization of 7 where
7 = g*h, g and h algebraic integers.
So that
b = (c/g) * (d/h)
where (c/g) and (d/h) are algebraic integers?
> Understand yet?
I understand that you don't seem to understand what I'm saying.
I understand that you don't know what Galois theory is.
>
> James Harris
MichaelW
>
> > 5*7/7 = 5*7/7
>
> 7/7 = 1 in the ring of integers.
>
> > Divide off the 7
>
> So you're dividing 5 by 7. Invalid in that ring as 7 is not a factor
> of 5.
>
> > 5/7 = 5/7
>
> > No longer in the ring of integers.
>
> Your fallacy is in believing that division is a ring operation. It's not
>
>
> Review ring operations.
>
> James Harris
Not my fallacy, it's yours. I just simplified it to make it clear. I
doubt if you will ever understand. Not that you are incompetent but
rather...
Regards, Michael W.
JSH
> > Understand yet?
>
> > James Harris
>
> What I understand is that you made a ridiculous claim
>
> > > > > > For VALID mathematics you must be able to divide back off what you
> > > > > > multiply.
>
> And I've shown you very simple examples in which you can't divide back
> off what you multiply.
>
> > > > i.e. 7x = 7x, therefore, x = x.
>
> Did you mean to say 7x = 7y, therefore, x = y?
Nope.
> > > 2 * 2 = 0 mod 4, but this doesn't mean that 2 = 0 mod 4.
>
> > That's not an identity!!!!!!!!!!!!!
Oh, my mistake. It is.
> Do you know what an "identity" is?
Yup.
> Are you telling me that
>
> 4 = 0 mod 4
>
> is not an identity?
Well I did not looking closely and I was wrong. It IS an identity.
> (2*2) = (0*1) (mod 4)
>
> is not an identity?
Yes it is an identity. I was wrong before.
>
> Good luck dividing off the 2 up there.
2*2 = 0 mod 4
2 = 0 mod 2. Easy.
I saw "mod" and didn't look more closely until this reply, so yes, 2*2
= 0 mod 4 is an identity, but you can divide off the 2.
You seem to believe that what's on the other side of "mod" is locked.
That's not true.
James Harris
Ki Song
... I don't think I understand what you mean by "dividing off."
You are allowed to switch rings?
>
> I saw "mod" and didn't look more closely until this reply, so yes, 2*2
> = 0 mod 4 is an identity, but you can divide off the 2.
>
> You seem to believe that what's on the other side of "mod" is locked.
Sigh.
Do you not understand that Z_4 and Z_2 are two different rings?
When you change from mod 4 to mod 2, you are jumping from one ring
into a different ring.
You were earlier making a huge fuss when MichaelW went from integers
to rational numbers, but you are okay going from Z_4 to Z_2?
Do you understand that in the example above, you had to go from a
ring, Z_4, into a different ring Z_2? So in fact, the division didn't
take place in Z_4?
Here's another example.
[ 0 1 ] ^2 = [ 0 0 ]
[ 0 0 ] [ 0 0 ]
Now, tell me if you can divide anything off here. There are many
other examples, but typing matrices is a pain, so this is what I'll
leave it as.
Ki Song
The problem with OP is that he thinks he understand more mathematics
than other people. Another big problem is that he has very strong
ideas as to what "mathematics" should consist of. According to the
OP, some rings that have a zero divisor is 'broken.' Any ring that is
not a unique factorization domain is 'broken.'
hagman
What does "mentioning" 7 mean? I almost guess your getting weird even
by oyur standards.
Let's see
7*(x^2 + 6x-7) = (7x+49)*(x-1)
Divide off the 7
(x^2 + 6x-7) = (x+7)*(x-1)
and the result still "mentions" 7. LOL
MichaelW
>
> The problem with OP is that he thinks he understand more mathematics
> than other people. Another big problem is that he has very strong
> ideas as to what "mathematics" should consist of. According to the
> OP, some rings that have a zero divisor is 'broken.' Any ring that is
> not a unique factorization domain is 'broken.'
I have used James' original argument in a slightly modified form to
"prove" that the ring of integers is broken. Presumably this means
that maths took a wrong turn 20,000 years ago when people started
counting! Any ring can be "proved" invalid using this logic since at
its core is a stealth illegal division.
BTW his claim in this thread that
2*2 = 0 mod 4
can be simplified to
2 = 0 mod 2
is so awful that I had to read several times before I could believe he
really wrote it. Definitely a keeper.
Regards, Michael W.
JSH
Oh, that was stupid. I'm still wrong!
> ... I don't think I understand what you mean by "dividing off."
>
> You are allowed to switch rings?
Nope. Made another mistake in reply.
> > I saw "mod" and didn't look more closely until this reply, so yes, 2*2
> > = 0 mod 4 is an identity, but you can divide off the 2.
>
> > You seem to believe that what's on the other side of "mod" is locked.
>
> Sigh.
>
> Do you not understand that Z_4 and Z_2 are two different rings?
>
> When you change from mod 4 to mod 2, you are jumping from one ring
> into a different ring.
Yup. I was wrong.
> You were earlier making a huge fuss when MichaelW went from integers
> to rational numbers, but you are okay going from Z_4 to Z_2?
I made a mistake.
> Do you understand that in the example above, you had to go from a
> ring, Z_4, into a different ring Z_2? So in fact, the division didn't
> take place in Z_4?
>
> Here's another example.
>
> [ 0 1 ] ^2 = [ 0 0 ]
> [ 0 0 ] [ 0 0 ]
>
> Now, tell me if you can divide anything off here. There are many
> other examples, but typing matrices is a pain, so this is what I'll
> leave it as.
I hate matrices. But in any event, I was wrong before. I made a
mistake, as I've noted repeatedly above.
James Harris
Bacle
Yet, instead of being cautious next time you make a claim, you will open your mouth , spout out nonsense with full vehemence. You seem to learn slowly, if at all.
>
> > > I saw "mod" and didn't look more closely until
> this reply, so yes, 2*2
> >