Algebraic integer result with quadratics
JSH
of algebraic integers where quadratics thankfully are useful for a
somewhat subtle result, and the distributive property is almost
bizarrely the linchpin of the proof.
Consider in an integral domain
P(x) = 175x^2 - 15x + 2
and
7*P(x) = (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 I have a non-monic quadratic with integer coefficients which is
multiplied times 7 and then factored using functions given by the
roots of a monic quadratic expression, so I've entangled one quadratic
with a quadratic generator in such a way as to allow myself to probe
at the ring of algebraic integers in a way never before done without
such remarkable tools.
Of course with a regular polynomial factorization like
7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
it's trivial to just divide the 7 off to get
(x^2 + 3x + 2) = (x + 1)*(x+2)
but by entangling the two expressions and making the functional values
roots of monic quadratics with integer coefficients when x is an
integer, I've managed to FORCE a division of the 7 across non-rational
solutions, and in that way remove the ability to in general divide it
off from
7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
in the ring of algebraic integers.
That was done by a clever re-grouping of terms:
7*P(x) = 1225x^2 - 105x + 14
and with the re-grouping I have
7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2
to get
7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)
which clearly gives
a^2 - (7x-1)a + (49x^2 - 14x) = 0,
But what's the point of the exercise?
Well by entangling the function into the roots of a quadratic
generator I can put the distributive property to the test! As there
isn't a lot of difference really between
7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
and
7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, because
the distributive property doesn't care what is being multiplied, so in
the first case you have (x+1)*(x+2), while in the second you have some
weird unseen factorization where more complicated than linear
functions are involved as yes, x+1 and x+2 are just linear functions.
So in EACH case you have FUNCTIONS involved but in the more
complicated case I forced the functions to be more complicated by
making them roots of a quadratic generator.
So the two examples are the same mathematically except in the second
case the functions are not linear, which is why it's a non-polynomial
factorization, which allows non-rational solutions with rational x.
So as they are mathematically the same in all key ways, not
surprisingly, in each case you simply have 7 multiplied rather simply,
which can be seen with the more complex example by letting x=0 as then
a^2 - (7x-1)a + (49x^2 - 14x) = 0
gives
a^2 + a = 0
so one of the a's is 0 and the other is -1, so I have from
7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
that
7*P(0) = (7)*(2) or (2)*(7).
So one of (5a_1(x) + 7) or (5a_2(x)+ 7) is coprime to 7 at that value
just like with
7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
one of them is coprime to 7 at x=0, and since I've mentioned
coprimeness I'll go ahead and say that we'll try to be in the ring of
algebraic integers.
I say try because you may already know that now there is a HUGE
problem as if integer x generates a quadratic that is irreducible over
Q, then you already know that NEITHER of the a's can have 7 as a
factor in that ring!!!
For example with x=1, the result would need that for
a^2 - 6a + 35 = 0
one of the roots has 7 as a factor, but provably NEITHER of them do in
the ring of algebraic integers.
And that is the quick demonstration of the major issue over which so
many arguments have raged in the roughly six years or so since I
discovered the problem.
After pondering things for a while I concluded that the problem was
not with my understanding of the distributive property--which is after
all rather basic--but with the ring of algebraic integers itself,
which lead me to wondering what ring might be ok, and years ago I came
up with what I call the ring of objects:
The object ring is defined by two conditions, and includes all numbers
such that these conditions are true:
1. 1 and -1 are the only rationals that are units in the ring.
2. Given a member m of the ring there must exist a non-zero member n
such that mn is an integer, and if mn is not a factor of m, then n
cannot be a unit in the ring.
In that ring I found I didn't have a problem! And can simply divide
the 7 off again like before:
P(x) = (5b_1(x) + 1)(5a_2(x)+ 7) or (5a_1(x) + 7)(5b_2(x)+ 1)
where a_1(x) = 7*b_1(x) or a_2(x) = 7*b_2(x),
where you have an ambiguity as to which one unless you have rational
values for the functions.
AND I went ahead and figured out how the ring of algebraic integers
can allow dividing off of the 7 as well, using what I call wrappers,
and you can find my research on that subject by searching in Google on
"wrapper theorem".
The problem though with this issue is that it shows that the ring of
algebraic integers has additional properties beyond those previously
known, and you can do some odd things with it.
Unfortunately some of those things include coming up with mathematical
arguments that are not actually proofs as they rely on the quirks of
the ring to imply something false.
James Harris
Rotwang
>
> [...]
>
> After pondering things for a while I concluded that the problem was
> not with my understanding of the distributive property--which is after
> all rather basic--but with the ring of algebraic integers itself,
> which lead me to wondering what ring might be ok, and years ago I came
> up with what I call the ring of objects:
>
> The object ring is defined by two conditions, and includes all numbers
> such that these conditions are true:
>
> 1. 1 and -1 are the only rationals that are units in the ring.
>
> 2. Given a member m of the ring there must exist a non-zero member n
> such that mn is an integer, and if mn is not a factor of m, then n
> cannot be a unit in the ring.
>
> In that ring I found I didn't have a problem! And can simply divide
> the 7 off again like before:
>
> P(x) = (5b_1(x) + 1)(5a_2(x)+ 7) or (5a_1(x) + 7)(5b_2(x)+ 1)
>
> where a_1(x) = 7*b_1(x) or a_2(x) = 7*b_2(x),
>
> where you have an ambiguity as to which one unless you have rational
> values for the functions.
James, from reading your posts in the "Distributive property and key
result" thread, my understanding of your concept of "wrappers" is
this: when Rick and Dale provide alternative ways to factor 7 out of
your product
7 * P(x) = (5a_1(x) + 7)(5a_2(x) + 7)
in such a way that all coefficients remain in the algebraic integers,
you say that in fact the factorisations 7 = w_1(x)*w_2(x) they found
differ from the factorisation 7 = 1*7 (or 7*1) by numbers which are
not units in the algebraic integers, but which are units in the Object
ring, is this correct?
If so, we have the following theorem: the ring of Objects contains non-
integer rationals. Proof:
I will first quote from one of your posts in the above mentioned
thread:
On 29 Aug, 01:34, JSH <jst...@gmail.com> wrote:
> On Aug 28, 2:32 am, "W. Dale Hall"
>
> <wdunderscorehallatpacbelldotnet@last> wrote:
> > Rick Decker wrote:
> > > marcus_b wrote:
> > >> On Aug 26, 8:06 pm, JSH <jst...@gmail.com> wrote:
>
> > > <snip>
>
> > >>> Of course I've given an example, which I'll put here so that readers
> > >>> can understand what you're saying:
>
> > >>> P(x) = 175x^2 - 15x + 2
>
> > >>> and readers can notice that there is no w_1(x) or w_2(x) where 7 =
> > >>> w_1(x)*w_2(x) there at all.
>
> > >> Wrong. 7 CAN be written in the form 7 = w_1(x)*w_2(x).
>
> > > Part of James' problem is that the w's aren't
> > > exactly obvious. For example, with x = 1 we can find
> > > that
>
> > > a_1(1) = 3 + sqrt(-26)
> > > a_2(1) = 3 - sqrt(-26)
>
> > > and we can find (with some help from Mathematica)
>
> > > w_1(1) = (307 + 30 * sqrt(-26))^{1/6}
> > > w_2(1) = (307 - 30 * sqrt(-26))^{1/6}
>
> > > with w_1(1) | a_1(1), w_2(1) | a_2(1) and
>
> > > 7 = w_1(1) * w_2(1)
>
> > > He seems to think that if he can't find the factors of
> > > 7 then they don't exist. As you said, they will
> > > always be there for any integer x, even though
> > > they might not be obvious.
>
> > A couple more non-obvious values for these factors:
>
> > If x = 2:
>
> > w(2) = (109173808 +/- 32965641 sqrt(-26))^(1/21)
>
> > If x = 3:
>
> > w(3) = (745 +/- 132 sqrt(-299))^(1/8)
>
> Those are what I call wrappers.
Consider the values of w(1) and w(2) given above, namely
w_1(1) = (307 + 30 * sqrt(-26))^{1/6}
w_2(1) = (307 - 30 * sqrt(-26))^{1/6}
w_1(2) = (109173808 + 32965641 sqrt(-26))^(1/21)
w_2(2) = (109173808 - 32965641 sqrt(-26))^(1/21)
Now, assuming that I have correctly understood what a "wrapper" is,
you are saying that one of the two values for w(1)/7 is a unit in the
object ring, and similarly for one of the two values of w(2)/7. Let's
suppose for the sake of argument that we may take the plus sign in
both cases (so that our value of w(1)^6 has positive imaginary part,
and similarly for w(2)^21). Since the ring of objects contains the
numbers
x = (w_1(1)/7)^6 = 1/7^6 * (307 + 30 * sqrt(-26))
y = (w_1(2)/7)^21 = 1/7^21 * (109173808 + 32965641 * sqrt(-26))
It also contains the number
10988547 * x - 10* 7^15 * y = 1/7^6 * 2281745849
It is easy to check that this is not an integer. There are three other
cases to consider, namely the cases where we assume that the pairs
(w_1(1)/7, w_2(2)/7), (w_2(1)/7, w_1(2)/7) and (w_2(1)/7, w_2(2)/7)
are in the object ring. It is easy to check that each of these
assumptions leads, by the same method, to a proof that the Object ring
contains non-integer rationals.
Note that if it contains any non-integer rational then necessarily it
contains a integer units other than 1 or -1.
JSH
You can't so assume.
A problem here is that mathematicians have long said things like,
sqrt(4) is 2. But that is false, as the sqrt(4) is 2 or -2.
The ambiguity CANNOT BE REMOVED but students learn that sqrt(4) is
defined to be 2, which is like legislating the ambiguity away, but -2
still works even if you say sqrt(4) is 2.
That's because mathematically sqrt(x) is saying that the given number
when multiplied times itself gives x. And -2 works just fine as (-2)
(-2) = 4.
Oddly enough from my reading into the history, part of the reason for
the attempt to remove the negative solution was a notion back in math
history that functions should only return a single answer, so
mathematicians at that time tried to force all functions to do so,
which just doesn't work.
So it was just kind of a fashion trend in mathematics that went away
but left vestiges like people saying that the sqrt() function only
returns the positive value.
James Harris
Rotwang
And I didn't. Had you bothered to read the rest of my article you
would have seen that I consider the remaining possibilities for the
square roots later, and I pointed out that no matter which combination
of square roots with positive or negative imaginary part you take you
still end up with non-integer rationals in your ring.
>
> A problem here is that mathematicians have long said things like,
> sqrt(4) is 2. But that is false, as the sqrt(4) is 2 or -2.
>
> The ambiguity CANNOT BE REMOVED but students learn that sqrt(4) is
> defined to be 2, which is like legislating the ambiguity away, but -2
> still works even if you say sqrt(4) is 2.
>
> That's because mathematically sqrt(x) is saying that the given number
> when multiplied times itself gives x.
No, that isn't what sqrt(x) is saying. That may be what /you/ mean
when you write sqrt(x), but the fact that you insist on using a
private language is your problem, not that of mathematics. If you call
a tail a leg, how many legs does a dog have?
Joshua Cranmer
> A problem here is that mathematicians have long said things like,
> sqrt(4) is 2. But that is false, as the sqrt(4) is 2 or -2.
The sqrt, or its equivalent, the radical, unambiguously refers to the
positive solution, the principal . What you mean to say is that the
solution to the equation x^2 = 4 is x = +/- 2.
From Wikipedia:
Every non-negative real number x has a unique non-negative square root,
called the principal square root and denoted with a radical symbol as √x.
From PlanetMath:
The square root of a nonnegative real number $x$, written as $\sqrt{x}$,
is the unique nonnegative real number $y$ such that $y^2=x$.
From Mathworld:
Any nonnegative real number x has a unique nonnegative square root r;
this is called the principal square root and is written r=x^(1/2) or r=√x.
sqrt(x) is the version of √x that can actually be written in ASCII text.
> Oddly enough from my reading into the history, part of the reason for
> the attempt to remove the negative solution was a notion back in math
> history that functions should only return a single answer, so
> mathematicians at that time tried to force all functions to do so,
> which just doesn't work.
It's not that a function "should" return a single answer, it's that a
function must return a single answer. Definitions:
From Wikipedia:
The mathematical concept of a function expresses dependence between two
quantities, one of which is given (the independent variable, argument of
the function, or its "input") and the other produced (the dependent
variable, value of the function, or "output"). A function associates a
single output to each input element drawn from a fixed set, such as the
real numbers.
From PlanetMath:
A function is a triplet $(f,A,B)$ where:
1. $A$ is a set (called the domain of the function).
2. $B$ is a set (called the codomain of the function).
3. $f$ is a binary relation between $A$ and $B$.
4. For every $a \in A$, there exists $b \in B$ such that $(a,b) \in f$.
5. If $a \in A$, $b_1,b_2 \in B$, and $(a,b_1) \in f$ and $(a,b_2)
\in f$, then $b_1 = b_2$.
From Mathworld:
A function is a relation that uniquely associates members of one set
with members of another set. More formally, a function from A to B is an
object f such that every a in A is uniquely associated with an object
f(a) in B.
Where does it imply that a function can have two outputs?
[Quoting from multiple sources to avoid people screaming at me that WP
is unreliable. I could dredge up more if you really want to.]
Joshua Cranmer
Joshua Cranmer wrote:
> It's not that a function *should* return a single answer, it's that a
> function *must* return a single answer. Definitions:
>
> From Wikipedia:
> The mathematical concept of a function expresses dependence between two
> quantities, one of which is given (the independent variable, argument of
> the function, or its "input") and the other produced (the dependent
> variable, value of the function, or "output"). A function associates a
> *single* output to each input element drawn from a fixed set, such as the
> real numbers.
>
> From PlanetMath:
> A function is a triplet $(f,A,B)$ where:
>
> 1. $A$ is a set (called the domain of the function).
> 2. $B$ is a set (called the codomain of the function).
> 3. $f$ is a binary relation between $A$ and $B$.
> 4. For every $a \in A$, there exists $b \in B$ such that $(a,b) \in f$.
> ** 5. If $a \in A$, $b_1,b_2 \in B$, and $(a,b_1) \in f$ and $(a,b_2)
> \in f$, then $b_1 = b_2$.
>
> From Mathworld:
> A function is a relation that *uniquely* associates members of one set
> with members of another set. More formally, a function from A to B is an
> object f such that every a in A is *uniquely* associated with an object
> f(a) in B.
JSH
Same numbers.
For example, there is no difference between 1+sqrt(4) and 1-sqrt(4) as
in both cases you have
-1 or 3.
> w_1(2) = (109173808 + 32965641 sqrt(-26))^(1/21)
> w_2(2) = (109173808 - 32965641 sqrt(-26))^(1/21)
Same numbers.
w_1(x) is indistinguishable from w_2(x)
> Now, assuming that I have correctly understood what a "wrapper" is,
> you are saying that one of the two values for w(1)/7 is a unit in the
> object ring, and similarly for one of the two values of w(2)/7. Let's
> suppose for the sake of argument that we may take the plus sign in
> both cases (so that our value of w(1)^6 has positive imaginary part,
> and similarly for w(2)^21). Since the ring of objects contains the
> numbers
>
> x = (w_1(1)/7)^6 = 1/7^6 * (307 + 30 * sqrt(-26))
> y = (w_1(2)/7)^21 = 1/7^21 * (109173808 + 32965641 * sqrt(-26))
you cannot distinguish w_1(x) from w_2(x)
> It also contains the number
>
> 10988547 * x - 10* 7^15 * y = 1/7^6 * 2281745849
So that is incorrect.
I tried to explain in another reply where I noted that sqrt(4) has TWO
VALUES.
But you CAN eliminate the radicals with your example, and hold to the
assumption that if 7 is a factor of one of
307 +/- 30 * sqrt(-26)
and is also a factor of one of
109173808 -/+ 32965641 sqrt(-26)
then it must be a factor of one of the results when sqrt(-26) is
eliminated.
Actually, you can one-up that and say that 7^6 must be a factor I
think.
I don't feel like checking, as if it doesn't work then I need to be
sure you have the correct solutions for the w's, and it's kind of a
tedious exercise anyway.
James Harris
Rotwang
Only according to your private definition. I'm not willing to indulge
that. But it makes no difference to the argument at hand, so let's
ignore that.
>
> For example, there is no difference between 1+sqrt(4) and 1-sqrt(4) as
> in both cases you have
>
> -1 or 3.
>
> > w_1(2) = (109173808 + 32965641 sqrt(-26))^(1/21)
> > w_2(2) = (109173808 - 32965641 sqrt(-26))^(1/21)
>
> Same numbers.
>
> w_1(x) is indistinguishable from w_2(x)
>
> > Now, assuming that I have correctly understood what a "wrapper" is,
> > you are saying that one of the two values for w(1)/7 is a unit in the
> > object ring, and similarly for one of the two values of w(2)/7. Let's
> > suppose for the sake of argument that we may take the plus sign in
> > both cases (so that our value of w(1)^6 has positive imaginary part,
> > and similarly for w(2)^21). Since the ring of objects contains the
> > numbers
>
> > x = (w_1(1)/7)^6 = 1/7^6 * (307 + 30 * sqrt(-26))
> > y = (w_1(2)/7)^21 = 1/7^21 * (109173808 + 32965641 * sqrt(-26))
>
> you cannot distinguish w_1(x) from w_2(x)
Sure you can. One has positive imaginary part when raised to the 6th
power, the other has negative imaginary part when raised to the 6th
power. But that's not important here.
>
> > It also contains the number
>
> > 10988547 * x - 10* 7^15 * y = 1/7^6 * 2281745849
>
> So that is incorrect.
>
> I tried to explain in another reply where I noted that sqrt(4) has TWO
> VALUES.
Yes, you tried to explain twice. And both times you deleted the part
of my post where I explained why your explanation makes no difference.
Here it is again:
> There are three other
> cases to consider, namely the cases where we assume that the pairs
> (w_1(1)/7, w_2(2)/7), (w_2(1)/7, w_1(2)/7) and (w_2(1)/7, w_2(2)/7)
> are in the object ring. It is easy to check that each of these
> assumptions leads, by the same method, to a proof that the Object ring
> contains non-integer rationals.
If you don't like my referring to w_1 and w_2 as different numbers,
you can instead rewrite the above as referring to the four possible
combinations of values of w_1(1)/7 and w_1(2)/7. Either way, the
alleged ambiguity of the square root makes no difference: WHICHEVER of
the possible values of w(1)/7 and w(2)/7 are in the Object ring, the
Object ring is shown to contain non-integer rationals.
>
> But you CAN eliminate the radicals with your example, and hold to the
> assumption that if 7 is a factor of one of
>
> 307 +/- 30 * sqrt(-26)
>
> and is also a factor of one of
>
> 109173808 -/+ 32965641 sqrt(-26)
>
> then it must be a factor of one of the results when sqrt(-26) is
> eliminated.
>
> Actually, you can one-up that and say that 7^6 must be a factor I
> think.
Yes, that's exactly right. This is my point: you can eliminate the
radical, in such a way that you are left with an integer which has 7^6
as a factor in your ring. But that same integer does not have 7^6 as a
factor in the ring of integers. So your wrapper theorem implies that
your Object ring contains non-integer rationals, in direct
contradiction with its definition.
marcus_b
Assuming a_1(0) = 0, you think the distributive property
gives you:
P(x) = (5a_1(x)/7 + 1)*(5a_2(x) + 7).
Here's what the distributive property actually says:
a * (b + c) = a*b + a*c.
Notice: no division involved.
You are not trying to distribute 7 across your factors.
You are trying to distribute 1/7. The problem being, 1/7
is not in your ring. It is not even in your "object ring".
The distributive property is a statement about multiplication,
not division.
When you write 5a_1(x)/7, and declare that that MUST be
an element of your ring, you are just assuming what you
want to prove: that a_1(x) is divisible by 7. True, a_1(0)
is divisible by 7, because a_1(0) = 0 and 0 is divisible by
anything. But that does not by any means imply a_1(x) is
divisible by 7 for x <> 0.
Your real mistake is, you think 7 must divide out of
the product
(5a_1(x) + 7)*(5a_2(x) + 7)
in a constant way. That is, you think that how 7 divides
out of that product must be independent of x. The
distributive property does NOT tell you that. Again,
the distributive property is
a * (b + c) = a*b + a*c.
The distributive property in rings does NOT say
(b + c)/a = b/a + c/a.
That sounds wrong, doesn't it? After all, the
above equation is true in the real numbers when
a <> 0. It is true in the complex numbers. Those
are both rings. Why isn't it true in other rings?
Because b/a and c/a in general are not elements of
those rings. Division is not a ring operation. Take
the ring of integers. In that ring, is it true that
(2 + 3)/7 = 2/7 + 3/7 ?
No! 2/7 and 3/7 are not integers! The equation doesn't
make any sense in the ring of integers! It doesn't even
EXIST in that ring.
So when you say
(5a_1(x) + 7)/7 = 5a_1(x)/7 + 1,
is true in your ring, you are really saying that
5a_1(x) / 7 is an element of your ring. This
means you are ASSUMING that a_1(x) is divisible by 7.
I.e., you are assuming what you want to prove.
So you DEFINE your ring to include a_1(x)/7. That's
your object ring. That, you think, enables you to
factor things the way you think they should be
factored. But the fact is, you can factor 7
out of
(5a_1(x) + 7)*(5a_2(x) + 7),
that is, you CAN compute
(5a_1(x) + 7)*(5a_2(x) + 7)/7
and have the result expressed only in terms of
algebraic integers. You don't NEED the object
ring. There is already a perfectly good way to
factor 7 out of
(5a_1(x) + 7)*(5a_2(x) + 7)
without invoking some new ring. That is what
you say is impossible. It is also what Dale
Hall showed you how to do explicitly for x = 1,
2, and 3. It can be done for any integer, but
the computations are difficult. That is why Hall
had to use a program written exactly to carry out
computations in algebraic number theory. Gee,
computer verification of math results! Isn't that
something you're in favor of? And of course
something you haven't EVER done with your own
claims.
> in the ring of algebraic integers.
>
> That was done by a clever re-grouping of terms:
>
> 7*P(x) = 1225x^2 - 105x + 14
>
> and with the re-grouping I have
>
> 7*P(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 7^2
>
> to get
>
> 7*P(x) = (5a_1(x) + 7)(5a_2(x)+ 7)
>
That is, you are regarding 7*P(x) as if it were
a polynomial in the symbol "5".
> which clearly gives
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0,
>
> But what's the point of the exercise?
>
> Well by entangling the function into the roots of a quadratic
> generator I can put the distributive property to the test! As there
> isn't a lot of difference really between
>
> 7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
>
> and
>
> 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
>
No there is a very CRUCIAL difference. The
first polynomial is reducible. The second in general
is not. This makes an enormous difference. Do not
say they are the same.
> where the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0, because
> the distributive property doesn't care what is being multiplied, so in
> the first case you have (x+1)*(x+2), while in the second you have some
> weird unseen factorization where more complicated than linear
> functions are involved as yes, x+1 and x+2 are just linear functions.
>
> So in EACH case you have FUNCTIONS involved but in the more
> complicated case I forced the functions to be more complicated by
> making them roots of a quadratic generator.
>
> So the two examples are the same mathematically
NOT true as noted above.
> except in the second
> case the functions are not linear, which is why it's a non-polynomial
> factorization, which allows non-rational solutions with rational x.
>
> So as they are mathematically the same in all key ways,
They are NOT the same. There is a very important difference
which you seem unable to recognize between reducible and irreducible
polynomials.
Saying two things are the same when they are not - what is
the label for that logical error? Just plain stupid?
> not
> surprisingly, in each case you simply have 7 multiplied rather simply,
> which can be seen with the more complex example by letting x=0 as then
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0
>
> gives
>
> a^2 + a = 0
>
> so one of the a's is 0 and the other is -1, so I have from
>
> 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
>
> that
>
> 7*P(0) = (7)*(2) or (2)*(7).
>
> So one of (5a_1(x) + 7) or (5a_2(x)+ 7) is coprime to 7 at that value
Yes. AT THAT VALUE, i.e., when x = 0. That is all you have proved.
> just like with
>
> 7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
>
No, NOT just like with. Factorizations of reducible and
irreducible polynomials are not the same.
> one of them is coprime to 7 at x=0, and since I've mentioned
> coprimeness I'll go ahead and say that we'll try to be in the ring of
> algebraic integers.
>
> I say try because you may already know that now there is a HUGE
> problem as if integer x generates a quadratic that is irreducible over
> Q, then you already know that NEITHER of the a's can have 7 as a
> factor in that ring!!!
>
> For example with x=1, the result would need that for
>
> a^2 - 6a + 35 = 0
>
> one of the roots has 7 as a factor, but provably NEITHER of them do in
> the ring of algebraic integers.
>
Correct.
So it's clear what you are thinking. You know that neither root
has 7 as a factor and neither root is coprime to 7 in the ring of
algebraic numbers. Which means that within that ring, there is
a way to factor 7 out of
(5a_1(x) + 7)*(5a_2(x) + 7),
WITHOUT factoring it out of either factor individually. OK. But
you don't LIKE that factorization. You want the factorization to
be like what happens with reducible quadratics. So you say: there
is another ring where I get the kind of factorization I like. To
which the right response is: SO WHAT? There is no need to invoke
a new ring. There is perfectly good factorization in the ring
of algebraic integers. The part you don't like is, that factorization
is not constant. It is a function of x.
But that does not make it wrong, or illogical, or mathematically
incorrect. That just makes it something other than what you want.
But guess what, wienie.
THE MATH DOESN'T CARE WHAT YOU WANT.
Sound familiar?
> And that is the quick demonstration of the major issue over which so
> many arguments have raged in the roughly six years or so since I
> discovered the problem.
>
> After pondering things for a while I concluded that the problem was
> not with my understanding of the distributive property--
No, that is exactly where the problem is. Incredible as it may
seem, you do not understand the distributive property. It is not
about division. It is about multiplication.
> which is after
> all rather basic--but with the ring of algebraic integers itself,
> which lead me to wondering what ring might be ok, and years ago I came
> up with what I call the ring of objects:
>
> The object ring is defined by two conditions, and includes all numbers
> such that these conditions are true:
>
> 1. 1 and -1 are the only rationals that are units in the ring.
>
> 2. Given a member m of the ring there must exist a non-zero member n
> such that mn is an integer, and if mn is not a factor of m, then n
> cannot be a unit in the ring.
>
Consider
a^2 - 6a + 35.
Let r_1 and r_2 be the roots.
Either r_1/7 or r_2/7 must be in your 'object ring'.
Which one? Are they both in it? You have just given a
"definition" of the 'object ring'. You must be able to tell from
that which numbers are in it and which are not. Is it r_1, or r_2,
or both?
> In that ring I found I didn't have a problem! And can simply divide
> the 7 off again like before:
>
> P(x) = (5b_1(x) + 1)(5a_2(x)+ 7) or (5a_1(x) + 7)(5b_2(x)+ 1)
>
> where a_1(x) = 7*b_1(x) or a_2(x) = 7*b_2(x),
>
> where you have an ambiguity as to which one unless you have rational
> values for the functions.
>
To put it another way, you just define a new ring where
the expression factors the way you want. Never mind that
there is already a perfectly good factorization - which you
think either cannot or should not exist - in the ring of
algebraic numbers. Is that ALL you have done? Define a new
ring which has a property that you want? Does that somehow
remove what you claim is a "contradiction" involving the ring
of algebraic integers? You have claimed to have found a
"core error", and that mathematicians having been doing
things wrong for over 100 years. What is the core error?
Is there a contradiction or not? Other than not giving
factorizations of the form that you prefer, what is WRONG
with the ring of algebraic integers?
> AND I went ahead and figured out how the ring of algebraic integers
> can allow dividing off of the 7 as well, using what I call wrappers,
> and you can find my research on that subject by searching in Google on
> "wrapper theorem".
>
Look, you either have produced a contradiction or you haven't.
If you haven't, i.e., if you admit that Dale Hall is right that
factorizations exist where 7 is split up into w_1(x) and w_2(x),
then there is really no problem. If you HAVE produced a
contradiction,
then no 'wrapper theorem' is going to fix it.
But it's clear now. There are factorizations as Dale Hall
describes. The problem for you being, you don't LIKE them.
This is not about logic or math. It is only about your
preferences.
> The problem though with this issue is that it shows that the ring of
> algebraic integers has additional properties beyond those previously
> known, and you can do some odd things with it.
>
> Unfortunately some of those things include coming up with mathematical
> arguments that are not actually proofs as they rely on the quirks of
> the ring to imply something false.
>
Now what are you saying? That what you have said is a
proof is not actually a proof? Doesn't this pretty much
undermine everything you have claimed?
Marcus.
> James Harris
W. Dale Hall
So, what's to prevent you from using the convention that for positive x,
sqrt(x) refers to the positive real number r for which r^2 = x? That
number surely exists, and is unique. If one were simply to rewrite
sqrt(x) as |sqrt(x)|, then that value would be obtained. It is not
a mathematical error to impose such a convention, if only to simplify
discussion.
BTW, I don't notice you having expressed similar qualms about any
odd-order roots, but you surely must be aware that any complex number
(so including reals) has n roots of order n.
>
>> w_1(2) = (109173808 + 32965641 sqrt(-26))^(1/21)
>> w_2(2) = (109173808 - 32965641 sqrt(-26))^(1/21)
A correction: Rick Decker has noted (correctly that my values for
w's here are in error: the sqrt() radicand should be -503, and
not -26. He was kind enough to observe this as the typographical
error that it was.
>
> Same numbers.
>
> w_1(x) is indistinguishable from w_2(x)
That is just not true. There is a convention that we observe with
the definition of the square root. You are aware of that convention,
but refuse to admit it. Fine. There are two values of w here, one
of which uses the positive number whose square is 503, and the other
uses the negative number whose square is 503. All I have done is to
have spelled out the convention in unambiguous terms.
Maybe you want to say that there is no positive number whose square
is 503, and that there is no negative number square is 503. If so,
you shouldn't have any trouble making that claim explicit.
That having been said, there are *two* numbers that have been
displayed. Their product is 7, as you could surely check, by
removing the (1/21) as exponent, multiplying the results, and
then comparing to 7^21.
>
>> Now, assuming that I have correctly understood what a "wrapper" is,
>> you are saying that one of the two values for w(1)/7 is a unit in the
>> object ring, and similarly for one of the two values of w(2)/7. Let's
>> suppose for the sake of argument that we may take the plus sign in
>> both cases (so that our value of w(1)^6 has positive imaginary part,
>> and similarly for w(2)^21). Since the ring of objects contains the
>> numbers
>>
>> x = (w_1(1)/7)^6 = 1/7^6 * (307 + 30 * sqrt(-26))
>> y = (w_1(2)/7)^21 = 1/7^21 * (109173808 + 32965641 * sqrt(-26))
>
> you cannot distinguish w_1(x) from w_2(x)
However, if one takes the convention regarding the square root function
one can produce numbers that are distinguishable. Don't be silly.
>
>> It also contains the number
>>
>> 10988547 * x - 10* 7^15 * y = 1/7^6 * 2281745849
>
> So that is incorrect.
>
> I tried to explain in another reply where I noted that sqrt(4) has TWO
> VALUES.
>
> But you CAN eliminate the radicals with your example, and hold to the
> assumption that if 7 is a factor of one of
^^^^^^^^^^ ... note it's an assumption
>
> 307 +/- 30 * sqrt(-26)
>
> and is also a factor of one of
>
> 109173808 -/+ 32965641 sqrt(-26)
>
> then it must be a factor of one of the results when sqrt(-26) is
> eliminated.
The second expression should have (-503) as the radicand, as
I mentioned above.
Of course, what you've said is true, because 7 is NOT a factor of
307 + 30 sqrt(-26). Since False ==> True, the implication is true.
To show that divisibility doesn't hold:
If I divide this number by 7, it satisfies this equation:
7 x^2 - 614 x + 16807 = 0
While 7 divides the leading coefficient and the constant term (= 7^5),
614 factors as 2 x 307, both of which are prime. In other words, this
is a non-monic polynomial that is primitive (gcd of its coefficients
is 1). It has discriminant -93600 = - 2^5 x 3^2 x 5^2 x 13, which is
not a square integer, and so there are no rational roots: thus, it is
irreducible over the rationals.
That tells us that w/7 is not an algebraic integer. In turn, this
implies that w/7^6 is not an algebraic integer.
>
> Actually, you can one-up that and say that 7^6 must be a factor I
> think.
>
> I don't feel like checking, as if it doesn't work then I need to be
> sure you have the correct solutions for the w's, and it's kind of a
> tedious exercise anyway.
>
I presume you have a computer. I have already given a source for the
program I use (Kash/Kant), but here it is again:
http://www.math.tu-berlin.de/~kant/kash.html
It's free, and it works under MS-Windows. It is a simple download
and installation, and the above page has a link to a documents page
with user manuals in .pdf format.
There's actually a link to a web-based version, in which you submit
lines of code & click a button to get it evaluated. It is limited to
20 seconds of run time, and it has the disadvantage that if you don't
quite know the syntax, it can be difficult to get something to run.
>
> James Harris
Dale
W. Dale Hall
Why wouldn't you regard this as a shortcoming in your presumed ring?
After all, you can't tell whether a given element is a multiple of 7.
You seem to be saying that your desired conclusion "irrationals should
have the same properties as rationals", still doesn't hold in your ring.
That problem simply does not exist in the algebraic integers. In fact,
given any two algebraic integers p and q, the question of whether one
is a multiple of the other (in the ring of algebraic integers) can
always be resolved. It's "just" a matter of establishing whether the
ideal <p q> is generated by a single element (i.e., whether the
ideal is principal), and if so, finding a generator g, and
determining whether either of the algebraic integers p/g or q/g is
a unit.
>
> AND I went ahead and figured out how the ring of algebraic integers
> can allow dividing off of the 7 as well, using what I call wrappers,
> and you can find my research on that subject by searching in Google on
> "wrapper theorem".
>
> The problem though with this issue is that it shows that the ring of
> algebraic integers has additional properties beyond those previously
> known, and you can do some odd things with it.
>
> Unfortunately some of those things include coming up with mathematical
> arguments that are not actually proofs as they rely on the quirks of
> the ring to imply something false.
>
This is an odd way of admitting that your earlier arguments were flawed.
BTW, the "quirks of the ring" as you term them, were seen over 100 years
ago, with the recognition that unique factorization is not a universal
property of rings. If you had taken a course in algebra, that would not
have come as any sort of surprise.
What I'm saying, of course, is that you could do worse than taking some
course in algebra that covers field theory & commutative rings. I'd have
to say, however, that there is a bit of machinery to be mastered, and
all you want might not be covered in a single course.
>
> James Harris
Dale
Rotwang
<wdunderscorehallatpacbelldotnet@last> wrote:
> > On Aug 30, 12:46 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
>
> >> w_1(2) = (109173808 + 32965641 sqrt(-26))^(1/21)
> >> w_2(2) = (109173808 - 32965641 sqrt(-26))^(1/21)
>
> A correction: Rick Decker has noted (correctly that my values for
> w's here are in error: the sqrt() radicand should be -503, and
> not -26. He was kind enough to observe this as the typographical
> error that it was.
Oh. That sinks my proof that James' ring contains non-integer
rationals. Do you know whether other values of x will give w's which
will salvage it?
Note incidentally that I don't need any other values of x if I assume
that the Object ring is closed under complex conjugation. Question for
James: is it?
JSH
Not possible. Besides W. Dale Hall has already posted that he had an
error in his w's, and he posted a correction.
Turns out that regardless of what you believe, or I believe, if you
eliminate the radicals then one of the two possible numbers left MUST
have 7 as a factor. Duh.
So it's a wasted exercise in futility.
If you do NOT eliminate the radicals then my argument applies that you
can't distinguish between square roots as sqrt(4), for instance is 2
or -2.
So there is an impasse.
I've tried the line of approach you're using, before and ran into the
same conclusion, years ago.
In contrast my original post steps through some basic algebra and
relies on the distributive property.
Maybe if you work at it for a few years actually trying to figure out
what's going on, you'll understand that's true, and then understand
the mathematical solution that removes the appearance of
contradiction.
Maybe.
James Harris
JSH
> On 31 Aug, 01:01, "W. Dale Hall"
>
> <wdunderscorehallatpacbelldotnet@last> wrote:
> > > On Aug 30, 12:46 pm, Rotwang <sg...@hotmail.co.uk> wrote:
>
> > >> w_1(2) = (109173808 + 32965641 sqrt(-26))^(1/21)
> > >> w_2(2) = (109173808 - 32965641 sqrt(-26))^(1/21)
>
> > A correction: Rick Decker has noted (correctly that my values for
> > w's here are in error: the sqrt() radicand should be -503, and
> > not -26. He was kind enough to observe this as the typographical
> > error that it was.
>
> Oh. That sinks my proof that James' ring contains non-integer
Ugh!!! Something false is NOT PROOF!!!!!!!!!!!!!
So you did NOT have a proof and it's a horribly dangerous error to
claim you had a proof that could be sunk as it shows a distressing
lack of understanding of what proof is.
Proof means you are right.
Mathematical proofs cannot be sunk, are not delicate and they cannot
be wrong.
Please Google "definition of mathematical proof" and read my math blog
post on the subject, which explains in detail what a proof is.
It is one of the worst things the modern math community did to teach
this insane view that proof can be wrong.
James Harris
Joshua Cranmer
> Ugh!!! Something false is NOT PROOF!!!!!!!!!!!!!
>
> So you did NOT have a proof and it's a horribly dangerous error to
> claim you had a proof that could be sunk as it shows a distressing
> lack of understanding of what proof is.
... You know, I have the same feeling. Shall we talk about your proof of
correctness for your Euclidean TSP algorithm? That was sunk in about ten
minutes of not-so-pensive thought?
JSH
Yeah, I figured it had to be wrong, as the radical ALWAYS must shift
as x shifts or it's trivial to prove that 7 must be a factor of just
one of the w's, which is just kind of one of those funny things.
I think math people say it's the class number or something or other
that has to shift based on Galois Theory, so "Rotwang" didn't realize
it, but he was arguing against GT.
James Harris
Rick Decker
<snip>
>
> The object ring is defined by two conditions, and includes all numbers
> such that these conditions are true:
>
> 1. 1 and -1 are the only rationals that are units in the ring.
>
> 2. Given a member m of the ring there must exist a non-zero member n
> such that mn is an integer, and if mn is not a factor of m, then n
> cannot be a unit in the ring.
Okay, let's look at this ring. Let's call it OR. Are these conclusions
correct in your understanding of OR?
A. Since 1 and -1 \in OR and any ring is closed under addition,
we can conclude immediately Z \subset OR.
B. Since OR is defined to include _all_ numbers satisfying your (2),
then for any nonzero m \in Z, 1/m \in OR, since m * (1/m) \in Z.
Consequently Q (the rationals) \subset OR. More generally,
if m \in OR and m != 0, then by your (2), 1/m \in OR. Thus,
your object ring is a field.
C. Let sqrt(2) represent, as usual, the positive square root of 2.
Then sqrt(2) is not a member of OR. Suppose, to the contrary,
that sqrt(2) \in OR. Since Z \subset OR and all rings are closed
under addition, we would have 1 + sqrt(2) \in OR and -1 + sqrt(2)
\in OR. However,
(1 + sqrt(2))(-1 + sqrt(2)) = 1
Contradicting the assumption that 1 and -1 are the only units in OR.
In fact, we can show that _no_ real quadratic irrational is in OR.
D. Suppose sqrt(-26) \in OR (if you're concerned about the ambiguity
of sqrt, take either root -- it's immaterial for this argument).
Then, as we've argued above, 307 + 30sqrt(-26) \in OR and
307 - 30sqrt(-26) \in OR. Thus by (B),
u_1 = (307 + 30sqrt(-26)) / 7^3 \in OR and
u_2 = (307 - 30sqrt(-26)) / 7^3 \in OR
However, u_1 * u_2 = 1, which is forbidden by (1). Thus,
3 + sqrt(-26) is not a member of OR which is going to make
your factorization
7 P(x) = (5a_1(x) + 7)(5a_2(x) + 7)
a bit problematic for x = 1, since then we have
a_i = 3 +/- sqrt(-26). In a similar way, we can show that
no imaginary quadratic irrational is in OR.
Regards,
Rick
hagman
> JSH wrote:
>
> <snip>
>
>
>
> > The object ring is defined by two conditions, and includes all numbers
> > such that these conditions are true:
>
> > 1. 1 and -1 are the only rationals that are units in the ring.
>
> > 2. Given a member m of the ring there must exist a non-zero member n
> > such that mn is an integer, and if mn is not a factor of m, then n
> > cannot be a unit in the ring.
>
> Okay, let's look at this ring. Let's call it OR. Are these conclusions
> correct in your understanding of OR?
>
> A. Since 1 and -1 \in OR and any ring is closed under addition,
> we can conclude immediately Z \subset OR.
>
> B. Since OR is defined to include _all_ numbers satisfying your (2),
> then for any nonzero m \in Z, 1/m \in OR, since m * (1/m) \in Z.
> Consequently Q (the rationals) \subset OR. More generally,
> if m \in OR and m != 0, then by your (2), 1/m \in OR. Thus,
> your object ring is a field.
I don't buy B. Rewrite the awfully formulated condition 2 as
2. Given m in OR there exists n in OR, n != 0, such that mn in Z;
if this n is a unit in OR, then mn divides m.
and note that the "if then" is trivial.
Thus we really have
(1') OR^* intersect Q = Z^*
and
(2') non-trivial principle ideals of OR intersect Z non-trivially.
It is however unclear what "and includes all numbers
such that these conditions are true" should mean because
a) from which "all" numbers do we select the members of OR? Probably
from the algebraic numbers Qbar ...
b) (1') and (2') are not conditions for an algebraic number (there's
no free variable in the condition).
Probably OR should be "the biggest" subring of Qbar with (1') and
(2').
Is there a biggest such ring?
Call a subring R of Qbar object-like if (1') and (2') hold for R, i.e.
R^* intersect Q = Z^* and for all non-trivial priciple ideals of R
intersect Z non-trivially.
Clearly Z is object-like; and if we have a chain of object-like rings
then their union is an object like ring. By Zorn's lemma a maximal
object-like ring exists -- however, it may not be uniquely determined.
> C. Let sqrt(2) represent, as usual, the positive square root of 2.
> Then sqrt(2) is not a member of OR. Suppose, to the contrary,
> that sqrt(2) \in OR. Since Z \subset OR and all rings are closed
> under addition, we would have 1 + sqrt(2) \in OR and -1 + sqrt(2)
> \in OR. However,
>
> (1 + sqrt(2))(-1 + sqrt(2)) = 1
>
> Contradicting the assumption that 1 and -1 are the only units in OR.
1 and -1 are the only *rational* units in OR.
This does not prevent irrational numbers like 1+-sqrt(2) from being
units.
Rotwang
> > w's here are in error: the sqrt() radicand should be -503, and
> > not -26. He was kind enough to observe this as the typographical
> > error that it was.
>
> Yeah, I figured it had to be wrong, as the radical ALWAYS must shift
> as x shifts or it's trivial to prove that 7 must be a factor of just
> one of the w's, which is just kind of one of those funny things.
>
> I think math people say it's the class number or something or other
> that has to shift based on Galois Theory, so "Rotwang" didn't realize
> it, but he was arguing against GT.
It's quite possible. If somebody who actually knows something about
Galois theory knows that what I'm trying to do is impossible, by all
means speak up. In the mean time here's another stab at the Object
ring. I don't know how to compute algebraic integer GCD's myself[1],
so let's instead consider another polynomial, which shares many of the
features with your original one, but is simple enough that some of the
w's can be found by inspection:
Q(x) = 25x^2 + 5x + 2
Like your original polynomial this is primitive and irreducible over
the rationals. Also like your original polynomial, we can factor
7*Q(x) as
7*Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)
where the a's are now the roots of
a^2 - (x - 1)a + 7x^2 = 0.
Like your original polynomial, at x = 0 we have that one of the a's is
equal to zero, so that at x = 0 we can divide the 7 out of one term.
So it looks the argument you used with your original polynomial, if
correct, should apply to this polynomial also. Is this the case? If
not, what is the difference which means that your argument no longer
applies?
Assuming that your argument does apply to Q(x), consider now x = 1.
Here we have
a = +/- sqrt(-7)
so at this value of x, 7 is not a factor of either of the terms
(5a_i(1) + 7). But we can find w_1(1) and w_2(1) such that
w_1(1)*w_2(1) = 7 and w_i divides (5a_i(x) + 7) such that all
coefficients remain in the algebraic integers, namely
w_1(1) = w_2(1) = sqrt(7)
Is this correct so far? Assuming it is, I believe now that your claim
is that the w's must "wrappers", meaning that the factorisation 7 =
w_1*w_2 is equivalent up to Object ring units to 7 = 7*1; in other
words, one of the w's has 7 as a factor in the Object ring. But then
the Object ring contains sqrt(7)/7 = 1/sqrt(7), and therefore also
contains 1/7.
[1] I had a quick look at the Kash documentation but I found myself
pretty far out of my depth. If someone can refer me to somewhere I can
learn what I need to know to calculate the GCD's I'll be grateful.
William Hale
<c757dc9a-40aa-425d...@k37g2000hsf.googlegroups.com>,
Rotwang <sg...@hotmail.co.uk> wrote:
[cut]
> I don't know how to compute algebraic integer GCD's myself[1],
> [1] I had a quick look at the Kash documentation but I found myself
> pretty far out of my depth. If someone can refer me to somewhere I can
> learn what I need to know to calculate the GCD's I'll be grateful.
I'm not sure myself, but I think it goes something like this.
Suppose we want to find the GCD of the two algebraic integers a and b.
Consider the ideal (a, b) generated by a and b. By the finiteness of the
class number, there is a finite natural number h and an algebraic
integer c such that the ideal (c) generated by c is equal to (a, b)^h.
I am being sloppy here, since I am not specifying the ring I am working
in.
For simplicity, I will assume that the class number h of the ideal
generated by a and b is equal to 2. Thus, (a, b)^2 = (c). Find an
algebraic integer e such that e^2 = c. I will again be sloppy here,
since I am now working in an even larger ring containing a, b, c and e.
Thus, I have (a, b)^2 = (e^2). I claim that e is a GCD of a and b. That
is, I claim that GCD(a, b) = e.
Note that (a, b)^2 = (a^2, a*b, b^2). Thus, a^2 and b^2 are multiples of
e^2. Thus, a and b are multiples of e. I don't know if this is obvious
or not, but at least it follows from the unique factorization of ideals
(I think). Hence, e is a common divisor of a and b. [In more detail: the
ideal (a^2) is a subset of the ideal (e^2); so, the ideal (a) times the
ideal (a) is a subset of the ideal (e) times the ideal (e); so, the
ideal (a)*(a) = (e)*(e)*J for some ideal J; by unique factorization of
ideals, the number of times a prime ideal factor occurs in the ideal (e)
must be at least the number of times that prime ideal occurs as a prime
ideal factor of the ideal (a); so it follows that the ideal (a) is a
subset of the ideal (e).]
Conversely, let d be a common divisor of a and b. Then, the ideals
(a^2), (a*b), and (b^2) are subsets of the ideal (d^2). Since (e^2)
equals the ideal (a^2, a*b, b^2), the ideal (e^2) is a subset of the
ideal (d^2). Like above, it follows that the ideal (e) is a subset of
the ideal (d). Thus, e is a multiple of d. Hence, e is a greatest common
divisor of a and b.
I guess all the above can be shortened to just noting that (a, b) = (e)
follows from (a, b)^2 = (e)^2 by using the unique factorization of
ideals.
The hard part is figuring out what rings you are working in. Also, I
think there was some discussion of what the definition of GCD is in this
situation.
Also note that I am using the unique factorization of ideals into prime
ideals (which is relatively easy to prove, although there is one very
tricky step in doing so) and the finiteness of the class number (which
is a hard thing to prove).
-- Bill Hale
Rotwang
> In article
> <c757dc9a-40aa-425d-9a6d-4fd7c7f7b...@k37g2000hsf.googlegroups.com>, Rotwang <sg...@hotmail.co.uk> wrote:
>
> [cut]
>
> > I don't know how to compute algebraic integer GCD's myself[1],
> [cut]
> > [1] I had a quick look at the Kash documentation but I found myself
> > pretty far out of my depth. If someone can refer me to somewhere I can
> > learn what I need to know to calculate the GCD's I'll be grateful.
>
> I'm not sure myself, but I think it goes something like this.
Thanks for your reply. This stuff is new to me (as in I've just looked
up "Ideal class" on planetmath) so I have a few questions and I hope
you will forgive me if I say anything thick.
>
> Suppose we want to find the GCD of the two algebraic integers a and b.
> Consider the ideal (a, b) generated by a and b. By the finiteness of the
> class number, there is a finite natural number h and an algebraic
> integer c such that the ideal (c) generated by c is equal to (a, b)^h.
> I am being sloppy here, since I am not specifying the ring I am working
> in.
At this point I suppose we can take the ring R to be the ring of
algebraic integers in Q(a,b), which is a finite extension so that the
class number is finite.
According to what I've read, the class number is the number of
equivalence classes of ideals, where two ideals I and J are equivalent
(written I~J) if there exist non-zero ring elements alpha and beta
such that
(alpha)I = (beta)J.
I think it is easy to show that an ideal I is principal iff I is
equivalent to the whole ring (1). The statement you give above then
follows from the finiteness of the class number together with the fact
that the ideal class group is a group, since we must have (a, b)^n =
(a, b)^m for some n > m from finiteness, and therefore that (a, b)^{n
- m) = (1) from general properties of groups. Is it easy to show that
the ideal class group is indeed a group?
> For simplicity, I will assume that the class number h of the ideal
> generated by a and b is equal to 2. Thus, (a, b)^2 = (c). Find an
> algebraic integer e such that e^2 = c. I will again be sloppy here,
> since I am now working in an even larger ring containing a, b, c and e.
I think this is OK. Let's now consider the whole ring A of algebraic
integers, and define a map f from the set of ideals of R to those of
A, such that f(I) is the smallest ideal of A containing I. Then we can
show that
f((x,y,z,...)_R) = (x,y,z,...)_A
(using a notation whose meaning I hope is obvious), and
f(I)f(J) = f(IJ)
so that the relation (a, b)^2 = (c) still holds in A.
> Thus, I have (a, b)^2 = (e^2). I claim that e is a GCD of a and b. That
> is, I claim that GCD(a, b) = e.
>
> Note that (a, b)^2 = (a^2, a*b, b^2). Thus, a^2 and b^2 are multiples of
> e^2. Thus, a and b are multiples of e. I don't know if this is obvious
> or not, but at least it follows from the unique factorization of ideals
> (I think).
Much simpler than that, I think - the algebraic integers are closed
under extraction of roots, so if e.g. a^2/e^2 is in A then so is a/e.
> Hence, e is a common divisor of a and b. [In more detail: the
> ideal (a^2) is a subset of the ideal (e^2); so, the ideal (a) times the
> ideal (a) is a subset of the ideal (e) times the ideal (e); so, the
> ideal (a)*(a) = (e)*(e)*J for some ideal J; by unique factorization of
> ideals, the number of times a prime ideal factor occurs in the ideal (e)
> must be at least the number of times that prime ideal occurs as a prime
> ideal factor of the ideal (a); so it follows that the ideal (a) is a
> subset of the ideal (e).]
>
> Conversely, let d be a common divisor of a and b. Then, the ideals
> (a^2), (a*b), and (b^2) are subsets of the ideal (d^2). Since (e^2)
> equals the ideal (a^2, a*b, b^2), the ideal (e^2) is a subset of the
> ideal (d^2). Like above, it follows that the ideal (e) is a subset of
> the ideal (d). Thus, e is a multiple of d. Hence, e is a greatest common
> divisor of a and b.
Looks good.
> I guess all the above can be shortened to just noting that (a, b) = (e)
> follows from (a, b)^2 = (e)^2 by using the unique factorization of
> ideals.
>
> The hard part is figuring out what rings you are working in. Also, I
> think there was some discussion of what the definition of GCD is in this
> situation.
My understanding is that a GCD of a and b is defined to be an e such
that e|a, e|b and ((d|a & d|b) -> d|e). I thought this was standard.
W. Dale Hall
Here's how to get started with Kash. I'll compute the w's for x = 4,
and give a transcription of my session. Then I'll go through and try
to provide some indication of what on Earth I just did.
A min[aj]or caveat: I'm no algebraist. Although I've taken adequate
coursework in algebra and commutative algebra to be able to decipher
the instructions, I've found I'm somewhat behind the curve. Most any
practicing algebraist will be able to correct my misconceptions, after
recovering from his laughing fit.
A couple of key things.
X must be capitalized for KASH to recognize the indeterminate
in a polynomial.
Lists are [thing1, thing2, ..., thingn], where thing.. must
have been defined already. Brackets & commas are compulsory,
ellipses are not allowed.
Help works like this:
?Keyword
gives you help on "Keyword" if that is an exact & completet
keyword.
?*eyword gives you help on all completions of "eyword"
It's a good idea to set the window where KASH starts to
have a HUGE scrollback range.
Be prepared for KASH to complain lots & lots.
The .pdf files are more useful if you have a function in mind,
and do serve to spell out a few examples explicitly. I'd bet
they're a heap more useful to real algebraists.
If you have questions, I'll be glad to look into them, but
only realize I'm a rank amateur at this program.
I'll sign of here and let you continue your reading program...
Dale
Here we go:
*** REMARK: THIS IS THE INTRO TO THE SESSION ***
ooooqp
oooo oooo .o. .oooooo..o ooooo ooooo .dP'
`888 .8P' .888. d8P' `Y8 `888' `888' d88b.
888 d8' .8'888. Y88bo. 888 888 o. )8
88888K .8' `888. `'Y8888o. 888ooooo888 `888P'
888`88b. .88ooo8888. `'Y88b 888 888
888 `88b. .8' `888. oo .d8P 888 888
o888o o888o o88o o8888o 8''88888P' o888o o888o
Shell of the KANT V4 Software, Version 3, build: Windows-2005-11-19
Copyright (c) 1994-2005 Prof. Dr. M. E. Pohst,
Technische Universitaet Berlin. All rights reserved.
For registration and support send an email to ka...@math.tu-berlin.de
--------------------------------------------------------------------
KANT V4 is based on Magma developed by Prof. J. Cannon,
Copyright (c) 2002 Prof. J. Cannon, University of Sydney.
Shell is based on GAP developed by Lehrstuhl D Mathematik, RWTH
Aachen,
Copyright (c) 1992 Lehrstuhl D Mathematik, RWTH Aachen.
Enter "?" for help and "quit;" to leave KASH
list [0] alist [0] term [0] __DOC [0.148] doc [0.002] docui [0.123]
docui [0.0\
02] method [0.002] init-methods [0.117] constants [0.007] kash [0.002]
matrix \
[0.005] map [0.008] qaos [0.004] locFact [0.01]
*** REMARK: END OF SESSION INTRO ***
*** REMARK: First I'll define the coefficients of
*** REMARK: a^2 - (7 x - 1)a + (49 x^2 - 14 x):
kash% q1 := -(7*X - 1);
-7*X + 1
Time: 0 s
kash% q0 := 49*X^2 - 14*X;
49*X^2 - 14*X
Time: 0 s
*** REMARK: set x equal to 4:
kash% c1 := Evaluate(q1,4);
-27
Time: 0 s
kash% c0 := Evaluate(q0,4);
728
Time: 0 s
*** REMARK: define the polynomial that sets a's value:
kash% p := X^2 + c1*X + c0;
X^2 - 27*X + 728
Time: 0.044 s
*** REMARK: an "order" of a number field is a free abelian group ord
*** REMARK: that spans the field over Q; in Kash, the "equation order"
*** REMARK: of the polynomial p is the basis for Q[x]/<p> generated
*** REMARK: by the powers of x from x^0 to x^(deg(p)-1). The element
*** REMARK: corresponding to x is one root of the polynomial p
*** REMARK:
*** REMARK: a link for "order"
*** REMARK: http://en.wikipedia.org/wiki/Maximal_order
kash% o := EquationOrder(p);
Equation Order with defining polynomial X^2 - 27*X + 728 over Z
Time: 0.124 s
*** REMARK: The maximal order associated to the order ord is an
*** REMARK: order Ord, that spans the integers of the field over Z
*** REMARK: It may be the same as the order ord, but may be strictly
*** REMARK: larger, as it turns out in this case.
kash% O := MaximalOrder(o);
Maximal Equation Order with defining polynomial X^2 - 27*X + 728 over Z
Time: 0.129 s
*** REMARK: We need the maximal order to compute the class group
*** REMARK: which we do here:
kash% ClassGroup(O);
Abelian Group isomorphic to Z/42
Defined on 1 generator
Relations:
42*_HE.1 = 0, extended by:
ext1 := Mapping from: grp^abl: _HE to ids/ord^num: _EJ
Time: 0.752 s
*** REMARK: Here, I'm trying to find out about the
*** REMARK: ideal generated by the root (a) and 7
kash% e1 := Element(O,[0,1]);
[0, 1]
Time: 0.032 s
*** REMARK: First to test whether this is (a)
kash% Evaluate(p,e1);
[0, 0]
Time: 0.002 s
*** REMARK: It is. Next to find the ideal <7, a>:
kash% Ia := Ideal(O,[7, e1]);
Ideal of O
Two element generators:
[7, 0]
[0, 1]
Time: 0.091 s
*** REMARK: Is there a common divisor here?
kash% IsPrincipal(Ia);
FALSE, extended by:
ext1 := Unassign
Time: 0.02 s
*** REMARK: nope. Next, I raise Ia to powers that divide 42,
*** REMARK: the order of the ideal class group (Lagrange's
*** REMARK: theorem says these are the only possible orders
*** REMARK: (different meaning of "order") of elements of
*** REMARK: the class group.
kash% IsPrincipal(Ia^2);
FALSE, extended by:
ext1 := Unassign
Time: 0.087 s
kash% IsPrincipal(Ia^3);
FALSE, extended by:
ext1 := Unassign
Time: 0.001 s
kash% IsPrincipal(Ia^6);
FALSE, extended by:
ext1 := Unassign
Time: 0.001 s
kash% IsPrincipal(Ia^7);
FALSE, extended by:
ext1 := Unassign
Time: 0 s
kash% IsPrincipal(Ia^14);
FALSE, extended by:
ext1 := Unassign
Time: 0.027 s
kash% IsPrincipal(Ia^21);
TRUE, extended by:
ext1 := [601438915, -31021218]
Time: 0.001 s
*** REMARK: Oh boy. I only have to take the 21st power
*** REMARK: to get a common divisor!
*** REMARK: Here's the divisor, in terms of 1 and a:
kash% a_gen := IsPrincipal(Ia^21).ext1;
[601438915, -31021218]
Time: 0.001 s
kash% a_gen - (601438915 - 31021218*e1);
[0, 0]
Time: 0.001 s
*** REMARK: Oops, I wanted this in terms of 1 and
*** REMARK: sqrt(discriminant(p)):
kash% Discriminant(p);
-2183
Time: 0 s
*** REMARK: Just curious. How does discriminant(p) factor?
kash% Factorization(2183);
[ <37, 1>, <59, 1> ], extended by:
ext1 := 1,
ext2 := Unassign
Time: 0.042 s
*** REMARK: Remind me what p is:
kash% p;
X^2 - 27*X + 728
Time: 0.001 s
*** REMARK: Let's take the extension by just including
*** REMARK: the square root of the discriminant (-2183):
kash% ox := EquationOrder(X^2 + 2183);
Equation Order with defining polynomial X^2 + 2183 over Z
Time: 0 s
kash% Ox := MaximalOrder(ox);
Maximal Order of ox
Time: 0.005 s
*** REMARK: And here's its class group. We hope I find
*** REMARK: the same class group as before:
kash% ClassGroup(Ox);
Abelian Group isomorphic to Z/42
Defined on 1 generator
Relations:
42*_TD.1 = 0, extended by:
ext1 := Mapping from: grp^abl: _TD to ids/ord^num: _VY
Time: 0.242 s
*** REMARK: Goody. Now to write (a) in terms of 1 and sqrt(disc(p)):
*** REMARK: First, the generator [0,1] of the order Ox:
kash% x1 := Element(Ox, [0,1]);
[0, 1]
Time: 0 s
*** REMARK: Is it really sqrt(-2183)?
kash% Evaluate(X^2 + 2183,x1);
[1637, 1]
Time: 0 s
*** REMARK: Sheeeit. Now I gotta dig.
*** REMARK: Let's find out what on earth it is:
kash% MinimalPolynomial(x1);
X^2 - X + 546
Time: 0.041 s
*** REMARK: OK, so it's (1 + sqrt(disc(thisthing)))/2
kash% Discriminant(MinimalPolynomial(x1));
-2183
Time: 0 s
*** REMARK: or, (1 + sqrt(-2183))/2.
*** REMARK: Solving (by hand), I see that
*** REMARK: sqrt(-2183) = -1 + 2*root:
kash% Evaluate(X^2 + 2183, -1 + 2*x1);
[0, 0]
Time: 0 s
*** REMARK: Yay. Let's give it a name and proceed:
kash% r1 := -1 + 2*x1;
[-1, 2]
Time: 0 s
kash% MinimalPolynomial(r1);
X^2 + 2183
Time: 0 s
*** REMARK: I forgot what p was again, duh:
kash% p;
X^2 - 27*X + 728
Time: 0.001 s
kash% Discriminant(p);
-2183
Time: 0 s
*** REMARK: and I know r1 = sqrt(-2183), so I
*** REMARK: can write (a) out directly:
kash% a_1 := (27 + r1)/2;
13/1*_GT.1 + _GT.2
Time: 0.001 s
kash% MinimalPolynomial(a_1);
_SM.1^2 - 27*_SM.1 + 728
Time: 0 s
*** REMARK: OK, it's got the right minimal polynomial
*** REMARK: Next check on p itself:
kash% Evaluate(p, a_1);
0
Time: 0.001 s
*** REMARK: and I proceed to define that ideal again:
kash% Ia_1 := Ideal(Ox, [7, a_1]);
Ideal of Ox
Two element generators:
[7, 0]
[13, 1]
Time: 0.001 s
*** REMARK: I have to raise it to the 21st power to
*** REMARK: get a principal ideal:
*** REMARK:
kash% IsPrincipal(Ia_1^21);
TRUE, extended by:
ext1 := [198163081, -31021218]
Time: 0.001
*** REMARK: Are these coefficients the same
*** REMARK: I got earlier?
kash% a_gen
% ;
[601438915, -31021218]
*** REMARK: Not entirely. the second one is OK, but
*** REMARK: not the first. I shouldn't have expected
*** REMARK: more, since the two maximal orders have
*** REMARK: different generating sets.
Time: 0 s
kash% a1_gen := IsPrincipal(Ia_1^21).ext1;
[198163081, -31021218]
Time: 0.001 s
*** REMARK: I'll copy the coefficients, since I
*** REMARK: need to do some arithmetic:
kash% coeff1 := 198163081;
198163081
Time: 0.001 s
kash% coeff2 := -31021218;
-31021218
Time: 0 s
*** REMARK: I'll rewrite a1_gen:
kash% coeff1 + coeff2*x1;
[198163081, -31021218]
Time: 0 s
kash% coeff1 + coeff2*x1 - a1_gen;
[0, 0]
Time: 0.001 s
*** REMARK: I forgot how r1 ( = sqrt(-2183))
*** REMARK: came out in its basis
kash% r1 := -1 + 2*x1;
[-1, 2]
Time: 0 s
*** REMARK: so I adjust coefficients to write
*** REMARK: the generator in terms of 1 & sqrt(-2183):
kash% k2 := coeff2/2;
-15510609
Time: 0 s
kash% k1 := coeff1 + k2;
182652472
Time: 0.001 s
*** REMARK: here it should be:
kash% k1 + k2*r1;
198163081/1*_GT.1 - 31021218/1*_GT.2
Time: 0.001 s
*** REMARK: Let's check:
kash% k1 + k2*r1 - a1_gen;
0
Time: 0 s
*** REMARK: I'll give it a new name:
kash% gen1 = k1 + k2*r1;
Error, the variable 'gen1' must have a value
*** REMARK: Darn it. definitions are :=, not =
kash% gen1 := k1 + k2*r1;
198163081/1*_GT.1 - 31021218/1*_GT.2
Time: 0 s
*** REMARK: And I'll find the complex conjugate:
kash% gen2 := k1 - k2*r1;
167141863/1*_GT.1 + 31021218/1*_GT.2
Time: 0 s
*** REMARK: Do these numbers give the correct product?
kash% gen1*gen2;
558545864083284007/1*_GT.1
Time: 0 s
kash% 7^21;
558545864083284007
Time: 0 s
*** REMARK: Looks OK, but I never trust my eyes any more:
kash% gen1*gen2 - 7^21;
0
Time: 0 s
kash%
*** REMARK: Looks like my work is finished for now.
*** REMARK: END OF KASH SESSION
And there we have it: for x = 4, the factorization of 7 comes out like
this:
w(4) = (182652472 +/- 15510609 sqrt(-2183))^(1/21)
That's it.
I'm sure it would have been simpler if I knew what I was doing.
Dale
W. Dale Hall
Yes, but if the class number is greater than 1, then unique
factorization can fail, and there may be no GCD. It's when <a b> is
not principal, that you go to powers of the ideal <a b>, trying to
reach a power that is principal.
Some folks may use GCD(a,b) to denote the ideal <a,b> which, when
principal, does yield the older notion [as any of its generators].
>> Also note that I am using the unique factorization of ideals into prime
>> ideals (which is relatively easy to prove, although there is one very
>> tricky step in doing so) and the finiteness of the class number (which
>> is a hard thing to prove).
>>
>> -- Bill Hale
Dale
Rotwang
wrote:
> Rotwang wrote:
>
> > [1] I had a quick look at the Kash documentation but I found myself
> > pretty far out of my depth. If someone can refer me to somewhere I can
> > learn what I need to know to calculate the GCD's I'll be grateful.
>
> Here's how to get started with Kash. I'll compute the w's for x = 4,
> and give a transcription of my session. Then I'll go through and try
> to provide some indication of what on Earth I just did.
>
Thanks very much, I appreciate it. It'll take me a while to digest
your post, so I may have some questions later.
JSH
Really the key points are covered at that point. So it's worth
clearly showing how arguments have raged for years when you ignore the
possibility of unit factors.
After all, it's trivial to note that if you have a polynomial P(x)
with linear functions f_1(x) and f_2(x) such that
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
that the 7 must divide through only one of the f's, which of course,
posters have routinely acknowledged as then you have a polynomial
factorization, so there's no denial possible.
So you have
P(x) = (5f_1(x)/7 + 1)*(5f_2(x)+ 7)
or
P(x) = (5f_1(x) + 7)*(5f_2(x)/7 + 1).
But change the functions to non-linear ones by making them roots of a
quadratic generator as I did with the a's and the arguments start up
again, but what changed?
The TYPE of the functions is what changed!!!
But if a*(f(x) + b) = a*f(x) + a*b and the value of the function
doesn't matter, then how can the type of function change the
distributive property?
Of course it cannot. For years posters dance around this issue, but
it is a mathematical fact.
A function is a function. If one function works one way based on the
distributive property then how can mathematics be consistent and
changing the type of FUNCTION shift the distributive property? It
cannot, if mathematics is consistent.
So what actually happens?
Well, I don't have a case where you have something like
a*(f(x) + b) = a*f(x) + a*b
but instead have cases like
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
where you can have unit factors u_1*u_2 = 1, and get
P(x) = (5u_1*f_1(x)/7 + u_1)*(5*u_2*f_2(x)+ 7*u_2)
or
P(x) = (5u_1*f_1(x) + 7*u_1)*(5*u_2*f_2(x)/7 + u_2)
where the u's are not units in the ring of algebraic integers because
they are not roots of a monic polynomial with integer coefficients
with a unit last coefficient.
And mathematical consistency remains.
But those factorization cannot exist in the ring of algebraic integers
in that form as 7*u_1 is not available in the ring of algebraic
integers which does not allow that 7 be a factor in that ring, and u_1
and u_2 are NOT units in the ring of algebraic integers.
Notice here that the problem is, how can the distributive property
vary depending on the TYPE of function?
The answer is it cannot.
But then you have to resolve the fact that 7 cannot be a factor in the
ring of algebraic integers of the a's when
7*P(x) = (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 for those looking for an absolute crux point where posters fail in
their reasoning then it is in their position that the type of function
can shift the distributive property.
Once you dismiss that illogical belief then you have a mystery to
resolve.
To preserve mathematical consistency the resolution of the mystery
requires unit factors, but you can quickly find that units are not
available in the ring of algebraic integers.
Which proves that the ring is incomplete, or mathematical consistency
fails.
Disputes here no matter what posters say, or what they delete out in
reply reduce to asserting that the TYPE of function shifts
a*(f(x) + b) = a*f(x) + a*b
or the result that is clearly correct with linear functions would not
be an issue with non-linear ones cleverly forced to be the roots of a
quadratic generator as the a's are with
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
It is frustrating to have a logical argument, which is being refuted
by people who are arguing against the distributive property who simply
say they are not, and get accepted by a society that claims it is
logical.
But actions speak louder than words.
James Harris
marcus_b
Right there, in both equations, you are not using the
distributive property.
The distributive property is about multiplication. It
says:
a * (b + c) = a*b + a*c.
But you are DIVIDING by 7. Division is not a ring
operation. Division works in fields. In general, you
cannot talk about a/b in a ring. The ring may not contain
a/b. The "version" of the distributive property that you want
is
(b + c) / a = b/a + c/a.
This is NOT a statement of the distributive property
in rings. Look it up.
In your case, "b/a" is 5a_1(x)/7. But you don't even
know that 5a_1(x)/7 is in your ring. If you assume it is,
then you are assuming what you want to prove.
> where the u's are not units in the ring of algebraic integers because
> they are not roots of a monic polynomial with integer coefficients
> with a unit last coefficient.
>
> And mathematical consistency remains.
>
> But those factorization cannot exist in the ring of algebraic integers
> in that form as 7*u_1 is not available in the ring of algebraic
> integers
Of course it is. You can let u_1 = 1.
> which does not allow that 7 be a factor in that ring, and u_1
> and u_2 are NOT units in the ring of algebraic integers.
>
> Notice here that the problem is, how can the distributive property
> vary depending on the TYPE of function?
>
> The answer is it cannot.
>
Of course it can. Look: a_1(x) is a variable function. Right?
What
you need is the greatest common divisor of a_1(x) and 7. Since a_1(x)
is a variable function of x, there is EVERY REASON to believe that the
greatest common divisor of a_1(x) and 7 must ALSO be a variable
function of x.
You SHOULD assume it is a function of x until you have proven
otherwise. And for sure here, wiener, you have not proven
otherwise. And in fact, as shown dramatically by Dale Hall's
computations, the correct factors DO depend on x.
> But then you have to resolve the fact that 7 cannot be a factor in the
> ring of algebraic integers of the a's when
>
> 7*P(x) = (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 side of the resolution is YOUR problem. It you who is
claiming this, based on your incorrect logic where you regard
the distributive property as a statement about division rather
than about multiplication.
> So for those looking for an absolute crux point where posters fail in
> their reasoning then it is in their position that the type of function
> can shift the distributive property.
>
Assuming that 7 = w_1(x) * w_2(x), etc., does not "shift" or
otherwise violate the distributive property. It is safely preserved.
> Once you dismiss that illogical belief then you have a mystery to
> resolve.
>
No, the mystery is all on your side. You erroneously think your
flawed logic implies something which is not true.
> To preserve mathematical consistency the resolution of the mystery
> requires unit factors, but you can quickly find that units are not
> available in the ring of algebraic integers.
>
Units have nothing to do with the central problem. The central
problem is that you think there is only one, CONSTANT way to divide 7
out of
(5a_1(x) + 7)*(5a_2(x) + 7),
even though both a_1(x) and a_2(x) are ***variable*** functions of x.
> Which proves that the ring is incomplete, or mathematical consistency
> fails.
>
> Disputes here no matter what posters say, or what they delete out in
> reply reduce to asserting that the TYPE of function shifts
>
> a*(f(x) + b) = a*f(x) + a*b
>
> or the result that is clearly correct with linear functions would not
> be an issue with non-linear ones cleverly forced to be the roots of a
> quadratic generator as the a's are with
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0.
>
> It is frustrating to have a logical argument,
You are in no position to know that. You don't have one.
Marcus.
William Hughes
> After all, it's trivial to note that if you have a polynomial P(x)
> with linear functions f_1(x) and f_2(x) such that
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> that the 7 must divide through only one of the f's, which of course,
is not the distributive property. The distributive property does
not tell you if the 7 divides through one f or two.
> posters have routinely acknowledged as then you have a polynomial
> factorization, so there's no denial possible.
>
> So you have
>
> P(x) = (5f_1(x)/7 + 1)*(5f_2(x)+ 7)
>
> or
>
> P(x) = (5f_1(x) + 7)*(5f_2(x)/7 + 1).
>
> But change the functions to non-linear ones by making them roots of a
> quadratic generator as I did with the a's and the arguments start up
> again, but what changed?
The 7 divides through two f's rather than one.
Note that this change has nothing to do with the
distibutive property.
- William Hughes
JSH
Yes, I am.
If you have
7*(x+3x+2) = (7x + 7)*(x+2)
and divide off the 7, are you not using the distributive property?
Given that a*(b+c) = a*b + a*c it must be true that 'a' is a factor.
So the point is that 7 is a factor based on the distributive property.
If it is a factor then it can be removed, which is normally described
as "dividing it off".
> The distributive property is about multiplication. It
> says:
>
> a * (b + c) = a*b + a*c.
Yes.
> But you are DIVIDING by 7. Division is not a ring
> operation. Division works in fields. In general, you
> cannot talk about a/b in a ring. The ring may not contain
> a/b. The "version" of the distributive property that you want
> is
>
> (b + c) / a = b/a + c/a.
No.
The reality can be seen with linear functions:
7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
where dividing off the 7 is trivial.
I pointed out that there is no critical difference mathematically
between that example and
P(x) = 175x^2 - 15x + 2
7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0,
while there is the difference of linear versus non-linear function
being used, as 7x is a linear function, or do you disagree?
If
a*(f(x) + b) = a*f(x) + a*b
without regard to whether you have a linear or a non-linear function
how can the behavior change just because the a's are non-linear
versus
7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
where they are?
> This is NOT a statement of the distributive property
> in rings. Look it up.
>
> In your case, "b/a" is 5a_1(x)/7. But you don't even
> know that 5a_1(x)/7 is in your ring. If you assume it is,
> then you are assuming what you want to prove.
The full statement is
7*P(x) = (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 7 is multiplied times P(x), so it's not about assuming.
> > where the u's are not units in the ring of algebraic integers because
> > they are not roots of a monic polynomial with integer coefficients
> > with a unit last coefficient.
>
> > And mathematical consistency remains.
>
> > But those factorization cannot exist in the ring of algebraic integers
> > in that form as 7*u_1 is not available in the ring of algebraic
> > integers
>
> Of course it is. You can let u_1 = 1.
Nope, as then 7 would be a factor in that ring, and it is not.
The crucial logical point here is that
a*(f(x) + b) = a*f(x) + a*b
WITHOUT REGARD to the TYPE of function.
While your argument and that of Arturo Magidin, and W. Dale Hall, and
Rick Decker among others is that if f(x) is a non-linear function then
with
7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
somehow, someway the value of FUNCTIONS shift where factors of 7 go,
which is in direct contradiction to the reality that
a*(f(x) + b) = a*f(x) + a*b
WITHOUT REGARD to the TYPE of function.
So your position can be shown to directly contradict basic algebra.
You are arguing that functional type defines the output of the
distributive property when that is trivially wrong.
It is a bad math mistake.
James Harris
JSH
> On Sep 1, 4:35 pm, JSH <jst...@gmail.com> wrote:
>
> > After all, it's trivial to note that if you have a polynomial P(x)
> > with linear functions f_1(x) and f_2(x) such that
>
> > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> > that the 7 must divide through only one of the f's, which of course,
>
> is not the distributive property. The distributive property does
> not tell you if the 7 divides through one f or two.
Correct.
Because it's possible that each has sqrt(7) as a factor or an infinity
of other variations, even with linear functions.
But how could you determine whether or not that is true, rigorously?
> > posters have routinely acknowledged as then you have a polynomial
> > factorization, so there's no denial possible.
>
> > So you have
>
> > P(x) = (5f_1(x)/7 + 1)*(5f_2(x)+ 7)
>
> > or
>
> > P(x) = (5f_1(x) + 7)*(5f_2(x)/7 + 1).
>
> > But change the functions to non-linear ones by making them roots of a
> > quadratic generator as I did with the a's and the arguments start up
> > again, but what changed?
>
> The 7 divides through two f's rather than one.
> Note that this change has nothing to do with the
> distibutive property.
>
> - William Hughes
Incorrect. It's not determined.
The functional type does NOT impact the statement
a*(f(x) + b) = a*f(x) + a*b
which is true regardless of whether or not the function is linear or
non-linear.
Possibly understanding this point will help you understand the full
result?
Clearly you and other posters believe that linear versus non-linear
functions make a difference.
James Harris
William Hughes
> On Sep 1, 4:09 pm, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Sep 1, 4:35 pm, JSH <jst...@gmail.com> wrote:
>
> > > After all, it's trivial to note that if you have a polynomial P(x)
> > > with linear functions f_1(x) and f_2(x) such that
>
> > > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> > > that the 7 must divide through only one of the f's, which of course,
>
> > is not the distributive property. The distributive property does
> > not tell you if the 7 divides through one f or two.
>
> Correct.
So, the important point, "does 7 divide through one f or two?" has
nothing to do with the distributive property. In particular,
something can change whether 7 divides through one f or
two without changing the distributive property.
- William Hughes
Rotwang
>
> Really the key points are covered at that point. So it's worth
> clearly showing how arguments have raged for years when you ignore the
> possibility of unit factors.
>
> After all, it's trivial to note that if you have a polynomial P(x)
> with linear functions f_1(x) and f_2(x) such that
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> that the 7 must divide through only one of the f's, which of course,
> posters have routinely acknowledged as then you have a polynomial
> factorization, so there's no denial possible.
>
> So you have
>
> P(x) = (5f_1(x)/7 + 1)*(5f_2(x)+ 7)
>
> or
>
> P(x) = (5f_1(x) + 7)*(5f_2(x)/7 + 1).
>
>
> So what actually happens?
>
> Well, I don't have a case where you have something like
>
> a*(f(x) + b) = a*f(x) + a*b
>
> but instead have cases like
>
>
> where you can have unit factors u_1*u_2 = 1, and get
>
> P(x) = (5u_1*f_1(x)/7 + u_1)*(5*u_2*f_2(x)+ 7*u_2)
>
> or
>
>
> where the u's are not units in the ring of algebraic integers because
> they are not roots of a monic polynomial with integer coefficients
> with a unit last coefficient.
James, yesterday I posted a new polynomial and a factorisation
satisfying the equation I have labelled (1), and showed that at x = 1
the assumption that the u's in (2) are units in the object ring means
that the object ring contains 1/7. Did you see that post? If not, it
can be read here:
http://groups.google.co.uk/group/sci.math/msg/b04a1231f4c0c85c?hl=en
It looks to me like I followed your method exactly. Can you (or anyone
else) explain where, if anywhere, I went wrong?
JSH
Incorrect.
The proper answer is that you can simply check at some value of the
functions.
For instance, with *linear* functions, if at x=6 in the ring of
algebraic integers, you have that with
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
(5f_2(6)+ 7)
has 7 as a factor then you have that 7 is the factor for all x.
Understand?
James Harris
William Hughes
> For instance, with *linear* functions, if at x=6 in the ring of
> algebraic integers, you have that with
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> (5f_2(6)+ 7)
>
> has 7 as a factor then you have that 7 is the factor for all x.
>
> Understand?
Yes. For *linear* functions 7 must divide through one f for all x.
If the functions are not linear, 7 may divide through one f for some
values of x and divide
through two f's for other values of x. However, for non-linear
functions the
distributive property must still hold. The question of whether 7
divides
through one f or two f's has nothing to do with the distibutive
property.
- William Hughes
JSH
> On Sep 1, 8:27 pm, JSH <jst...@gmail.com> wrote:
>
> > For instance, with *linear* functions, if at x=6 in the ring of
> > algebraic integers, you have that with
>
> > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> > (5f_2(6)+ 7)
>
> > has 7 as a factor then you have that 7 is the factor for all x.
>
> > Understand?
>
> Yes. For *linear* functions 7 must divide through one f for all x.
Wrong. What I said was incorrect as maybe at x=6, P(6) has 7 as a
factor.
Correctly stated it should be that given
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
if at x=6, (5f_2(6)+ 7) has 7 as a factor and 5f_1(6) + 7 is coprime
to 7, then
5f_1(x) + 7
has 7 as a factor for all x.
> If the functions are not linear, 7 may divide through one f for some
> values of x and divide
> through two f's for other values of x. However, for non-linear
Incorrect.
Consider again
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
where P(x) is a polynomial function, does the distributive property
care about the type of the function?
a*(f(x) + b) = a*f(x) + a*b
Does the TYPE of the function have any impact?
Think carefully.
You've already answered more than one question wrong.
James Harris
William Hughes
>
> Consider again
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> where P(x) is a polynomial function, does the distributive property
> care about the type of the function?
The important question of whether 7 divides through one f
or two f's does care about the type of the function.
As you have noted, this question has nothing to do with
JSH
> On Sep 1, 9:15 pm, JSH <jst...@gmail.com> wrote:
>
>
>
> > Consider again
>
> > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> > where P(x) is a polynomial function, does the distributive property
> > care about the type of the function?
>
> The important question of whether 7 divides through one f
> or two f's does care about the type of the function.
Nope. Let's say that
P(x) = x^2 + 3x + 2 = (x+1)(x+2)
so you multiply 7*P(x), how many ways can you do that in the ring of
algebraic integers?
Ans: An infinity of ways.
Is there any way given
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
to determine *which* way was used.
> As you have noted, this question has nothing to do with
> the distributive property.
>
> - William Hughes
Of course it does because if you have
7*(x+1)*(x+2) = (7x + 7)(x + 2)
that's all about the distributive property as you *distribute* the 7
through the factor.
Now it's trivial algebra but the problem here is that posters like you
deny trivial algebra and then just claim that's not what you're doing
because the stakes are high, and you actually clearly suck at basic
math.
Now if
P(x) = x^2 + 3x + 2 = (x+1)(x+2)
and
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
there are an infinity of ways to distribute factors of 7 across the
factorization of P(x), but there's one clear way to determine which
one was used as let's say, where still the f's are linear functions,
that f_1(0) = 0.
In that case HOW did the 7 distribute?
Not a trick question but answering honestly will reveal that yes, it
is about the distributive property.
Given that answer with linear functions can you deny that your claims
that non-linear functions are somehow different are non-mathematical
and completely bogus efforts to deny very basic mathematics?
James Harris
Tim Smith
<6908f531-b4d0-4519...@b2g2000prf.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> A rather remarkable bit of mathematics casts a darker view on the ring
> of algebraic integers where quadratics thankfully are useful for a
> somewhat subtle result, and the distributive property is almost
> bizarrely the linchpin of the proof.
>
> Consider in an integral domain
>
> P(x) = 175x^2 - 15x + 2
>
> and
>
> 7*P(x) = (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,
a1_(x) = (7x-1+sqrt(1+42x-147x^2))/2 [1]
and
a2_(x) = (7x-1-sqrt(1+42x-147x^2))/2 [2]
....
> So as they are mathematically the same in all key ways, not
> surprisingly, in each case you simply have 7 multiplied rather simply,
> which can be seen with the more complex example by letting x=0 as then
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0
>
> gives
>
> a^2 + a = 0
>
> so one of the a's is 0 and the other is -1, so I have from
>
> 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
>
> that
>
> 7*P(0) = (7)*(2) or (2)*(7).
Or, using [1] and [2], plugging in 0 gives 0 and -1, which gives 7*2 for
7*P(0).
>
> So one of (5a_1(x) + 7) or (5a_2(x)+ 7) is coprime to 7 at that value
> just like with
>
> 7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
>
> one of them is coprime to 7 at x=0, and since I've mentioned
> coprimeness I'll go ahead and say that we'll try to be in the ring of
> algebraic integers.
>
> I say try because you may already know that now there is a HUGE
> problem as if integer x generates a quadratic that is irreducible over
> Q, then you already know that NEITHER of the a's can have 7 as a
> factor in that ring!!!
>
> For example with x=1, the result would need that for
Plugging 1 into [1] and [2] gives 3+sqrt(-26) and 3-sqrt(-26). That
gives us
7*P(1) = 7*162 = (22 + 5*sqrt(-26)) * (22 - 5*sqrt(-26)).
Why do you think there is something broken about being able to write
7*162 as the product of two irrational complex numbers?
--
--Tim Smith
W. Dale Hall
Not really. It has to do with the ring where the factorization occurs.
The distributive property has nothing to say about the distribution
of factors over a product. Instead, it has everything to say about the
distribution of a factor over a sum.
In the case you've described
P(x) = (x + 1)(x + 2),
if you write
7P = (ax + a)(bx + 2b)
the factors a and b multiply to give 7, of course, and if your
factorization is over the integers, or the rational numbers,
there are only two options (disregarding sign). If you are
factoring over Z[sqrt(7)], you have three options (again
disregarding sign), the additional option being a = b = sqrt(7).
Factoring over the full ring of algebraic integers, you get
an infinite array of options, amounting to the full set of
factorizations of 7 that are available in that ring.
Now, having specified a particular form of factorization:
(*) 7 P = (5 f_1 + 7)(5 f_2 + 7)
I run into a bit of a problem as to how you maintain that
a factorization of this sort exists. In the preceding case,
you held correctly that your a_1 & a_2 satisfied an
algebraic equation, namely the quadratic equation
a^2 - (7 x - 1)a + (49 x^2 - 14 x) = 0.
I note that for integer x, this has integer coefficients,
and so (being monic) resulted in algebraic integer solutions.
One might well say that your a's are integral over the ring
Z[x] of polynomials with integer coefficients. At least I'd
say that.
In this case, I don't see a quadratic polynomial that the f's
satisfy, in the same fashion. Expanding the product on the
right side of the above equation (*), we find
25 f_1 f_2 + 35 (f_1 + f_2) + 49
and if the f's satisfy a monic quadratic polynomial
f^2 + p_1 f + p_0
with the p's being polynomials in x, for the above
product to hold, you'd need
f_1 f_2 = p_0
f_1 + f_2 = -p_1
Since I've assumed the p's to be polynomials, the equation
7 x^2 + 21 x + 14 = 25 p_0 + 35 p_1 + 49
suggests that they be of no greater than 2nd degree.
Let
p_0 = r x^2 + s x + t
p_1 = u x^2 + v x + w
and we'll get
7 x^2 + 21 x + 14 = 25 (r x^2 + s x + t)
+ 35 (u x^2 + v x + w)
+ 49
This is supposed to be an identity, so we equate powers of x:
x^2: 25 r + 35 u = 7
x : 25 s + 35 v = 14
1: 25 t + 35 w + 49 = 14
or 25 t + 35 w = -35
So, we have these constraints on the coefficients:
25 r + 35 u = 7,
25 s + 35 v = 14,
25 t + 35 w = -35.
Integrality of these coefficients leads to the
following:
25 r = 7(1 - 5u)
Note that for integer r,u, the right side must be congruent
to 2 mod 5, while the left side is divisible by 5.
I obtain similar incongruities for s and v:
25 s = 7(2 - 5 v)
in which the right side is congruent to 4 mod 5,
the left divisible by 5.
For the final coefficients, t and w:
25 t = - 35( 1 + w)
I note that setting w congruent to 4 mod 5
(e.g., w = -1, 4, 9, ...) is sufficient to
obtain a solution for t.
Did you intend your functions f_1, f_2 to satisfy
a quadratic equation with polynomial coefficients?
Was that polynomial intended to be monic?
What else have I assumed that is not the case here?
> Given that answer with linear functions can you deny that your claims
> that non-linear functions are somehow different are non-mathematical
> and completely bogus efforts to deny very basic mathematics?
>
>
> James Harris
Again, it's not the linearity vs. non-linearity that is at issue.
I could (and did) factor the above polynomial over Z[sqrt(7)]
with a different distribution of factors than having 7 residing
in one factor or the other.
In exactly the same fashion, in the non-polynomial factorization
example, the w's that are obtained are really only the "minimal
examples", because one could also extend the ring in which the
factors reside, to further vary the range of factors of 7 available
for the factorization.
In closing, it's not necessary to castigate the efforts of others
in understanding the situation. Calling these efforts "non-mathematical"
and "completely bogus" does nothing to enhance your argument, and
merely suggests your own case to be based on emotion. Given the
fact that I have exhibited factors that you maintained to be
impossible (and disregarding your so-called "wrapper theorem", which
I have examined and found to be somewhat incoherent, followed by
an argument that is, if read literally, patently false), it is
erroneous to suggest that my correct computations are either
"non-mathematical" or "bogus".
Dale
William Hughes
> Now if
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> and
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> there are an infinity of ways to distribute factors of 7 across the
> factorization of P(x),
and for different values of x different ways may be needed.
> but there's one clear way to determine which
> one was used
Nope. There is no "which one" to determine. For every
value of x there can be a different way.
- William Hughes
marcus_b
No. You are dividing, not multiplying. You do not
know a priori that 5a_2(x)/7 is in whatever ring you
start with.
> If you have
>
> 7*(x+3x+2) = (7x + 7)*(x+2)
>
> and divide off the 7, are you not using the distributive property?
>
It's OK in that case, because 7x is obviously divisible by 7. But
as usual the analogy doesn't do you any good, because you don't
know that 5a_1(x) is divisible by 7. The fact that 5a_1(0) is 0
does not tell you about other values of x. And the distributive
property does not tell you about divisibility in general either.
> Given that a*(b+c) = a*b + a*c it must be true that 'a' is a factor.
>
> So the point is that 7 is a factor based on the distributive property.
>
Works with 7x + 7 because 7x is clearly divisible by 7.
Doesn't work with 5a_1(x) + 7 because you do NOT know that
5a_1(x) is divisible by 7. And you cannot just assume that.
> If it is a factor then it can be removed, which is normally described
> as "dividing it off".
>
Tiresome. Clearly 7 divides 7x + 7. You have to PROVE that
7 divides 5a_1(x) + 7. You haven't.
> > The distributive property is about multiplication. It
> > says:
>
> > a * (b + c) = a*b + a*c.
>
> Yes.
>
So why do you think you can use it with division, in a
ring where division is not even an operation?
> > But you are DIVIDING by 7. Division is not a ring
> > operation. Division works in fields. In general, you
> > cannot talk about a/b in a ring. The ring may not contain
> > a/b. The "version" of the distributive property that you want
> > is
>
> > (b + c) / a = b/a + c/a.
>
> No.
>
Sorry, that's what you are doing: DIVIDING by 7.
> The reality can be seen with linear functions:
>
> 7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
>
> where dividing off the 7 is trivial.
>
Right. Why? Because 7x is clearly divisible by 7.
Stop thinking in oversimplified analogies. They do
not apply. 5a_1(x) is NOT divisible by 7 in general.
> I pointed out that there is no critical difference mathematically
> between that example and
>
> P(x) = 175x^2 - 15x + 2
>
> 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
>
> where the a's are roots of
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0,
>
> while there is the difference of linear versus non-linear function
> being used, as 7x is a linear function, or do you disagree?
>
No critical difference mathematically, between dividing
7x + 7
by 7, and dividing
5a_1(x) + 7
by 7? Of course there is. 7x is CLEARLY, OBVIOUSLY divisible
by 7. a_1(x) is NOT. All the difference in the world.
> If
>
> a*(f(x) + b) = a*f(x) + a*b
>
> without regard to whether you have a linear or a non-linear function
> how can the behavior change just because the a's are non-linear
> versus
>
> 7*(x^2 + 3x + 2) = (7x + 7)*(x+2)
>
> where they are?
>
Ultra tiresome. See above.
> > This is NOT a statement of the distributive property
> > in rings. Look it up.
>
> > In your case, "b/a" is 5a_1(x)/7. But you don't even
> > know that 5a_1(x)/7 is in your ring. If you assume it is,
> > then you are assuming what you want to prove.
>
Response to this???
> The full statement is
>
> 7*P(x) = (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 7 is multiplied times P(x), so it's not about assuming.
>
So you just somehow KNOW that 5a_1(x) is divisible by 7.
The distributive property tells you this??? No.
> > > where the u's are not units in the ring of algebraic integers because
> > > they are not roots of a monic polynomial with integer coefficients
> > > with a unit last coefficient.
>
> > > And mathematical consistency remains.
>
> > > But those factorization cannot exist in the ring of algebraic integers
> > > in that form as 7*u_1 is not available in the ring of algebraic
> > > integers
>
> > Of course it is. You can let u_1 = 1.
>
> Nope, as then 7 would be a factor in that ring, and it is not.
>
> The crucial logical point here is that
>
> a*(f(x) + b) = a*f(x) + a*b
>
> WITHOUT REGARD to the TYPE of function.
>
But this is not what you are doing! You are not dividing
7 out of just one term, but the product of two terms. It can
easily happen that you HAVE TO split 7 into two pieces to
accomplish such division. In fact this is exactly what
happens. See Hall's example for x = 1, 2, 3.
> While your argument and that of Arturo Magidin, and W. Dale Hall, and
> Rick Decker among others is that if f(x) is a non-linear function then
> with
>
> 7*P(x) = (5a_1(x) + 7)*(5a_2(x)+ 7)
>
> where the a's are roots of
>
> a^2 - (7x-1)a + (49x^2 - 14x) = 0
>
> somehow, someway the value of FUNCTIONS shift where factors of 7 go,
Yep, right.
> which is in direct contradiction to the reality that
>
> a*(f(x) + b) = a*f(x) + a*b
>
Nope, wrong. The latter equation involves factoring out
of only one term. What you have in reality involves two
terms. Ultimately, you do NOT know that 5a_1(x) is
divisible by 7. This is not analogous to a*f(x), which
is clearly divisible by a.
> WITHOUT REGARD to the TYPE of function.
>
> So your position can be shown to directly contradict basic algebra.
>
Not all all. No violation of any law of basic algebra
is involved.
> You are arguing that functional type defines the output of the
> distributive property when that is trivially wrong.
>
You are really incapable of getting it, aren't you?
> It is a bad math mistake.
>
You would instantly fail the first real math course
you ever took.
Marcus.
> James Harris
Michael Press
<1f19fb23-31c4-4de9...@b2g2000prf.googlegroups.com>,
JSH <jst...@gmail.com> wrote:
> The functional type does NOT impact the statement
>
> a*(f(x) + b) = a*f(x) + a*b
>
> which is true regardless of whether or not the function is linear or
> non-linear.
The word `impact' is not a verb.
Learn the difference between `affect' and `effect',
then eschew that abomination.
--
Michael Press
marcus_b
> On Sep 1, 7:23 pm, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On Sep 1, 9:15 pm, JSH <jst...@gmail.com> wrote:
>
> > > Consider again
>
> > > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> > > where P(x) is a polynomial function, does the distributive property
> > > care about the type of the function?
>
> > The important question of whether 7 divides through one f
> > or two f's does care about the type of the function.
>
> Nope. Let's say that
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> so you multiply 7*P(x), how many ways can you do that in the ring of
> algebraic integers?
>
> Ans: An infinity of ways.
Correct.
>
> Is there any way given
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> to determine *which* way was used.
>
Yes. If only one of f_1(x) or f_2(x) is divisible by
7, then you know that 7 factors out of the corresponding
term in parentheses.
Of course it may happen that neither f_1(x) nor f_2(x)
is divisible by 7. In that case each of f_1(x) and f_2(x)
must contain "part" of 7. That is, it must be the case
that 7 = w_1(x) * w_2(x), and
f_1(x) is divisible by w_1(x), and
f_2(x) is divisible by w_2(x).
> > As you have noted, this question has nothing to do with
> > the distributive property.
>
> > - William Hughes
>
> Of course it does because if you have
>
> 7*(x+1)*(x+2) = (7x + 7)(x + 2)
>
> that's all about the distributive property as you *distribute* the 7
> through the factor.
>
Or you could write:
7*(x+1)*(x+2) = (x + 1)(7x + 14).
Or:
7*(x+1)*(x+2) = (sqrt(7)x + sqrt(7))(sqrt(7)x + 2sqrt(7)).
Or any number of other ways. The point being, the distributive
property allows all sorts of factorizations in this case, all
of which have coefficients in the ring of algebraic integers.
The distributive property does NOT tell you that it has to be
7*(x+1)*(x+2) = (7x + 7)(x + 2).
But you keep saying that the factorization of
7*P(x) = (5a_1(x) + 7)(5a_2(x) + 7)
can be done in only one way, and that way is determined by
the distributive property.
That's wrong. Factorization of 7 out of
(5a_1(x) + 7)(5a_2(x) + 7)
is determined by w_1(x) = GCD(a_1(x), 7) and
w_2(x) = GCD(a_2(x), 7).
This is actually exactly the same as your reducible case
factorization: in
(7x + 7)*(x + 2),
the coefficient corresponding to a_1(x) is 7, and the
coefficient corresponding to a_2(x) is 1.
Note that GCD(7, 7) = 7 and GCD(1, 7) = 1.
That is, it works out in your example exactly the way
it works out in
(5a_1(x) + 7)*(5a_2(x) + 7).
You find the correct factorization by computing GCDs in
both cases!
> Now it's trivial algebra but the problem here is that posters like you
> deny trivial algebra and then just claim that's not what you're doing
> because the stakes are high, and you actually clearly suck at basic
> math.
>
See above.
> Now if
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> and
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> there are an infinity of ways to distribute factors of 7 across the
> factorization of P(x), but there's one clear way to determine which
> one was used as let's say, where still the f's are linear functions,
> that f_1(0) = 0.
>
> In that case HOW did the 7 distribute?
>
> Not a trick question but answering honestly will reveal that yes, it
> is about the distributive property.
>
Obviously in that case 7 factors out of (5a_1(0) + 7).
Why? Because GCD(a_1(0), 7) = GCD(0, 7) = 7.
But GCD(a_1(x), 7) in general is not 7. That is completely
clear from Dale Hall's calculation. What he was doing was
computing GCD(a_1(x), 7) for different values of x: x = 1,
x = 2, x = 3. He used a well-validated algebraic-number-theory
computer program to compute the GCDs.
Now, who should we believe when it comes to computing GCDs:
You, or a computer program that gets it right every time?
So tell us how YOU would compute GCD(a_1(1), 7) when
a_1(x) = 3 + sqrt(-26).
See, this GCD(a_1(1), 7) HAS to be the right factor.
You are trying to tell us that GCD(a_1(1), 7) = 7. But
that is not what the computer program gets!
Excellent example of where your claims can be checked
by a computer program. Isn't that you keep asking for?
And here, it gives a different answer from what you get.
How do you explain this? Are you understanding this?
Marcus.
[snip repetition]
>
> James Harris
Jesse F. Hughes
> In article
> <1f19fb23-31c4-4de9...@b2g2000prf.googlegroups.com>,
> JSH <jst...@gmail.com> wrote:
>
>> The functional type does NOT impact the statement
>>
>> a*(f(x) + b) = a*f(x) + a*b
>>
>> which is true regardless of whether or not the function is linear or
>> non-linear.
>
> The word `impact' is not a verb.
It has been since 1601.
impact (v.)
1601, "press closely into something," from L. impactus, pp. of
impingere "to push into, dash against" (see impinge). Originally sense
preserved in impacted teeth (1876). Sense of "strike forcefully
against something" first recorded 1916. Figurative sense began with
use as a noun (1817, first in Coleridge) meaning "effect of coming
into contact with a thing or person."
I'm not saying he's using it right, but impact is a verb according to
http://www.etymonline.com/index.php?term=impact.
> Learn the difference between `affect' and `effect',
> then eschew that abomination.
--
Jesse F. Hughes
"Mathematicians don't fit in with a consistent view, unless you accept
that to a strangely large extent they are acting under the influence
of something very powerful, dark, and negative." -- James S. Harris
W. Dale Hall
> Michael Press <rub...@pacbell.net> writes:
>
>> In article
>> <1f19fb23-31c4-4de9...@b2g2000prf.googlegroups.com>,
>> JSH <jst...@gmail.com> wrote:
>>
>>> The functional type does NOT impact the statement
>>>
>>> a*(f(x) + b) = a*f(x) + a*b
>>>
>>> which is true regardless of whether or not the function is linear or
>>> non-linear.
>> The word `impact' is not a verb.
>
> It has been since 1601.
>
> impact (v.)
>
> 1601, "press closely into something," from L. impactus, pp. of
> impingere "to push into, dash against" (see impinge). Originally sense
> preserved in impacted teeth (1876). Sense of "strike forcefully
> against something" first recorded 1916. Figurative sense began with
> use as a noun (1817, first in Coleridge) meaning "effect of coming
> into contact with a thing or person."
>
> I'm not saying he's using it right, but impact is a verb according to
> http://www.etymonline.com/index.php?term=impact.
>
>> Learn the difference between `affect' and `effect',
>> then eschew that abomination.
Bummer. The one thing this thread is missing is a grammar brawl.
Joshua Cranmer
Right. Shall I effect one right now?
JSH
Wrong. There is NO way to have
P(x) = x^2 + 3x + 2 = (x+1)(x+2)
and
7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
in general in the ring of algebraic integers with linear functions.
___JSH
W. Dale Hall
Harrumph! Your affect!
William Hughes
> On Sep 2, 4:16 am, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > On Sep 2, 12:51 am, JSH <jst...@gmail.com> wrote:
>
> > > Now if
>
> > > P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> > > and
>
> > > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
At this point we know that f_1(x) and f_2(x)
are not linear functions
>
> > > there are an infinity of ways to distribute factors of 7 across the
> > > factorization of P(x),
>
> > and for different values of x different ways may be needed.
>
> > > but there's one clear way to determine which
> > > one was used
>
> > Nope. There is no "which one" to determine. For every
> > value of x there can be a different way.
>
> > - William Hughes
>
<snip stuff about linear functions since
we know that f_1 and f_2 are not linear functions>
Try again.
- William Hughes
JSH
> On Sep 2, 8:44 pm, JSH <jst...@gmail.com> wrote:
>
> > On Sep 2, 4:16 am, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On Sep 2, 12:51 am, JSH <jst...@gmail.com> wrote:
>
> > > > Now if
>
> > > > P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> > > > and
>
> > > > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> At this point we know that f_1(x) and f_2(x)
> are not linear functions
Nope. They are. If you thought they weren't or I said they weren't
that was a mistake!
I'll make things simpler by giving an example that CAN work:
P(x) = x^2 + 3x + 2 = (x+1)(x+2)
and
7*P(x) = (f_1(x) + 7)*(f_2(x)+ 7)
where the f's are LINEAR functions still. The ring is still the ring
of algebraic integers.
Now then, let's say that only one of them has sqrt(7) as a factor.
Can you give an example for actual f's that are linear functions that
will work?
Of course you can. And you can do the same if the factor is
1+sqrt(-6) or is still 7 itself.
Let's keep thing simple and say that one of the f's has 7 as a factor,
what can you do to know WHICH of them, if you have a single set of
values at say, x=1? That is, at that value you have P(1), f_1(1) and
f_2(1).
Do you now know which of the f's has 7 as a factor if P(1) is coprime
to 7 if you have values for f_1(1) and f_2(2)?
Yes. You do. But how?
Ans: The distributive property.
James Harrs
W. Dale Hall
Check my response earlier in this branch of the thread:
It appears that there is no factorization with f's satisfying
a quadratic equation over Z[x]. I don't see any factorization
using a polynomial of higher degree, but that doesn't mean
there isn't one.
Dale
William Hughes
>
> I'll make things simpler by giving an example that CAN work:
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> and
>
> 7*P(x) = (f_1(x) + 7)*(f_2(x)+ 7)
so for every x, 7 divides through (f_1(x) + 7)*(f_2(x)+ 7)
in some way
>
> where the f's are LINEAR functions still.
If the f's are LINEAR then 7 divides through
(f_1(x) + 7)*(f_2(x)+ 7) in the same way for every x.
We can use a single value of x to determine
this one way.
If the f's are NON-LINEAR then 7 may divide through
(f_1(x) + 7)*(f_2(x)+ 7) in different ways for differetn
x's. We cannot use a single value of x, because there
is more than one way to determine.
- William Hughes
Joshua Cranmer
> If the f's are LINEAR then 7 divides through
> (f_1(x) + 7)*(f_2(x)+ 7) in the same way for every x.
> We can use a single value of x to determine
> this one way.
Correct me if I am wrong, but it seems that we could not use x = 7 here?
William Hughes
[From context LINEAR must mean a linear polynomial (ax +b) and not a
linear
function (kx) as we need solutions to
7P(x)= 7x^2+ 21x +14 = (f_1(x) + 7)*(f_2(x)+ 7) ]
Any value of x for which P(x) is coprime to 7 will suffice to
determine
how 7 divides through (f_1(x) + 7)*(f_2(x)+ 7).
- William Hughes
gjedwards
Maybe the reviewers comments would help your side of the argument -
after all they must have thought it was right to publish it. Why not
post them since you keep emphasizing that your paper was 'peer-
reviewed'. Still waiting....
marcus_b
>
> > On Sep 1, 9:15 pm, JSH <jst...@gmail.com> wrote:
>
> > > Consider again
>
> > > 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> > > where P(x) is a polynomial function, does the distributive property
> > > care about the type of the function?
>
> > The important question of whether 7 divides through one f
> > or two f's does care about the type of the function.
>
> Nope. Let's say that
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> so you multiply 7*P(x), how many ways can you do that in the ring of
> algebraic integers?
>
> Ans: An infinity of ways.
>
> Is there any way given
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> to determine *which* way was used.
>
Yes. But let's start first with your
other example,
7 P(x) = (7x + 7)*(x + 2)
The coefficients of x on the right side are 7 and 1.
Note that GCD(7, 7) = 7 and GCD(1, 7) = 1.
So you should factor 7 out of the first parenthesis. OK,
this is obvious without using the GCD. The point is, using
the GCD to figure out how the factorization should go
GENERALIZES to other situations. In particular, it
generalizes to
(5a_1(x) + 7)*(5a_2(x) + 7).
For example, let x = 0. Then a_1(x) = 0, and GCD(0, 7) = 7.
So the factorization when x = 0 is
(0 + 7*1)(5a_2(0) + 7) = 7 * (-5 + 7) = 7 * 2.
When x = 1, the computation of GCD(a_1(1), 7) is difficult.
Dale Hall used a computer program to find it. The answer, w_1(x),
is NOT 7, which is what you think it should be. Similarly,
Dale gave answers from the same program for x = 2 and x = 3.
Neither of those answers equalled 7 either.
Dale's well-verified computer program gives the answer. It
is different from your answer.
According to the computer program, your answer is WRONG.
Sound familiar? You have argued long and hard for computer
verification of maths. Here someone actually did it and it
proved you wrong.
If you want a formula for the appropriate factorization of
7, it's easy to write down. 7 = w_1(x) * w_2(x), where
w_1(x) = GCD(a_1(x), 7) and
w_2(x) = GCD(a_2(x), 7).
You may say, "That's not a formula. It's just a string of
symbols. You're not giving the GCD explicitly."
The GCD function is perfectly well-defined, but even in the
ring of integers, it does not have a "formula" expression.
Examples:
GCD(14, 7) = 7
GCD(30, 35) = 5
GCD(210, 2940) = 70
GCD(85, 306) = 17
etc. You see any way to derive a "formula" for those? But
they are all well-defined and correct.
But there IS a way compute the GCD. There is an algorithm.
In the integers, it is called the Euclidean algorithm. In a
finite number of steps you can always compute the GCD of any
two integers.
The same is true in the algebraic integers. There is an
algorithm. It depends on the fact that the class number of any
finite-degree extension of the rationals being finite. The algorithm
is arduous, but it is well-defined and it is guaranteed to always
work, and it can be programmed. Dale Hall made use of one
program that can compute the GCD of two algebraic integers.
Again: The right factorization is defined by the GCD
function. A computer program proves that your factorization
is NOT right: GCD(a_1(x), 7) in general is NOT 7, as you keep
insisting. No violation of the distributive property is
involved. Your argument is just plain wrong.
> > As you have noted, this question has nothing to do with
> > the distributive property.
>
> > - William Hughes
>
> Of course it does because if you have
>
> 7*(x+1)*(x+2) = (7x + 7)(x + 2)
>
> that's all about the distributive property as you *distribute* the 7
> through the factor.
>
Works because GCD(7, 7) = 7.
> Now it's trivial algebra but the problem here is that posters like you
> deny trivial algebra and then just claim that's not what you're doing
> because the stakes are high, and you actually clearly suck at basic
> math.
>
Shoe is on the other foot.
> Now if
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> and
>
> 7*P(x) = (5f_1(x) + 7)*(5f_2(x)+ 7)
>
> there are an infinity of ways to distribute factors of 7 across the
> factorization of P(x), but there's one clear way to determine which
> one was used as let's say, where still the f's are linear functions,
> that f_1(0) = 0.
>
But that only works when x = 0.
> In that case HOW did the 7 distribute?
>
As you say. But the answer is different if x <> 0.
It depends on GCD(f_1(x), 7).
> Not a trick question but answering honestly will reveal that yes, it
> is about the distributive property.
>
Nope. You are, to use your own words, overinterpreting it.
There is a simple way to give the right answer and it can be
written as a computer program. No violation of the distributive
property occurs.
> Given that answer with linear functions can you deny that your claims
> that non-linear functions are somehow different are non-mathematical
> and completely bogus efforts to deny very basic mathematics?
>
No basic CORRECT mathematics is being denied. Only your bogus
argument. You are proven wrong by a computer program.
Marcus.
> James Harris
Michael Press
And now, back to our regularly scheduled programming.
--
Michael Press
junoexpress
> post on the subject, which explains in detail what a proof is.
>
> James Harris
I would advise against it unless you have a strong stomach and a barf
bag handy.
Face it, your "definition" is, at best, very poor. Not considering a
host of
technical points that are not addressed (such as a "truth", "logical
steps", "proceeds"), the "definition" ends with the phrase "which
then
must be true" (in reference to the conclusion). ROFLOL.
Shooting holes in your so-called defn takes little effort:
for ex., what you would defin as a "proof" would not permit a proof
by
contradition (since you are not starting with a truth, but actually a
falsehood). And yet, with this sort of crap that you're posting,
everyone is to
believe that you are the Messiah of math and everyone else is wrong.
Yeah, right.
It's clear from such simple observations like this that your high
ratings are not because people are using your definitions (since you
don't have a single reference, besides your own to show for), but
rather simply people linking to your site for a good laugh. In short,
you've become an Internet laughingstock.
Congratulations!
M
Edward
[Mr.] Lynn Kurtz
>Hello i am new to this forum and i want to ask about card Probability me and my friends had a game of cards with 7 cards we dealt them every time and than reshuffled them again one of my friends got seven times two cards that were marked "as special cards" now that friend got those same two cards seven time now what i ask is if the Probability for him to get those two cards is about 28%. is the Probability for him to get it seven times in a row in seven diffrent games does it still mean that the Probability stays about 28%?
Since you are new to this forum, I will suggest that you will receive
nicer responses if you put a little effort into your posts to make
them readable and literate. The personal pronoun is "I", not "i", and
sentences begin with capital letters and end with periods, for
starters.
You need to be more specific about your problem. How many cards are
being dealt. Is it two? In other words are you asking about the
probability that a particular person "Joe" get the same two cards
seven times in a row, when being dealt 2 cards from a pack of 7? I
will assume, perhaps incorrectly, that is what you intended to ask. It
is the same as the question what is the probablility that Joe gets the
two marked cards on a single draw, which is 1 / C(7,2) = 1 / 21, a
little under 5%. The probability of him doing that seven times in a
row is (1/21)^7 = 1 / 1801088541 = .0000000005552197892. Not very
likely.
--Lynn
Edward
The question here is that we played a game with 7 cards,Two cards had the same meaning we dealt it, played then gave the cards for a reshuffle.Now the thing here is as you called him "Joe" got one of probability change every time,Or because its diffrent case it stays the same probability?
I thank everyone for the rsponse and I am sorry for my lousy english.
Edward
Now what i ask is does the probability stay the same becuse every time we deal its a new case,which means the probability stays the same, or does the probability for him to get those two cards 7 times in a row is smaller than just getting one of them once ?
Thanks for any response and sorry for my lousy english.