The Coverage Problem - thread 1

[Haydon Berrow]

What is the coverage problem? Not with constructions to illustrate it, but a simple statement such as

Example 1) There exists a complex number x and an integer n such that x is not an algebraic integer but n*x is an algebraic integer

or

Example 2) There exists a complex number x such that n*x is an algebraic integer with degree 2 for ever integer n>1.

I know that neither of these are correct, but what is the correct statement?

[JSH]

On Monday, November 13, 2017 at 6:00:56 AM UTC-5, Haydon Berrow wrote:

What is the coverage problem? Not with constructions to illustrate it, but a simple statement such as

Example 1) There exists a complex number x and an integer n such that x is not an algebraic integer but n*x is an algebraic integer

Given a complex number x and an integer n, which is not zero nor a unit, where x is NOT an algebraic integer, there exists numbers such that nx is an algebraic integer, regardless otherwise of n.

Now 1/2 is not an algebraic integer, but n/2 is for any even n. But am saying for ANY non-unit non-zero integer n. Weird, eh? These numbers are NOT in any way fractions. 

[Rotwang]

Are you saying that there is some number x, not an algebraic integer, such that n*x is an algebraic integer for any non-zero, non-unit natural number n? Because if such an x existed we would have x = 3x - 2x, where 3x and -2x are both algebraic integers. So the algebraic integers would not be closed under addition, and therefore not a ring. Is that what you're saying?

[Haydon Berrow]

On Monday, November 13, 2017 at 12:45:30 PM UTC, JSH wrote:

Ah, thank you, but that doesn't make sense You start with 'given .. x and n' and then follow with 'there exists numbers ...'. ie we're given x and n and some other number exists but it's not clear from your statement what these further numbers are. Do you mean

Example 3) There exists a complex number x such that x is not an algebraic integer but n*x is an algebraic integer for all integers n that aren't zero or a unit.

Of course this is totally equivalent to

Example 3a) There exists complex numbers x such that x is not an algebraic integer but n*x is an algebraic integer for all integers n that aren't zero or a unit.

and

Example 3b) There exists a complex number x such that x is not an algebraic integer but n*x is an algebraic integer for all integers n except 0 and 1.

Any of them can be proved from the other so we'll settle on whichever is the more useful.

Of course I'll immediately respond by pointing out that if 3*x and 2*x are both algebraic integers then so is the difference.

[JSH]

Interesting way of putting it. The proof is easy enough that such numbers exist. And have posted in this group. But here's the link to it on my blog:

http://somemath.blogspot.com/2015/11/non-polynomial-factorization-short.html

What I LOVE about THAT argument is use of a generalized quadratic factorization. Makes it possible to solve all of the variables. You can see a variable k, which you can set how you wish, and note that setting things up for algebraic integers for some functions, with integer k, you can push one in or out of the ring of algebraic integers depending on how you change k. Play with k = 1 for a bit, to get a handle though. HELPS IMMENSELY.

So yeah, if YOUR assertion were agreed upon by others then yes, could establish is not a ring. You might re-evaluate arguments now where mathematicians tested that the addition of algebraic integers is an algebraic integer. What did they miss? 

We can guess. If you ASSUME that for any complex number x NOT an algebraic integer that if nx is then, is some kind of fraction, then yeah, can miss it.

Result shows that some numbers are simply NOT in the ring of algebraic integers, but multiply times a non-unit, non-zero integer, and result is. Is wild.

___JSH

[JSH]

You think then the difference between 3x - 2x is an algebraic integer, but how do you prove? Rotwang put up same argument in reply by the way. You can look at my response there.

___JSH 

[Rotwang]

On Monday, November 13, 2017 at 2:21:37 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 8:54:22 AM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 12:45:30 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 6:00:56 AM UTC-5, Haydon Berrow wrote:

What is the coverage problem? Not with constructions to illustrate it, but a simple statement such as

Example 1) There exists a complex number x and an integer n such that x is not an algebraic integer but n*x is an algebraic integer

Given a complex number x and an integer n, which is not zero nor a unit, where x is NOT an algebraic integer, there exists numbers such that nx is an algebraic integer, regardless otherwise of n.

Now 1/2 is not an algebraic integer, but n/2 is for any even n. But am saying for ANY non-unit non-zero integer n. Weird, eh? These numbers are NOT in any way fractions.

Are you saying that there is some number x, not an algebraic integer, such that n*x is an algebraic integer for any non-zero, non-unit natural number n? Because if such an x existed we would have x = 3x - 2x, where 3x and -2x are both algebraic integers. So the algebraic integers would not be closed under addition, and therefore not a ring. Is that what you're saying?

Interesting way of putting it. The proof is easy enough that such numbers exist. And have posted in this group. But here's the link to it on my blog:

http://somemath.blogspot.com/2015/11/non-polynomial-factorization-short.html

What I LOVE about THAT argument is use of a generalized quadratic factorization. Makes it possible to solve all of the variables. You can see a variable k, which you can set how you wish, and note that setting things up for algebraic integers for some functions, with integer k, you can push one in or out of the ring of algebraic integers depending on how you change k. Play with k = 1 for a bit, to get a handle though. HELPS IMMENSELY.

So yeah, if YOUR assertion were agreed upon by others then yes, could establish is not a ring.

You seem to be hedging your bets. The assertion that there exists a non-algebraic integer x such that n*x is an algebraic integer for all non-zero, non-unit integers n is how I interpreted /your/ assertion. Was it a correct paraphrase of your assertion, or not? Do /you/ believe that the algebraic integers are not closed under addition?

You might re-evaluate arguments now where mathematicians tested that the addition of algebraic integers is an algebraic integer. What did they miss?

I'm willing to look up the proof that the algebraic integers are closed under addition and post it here in my own words if you want (I'd need to remind myself how it goes - IIRC it isn't difficult, but it isn't elementary either). But I'd like to be clear about whether you are claiming they aren't before put the time in, because otherwise there's not much point.

[Haydon Berrow]

On Monday, November 13, 2017 at 2:23:21 PM UTC, JSH wrote:

You think then the difference between 3x - 2x is an algebraic integer, but how do you prove? Rotwang put up same argument in reply by the way. You can look at my response there.

You didn't confirm that Example 3) is what your statement of the coverage problem is. Do I have it correctly and do you prefer 3, 3a, or 3b ?

I want to be sure of my understanding before I try to understand more.

I agree that it's not obvious that the difference between two algebraic integers is another one and I'd have to look the proof up if I wanted to. When we get to your construction, and I've said before that I didn't understand it, I'm going to tax your patience by looking at these numbers that you get from a quadratic that have the strange property that 3*x and 2*x are algebraic integers. I *think* that they have degree 2, ie there are quadratics f₂() and  f₃() with integer coefficients such f₂(2x) and  f₃(3x) equal zero, but I don't see how this can be possible.

 

[JSH]

On Monday, November 13, 2017 at 9:37:36 AM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 2:21:37 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 8:54:22 AM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 12:45:30 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 6:00:56 AM UTC-5, Haydon Berrow wrote:

What is the coverage problem? Not with constructions to illustrate it, but a simple statement such as

Example 1) There exists a complex number x and an integer n such that x is not an algebraic integer but n*x is an algebraic integer

Given a complex number x and an integer n, which is not zero nor a unit, where x is NOT an algebraic integer, there exists numbers such that nx is an algebraic integer, regardless otherwise of n.

Now 1/2 is not an algebraic integer, but n/2 is for any even n. But am saying for ANY non-unit non-zero integer n. Weird, eh? These numbers are NOT in any way fractions.

Are you saying that there is some number x, not an algebraic integer, such that n*x is an algebraic integer for any non-zero, non-unit natural number n? Because if such an x existed we would have x = 3x - 2x, where 3x and -2x are both algebraic integers. So the algebraic integers would not be closed under addition, and therefore not a ring. Is that what you're saying?

Interesting way of putting it. The proof is easy enough that such numbers exist. And have posted in this group. But here's the link to it on my blog:

http://somemath.blogspot.com/2015/11/non-polynomial-factorization-short.html

What I LOVE about THAT argument is use of a generalized quadratic factorization. Makes it possible to solve all of the variables. You can see a variable k, which you can set how you wish, and note that setting things up for algebraic integers for some functions, with integer k, you can push one in or out of the ring of algebraic integers depending on how you change k. Play with k = 1 for a bit, to get a handle though. HELPS IMMENSELY.

So yeah, if YOUR assertion were agreed upon by others then yes, could establish is not a ring.

You seem to be hedging your bets. The assertion that there exists a non-algebraic integer x such that n*x is an algebraic integer for all non-zero, non-unit integers n is how I interpreted /your/ assertion. Was it a correct paraphrase of your assertion, or not? Do /you/ believe that the algebraic integers are not closed under addition?

I noted that there can exist a complex number x, which is NOT an algebraic integer, where ANY non-unit, nonzero integer k times that x will give an algebraic integer. That is, where kx is an algebraic integer for ANY non-unit, nonzero k, when x is not. You noted that x = 3x - 2x, and YOU noted that would indicate a failure of the ring of algebraic integers. And I think that's interesting to assert.

Makes you ponder the argument that the ring of algebraic integer IS so closed. And have looked it over, in the past. 

You might re-evaluate arguments now where mathematicians tested that the addition of algebraic integers is an algebraic integer. What did they miss?

I'm willing to look up the proof that the algebraic integers are closed under addition and post it here in my own words if you want (I'd need to remind myself how it goes - IIRC it isn't difficult, but it isn't elementary either). But I'd like to be clear about whether you are claiming they aren't before put the time in, because otherwise there's not much point.

Cool. The proof is EASY that these numbers exist. But they are not intuitive, which is why I guess were left for me to find. And I got lucky.

Since there is a mathematical proof that they exist, is an intellectual exercise to see how mathematicians who thought they had proof about the ring of algebraic integers, messed up. Have fun.

___JSH

[JSH]

On Monday, November 13, 2017 at 9:46:22 AM UTC-5, Haydon Berrow wrote:

On Monday, November 13, 2017 at 2:23:21 PM UTC, JSH wrote:

You think then the difference between 3x - 2x is an algebraic integer, but how do you prove? Rotwang put up same argument in reply by the way. You can look at my response there.

You didn't confirm that Example 3) is what your statement of the coverage problem is. Do I have it correctly and do you prefer 3, 3a, or 3b ?

I realized your FINAL statement encompassed the above so answered it, to be more efficient.

Answer is as given. Relevance of the prior examples was suspect. That is, they did not actually really separate in a way that justified that spread, in my opinion. Kind of like asking same thing slightly differently. So I simply addressed most relevant point, which still stands by the way.

Have made progress with Rotwang.

 

I want to be sure of my understanding before I try to understand more.

Makes sense. 

I agree that it's not obvious that the difference between two algebraic integers is another one and I'd have to look the proof up if I wanted to. When we get to your construction, and I've said before that I didn't understand it, I'm going to tax your patience by looking at these numbers that you get from a quadratic that have the strange property that 3*x and 2*x are algebraic integers. I *think* that they have degree 2, ie there are quadratics f₂() and  f₃() with integer coefficients such f₂(2x) and  f₃(3x) equal zero, but I don't see how this can be possible.

 

To be a member of the ring of algebraic integers, a number MUST be the root of SOME monic polynomial with integer coefficients.

Given the assertion that 2x is an algebraic integer, then by definition that assertion is being made. And you see?

By mathematical proof, always. And you say you struggle with the construction? Ok. Is an abstraction, which relies upon quadratics.

Still could be difficult to fully grasp that abstraction, which is true for any mathematics. Is like how some struggle with calculus.

I LIKE calculus. But I struggle in many areas with it. Is just reality of mathematics. You can struggle, but if you do not get to an understanding, then you will not be able to understand the proof.

Is like if am struggling with calculus and you keep repeating to me the calculus argument, will not go anywhere, right?

___JSH 

[Rotwang]

On Monday, November 13, 2017 at 3:01:43 PM UTC, JSH wrote:

[...]

I noted that there can exist a complex number x, which is NOT an algebraic integer, where ANY non-unit, nonzero integer k times that x will give an algebraic integer. That is, where kx is an algebraic integer for ANY non-unit, nonzero k, when x is not. You noted that x = 3x - 2x, and YOU noted that would indicate a failure of the ring of algebraic integers. And I think that's interesting to assert.

Makes you ponder the argument that the ring of algebraic integer IS so closed. And have looked it over, in the past. 

You might re-evaluate arguments now where mathematicians tested that the addition of algebraic integers is an algebraic integer. What did they miss?

I'm willing to look up the proof that the algebraic integers are closed under addition and post it here in my own words if you want (I'd need to remind myself how it goes - IIRC it isn't difficult, but it isn't elementary either). But I'd like to be clear about whether you are claiming they aren't before put the time in, because otherwise there's not much point.

Cool. The proof is EASY that these numbers exist. But they are not intuitive, which is why I guess were left for me to find. And I got lucky.

Since there is a mathematical proof that they exist, is an intellectual exercise to see how mathematicians who thought they had proof about the ring of algebraic integers, messed up. Have fun.

You still seem to be hedging your bets. Do you believe that the algebraic integers are not closed under addition? Yes, or no?
 

[JSH]

The ring of algebraic integer is NOT closed under addition, for example I know that 3x can be an algebraic integer and 2x can be an algebraic integer, when x is NOT an algebraic integer. So yeah, add negative 2x to 3x and can NOT be an algebraic integer when they both are.

Review the literature, and see if that case is handled. Reviewed what was done in the late 1800's myself over a decade ago.

Is is an interesting subject.

___JSH

[Rotwang]

On Monday, November 13, 2017 at 4:03:11 PM UTC, JSH wrote:

[...]

The ring of algebraic integer is NOT closed under addition, for example I know that 3x can be an algebraic integer and 2x can be an algebraic integer, when x is NOT an algebraic integer.

OK, thanks. Here's a nice proof that the algebraic integers are closed under addition:

First, a lemma: a complex number x is an algebraic integer if and only if it is an eigenvalue of a square matrix with integer coefficients. For if

M u = x u

for some square matrix M and vector u, then

(M - x I) u = 0 and so
|M - x I| = 0

where I is the identity matrix and |.| denotes the determinant. But the determinant of M - x I is a monic polynomial in x, so x is an algebraic integer. Conversely suppose that x is an algebraic integer, and therefore the root of a monic polynomial with integer coefficients, i.e.

x^m + a_{m - 1} x^{m - 1} + ... + a_0 = 0

for some integers a_0, a_1, ... a_{m - 1}. Let

u = (1, x, x^2, ... x^{m - 1})

and let M be an nxn matrix whose bottom row is given by (-a0, -a1, ... -a_{m - 1}) and whose upper m - 1 rows satisfy

M_{i, j} = delta_{i + 1, j}

(here delta is the Kronecker delta symbol). So e.g. in the case where m = 3 we have

    /   0   1   0 \
M = |   0   0   1 |
    \ -a0 -a1 -a2 /

Then it's easy to see that M u = x u, and the lemma is proved.

Suppose now that x and y are algebraic integers. Then there exist natural numbers m and n, an mxm matrix M, an nxn matrix N, an m-vector u and an n-vector v such that

M u = x u
N v = y v

We define an (n m)x(n m) matrix P and and (n m)-vector w as follows:

P_{m i + j, m k + l} = M_{j, l} delta_{i, k} + N_{i, k} delta_{j, l}
w_{m i + j} = u_j v_i

for i and k in {0, 1, ... n - 1}, j and l in {0, 1, ... m - 1} (I am using the fact that every natural number less than n m may be uniquely represented in the form m i + j with i, j in the aforementioned sets). Alternatively, if you're familiar with the tensor product of vector spaces, the above definitions simply say that

P = M \otimes I + I \otimes N
w = u \otimes w

We then find that

(P w)_{m i + j} = sum_{k, l} P_{m i + j, m k + l} w_{m k + l}
                = sum_{k, l} (  M_{j, l} delta_{i, k} u_l v_k
                              + N_{i, k} delta_{j, l} u_l v_k)
                =   sum_l M_{j, l} u_l v_i
                  + sum_k N_{i, k} u_j v_k
                = (M u)_j v_i + (N v)_i u_j
                = x u_j v_i + y v_i u_j
                = (x + y) w_{m i + j}

where I used the facts that M u = x u and N v = y v in the penultimate step. Thus we see that w is an eigenvector of P with eigenvalue x + y, so that x + y is an algebraic integer from the lemma.

[JSH]

Admittedly skimmed as NOW go through and replace x with 3x and y with -2x, please.

Report back.

___JSH 

[Rotwang]

On Monday, November 13, 2017 at 7:08:49 PM UTC, JSH wrote:

[...]

Admittedly skimmed as NOW go through and replace x with 3x and y with -2x, please.

Sure, I'll do it when I get home in a couple of hours. Will it be OK if I assume, for the purpose of illustration, that the monic polynomials of which 3x and -2x are roots are both quadratics (or I could assume one is a quadratic and another a cubic)? That will enable me to write the various matrices more explicitly, rather than defining them by indices.

[JSH]

You are simply checking if there is a mathematical reason that given 3x and -2x being algebraic integers, there is a mathematical reason that would say that x must be as well, when I'm like, I can prove it might not be. And proofs don't fight.

That is, when you have mathematical truth, it is consistent. Given that I KNOW that I can prove the reality, is not really that interesting to me your attempts but might help you. I reviewed arguments that ring of algebraic integers was ok, over a decade ago. Was surprised. Gaping gaps in reasoning, when you're looking for them.

But yeah, proofs don't fight.

So yeah, you're doing something for yourself. So you can approach as you wish. 

James Harris 

[Rotwang]

On Monday, November 13, 2017 at 7:35:31 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 2:22:57 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:08:49 PM UTC, JSH wrote:

[...]

Admittedly skimmed as NOW go through and replace x with 3x and y with -2x, please.

Sure, I'll do it when I get home in a couple of hours. Will it be OK if I assume, for the purpose of illustration, that the monic polynomials of which 3x and -2x are roots are both quadratics (or I could assume one is a quadratic and another a cubic)? That will enable me to write the various matrices more explicitly, rather than defining them by indices.

You are simply checking if there is a mathematical reason that given 3x and -2x being algebraic integers, there is a mathematical reason that would say that x must be as well, when I'm like, I can prove it might not be.

How do you know that? There have been many previous occasions when you were convinced you had proven something, only to subsequently discover you were mistaken.

 

And proofs don't fight.

That is, when you have mathematical truth, it is consistent. Given that I KNOW that I can prove the reality, is not really that interesting to me your attempts but might help you.

Just to be clear, are you saying you have no intention of actually checking any proof I provide with the specific substitutions you asked for? Because if that's the case then I won't bother.

 

I reviewed arguments that ring of algebraic integers was ok, over a decade ago. Was surprised. Gaping gaps in reasoning, when you're looking for them.

Is it possible that you have misremembered? Here's a post from just over a decade ago, in which you explicitly state that the algebraic integers indeed form a ring:

https://groups.google.com/forum/#!original/sci.math/3fG6ReIrmao/B2NsXeUeNygJ

Anyway, if the gaps in reasoning are so gaping then how about you point one out in the proof I posted?

[JSH]

On Monday, November 13, 2017 at 7:21:00 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:35:31 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 2:22:57 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:08:49 PM UTC, JSH wrote:

[...]

Admittedly skimmed as NOW go through and replace x with 3x and y with -2x, please.

Sure, I'll do it when I get home in a couple of hours. Will it be OK if I assume, for the purpose of illustration, that the monic polynomials of which 3x and -2x are roots are both quadratics (or I could assume one is a quadratic and another a cubic)? That will enable me to write the various matrices more explicitly, rather than defining them by indices.

You are simply checking if there is a mathematical reason that given 3x and -2x being algebraic integers, there is a mathematical reason that would say that x must be as well, when I'm like, I can prove it might not be.

How do you know that? There have been many previous occasions when you were convinced you had proven something, only to subsequently discover you were mistaken.

Yup. But testing YOUR point that if 3x - 2x = x, means that if 3x is an algebraic integer, and 2x is, and x is not, then there is a problem. You brought that up, not me. 

 

 

And proofs don't fight.

That is, when you have mathematical truth, it is consistent. Given that I KNOW that I can prove the reality, is not really that interesting to me your attempts but might help you.

Just to be clear, are you saying you have no intention of actually checking any proof I provide with the specific substitutions you asked for? Because if that's the case then I won't bother.

I'm saying you cannot prove. Proofs do not fight.

You cannot. That is my point. Proofs do not fight.

Proofs do not fight. Human beings can say things though.

The math does not care what you say. Proofs do NOT fight.

 

 

I reviewed arguments that ring of algebraic integers was ok, over a decade ago. Was surprised. Gaping gaps in reasoning, when you're looking for them.

Is it possible that you have misremembered? Here's a post from just over a decade ago, in which you explicitly state that the algebraic integers indeed form a ring:

https://groups.google.com/forum/#!original/sci.math/3fG6ReIrmao/B2NsXeUeNygJ

Anyway, if the gaps in reasoning are so gaping then how about you point one out in the proof I posted?

Wow was really talkative in that thing! But was mostly on point. AM impressed, with myself. Regardless at THAT time was NOT considering:

3x - 2x = x

That is YOUR example, given in THIS thread, in the year 2017.

YOU presented something with x+y, and I merely asked you to replace x with 3x and y with -2x, and report back.

___JSH 

[Rotwang]

On Tuesday, November 14, 2017 at 12:52:29 AM UTC, JSH wrote:

On Monday, November 13, 2017 at 7:21:00 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:35:31 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 2:22:57 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:08:49 PM UTC, JSH wrote:

[...]

Admittedly skimmed as NOW go through and replace x with 3x and y with -2x, please.

Sure, I'll do it when I get home in a couple of hours. Will it be OK if I assume, for the purpose of illustration, that the monic polynomials of which 3x and -2x are roots are both quadratics (or I could assume one is a quadratic and another a cubic)? That will enable me to write the various matrices more explicitly, rather than defining them by indices.

You are simply checking if there is a mathematical reason that given 3x and -2x being algebraic integers, there is a mathematical reason that would say that x must be as well, when I'm like, I can prove it might not be.

How do you know that? There have been many previous occasions when you were convinced you had proven something, only to subsequently discover you were mistaken.

Yup. But testing YOUR point that if 3x - 2x = x, means that if 3x is an algebraic integer, and 2x is, and x is not, then there is a problem. You brought that up, not me. 

Yes, I brought it up because it's an immediate consequence of an assertion you made earlier in this thread. What bearing does this have on my question?

 

 

And proofs don't fight.

That is, when you have mathematical truth, it is consistent. Given that I KNOW that I can prove the reality, is not really that interesting to me your attempts but might help you.

Just to be clear, are you saying you have no intention of actually checking any proof I provide with the specific substitutions you asked for? Because if that's the case then I won't bother.

I'm saying you cannot prove. Proofs do not fight.

You cannot. That is my point. Proofs do not fight.

Proofs do not fight. Human beings can say things though.

The math does not care what you say. Proofs do NOT fight.

Right. Which means that at least one of our alleged proofs is not, in fact, a proof. Again: what makes you so certain that my claim must be in error, rather than yours? You realise that you don't exactly have a perfect track record when it comes to spotting your errors, yes?

I reviewed arguments that ring of algebraic integers was ok, over a decade ago. Was surprised. Gaping gaps in reasoning, when you're looking for them.

Is it possible that you have misremembered? Here's a post from just over a decade ago, in which you explicitly state that the algebraic integers indeed form a ring:

https://groups.google.com/forum/#!original/sci.math/3fG6ReIrmao/B2NsXeUeNygJ

Anyway, if the gaps in reasoning are so gaping then how about you point one out in the proof I posted?

Wow was really talkative in that thing! But was mostly on point. AM impressed, with myself.

But it flatly contradicts what you've written in this thread! 2017 James says that the algebraic integers are not closed under addition, but 2007 James agrees with me that they are. You brought up the fact that you reviewed arguments over a decade ago, apparently as a reason to dismiss the proof I posted without reading it, but 2007 James - who had presumably reviewed those arguments in the recent past - agreed with its conclusion. Why do you suppose that is?

[JSH]

On Wednesday, November 15, 2017 at 8:24:49 PM UTC-5, Rotwang wrote:

On Tuesday, November 14, 2017 at 12:52:29 AM UTC, JSH wrote:

On Monday, November 13, 2017 at 7:21:00 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:35:31 PM UTC, JSH wrote:

On Monday, November 13, 2017 at 2:22:57 PM UTC-5, Rotwang wrote:

On Monday, November 13, 2017 at 7:08:49 PM UTC, JSH wrote:

[...]

Admittedly skimmed as NOW go through and replace x with 3x and y with -2x, please.

Sure, I'll do it when I get home in a couple of hours. Will it be OK if I assume, for the purpose of illustration, that the monic polynomials of which 3x and -2x are roots are both quadratics (or I could assume one is a quadratic and another a cubic)? That will enable me to write the various matrices more explicitly, rather than defining them by indices.

You are simply checking if there is a mathematical reason that given 3x and -2x being algebraic integers, there is a mathematical reason that would say that x must be as well, when I'm like, I can prove it might not be.

How do you know that? There have been many previous occasions when you were convinced you had proven something, only to subsequently discover you were mistaken.

Yup. But testing YOUR point that if 3x - 2x = x, means that if 3x is an algebraic integer, and 2x is, and x is not, then there is a problem. You brought that up, not me. 

Yes, I brought it up because it's an immediate consequence of an assertion you made earlier in this thread. What bearing does this have on my question?

 

I had not considered that, and DO agree: 3x - 2x = x is a GREAT example to test certain concepts. As if 3x is an algebraic integer and 2x is, while x is not, it shows lack of closure under addition. Which is YOUR point. And I think is a great one!!!

Thanks!!!

It puzzles me why this weird back-and-forth where I'm just agreeing.

 

 

And proofs don't fight.

That is, when you have mathematical truth, it is consistent. Given that I KNOW that I can prove the reality, is not really that interesting to me your attempts but might help you.

Just to be clear, are you saying you have no intention of actually checking any proof I provide with the specific substitutions you asked for? Because if that's the case then I won't bother.

I'm saying you cannot prove. Proofs do not fight.

You cannot. That is my point. Proofs do not fight.

Proofs do not fight. Human beings can say things though.

The math does not care what you say. Proofs do NOT fight.

Right. Which means that at least one of our alleged proofs is not, in fact, a proof. Again: what makes you so certain that my claim must be in error, rather than yours? You realise that you don't exactly have a perfect track record when it comes to spotting your errors, yes?

If I believe I have a proof, and YOU make a claim that would prove is not, then of course am curious. However, if I know IS a proof, am also confident that you probably have a mistake, especially if relying on something VASTLY more complicated, and if you degenerate into a verbal back-and-forth when given simple requests, and facing agreement on basic points. I have no reason to argue. For me is calm.

Proofs do not fight, is a basic statement which notes that you cannot have two different proofs, which contradict each other. Is just a statement of mathematical consistency.

And yeah, when you have two people with dueling claims, tend to go with the simpler argument--first. One must be in error, and is more likely with greater complexity, one person has either made accidental mistake, or hidden or tried to hide something more subtle.

Is common sense. 

I reviewed arguments that ring of algebraic integers was ok, over a decade ago. Was surprised. Gaping gaps in reasoning, when you're looking for them.

Is it possible that you have misremembered? Here's a post from just over a decade ago, in which you explicitly state that the algebraic integers indeed form a ring:

https://groups.google.com/forum/#!original/sci.math/3fG6ReIrmao/B2NsXeUeNygJ

Anyway, if the gaps in reasoning are so gaping then how about you point one out in the proof I posted?

Wow was really talkative in that thing! But was mostly on point. AM impressed, with myself.

But it flatly contradicts what you've written in this thread! 2017 James says that the algebraic integers are not closed under addition, but 2007 James agrees with me that they are. You brought up the fact that you reviewed arguments over a decade ago, apparently as a reason to dismiss the proof I posted without reading it, but 2007 James - who had presumably reviewed those arguments in the recent past - agreed with its conclusion. Why do you suppose that is?

You should have noted 3x - 2x = x before, as a definitive example demonstrating lack of closure if is proven that 3x is an algebraic integer, and 2x is an algebraic integer when x is not, which follows as possible from my research. To me is just cool.

Am glad you noted that in 2017.

Oh and said SKIMMED what you presented where I noticed x and y, and simply requested you replace x with 3x and y with -2x in your argument. Next thing I know there is endless debate.

Why might that make a difference? Because proofs do NOT fight. The math will adjust IF you presented an otherwise correct argument, and if you simply accept truth, will see it allows the possibility. I've watched that happen with my own research and is like, dealing with an intelligence.

The math is infinite intelligence.

___JSH 

[Rotwang]

On Thursday, November 16, 2017 at 11:51:08 AM UTC, JSH wrote:

[...]

Yes, I brought it up because it's an immediate consequence of an assertion you made earlier in this thread. What bearing does this have on my question?

I had not considered that, and DO agree: 3x - 2x = x is a GREAT example to test certain concepts. As if 3x is an algebraic integer and 2x is, while x is not, it shows lack of closure under addition. Which is YOUR point. And I think is a great one!!!

Thanks!!!

It puzzles me why this weird back-and-forth where I'm just agreeing.

 

 

And proofs don't fight.

That is, when you have mathematical truth, it is consistent. Given that I KNOW that I can prove the reality, is not really that interesting to me your attempts but might help you.

Just to be clear, are you saying you have no intention of actually checking any proof I provide with the specific substitutions you asked for? Because if that's the case then I won't bother.

I'm saying you cannot prove. Proofs do not fight.

You cannot. That is my point. Proofs do not fight.

Proofs do not fight. Human beings can say things though.

The math does not care what you say. Proofs do NOT fight.

Right. Which means that at least one of our alleged proofs is not, in fact, a proof. Again: what makes you so certain that my claim must be in error, rather than yours? You realise that you don't exactly have a perfect track record when it comes to spotting your errors, yes?

If I believe I have a proof, and YOU make a claim that would prove is not, then of course am curious. However, if I know IS a proof, am also confident that you probably have a mistake, especially if relying on something VASTLY more complicated

The proof I posted isn't very complicated at all - just basic linear algebra.

Though, as luck would have it, I've just thought of a simpler proof for the special case we're discussing. Namely, suppose that 3x and -2x are both algebraic integers. Then there exist monic polynomials P(y) and Q(y) such that

P(3x)  = 0
Q(-2x) = 0

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root). So

P(3x) = (3x)^n + a_{n - 1} (3x)^{n - 1} + ... + a_0
      = 3^n x^n + 3^{n - 1} a_{n - 1} x^{n - 1} + ... + a0        (1)
Q(-2x) = (-3x)^n + b_{n - 1} (-2x)^{n - 1} + ... + b_0
       = (-2)^n x^n + (-2)^{n - 1} b_{n - 1} x^{n - 1} + ... + b0 (2)

for some integer coefficients a_0, b_0, a_1, b_1 and so on; the formulas I've labelled (1) and (2) are non-monic polynomials with integer coefficients which both have x as a root. But 3^n and (-2)^n are coprime, so there exist integers i and j with

3^n i + (-2)^n j = 1

But then

0 = i P(3x) + j Q(-2x)
  = (3^n i + (-2)^n j) x^n + ... + i a0 + j b0
  = x^n + ... + i a0 + j b0

and so i P(3x) + j Q(-2x) is a monic polynomial with integer coefficients which has x as a root. So x is an algebraic integer.

[Haydon Berrow]

On Thursday, November 16, 2017 at 1:19:14 PM UTC, Rotwang wrote:

The proof I posted isn't very complicated at all - just basic linear algebra.

Though, as luck would have it, I've just thought of a simpler proof for the special case we're discussing. Namely, suppose that 3x and -2x are both algebraic integers. Then there exist monic polynomials P(y) and Q(y) such that

P(3x)  = 0
Q(-2x) = 0

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root).

Can you expand on that? It's not obvious why it's still monic, but maybe it should be

So

P(3x) = (3x)^n + a_{n - 1} (3x)^{n - 1} + ... + a_0
      = 3^n x^n + 3^{n - 1} a_{n - 1} x^{n - 1} + ... + a0        (1)
Q(-2x) = (-3x)^n + b_{n - 1} (-2x)^{n - 1} + ... + b_0

there's a typo there, -3x should be -2x.

       = (-2)^n x^n + (-2)^{n - 1} b_{n - 1} x^{n - 1} + ... + b0 (2)

for some integer coefficients a_0, b_0, a_1, b_1 and so on; the formulas I've labelled (1) and (2) are non-monic polynomials with integer coefficients which both have x as a root. But 3^n and (-2)^n are coprime, so there exist integers i and j with

3^n i + (-2)^n j = 1

But then

0 = i P(3x) + j Q(-2x)
  = (3^n i + (-2)^n j) x^n + ... + i a0 + j b0
  = x^n + ... + i a0 + j b0

and so i P(3x) + j Q(-2x) is a monic polynomial with integer coefficients which has x as a root. So x is an algebraic integer.

It did occur to me that P and Q can't both be quadratic, because then P-2Q would be monic, but that's the best I could do. I'g getting senile :-(

PS Did you that this medium is much better then sci.math, we can write in an html-editor and then copy and paste

P(3x) = (3x)ⁿ + a_n-1 (3x)^n-1 + ... + a₀
         = 3ⁿ xⁿ + 3^n-1 a_n-1 x^n-1 + ... + a₀ (1)
Q(-2x) = (-2x)ⁿ + b_n-1 (-2x)^n-1 + ... + b₀
          = (-2)ⁿ xⁿ + (-2)^n-1 b_n-1 x^n-1 + ... + b₀ (2)

[Haydon Berrow]

On Thursday, November 16, 2017 at 2:33:05 PM UTC, Haydon Berrow wrote:

On Thursday, November 16, 2017 at 1:19:14 PM UTC, Rotwang wrote:

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root).

Can you expand on that? It's not obvious why it's still monic, but maybe it should be

Oh sod! Of course it is. Perhaps an expansion would still be a good idea for people like me.

[Haydon Berrow]

No, 20 minutes later and it's not obvious again.

[Rotwang]

On Thursday, November 16, 2017 at 2:33:05 PM UTC, Haydon Berrow wrote:

On Thursday, November 16, 2017 at 1:19:14 PM UTC, Rotwang wrote:

The proof I posted isn't very complicated at all - just basic linear algebra.

Though, as luck would have it, I've just thought of a simpler proof for the special case we're discussing. Namely, suppose that 3x and -2x are both algebraic integers. Then there exist monic polynomials P(y) and Q(y) such that

P(3x)  = 0
Q(-2x) = 0

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root).

Can you expand on that? It's not obvious why it's still monic, but maybe it should be

I see you've already figured this out, but maybe an example would help for others: say P(y) is cubic and Q(y) is quadratic, then if

Q(y) = y² + b₁ y + b₀

we can define

Q'(y) = y³ + b₁ y² + b₀ y

and we now have a monic cubic which still has -2x as a root.

 

So

P(3x) = (3x)^n + a_{n - 1} (3x)^{n - 1} + ... + a_0
      = 3^n x^n + 3^{n - 1} a_{n - 1} x^{n - 1} + ... + a0        (1)
Q(-2x) = (-3x)^n + b_{n - 1} (-2x)^{n - 1} + ... + b_0

there's a typo there, -3x should be -2x.

Right, thanks.

 

       = (-2)^n x^n + (-2)^{n - 1} b_{n - 1} x^{n - 1} + ... + b0 (2)

for some integer coefficients a_0, b_0, a_1, b_1 and so on; the formulas I've labelled (1) and (2) are non-monic polynomials with integer coefficients which both have x as a root. But 3^n and (-2)^n are coprime, so there exist integers i and j with

3^n i + (-2)^n j = 1

But then

0 = i P(3x) + j Q(-2x)
  = (3^n i + (-2)^n j) x^n + ... + i a0 + j b0
  = x^n + ... + i a0 + j b0

and so i P(3x) + j Q(-2x) is a monic polynomial with integer coefficients which has x as a root. So x is an algebraic integer.

It did occur to me that P and Q can't both be quadratic, because then P-2Q would be monic, but that's the best I could do.

I'm not sure whether you're disagreeing with me or not? If P and Q are both quadratic (i.e. n = 2) then 3² - 2(-2)² = 1 so i = 1 and j = -2. Then

i P(3x) + j Q(-2x) = 9 x² + 3 a₁ x + a₀ - 2 (4 x² - 2 b₁ x + b₀)

                   = (9 - 8) x² + (3 a₁ + 4 b₁) x + a₀ - 2 b₀

                   = x² + (3 a₁ + 4 b₁) x + a₀ - 2 b₀

And this generalises to all other degrees - we can always find i and j such that 3ⁿ i + (-2)ⁿ j = 1, using the extended Euclidean algorithm.

PS Did you that this medium is much better then sci.math, we can write in an html-editor and then copy and paste

P(3x) = (3x)ⁿ + a_n-1 (3x)^n-1 + ... + a₀
         = 3ⁿ xⁿ + 3^n-1 a_n-1 x^n-1 + ... + a₀ (1)
Q(-2x) = (-2x)ⁿ + b_n-1 (-2x)^n-1 + ... + b₀
          = (-2)ⁿ xⁿ + (-2)^n-1 b_n-1 x^n-1 + ... + b₀ (2)

Good idea, thanks. Old habits die hard.

[Rotwang]

If P(y) is a monic of degree n and Q(y) is a monic of degree m, with n > m, then let's say

Q(y) = y^m + b_{m - 1} y^{m - 1} + ... + b₀

so that

Q'(y) := Q(y) y^{n - m} = yⁿ + b_{m - 1} y^{n - 1} + ... + b₀ y^{n - m}

which is monic of degree n, and

Q'(-2x) = (-2x)^{n - m} Q(-2x) = 0

since Q(-2x) = 0.

[JSH]

Like: 3x² - 7x + 5 = 0, and 2x² + 6x + 3 = 0, why would I assume an intersection of solutions?

Actually have played with such things in the past. Single variable polynomials set to zero are determinate. 

To simplify for you, what you are trying is like saying x = 3 and x = 2.

Trivial error.

___JSH

[Haydon Berrow]

On Thursday, November 16, 2017 at 3:03:45 PM UTC, Rotwang wrote:

It did occur to me that P and Q can't both be quadratic, because then P-2Q would be monic, but that's the best I could do.

I'm not sure whether you're disagreeing with me or not?

Neither, I'm just saying I could only do the very simple case. Your use of the euclidean algorithm and trivial transformation of the lower-degree polynomial is far superior. I experimented with the lcm of the degrees and got absolutely nowhere.

[Rotwang]

On Thursday, November 16, 2017 at 3:29:00 PM UTC, JSH wrote:

On Thursday, November 16, 2017 at 8:19:14 AM UTC-5, Rotwang wrote:

[...]

Though, as luck would have it, I've just thought of a simpler proof for the special case we're discussing. Namely, suppose that 3x and -2x are both algebraic integers. Then there exist monic polynomials P(y) and Q(y) such that

P(3x)  = 0
Q(-2x) = 0

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root). So

P(3x) = (3x)^n + a_{n - 1} (3x)^{n - 1} + ... + a_0
      = 3^n x^n + 3^{n - 1} a_{n - 1} x^{n - 1} + ... + a0        (1)
Q(-2x) = (-3x)^n + b_{n - 1} (-2x)^{n - 1} + ... + b_0
       = (-2)^n x^n + (-2)^{n - 1} b_{n - 1} x^{n - 1} + ... + b0 (2)

for some integer coefficients a_0, b_0, a_1, b_1 and so on; the formulas I've labelled (1) and (2) are non-monic polynomials with integer coefficients which both have x as a root. But 3^n and (-2)^n are coprime, so there exist integers i and j with

3^n i + (-2)^n j = 1

But then

0 = i P(3x) + j Q(-2x)
  = (3^n i + (-2)^n j) x^n + ... + i a0 + j b0
  = x^n + ... + i a0 + j b0

and so i P(3x) + j Q(-2x) is a monic polynomial with integer coefficients which has x as a root. So x is an algebraic integer.

Like: 3x² - 7x + 5 = 0, and 2x² + 6x + 3 = 0, why would I assume an intersection of solutions?

I'm afraid I don't follow. Do you mean, why would I assume that P(3x) and Q(-2x) have a root x in common? I didn't assume that, it follows from the hypotheses. Remember, we're supposing that there is a complex number x such that both 3x and -2x are algebraic integers. P and Q are monic polynomials that have 3x and -2x as roots, for that x. So P(3x) and Q(-2x) are 0, and therefore so is i P(3x) + j Q(-2x), which turns out to be a monic polynomial in x.

By the way my proof is constructive, so if you give me monic polynomials P and Q that have 3x and -2x respectively as roots (where x is one of the numbers you claim to have proven exists, i.e. a non-algebraic integer with the property that nx is an algebraic integer for all non-unit integers n) then I can reply with a monic polynomial that has x as a root. Care to try it?

[Rotwang]

Ah, thanks.

[JSH]

On Thursday, November 16, 2017 at 1:07:53 PM UTC-5, Rotwang wrote:

On Thursday, November 16, 2017 at 3:29:00 PM UTC, JSH wrote:

On Thursday, November 16, 2017 at 8:19:14 AM UTC-5, Rotwang wrote:

[...]

Though, as luck would have it, I've just thought of a simpler proof for the special case we're discussing. Namely, suppose that 3x and -2x are both algebraic integers. Then there exist monic polynomials P(y) and Q(y) such that

P(3x)  = 0
Q(-2x) = 0

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root). So

P(3x) = (3x)^n + a_{n - 1} (3x)^{n - 1} + ... + a_0
      = 3^n x^n + 3^{n - 1} a_{n - 1} x^{n - 1} + ... + a0        (1)
Q(-2x) = (-3x)^n + b_{n - 1} (-2x)^{n - 1} + ... + b_0
       = (-2)^n x^n + (-2)^{n - 1} b_{n - 1} x^{n - 1} + ... + b0 (2)

for some integer coefficients a_0, b_0, a_1, b_1 and so on; the formulas I've labelled (1) and (2) are non-monic polynomials with integer coefficients which both have x as a root. But 3^n and (-2)^n are coprime, so there exist integers i and j with

3^n i + (-2)^n j = 1

But then

0 = i P(3x) + j Q(-2x)
  = (3^n i + (-2)^n j) x^n + ... + i a0 + j b0
  = x^n + ... + i a0 + j b0

and so i P(3x) + j Q(-2x) is a monic polynomial with integer coefficients which has x as a root. So x is an algebraic integer.

Like: 3x² - 7x + 5 = 0, and 2x² + 6x + 3 = 0, why would I assume an intersection of solutions?

I'm afraid I don't follow. Do you mean, why would I assume that P(3x) and Q(-2x) have a root x in common? I didn't assume that, it follows from the hypotheses. Remember, we're supposing that there is a complex number x such that both 3x and -2x are algebraic integers. P and Q are monic polynomials that have 3x and -2x as roots, for that x. So P(3x) and Q(-2x) are 0, and therefore so is i P(3x) + j Q(-2x), which turns out to be a monic polynomial in x.

Let P(3x) = 3x + 4, and Q(-2x) = -2x + 5. P(3x) = 0 = 3x + 4, and Q(-2x) = 0 = -2x + 5

As an example. Once you set to zero, you've set solutions. Like above x equals one thing for one and something else for another. So is equivalent to saying x  = 3 and x = 2, which is a direct contradiction. Why do I have to spell this out for you? Just use linear equations like I just did, with your idea.

Oh, but considering this area has been fun. Also, I finally remembered, the Wrapper Theorem. Here's source:

http://somemath.blogspot.com/2007/07/wrapper-theorem-and-ring-of-algebraic.html

The ring of algebraic integers can wrap up these numbers with unit factors which are NOT units in ring of algebraic integers.

Is so wild. They behave very much in a way that keeps you from looking at them directly. Where I've had over a decade to ponder.

What do they remind me of? Quarks. They, like quarks, resist attempts to look at them directly.

I find it fascinating. So yeah, thanks for bringing the up, about 3x - 2x = x.

By the way my proof is constructive, so if you give me monic polynomials P and Q that have 3x and -2x respectively as roots (where x is one of the numbers you claim to have proven exists, i.e. a non-algebraic integer with the property that nx is an algebraic integer for all non-unit integers n) then I can reply with a monic polynomial that has x as a root. Care to try it?

You shouldn't use word proof, when you don't have one. What you have is a persistence in making claims.

Demonstrate with quadratic examples, end to end. That is, present quadratics with INTEGERS, where can step through your complete argument.

That allows solving for things with quadratic formula.

Demonstrate your claims with a detailed quadratic example. And quit dancing. What good is it, really?

Isn't mathematical truth good enough for you? Or would you rather believe nonsense?

I CAN step through my proof with numbers with quadratic examples. Because it is math.

If you have math, act like it. What's with all this other?

___JSH

[Rotwang]

On Thursday, November 16, 2017 at 7:41:01 PM UTC, JSH wrote:

On Thursday, November 16, 2017 at 1:07:53 PM UTC-5, Rotwang wrote:

On Thursday, November 16, 2017 at 3:29:00 PM UTC, JSH wrote:

On Thursday, November 16, 2017 at 8:19:14 AM UTC-5, Rotwang wrote:

[...]

Though, as luck would have it, I've just thought of a simpler proof for the special case we're discussing. Namely, suppose that 3x and -2x are both algebraic integers. Then there exist monic polynomials P(y) and Q(y) such that

P(3x)  = 0
Q(-2x) = 0

We may assume that P and Q both have the same degree n (since if either of them has a smaller degree we can just multiply it by some power of y to get a polynomial of higher degree which still has the required root). So

P(3x) = (3x)^n + a_{n - 1} (3x)^{n - 1} + ... + a_0
      = 3^n x^n + 3^{n - 1} a_{n - 1} x^{n - 1} + ... + a0        (1)
Q(-2x) = (-3x)^n + b_{n - 1} (-2x)^{n - 1} + ... + b_0
       = (-2)^n x^n + (-2)^{n - 1} b_{n - 1} x^{n - 1} + ... + b0 (2)

for some integer coefficients a_0, b_0, a_1, b_1 and so on; the formulas I've labelled (1) and (2) are non-monic polynomials with integer coefficients which both have x as a root. But 3^n and (-2)^n are coprime, so there exist integers i and j with

3^n i + (-2)^n j = 1

But then

0 = i P(3x) + j Q(-2x)
  = (3^n i + (-2)^n j) x^n + ... + i a0 + j b0
  = x^n + ... + i a0 + j b0

and so i P(3x) + j Q(-2x) is a monic polynomial with integer coefficients which has x as a root. So x is an algebraic integer.

Like: 3x² - 7x + 5 = 0, and 2x² + 6x + 3 = 0, why would I assume an intersection of solutions?

I'm afraid I don't follow. Do you mean, why would I assume that P(3x) and Q(-2x) have a root x in common? I didn't assume that, it follows from the hypotheses. Remember, we're supposing that there is a complex number x such that both 3x and -2x are algebraic integers. P and Q are monic polynomials that have 3x and -2x as roots, for that x. So P(3x) and Q(-2x) are 0, and therefore so is i P(3x) + j Q(-2x), which turns out to be a monic polynomial in x.

Let P(3x) = 3x + 4, and Q(-2x) = -2x + 5. P(3x) = 0 = 3x + 4, and Q(-2x) = 0 = -2x + 5

As an example.

But those aren't examples of the P and Q I defined in the proof. Obviously if you ignore my definitions and come up with arbitrary P and Q then the things I say about the polynomials I defined will not, in general, be true of the ones you come up with. Remember, we're assuming that there exists an x such that 3x and -2x are both algebraic integers. and P and Q are then chosen to be monic polynomials with integer coefficients that have 3x and -2x respectively as roots.

 

Once you set to zero, you've set solutions. Like above x equals one thing for one and something else for another. So is equivalent to saying x  = 3 and x = 2, which is a direct contradiction. Why do I have to spell this out for you? Just use linear equations like I just did, with your idea.

OK, here's an example with linear equations. Let x = 5. Then we can take

P(y) = y - 15

Q(y) = y + 10

so that P(3x) = P(15) = 0 and Q(-2x) = Q(-10) = 0. The proof then proceeds by constructing the polynomial

P(3x) + Q(-2x) = 3x - 15 - 2x + 10

               = x - 5

which is a monic polynomial whose root is 5. This is what the construction does generally: given monic integer polynomials with 3x and -2x as roots, it defines a new monic integer polynomial with x as a root.

 

[...]

Demonstrate with quadratic examples, end to end. That is, present quadratics with INTEGERS, where can step through your complete argument.

That allows solving for things with quadratic formula.

Demonstrate your claims with a detailed quadratic example.

OK. Let's take x = 1 + sqrt(5). Since 3x and -2x are algebraic integers, there are polynomials P and Q  such that P(3x) = Q(-2x) = 0. For example, we can use

P(y) = y² - 6 y - 36

Q(y) = y² + 4 y - 16

(you can easily verify that 3*(1 + sqrt(5)) is a root of P and -2*(1 + sqrt(5)) is a root of Q). The proof then proceeds by constructing

P(3x) - 2 Q(-2x) = 9 x² - 18 x - 36 - 2 (4 x² - 8 x - 16)

                 = 9 x² - 18 x - 36 - 8 x² + 16 x + 32

                 = x² - 2 x - 4

and, lo and behold, the result is a monic polynomial with 1 + sqrt(5) as a root. Thus, starting with the fact that 3(1 + sqrt(5)) and -2(1 + sqrt(5)) are algebraic integers, we've shown that 1 + sqrt(5) is also an algebraic integer. The general proof shows that this can always be done with any x that has 3x and -2x being algebraic integers.

[JSH]

Or succinctly, given monic polynomials with integer coefficients P(y) and Q(y), with the substitutions 3x and -2x, P(3x) - jQ(2x), exists such that result is a monic polynomial as well with integer coefficients for some non-zero integer j. Could write how to calculate j, but why bother?

___JSH 

[Rotwang]

Not just any monic polynomials with integer coefficients - P must have 3x as a root and Q must have -2x as a root, and P and Q must have the same degree (I have shown that a pair of polynomials satisfying these requirements always exists, assuming that 3x and -2x are algebraic integers). Obviously if you ignore the definitions of key terms then that will make it harder to follow the argument.

 

with the substitutions 3x and -2x, P(3x) - jQ(2x),

No, you need i P(3x) + j Q(-2x), with i not necessarily equal to 1. For example, when P and Q are cubics you can take i = 3 and j = 10.

 

exists such that result is a monic polynomial as well with integer coefficients for some non-zero integer j.

Right, but it's not just any monic polynomial - it's a monic polynomial which has the number x that you started with as a root. This implies that x is an algebraic integer.

[JSH]

Oh yeah, my mistake. Need j and k, such that jP(3x) - kQ(-2x) is a monic, where of course they would exist as nonzero integers.

And yes, the root would definitely be an algebraic integer. However, you can't set P(3x) = 0 and Q(-2x) = 0, simultaneously.

That is like x = 3 and x = 2, which is a direct contradiction. BUT one can suppose that jP(3x) - kQ(-2x) = 0, exists. Which gives you:

jP(3x) = kQ(-2x) without contradiction. You seem focused on setting two different polynomials, which sets x, and then solving for another solution.

Like consider: x = 3 and x = 5, then, 2(x) = 2(3), and subtract, and 2x = 6, and x = 1. Contradictions abound.

See? Once you SET values for x with one equation, then you can't set to different values any more with higher order polynomials any more than you can with linear.

However can seem clever. Like suppose:  2x² + 8x + 8 = 0, and  x² + 5x + 6 = 0, and you subtract, erroneously:

 x² + 3x + 2 = 0 = (x+1)(x+2)

And it looks like, yay! Got the right answer IN there at least somewhere, right?

___JSH 

[Rotwang]

On Friday, November 17, 2017 at 12:00:57 AM UTC, JSH wrote:

[...]

Oh yeah, my mistake. Need j and k, such that jP(3x) - kQ(-2x) is a monic, where of course they would exist as nonzero integers.

And yes, the root would definitely be an algebraic integer. However, you can't set P(3x) = 0 and Q(-2x) = 0, simultaneously.

Yes, I can. Because P and Q are defined to be polynomials such that P(3x) = 0 and Q(-2x) = 0. I know that such polynomials exist because we're assuming that 3x and -2x are algebraic integers. By definition this means that there exists a polynomial which has 3x as a root, and another that has -2x as a root. Those two polynomials are what I'm calling P and Q. I can set P(3x) = 0 and Q(-2x) = 0 simultaneously because P and Q are specifically chosen so as to make those equations true.

 

That is like x = 3 and x = 2, which is a direct contradiction. BUT one can suppose that jP(3x) - kQ(-2x) = 0, exists. Which gives you:

jP(3x) = kQ(-2x) without contradiction. You seem focused on setting two different polynomials, which sets x, and then solving for another solution.

No, you have misunderstood. I am not solving for anything. And the polynomials don't set x; x determines the polynomials. Remember why I posted this proof in the first place? Earlier in this thread you wrote this:

    for example I know that 3x can be an algebraic integer and 2x can be

    an algebraic integer, when x is NOT an algebraic integer. So yeah, add

    negative 2x to 3x and can NOT be an algebraic integer when they both

    are.

That x is what my proof is about. I suppose that there exists a number x with the property that 3x is an algebraic integer and -2x is an algebraic integer, and I show that these facts imply that x is an algebraic integer. I am not just picking arbitrary polynomials P(3x) and Q(-2x) and assuming they have a shared root x; rather, I am defining P and Q to be those polynomials which have 3x and -2x as roots. That such P and Q exist is implied by the hypothesis that 3x and -2x are algebraic integers.

 

Like consider: x = 3 and x = 5, then, 2(x) = 2(3), and subtract, and 2x = 6, and x = 1. Contradictions abound.

See? Once you SET values for x with one equation,

But I don't set values for x with any equation. The x that appears in my proof simply refers to a number that you claim exists.

[JSH]

Is an interesting angle you presented. Have pondered these numbers for over a decade. Years ago spent so much time relentlessly trying to tease them out more until I logically proved the effort fruitless. DO like your observation though, that with 3x an algebraic integer and -2x an algebraic integer, that 3x - 2x = x, where x is not, reveals lack of closure of the ring under addition. Had not thought of that one before.

But imagine we were talking evens, how would you go about proving with evens that odds exist? Leaving you with that to ponder.

The mental exercise has been worth it. Had me reviewing plenty of things.

___JSH 

[Rotwang]

On Friday, November 17, 2017 at 1:52:45 PM UTC, JSH wrote:

[...]

Is an interesting angle you presented. Have pondered these numbers for over a decade. Years ago spent so much time relentlessly trying to tease them out more until I logically proved the effort fruitless. DO like your observation though, that with 3x an algebraic integer and -2x an algebraic integer, that 3x - 2x = x, where x is not, reveals lack of closure of the ring under addition.

But it's not true. I know you disagree with my argument but as far as I can tell none of your objections apply to what I've actually written. Maybe it will be more productive if I go through the argument step-by-step, and you can tell me at which point you first disagree with one of my statements.

You claim that there exist complex numbers that are not algebraic integers, but that give algebraic integers when multiplied by any non-unit integer. Let's pick such a number, and call it x. In everything that follows, "x" will always refer to this number.

Since 3x is an algebraic integer, there exist monic polynomials with integer coefficients that have 3x as a root. We pick one such polynomial and call it P. And since -2x is an algebraic integer, there exist monic polynomials with integer coefficients that have -2x as a root. We pick one such polynomial and call it Q. That is

P(3x)  = 0    (1)

Q(-2x) = 0    (2)

At this point, I will reiterate that x is not an unknown quantity to be solved for - it's just the number that we started with, one of the numbers you claim exist. Equations (1) and (2) are both true of that x.

Any objection so far?

[JSH]

On Saturday, November 18, 2017 at 9:07:06 AM UTC-5, Rotwang wrote:

On Friday, November 17, 2017 at 1:52:45 PM UTC, JSH wrote:

[...]

Is an interesting angle you presented. Have pondered these numbers for over a decade. Years ago spent so much time relentlessly trying to tease them out more until I logically proved the effort fruitless. DO like your observation though, that with 3x an algebraic integer and -2x an algebraic integer, that 3x - 2x = x, where x is not, reveals lack of closure of the ring under addition.

But it's not true. I know you disagree with my argument but as far as I can tell none of your objections apply to what I've actually written. Maybe it will be more productive if I go through the argument step-by-step, and you can tell me at which point you first disagree with one of my statements.

You are relying on direct contradiction which HAVE explained, but if you wish, will do so again. Good is a short post! Thanks.

As an exercise though, why not prove that odds exist, given only evens?

Can you? Is kind of similar, though MUCH easier. Even though will admit have been thinking of various ways without certainty they are valid.

Also might be kind of problem can just look up! To get the answer. 

You claim that there exist complex numbers that are not algebraic integers, but that give algebraic integers when multiplied by any non-unit integer. Let's pick such a number, and call it x. In everything that follows, "x" will always refer to this number.

Since 3x is an algebraic integer, there exist monic polynomials with integer coefficients that have 3x as a root. We pick one such polynomial and call it P. And since -2x is an algebraic integer, there exist monic polynomials with integer coefficients that have -2x as a root. We pick one such polynomial and call it Q. That is

P(3x)  = 0    (1)

Q(-2x) = 0    (2)

Ok, now what's the variable? Is like told you before. Say I have x = 3. Is that still a variable? No. It is set. Literally.

Now then, let's say I have: x + y = 3. You have two variables. Am explaining basic algebra to you.

 

At this point, I will reiterate that x is not an unknown quantity to be solved for - it's just the number that we started with, one of the numbers you claim exist. Equations (1) and (2) are both true of that x.

x = 3, is literally true for that x. x = 2 is also...um, now we have a direct contradiction.

 

Any objection so far?

Why don't you try to prove existence of odds, given evens only. Is a similar exercise. My point in offering it to you though is, if you struggle, which I suspect you may, as am explaining BASIC algebra to you, may rethink your confidence with something far more complex?

___JSH 

[JSH]

On Saturday, November 18, 2017 at 9:07:06 AM UTC-5, Rotwang wrote:

Have replied noting your error, however this approach is NOT without merit, if you recognize that you're looking for an intersection.

It is illogical to presume BOTH solutions are true for, say quadratics, but you CAN have one solution!

So, for instance, with quadratics you could find a solution for x, which is valid for BOTH quadratics!

Which is basic algebra, still. Continue in that direction though. But you still have one more flawed assumption.

___JSH

[Rotwang]

On Saturday, November 18, 2017 at 3:26:59 PM UTC, JSH wrote:

On Saturday, November 18, 2017 at 9:07:06 AM UTC-5, Rotwang wrote:

On Friday, November 17, 2017 at 1:52:45 PM UTC, JSH wrote:

[...]

Is an interesting angle you presented. Have pondered these numbers for over a decade. Years ago spent so much time relentlessly trying to tease them out more until I logically proved the effort fruitless. DO like your observation though, that with 3x an algebraic integer and -2x an algebraic integer, that 3x - 2x = x, where x is not, reveals lack of closure of the ring under addition.

But it's not true. I know you disagree with my argument but as far as I can tell none of your objections apply to what I've actually written. Maybe it will be more productive if I go through the argument step-by-step, and you can tell me at which point you first disagree with one of my statements.

You are relying on direct contradiction which HAVE explained, but if you wish, will do so again. Good is a short post! Thanks.

As an exercise though, why not prove that odds exist, given only evens?

Can you? Is kind of similar, though MUCH easier. Even though will admit have been thinking of various ways without certainty they are valid.

Also might be kind of problem can just look up! To get the answer. 

You claim that there exist complex numbers that are not algebraic integers, but that give algebraic integers when multiplied by any non-unit integer. Let's pick such a number, and call it x. In everything that follows, "x" will always refer to this number.

Since 3x is an algebraic integer, there exist monic polynomials with integer coefficients that have 3x as a root. We pick one such polynomial and call it P. And since -2x is an algebraic integer, there exist monic polynomials with integer coefficients that have -2x as a root. We pick one such polynomial and call it Q. That is

P(3x)  = 0    (1)

Q(-2x) = 0    (2)

Ok, now what's the variable?

What's what variable? Do you mean the variable in P and Q? If so you can call it whatever you want, except x (remember, I'm using x to denote a specific number that you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer). For example we could use y as a dummy variably, so that if e.g. P is a quadratic then

P(y)  = y² + a₁ y + a₀, for some integers a₁ and a₀ so that

P(3x) = (3x)² + a₁ (3x) + a₀ = 0

 

Is like told you before. Say I have x = 3. Is that still a variable? No. It is set. Literally.

Yes, x is set at the start of my proof. I set it to one of the numbers you claim exists.

Now then, let's say I have: x + y = 3. You have two variables.

No, as I've said, x is set at the start of my proof. It is not a variable.

 

Am explaining basic algebra to you.

 

At this point, I will reiterate that x is not an unknown quantity to be solved for - it's just the number that we started with, one of the numbers you claim exist. Equations (1) and (2) are both true of that x.

x = 3, is literally true for that x.

No, it isn't. Remember, x denotes one of the numbers you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer. P(3x) is true for that x, but x = 3 is not true for that x, unless you wish to claim that 3 is not an algebraic integer. Do you wish to claim that?

 

x = 2 is also

No, x = 2 is not true either of the x that appears in my argument - unless you wish to claim that 2 is not an algebraic integer.

 

...um, now we have a direct contradiction.

Right, but that contradiction arose from the false assumptions that x = 3 and x = 2, neither of which appeared in my argument - you just made those up. Can you point to an incorrect statement in my actual argument? Obviously, if you introduce additional false statements and derive a contradiction from those false statements, that doesn't show any problem with my argument; it just shows a problem with the statements you added.

Remember, in my argument the letter x refers to a particular number, of the sort you claim exists. Can you point out a flaw in my argument, without assuming that x also refers to something else?

Any objection so far?

Why don't you try to prove existence of odds, given evens only.

I don't really understand this question. It seems a bit like asking me to prove the existence of triangles, given circles only. Typically, mathematicians define the integers first, and then define the odd and even numbers as certain subsets of the integers. From that way of doing things, you don't even define the evens until after you've proved that both kinds of number exist. What does your question mean? What exactly am I allowed to assume in order to prove the existence of odds? Am I allowed to assume the usual axioms of set theory, for example? Because those can be used to prove that the natural numbers exist, so the assumption that evens exist wouldn't be needed at all. If not, then what am I supposed to assume?

[Rotwang]

On Saturday, November 18, 2017 at 4:27:41 PM UTC, JSH wrote:

On Saturday, November 18, 2017 at 9:07:06 AM UTC-5, Rotwang wrote:

On Friday, November 17, 2017 at 1:52:45 PM UTC, JSH wrote:

[...]

Is an interesting angle you presented. Have pondered these numbers for over a decade. Years ago spent so much time relentlessly trying to tease them out more until I logically proved the effort fruitless. DO like your observation though, that with 3x an algebraic integer and -2x an algebraic integer, that 3x - 2x = x, where x is not, reveals lack of closure of the ring under addition.

But it's not true. I know you disagree with my argument but as far as I can tell none of your objections apply to what I've actually written. Maybe it will be more productive if I go through the argument step-by-step, and you can tell me at which point you first disagree with one of my statements.

You claim that there exist complex numbers that are not algebraic integers, but that give algebraic integers when multiplied by any non-unit integer. Let's pick such a number, and call it x. In everything that follows, "x" will always refer to this number.

Since 3x is an algebraic integer, there exist monic polynomials with integer coefficients that have 3x as a root. We pick one such polynomial and call it P. And since -2x is an algebraic integer, there exist monic polynomials with integer coefficients that have -2x as a root. We pick one such polynomial and call it Q. That is

P(3x)  = 0    (1)

Q(-2x) = 0    (2)

At this point, I will reiterate that x is not an unknown quantity to be solved for - it's just the number that we started with, one of the numbers you claim exist. Equations (1) and (2) are both true of that x.

Have replied noting your error, however this approach is NOT without merit, if you recognize that you're looking for an intersection.

It is illogical to presume BOTH solutions are true for, say quadratics,

But I don't presume that. Equations (1) and (2) follow from the definitions of x, P and Q.

[Rotwang]

On Saturday, November 18, 2017 at 4:29:58 PM UTC, Rotwang wrote:

[...]

No, it isn't. Remember, x denotes one of the numbers you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer. P(3x) is true for that x,

Typo, I mean "P(3x) = 0 is true for that x".

[JSH]

On Saturday, November 18, 2017 at 11:29:58 AM UTC-5, Rotwang wrote:

On Saturday, November 18, 2017 at 3:26:59 PM UTC, JSH wrote:

On Saturday, November 18, 2017 at 9:07:06 AM UTC-5, Rotwang wrote:

On Friday, November 17, 2017 at 1:52:45 PM UTC, JSH wrote:

[...]

Is an interesting angle you presented. Have pondered these numbers for over a decade. Years ago spent so much time relentlessly trying to tease them out more until I logically proved the effort fruitless. DO like your observation though, that with 3x an algebraic integer and -2x an algebraic integer, that 3x - 2x = x, where x is not, reveals lack of closure of the ring under addition.

But it's not true. I know you disagree with my argument but as far as I can tell none of your objections apply to what I've actually written. Maybe it will be more productive if I go through the argument step-by-step, and you can tell me at which point you first disagree with one of my statements.

You are relying on direct contradiction which HAVE explained, but if you wish, will do so again. Good is a short post! Thanks.

As an exercise though, why not prove that odds exist, given only evens?

Can you? Is kind of similar, though MUCH easier. Even though will admit have been thinking of various ways without certainty they are valid.

Also might be kind of problem can just look up! To get the answer. 

You claim that there exist complex numbers that are not algebraic integers, but that give algebraic integers when multiplied by any non-unit integer. Let's pick such a number, and call it x. In everything that follows, "x" will always refer to this number.

Since 3x is an algebraic integer, there exist monic polynomials with integer coefficients that have 3x as a root. We pick one such polynomial and call it P. And since -2x is an algebraic integer, there exist monic polynomials with integer coefficients that have -2x as a root. We pick one such polynomial and call it Q. That is

P(3x)  = 0    (1)

Q(-2x) = 0    (2)

Ok, now what's the variable?

 

What's what variable? Do you mean the variable in P and Q? If so you can call it whatever you want, except

Yeah.

 

x (remember, I'm using x to denote a specific number that you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer). For example we could use y as a dummy variably, so that if e.g. P is a quadratic then

P(y)  = y² + a₁ y + a₀, for some integers a₁ and a₀ so that

P(3x) = (3x)² + a₁ (3x) + a₀ = 0

 

You're going in circles.

 

Is like told you before. Say I have x = 3. Is that still a variable? No. It is set. Literally.

Yes, x is set at the start of my proof. I set it to one of the numbers you claim exists.

Now then, let's say I have: x + y = 3. You have two variables.

No, as I've said, x is set at the start of my proof. It is not a variable.

Yeah, so what's the variable? Above you give y, then set. Going in circles. 

 

Am explaining basic algebra to you.

 

At this point, I will reiterate that x is not an unknown quantity to be solved for - it's just the number that we started with, one of the numbers you claim exist. Equations (1) and (2) are both true of that x.

x = 3, is literally true for that x.

No, it isn't. Remember, x denotes one of the numbers you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer. P(3x) is true for that x, but x = 3 is not true for that x, unless you wish to claim that 3 is not an algebraic integer. Do you wish to claim that?

No. Am noting that you are setting x, with solutions for polynomials. Then behaving as if you have NOT set x, by claiming solutions for two polynomials set to zero.

Have replied again, with more help for you. Is an interesting direction, turns out. But you need an INTERSECTION, which is a value valid between two expressions. Which reduces degree also. 

 

x = 2 is also

No, x = 2 is not true either of the x that appears in my argument - unless you wish to claim that 2 is not an algebraic integer.

 

...um, now we have a direct contradiction.

Right, but that contradiction arose from the false assumptions that x = 3 and x = 2, neither of which appeared in my argument - you just made those up. Can you point to an incorrect statement in my actual argument? Obviously, if you introduce additional false statements and derive a contradiction from those false statements, that doesn't show any problem with my argument; it just shows a problem with the statements you added.

Remember, in my argument the letter x refers to a particular number, of the sort you claim exists. Can you point out a flaw in my argument, without assuming that x also refers to something else?

Any objection so far?

Why don't you try to prove existence of odds, given evens only.

I don't really understand this question. It seems a bit like asking me to prove the existence of triangles, given circles only. Typically, mathematicians define the integers first,

Yeah. But here algebraic integers are defined, first. Get it? You wish to use established algebraic integers, with an argument noting the ring has a coverage problem. Evens are NOT a ring, but we know that odds exist.

But what if you were arguing with someone certain only evens existed? How would you prove existence of odds?

___JSH 

[JSH]

Which sets values for it, with your construction. Like if is quadratic? Has TWO values. But then you have Q also with two values. Contradiction.

You need an intersection. Values valid for P and Q at zero, without contradiction, which reduces degree.

Is a good direction. Continue. But as noted in another reply, you have ONE more flawed assumption.

___JSH 

[Rotwang]

On Saturday, November 18, 2017 at 4:40:54 PM UTC, JSH wrote:

[...]

Not much point replying to all of this; I'm going to concentrate on a single point of misunderstanding, because I believe that it will be easier to agree on the rest once we have agreement on this point.

x = 3, is literally true for that x.

No, it isn't. Remember, x denotes one of the numbers you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer. P(3x) is true for that x, but x = 3 is not true for that x, unless you wish to claim that 3 is not an algebraic integer. Do you wish to claim that?

No. Am noting that you are setting x, with solutions for polynomials.

No, I am not doing that. By the time I introduce the polynomials P and Q, x has already been set. I set it right at the start of the argument, remember? Here, I'll quote the start of the argument:

    You claim that there exist complex numbers that are not algebraic

    integers, but that give algebraic integers when multiplied by

    any non-unit integer. Let's pick such a number, and call it x.

    In everything that follows, "x" will always refer to this number.

You keep replying as if the equations I've labelled (1) and (2) are supposed to be solved, and that one of the solutions define x. But that isn't what I'm doing. By the time I introduce P and Q, x has already been defined. I then define P and Q, and from the definitions of x, P and Q it follows that (1) and (2) are true.

Maybe an analogy will help. Suppose that I were to claim that there there exist integers k such that 3k is even and -2k is even but k is not even. In order to prove to me that no such numbers exist, you might argue as follows:

    Suppose that there are such numbers; pick one and call it k. Since 3k

    is even, there exists an integer p such that 3k = 2p. And since -2k is

    even there exists an integer q such that -2k = 2q. So we have two identities:

    

     3k = 2p    (1)

    -2k = 2q    (2)

Obviously, if I accept that equations (1) and (2) hold then you will go on to argue that k = 3k - 2k = 2(p + q) is also even. But before you do so, I object like so:

    But why would I assume an intersection of solutions? Clearly solving

    3k = 2p will give us one value for k, but solving -2k = 2q will give

    us a different value. Contradiction.

You recognise that this is a spurious objection, right? But this is exactly the same objection you keep making to my argument! I start off by assuming the existence of a number x with certain properties you claim exists, just like the hypothetical James in the analogy starts off by assuming the the existence of a number k with certain properties that the hypothetical me claims exists. I then show that those properties imply the existence of polynomials P and Q such that my equations (1) and (2) are satisfied, just like the hypothetical James shows the existence of integers p and q such that his equations (1) and (2) are satisfied. In both cases there is no assumption that the equations (1) and (2) have a solution in common; it just follows from the hypotheses, because the objects that appear in them were specifically chosen to make it do so (and the fact that such choices are possible follows from the properties that x and k are claimed to have).

[JSH]

On Saturday, November 18, 2017 at 12:43:50 PM UTC-5, Rotwang wrote:

On Saturday, November 18, 2017 at 4:40:54 PM UTC, JSH wrote:

[...]

Not much point replying to all of this; I'm going to concentrate on a single point of misunderstanding, because I believe that it will be easier to agree on the rest once we have agreement on this point.

x = 3, is literally true for that x.

No, it isn't. Remember, x denotes one of the numbers you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer. P(3x) is true for that x, but x = 3 is not true for that x, unless you wish to claim that 3 is not an algebraic integer. Do you wish to claim that?

No. Am noting that you are setting x, with solutions for polynomials.

No, I am not doing that. By the time I introduce the polynomials P and Q, x has already been set. I set it right at the start of the argument, remember? Here, I'll quote the start of the argument:

    You claim that there exist complex numbers that are not algebraic

    integers, but that give algebraic integers when multiplied by

    any non-unit integer. Let's pick such a number, and call it x.

    In everything that follows, "x" will always refer to this number.

You keep replying as if the equations I've labelled (1) and (2) are supposed to be solved, and that one of the solutions define x. But that isn't what I'm doing. By the time I introduce P and Q, x has already been defined. I then define P and Q, and from the definitions of x, P and Q it follows that (1) and (2) are true.

Yet you literally have P(-3x) = 0, and Q(2x) = 0 in what you posted.

 

Maybe an analogy will help. Suppose that I were to claim that there there exist integers k such that 3k is even and -2k is even but k is not even. In order to prove to me that no such numbers exist, you might argue as follows:

    Suppose that there are such numbers; pick one and call it k. Since 3k

    is even, there exists an integer p such that 3k = 2p. And since -2k is

    even there exists an integer q such that -2k = 2q. So we have two identities:

    

     3k = 2p    (1)

    -2k = 2q    (2)

Obviously, if I accept that equations (1) and (2) hold then you will go on to argue that k = 3k - 2k = 2(p + q) is also even. But before you do so, I object like so:

    But why would I assume an intersection of solutions? Clearly solving

    3k = 2p will give us one value for k, but solving -2k = 2q will give

    us a different value. Contradiction.

Yet I actually replied noting using P(3x), and Q(-2x) without contradiction where at first forgot needed two variables, where you replied and corrected. 

You recognise that this is a spurious objection, right? But this is exactly the same objection you keep making to my argument! I start off by assuming the existence of a number x with certain properties you claim exists, just like the hypothetical James in the analogy starts off by assuming the the existence of a number k with certain properties that the hypothetical me claims exists. I then show that those properties imply the existence of polynomials P and Q such that my equations (1) and (2) are satisfied, just like the hypothetical James shows the existence of integers p and q such that his equations (1) and (2) are satisfied. In both cases there is no assumption that the equations (1) and (2) have a solution in common; it just follows from the hypotheses, because the objects that appear in them were specifically chosen to make it do so (and the fact that such choices are possible follows from the properties that x and k are claimed to have).

Let me help you further. Given two quadratics, as should be ok to keep simple with your approach, where you have P(3x) and Q(-2x), can you accept that P(3x) = 0 can have TWO solutions that will work? And Q(-2x) can have TWO solutions?

If so, consider the following logical statement:

IF two quadratics are distinct, and each has distinct solution as in different from each other, then for BOTH to be correct, for given supposition, it must be true that there is an intersection between a SINGLE value.

That is, if one quadratic is true, and another quadratic is true, for SAME variable, then MUST BE TRUE that the value is singular.

Like P(x) = (x+1)(x+2) = 0, and Q(x) = (x+2)(x+3) = 0 can BOTH be true, only if x = -2. Correct?

If you accept, then YOUR approach, should reduce degree, and give a solution for x with a LINEAR expression.

Maybe if spell out in that detail, progress?

___JSH 

[Rotwang]

On Saturday, November 18, 2017 at 6:17:13 PM UTC, JSH wrote:

On Saturday, November 18, 2017 at 12:43:50 PM UTC-5, Rotwang wrote:

On Saturday, November 18, 2017 at 4:40:54 PM UTC, JSH wrote:

[...]

Not much point replying to all of this; I'm going to concentrate on a single point of misunderstanding, because I believe that it will be easier to agree on the rest once we have agreement on this point.

x = 3, is literally true for that x.

No, it isn't. Remember, x denotes one of the numbers you claim exists, namely a number that isn't an algebraic integer but that gives an algebraic integer when multiplied by any non-unit integer. P(3x) is true for that x, but x = 3 is not true for that x, unless you wish to claim that 3 is not an algebraic integer. Do you wish to claim that?

No. Am noting that you are setting x, with solutions for polynomials.

No, I am not doing that. By the time I introduce the polynomials P and Q, x has already been set. I set it right at the start of the argument, remember? Here, I'll quote the start of the argument:

    You claim that there exist complex numbers that are not algebraic

    integers, but that give algebraic integers when multiplied by

    any non-unit integer. Let's pick such a number, and call it x.

    In everything that follows, "x" will always refer to this number.

You keep replying as if the equations I've labelled (1) and (2) are supposed to be solved, and that one of the solutions define x. But that isn't what I'm doing. By the time I introduce P and Q, x has already been defined. I then define P and Q, and from the definitions of x, P and Q it follows that (1) and (2) are true.

Yet you literally have P(-3x) = 0, and Q(2x) = 0 in what you posted.

Assuming you meant P(3x) = 0 and Q(-2x) = 0, yes, I literally have those equations in what I posted. Because they are literally a consequence of the definitions of x, P and Q. What of it? 

 

Maybe an analogy will help. Suppose that I were to claim that there there exist integers k such that 3k is even and -2k is even but k is not even. In order to prove to me that no such numbers exist, you might argue as follows:

    Suppose that there are such numbers; pick one and call it k. Since 3k

    is even, there exists an integer p such that 3k = 2p. And since -2k is

    even there exists an integer q such that -2k = 2q. So we have two identities:

    

     3k = 2p    (1)

    -2k = 2q    (2)

Obviously, if I accept that equations (1) and (2) hold then you will go on to argue that k = 3k - 2k = 2(p + q) is also even. But before you do so, I object like so:

    But why would I assume an intersection of solutions? Clearly solving

    3k = 2p will give us one value for k, but solving -2k = 2q will give

    us a different value. Contradiction.

Yet I actually replied noting using P(3x), and Q(-2x) without contradiction where at first forgot needed two variables, where you replied and corrected.

What are you talking about? Corrected what? P and Q are polynomials, P(3x) and Q(-2x) are evaluations thereof. Since I haven't yet given explicit formulas for P and Q there was no need to specify a variable, but if I did I could use any letter that doesn't already refer to something else in my argument. For example I could have

P(y) = yⁿ + a_{n - 1} y^{n - 1} + ... + a₀, or

P(z) = zⁿ + a_{n - 1} z^{n - 1} + ... + a₀

Both of those mean the same thing, since y and z are just dummy variables. x, on the other hand, is not - it's a specific number, which is assumed at the beginning of the argument. In any case, if P is as above then

P(3x) = (3x)ⁿ + a_{n - 1} (3x)^{n - 1} + ... + a₀ = 0

Note that this equation isn't supposed to be a definition of x (x has already been defined at the beginning of the argument). It's an equation that follows from the definitions of x and P.

 

You recognise that this is a spurious objection, right? But this is exactly the same objection you keep making to my argument! I start off by assuming the existence of a number x with certain properties you claim exists, just like the hypothetical James in the analogy starts off by assuming the the existence of a number k with certain properties that the hypothetical me claims exists. I then show that those properties imply the existence of polynomials P and Q such that my equations (1) and (2) are satisfied, just like the hypothetical James shows the existence of integers p and q such that his equations (1) and (2) are satisfied. In both cases there is no assumption that the equations (1) and (2) have a solution in common; it just follows from the hypotheses, because the objects that appear in them were specifically chosen to make it do so (and the fact that such choices are possible follows from the properties that x and k are claimed to have).

Let me help you further. Given two quadratics, as should be ok to keep simple with your approach, where you have P(3x) and Q(-2x), can you accept that P(3x) = 0 can have TWO solutions that will work? And Q(-2x) can have TWO solutions?

If I pretend that x is a variable then yes, the equations P(3x) = 0 and Q(-2x) = 0 can each have two solutions in general. However, in my argument, x is a specific number. P(3x) = 0 and Q(-2x) = 0 are facts that are true of that number, as a consequence of the definitions. It's true that there may be some other number, say y, such that P(3y) = 0, and some other number, say z, such that Q(-2z) = 0. But that's neither here or there. At no point in my argument do I assume otherwise; it simply has no bearing on anything that follows. I only need to use the facts that P(3x) = 0 and Q(-2x) = 0, for the specific x that I started with.

Do you agree that, for the x that I defined at the start of my argument, and the P and Q that I subsequently defined, it is true that

P(3x)  = 0 and

Q(-2x) = 0

?

If it would make things easier, I could go through the argument with a specific, numerical value for x - though of course such an x won't have all of the properties I'm assuming in the proof (namely that x is not an algebraic integer but 3x and -2x are), because the point of my argument is that no such x exists. But what the heck, I'll do it now. I already posted this example earlier, but hopefully it will be a bit clearer this time how it relates to my argument. Let x = 1 + sqrt(5). So 3x = 3 + 3 sqrt(5) and -2x = -2 - 2 sqrt(5) are both algebraic integers. I will use these two facts to prove that x is also an algebraic integer (which of course is obvious, but the example will show how the proof works).

Because 3 + 3 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have 3 + 3 sqrt(5) as a root. One such polynomial is y² - 6 y - 36. Let's call this P(y). Also because -2 - 2 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have -2 - 2 sqrt(5) as a root. One such polynomial is y² + 4 y - 16. Let's call this Q(y). So we have two polynomials P and Q, which satisfy

P(3 (1 + sqrt(5))  = (3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5))) - 36   = 0    (3)

Q(-2 (1 + sqrt(5)) = (-2 (1 + sqrt(5)))² + 4 (-2 (1 + sqrt(5))) - 16 = 0    (4)

Are you with me so far? Do you agree that the equations (3) and (4) are true? That's where we've got to with my argument; (3) and (4) are just a special case of the equations I've been calling (1) and (2). I could have written (3) and (4) with x instead of 1 + sqrt(5) and they would mean the same thing because, in this example, we've defined x to be 1 + sqrt(5). In the general case I can't give a numerical expression for x, because I'm assuming, in addition to the facts that 3x and -2x are algebraic integers, that x is not an algebraic integer. I can't give a numerical expression for such an x because no such x exists, but the general case works by assuming that such an x is given and showing that equations that are similar to (3) and (4) must hold.

 

If so, consider the following logical statement:

IF two quadratics are distinct, and each has distinct solution as in different from each other, then for BOTH to be correct, for given supposition, it must be true that there is an intersection between a SINGLE value.

That is, if one quadratic is true, and another quadratic is true, for SAME variable, then MUST BE TRUE that the value is singular.

Like P(x) = (x+1)(x+2) = 0, and Q(x) = (x+2)(x+3) = 0 can BOTH be true, only if x = -2. Correct?

In that example it's true that x must be -2, yes. Note, however, that my equations (1) and (2) state that P(3x) = 0 and Q(-2x) = 0, not P(x) = 0 and Q(x) = 0

 

If you accept, then YOUR approach, should reduce degree, and give a solution for x with a LINEAR expression.

No, that doesn't follow. See the specific example above where x is 1 + sqrt(5), P(y) is y² - 6 y - 36 and Q(y) is y² + 4 y - 16. Then P(3x) = 0 and Q(-2x) = 0, but x is not the root of any linear equation with integer coefficients, if that's what you mean.

[JSH]

x is not a specific number until you give a value, which you DO below.

Am literally explaining algebra to you.

Just because you say, x is a number, is no different from any other algebra using x. Like x+y = 5. x and y are some numbers.

But x = 1, is set.

Algebraically these distinctions matter. As saying you have x as set in your mind, you go through an argument where it is a variable where you engage in direct contradiction, which have repeatedly explained to you.

Worse there is a correct path with this approach, which you refuse, which with quadratics reduces to a linear equation, easy to evaluate.

 

In any case, if P is as above then

P(3x) = (3x)ⁿ + a_{n - 1} (3x)^{n - 1} + ... + a₀ = 0

Note that this equation isn't supposed to be a definition of x (x has already been defined at the beginning of the argument). It's an equation that follows from the definitions of x and P.

You are trying to re-invent algebra into something that isn't logical.

 

If I pretend that x is a variable then yes, the equations P(3x) = 0 and Q(-2x) = 0 can each have two solutions in general.

Pretend x is a variable? That is the start of algebra. If x is a value, then you GIVE the value.

You cannot as a human being just DECLARE that expressions with variables are now just a value because you, a human, have decided that algebra no longer needs follow rules of algebra, but will now follow rules made up by you, on the fly.

Which are illogical and promptly lead to a direct contradiction, also by the way.

But yeah, you are re-inventing algebra to some strange, illogical version of your own. When there is a correct path available too.

 

However, in my argument, x is a specific number. P(3x) = 0 and Q(-2x) = 0 are facts that are true of that number, as a consequence of the definitions. It's true that there may be some other number, say y, such that P(3y) = 0, and some other number, say z, such that Q(-2z) = 0. But that's neither here or there. At no point in my argument do I assume otherwise; it simply has no bearing on anything that follows. I only

Algebra does not care what you think. You have expressions with variables? Is algebra. Follows algebraic rules.

 

need to use the facts that P(3x) = 0 and Q(-2x) = 0, for the specific x that I started with.

Do you agree that, for the x that I defined at the start of my argument, and the P and Q that I subsequently defined, it is true that

P(3x)  = 0 and

Q(-2x) = 0

?

Yeah! Is what have been saying! Is basic algebra. You have polynomials set to zero, which will have multiple solutions for your variable x.

Yet you insist that because YOU believe that is a set value, as you put it, that algebra must mysteriously now follow your arbitrary distinction, when it does not. Meaning you approach leads to a direct contradiction, which YOU claim simply does not exist because you SET x, which is a variable. 

If it would make things easier, I could go through the argument with a specific, numerical value for x - though of course such an x won't have all of the properties I'm assuming in the proof (namely that x is not an algebraic integer but 3x and -2x are), because the point of my argument is that no such x exists. But what the heck, I'll do it now. I already posted this example earlier, but hopefully it will be a bit clearer this time how it relates to my argument. Let x = 1 + sqrt(5). So 3x = 3 + 3 sqrt(5) and -2x = -2 - 2 sqrt(5) are both algebraic integers. I will use these two facts to prove that x is also an algebraic integer (which of course is obvious, but the example will show how the proof works).

Because 3 + 3 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have 3 + 3 sqrt(5) as a root. One such polynomial is y² - 6 y - 36. Let's call this P(y). Also because -2 - 2 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have -2 - 2 sqrt(5) as a root. One such polynomial is y² + 4 y - 16. Let's call this Q(y). So we have two polynomials P and Q, which satisfy

P(3 (1 + sqrt(5))  = (3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5))) - 36   = 0    (3)

Q(-2 (1 + sqrt(5)) = (-2 (1 + sqrt(5)))² + 4 (-2 (1 + sqrt(5))) - 16 = 0    (4)

Are you with me so far? Do you agree that the equations (3) and (4) are true?

Yup. Keep going. NO use of x!!! x is a variable. Continue with your numeric examples. Then at least will be logical, I'm guessing.

If your approach is correct, should trivially show that yup, you end up with algebraic integers. Continue with EXPLICIT.

No more use of x, ok?

___JSH 

[Rotwang]

On Sunday, November 19, 2017 at 12:49:47 PM UTC, JSH wrote:

[...]

x is not a specific number until you give a value, which you DO below.

Am literally explaining algebra to you.

Just because you say, x is a number, is no different from any other algebra using x. Like x+y = 5. x and y are some numbers.

But x = 1, is set.

Algebraically these distinctions matter. As saying you have x as set in your mind, you go through an argument where it is a variable where you engage in direct contradiction, which have repeatedly explained to you.

No, the "contradiction" you keep explaining is a consequence of things you're adding to my argument, not what's actually there.

Worse there is a correct path with this approach, which you refuse, which with quadratics reduces to a linear equation, easy to evaluate.

 

In any case, if P is as above then

P(3x) = (3x)ⁿ + a_{n - 1} (3x)^{n - 1} + ... + a₀ = 0

Note that this equation isn't supposed to be a definition of x (x has already been defined at the beginning of the argument). It's an equation that follows from the definitions of x and P.

You are trying to re-invent algebra into something that isn't logical.

 

If I pretend that x is a variable then yes, the equations P(3x) = 0 and Q(-2x) = 0 can each have two solutions in general.

Pretend x is a variable? That is the start of algebra. If x is a value, then you GIVE the value.

No, that isn't how algebra works. Look, you've seen a proof by contradiction that there is no integers whose ratio is the square root of 2, right? It starts by assuming to the contrary that there are integers m and n such that (m/n)² = 2, and works from there to derive a contradiction. Obviously one can't GIVE values for m and n, because no such integers exist. But one can assume to the contrary that they do exist, give them names, and then derive some consequences. My argument works the same way: I assume that a number exists with the properties you claim, give it a name, and then derive a contradiction.

 

You cannot as a human being just DECLARE that expressions with variables are now just a value because you, a human, have decided that algebra no longer needs follow rules of algebra,

No, what I'm doing is standard algebra.

  

However, in my argument, x is a specific number. P(3x) = 0 and Q(-2x) = 0 are facts that are true of that number, as a consequence of the definitions. It's true that there may be some other number, say y, such that P(3y) = 0, and some other number, say z, such that Q(-2z) = 0. But that's neither here or there. At no point in my argument do I assume otherwise; it simply has no bearing on anything that follows. I only

Algebra does not care what you think. You have expressions with variables? Is algebra. Follows algebraic rules.

 

need to use the facts that P(3x) = 0 and Q(-2x) = 0, for the specific x that I started with.

Do you agree that, for the x that I defined at the start of my argument, and the P and Q that I subsequently defined, it is true that

P(3x)  = 0 and

Q(-2x) = 0

?

Yeah! Is what have been saying! Is basic algebra. You have polynomials set to zero, which will have multiple solutions for your variable x.

It's apparent that you are using the word "variable" in a non-standard way. But never mind what you call x. You claim that there exist numbers with certain properties. Think of such a number and then, whenever you see x in my argument, mentally replace x with the number you're thinking of. That's what I mean by x. If you want to call that a "variable" just because I haven't specified a value myself (which I can't, because no such number exists) then feel free, but it just seems to be confusing you - in particular, you see an equation such as

P(3x) = 0

and seem to think that means that I must be planning on solving it for x. But I'm not, any more than I plan to "solve for 2" in the equation

2² - 3*2 + 2 = 0

In both cases they're just facts that follow from the definitions of the symbols involved.

Yet you insist that because YOU believe that is a set value, as you put it, that algebra must mysteriously now follow your arbitrary distinction, when it does not. Meaning you approach leads to a direct contradiction, which YOU claim simply does not exist because you SET x, which is a variable. 

If it would make things easier, I could go through the argument with a specific, numerical value for x - though of course such an x won't have all of the properties I'm assuming in the proof (namely that x is not an algebraic integer but 3x and -2x are), because the point of my argument is that no such x exists. But what the heck, I'll do it now. I already posted this example earlier, but hopefully it will be a bit clearer this time how it relates to my argument. Let x = 1 + sqrt(5). So 3x = 3 + 3 sqrt(5) and -2x = -2 - 2 sqrt(5) are both algebraic integers. I will use these two facts to prove that x is also an algebraic integer (which of course is obvious, but the example will show how the proof works).

Because 3 + 3 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have 3 + 3 sqrt(5) as a root. One such polynomial is y² - 6 y - 36. Let's call this P(y). Also because -2 - 2 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have -2 - 2 sqrt(5) as a root. One such polynomial is y² + 4 y - 16. Let's call this Q(y). So we have two polynomials P and Q, which satisfy

P(3 (1 + sqrt(5))  = (3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5))) - 36   = 0    (3)

Q(-2 (1 + sqrt(5)) = (-2 (1 + sqrt(5)))² + 4 (-2 (1 + sqrt(5))) - 16 = 0    (4)

Are you with me so far? Do you agree that the equations (3) and (4) are true?

Yup. Keep going. NO use of x!!! x is a variable. Continue with your numeric examples.

OK. So, since you agree that (3) and (4) are true - in particular, that P(3 (1 + sqrt(5)) = 0 and  Q(-2 (1 + sqrt(5)) = 0, you hopefully also agree that

P(3 (1 + sqrt(5)) - 2 Q(-2 (1 + sqrt(5)) = 0    (5)

 

Now, expanding the polynomials on the left hand side of (5) gives us

(3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5)) - 36 - 2 (-2 (1 + sqrt(5)))² - 8 (-2 (1 + sqrt(5))) + 32

= (1 + sqrt(5))² - 2 (1 + sqrt(5)) - 4 = 0

Which shows us that

R(1 + sqrt(5)) = 0

where

R(y) = y² - 2 y - 4

So, to recap, we started out with a number, namely 1 + sqrt(5), with the property that its products with 3 and -2 are both algebraic integers. Because of this, we were able to find monic polynomials with integer coefficients P and Q that have 3 (1 + sqrt(5)) and -2 (1 + sqrt(5)) as roots, respectively. Then, using those two polynomials, we constructed a new monic polynomial with integer coefficients R that has 1 + sqrt(5) - that is, the number we started with - as a root. Thus we have shown that 1 + sqrt(5) is also an algebraic integer.

There is nothing particularly special about 1 + sqrt(5) that made this possible; given any other complex number whose products with 3 and -2 are algebraic integers, if I have monic polynomials with integer coefficients that have those products as roots then I can construct another monic polynomial with integer coefficients that has the original number as a root. This shows that every complex number that yields an algebraic integer when multiplied by 3 and -2 is itself an algebraic integer.

[JSH]

On Saturday, November 25, 2017 at 5:21:46 PM UTC-5, Rotwang wrote:

On Sunday, November 19, 2017 at 12:49:47 PM UTC, JSH wrote:

[...]

x is not a specific number until you give a value, which you DO below.

Am literally explaining algebra to you.

Just because you say, x is a number, is no different from any other algebra using x. Like x+y = 5. x and y are some numbers.

But x = 1, is set.

Algebraically these distinctions matter. As saying you have x as set in your mind, you go through an argument where it is a variable where you engage in direct contradiction, which have repeatedly explained to you.

No, the "contradiction" you keep explaining is a consequence of things you're adding to my argument, not what's actually there.

Here it is simply for you. If P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+5) = 0. No contradiction because x+1 = 0 can be true for BOTH cases.

But if:  P(x) = (x+1)(x+2) = 0, and Q(x) = (x+3)(x+5) = 0. Contradiction. As no value of x can satisfy both.

THAT is the simple algebra am noting for you. So let's continue from there, with that established.

Worse there is a correct path with this approach, which you refuse, which with quadratics reduces to a linear equation, easy to evaluate.

 

In any case, if P is as above then

P(3x) = (3x)ⁿ + a_{n - 1} (3x)^{n - 1} + ... + a₀ = 0

Note that this equation isn't supposed to be a definition of x (x has already been defined at the beginning of the argument). It's an equation that follows from the definitions of x and P.

You are trying to re-invent algebra into something that isn't logical.

 

If I pretend that x is a variable then yes, the equations P(3x) = 0 and Q(-2x) = 0 can each have two solutions in general.

Pretend x is a variable? That is the start of algebra. If x is a value, then you GIVE the value.

No, that isn't how algebra works. Look, you've seen a proof by contradiction that there is no integers whose ratio is the square root of 2, right? It starts by assuming to the contrary that there are integers m and n such that (m/n)² = 2, and works from there to derive a contradiction. Obviously one can't GIVE values for m and n, because no such integers exist. But one can assume to the contrary that they do exist, give them names, and then derive some consequences. My argument works the same way: I assume that a number exists with the properties you claim, give it a name, and then derive a contradiction.

Yeah is how algebra works. All I did was note you kept making claims that would lead to a direct contradiction as explained IN DETAIL by me, more than once.

In algebra you have variables, where classic is, x.

Those variables can be set, like with x + 1 = 0, or (x+1)(x+2) = 0, which allows x to have TWO values, but still is set. 

 

You cannot as a human being just DECLARE that expressions with variables are now just a value because you, a human, have decided that algebra no longer needs follow rules of algebra,

No, what I'm doing is standard algebra.

Have explained my position.

 

  

However, in my argument, x is a specific number. P(3x) = 0 and Q(-2x) = 0 are facts that are true of that number, as a consequence of the definitions. It's true that there may be some other number, say y, such that P(3y) = 0, and some other number, say z, such that Q(-2z) = 0. But that's neither here or there. At no point in my argument do I assume otherwise; it simply has no bearing on anything that follows. I only

Algebra does not care what you think. You have expressions with variables? Is algebra. Follows algebraic rules.

 

need to use the facts that P(3x) = 0 and Q(-2x) = 0, for the specific x that I started with.

Do you agree that, for the x that I defined at the start of my argument, and the P and Q that I subsequently defined, it is true that

P(3x)  = 0 and

Q(-2x) = 0

?

Yeah! Is what have been saying! Is basic algebra. You have polynomials set to zero, which will have multiple solutions for your variable x.

It's apparent that you are using the word "variable" in a non-standard way. But never mind what you call x. You claim that there exist numbers with certain properties. Think of such a number and then, whenever you see x in my argument, mentally replace x with the number you're thinking of. That's what I mean by x. If you want to call that a "variable" just because I haven't specified a value myself (which I can't, because no such number exists) then feel free, but it just seems to be confusing you - in particular, you see an equation such as

P(3x) = 0

and seem to think that means that I must be planning on solving it for x. But I'm not, any more than I

No. Am TELLING you that in so doing you are solving for x. If polynomial is of degree 2, you are giving two possible values.

 

plan to "solve for 2" in the equation

2² - 3*2 + 2 = 0

In both cases they're just facts that follow from the definitions of the symbols involved.

Yet you insist that because YOU believe that is a set value, as you put it, that algebra must mysteriously now follow your arbitrary distinction, when it does not. Meaning you approach leads to a direct contradiction, which YOU claim simply does not exist because you SET x, which is a variable. 

If it would make things easier, I could go through the argument with a specific, numerical value for x - though of course such an x won't have all of the properties I'm assuming in the proof (namely that x is not an algebraic integer but 3x and -2x are), because the point of my argument is that no such x exists. But what the heck, I'll do it now. I already posted this example earlier, but hopefully it will be a bit clearer this time how it relates to my argument. Let x = 1 + sqrt(5). So 3x = 3 + 3 sqrt(5) and -2x = -2 - 2 sqrt(5) are both algebraic integers. I will use these two facts to prove that x is also an algebraic integer (which of course is obvious, but the example will show how the proof works).

Because 3 + 3 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have 3 + 3 sqrt(5) as a root. One such polynomial is y² - 6 y - 36. Let's call this P(y). Also because -2 - 2 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have -2 - 2 sqrt(5) as a root. One such polynomial is y² + 4 y - 16. Let's call this Q(y). So we have two polynomials P and Q, which satisfy

P(3 (1 + sqrt(5))  = (3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5))) - 36   = 0    (3)

Q(-2 (1 + sqrt(5)) = (-2 (1 + sqrt(5)))² + 4 (-2 (1 + sqrt(5))) - 16 = 0    (4)

Are you with me so far? Do you agree that the equations (3) and (4) are true?

Yup. Keep going. NO use of x!!! x is a variable. Continue with your numeric examples.

OK. So, since you agree that (3) and (4) are true - in particular, that P(3 (1 + sqrt(5)) = 0 and  Q(-2 (1 + sqrt(5)) = 0, you hopefully also agree that

P(3 (1 + sqrt(5)) - 2 Q(-2 (1 + sqrt(5)) = 0    (5)

 

Now, expanding the polynomials on the left hand side of (5) gives us

(3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5)) - 36 - 2 (-2 (1 + sqrt(5)))² - 8 (-2 (1 + sqrt(5))) + 32

= (1 + sqrt(5))² - 2 (1 + sqrt(5)) - 4 = 0

Which shows us that

R(1 + sqrt(5)) = 0

where

R(y) = y² - 2 y - 4

So, to recap, we started out with a number, namely 1 + sqrt(5), with the property that its products with 3 and -2 are both algebraic integers. Because of this, we were able to find monic polynomials with integer coefficients P and Q that have 3 (1 + sqrt(5)) and -2 (1 + sqrt(5)) as roots, respectively. Then, using those two polynomials, we constructed a new monic polynomial with integer coefficients R that has 1 + sqrt(5) - that is, the number we started with - as a root. Thus we have shown that 1 + sqrt(5) is also an algebraic integer.

There is nothing particularly special about 1 + sqrt(5) that made this possible; given any other complex number whose products with 3 and -2 are algebraic integers, if I have monic polynomials with integer coefficients that have those products as roots then I can construct another monic polynomial with integer coefficients that has the original number as a root. This shows that every complex number that yields an algebraic integer when multiplied by 3 and -2 is itself an algebraic integer.

You picked monic polynomials with an excess factor. So with first 3 divides across, and gives a monic, and with second 2 divides across. THEN you still solve as if quadratics, when equations now only allow one solution.

Trivial algebra.

___JSH 

[Rotwang]

On Sunday, November 26, 2017 at 4:38:41 PM UTC, JSH wrote:

On Saturday, November 25, 2017 at 5:21:46 PM UTC-5, Rotwang wrote:

On Sunday, November 19, 2017 at 12:49:47 PM UTC, JSH wrote:

[...]

x is not a specific number until you give a value, which you DO below.

Am literally explaining algebra to you.

Just because you say, x is a number, is no different from any other algebra using x. Like x+y = 5. x and y are some numbers.

But x = 1, is set.

Algebraically these distinctions matter. As saying you have x as set in your mind, you go through an argument where it is a variable where you engage in direct contradiction, which have repeatedly explained to you.

No, the "contradiction" you keep explaining is a consequence of things you're adding to my argument, not what's actually there.

Here it is simply for you. If P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+5) = 0. No contradiction because x+1 = 0 can be true for BOTH cases.

But if:  P(x) = (x+1)(x+2) = 0, and Q(x) = (x+3)(x+5) = 0. Contradiction. As no value of x can satisfy both.

But those aren't examples of the P and Q I use in my argument. Remember, this is how I introduced P and Q:

    Since 3x is an algebraic integer, there exist monic polynomials with

    integer coefficients that have 3x as a root. We pick one such

    polynomial and call it P. And since -2x is an algebraic integer, there

    exist monic polynomials with integer coefficients that have -2x as a

    root. We pick one such polynomial and call it Q.

As you can see, the P and Q I use in my argument aren't just any old monic polynomials with integer coefficients - they are polynomials that are chosen so as to have particular roots. Obviously, if you ignore the part of my argument where I say what P and Q are, and instead just make up arbitrary polynomials and pretend I'm talking about those, the things I write about the P and Q I use in my argument will not in general be true of the P and Q you made up. That doesn't mean there's anything wrong with my argument, though.

 

[...]

If it would make things easier, I could go through the argument with a specific, numerical value for x - though of course such an x won't have all of the properties I'm assuming in the proof (namely that x is not an algebraic integer but 3x and -2x are), because the point of my argument is that no such x exists. But what the heck, I'll do it now. I already posted this example earlier, but hopefully it will be a bit clearer this time how it relates to my argument. Let x = 1 + sqrt(5). So 3x = 3 + 3 sqrt(5) and -2x = -2 - 2 sqrt(5) are both algebraic integers. I will use these two facts to prove that x is also an algebraic integer (which of course is obvious, but the example will show how the proof works).

Because 3 + 3 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have 3 + 3 sqrt(5) as a root. One such polynomial is y² - 6 y - 36. Let's call this P(y). Also because -2 - 2 sqrt(5) is an algebraic integer, there exist monic polynomials with integer coefficients that have -2 - 2 sqrt(5) as a root. One such polynomial is y² + 4 y - 16. Let's call this Q(y). So we have two polynomials P and Q, which satisfy

P(3 (1 + sqrt(5))  = (3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5))) - 36   = 0    (3)

Q(-2 (1 + sqrt(5)) = (-2 (1 + sqrt(5)))² + 4 (-2 (1 + sqrt(5))) - 16 = 0    (4)

Are you with me so far? Do you agree that the equations (3) and (4) are true?

Yup. Keep going. NO use of x!!! x is a variable. Continue with your numeric examples.

OK. So, since you agree that (3) and (4) are true - in particular, that P(3 (1 + sqrt(5)) = 0 and  Q(-2 (1 + sqrt(5)) = 0, you hopefully also agree that

P(3 (1 + sqrt(5)) - 2 Q(-2 (1 + sqrt(5)) = 0    (5)

 

Now, expanding the polynomials on the left hand side of (5) gives us

(3 (1 + sqrt(5)))² - 6 (3 (1 + sqrt(5)) - 36 - 2 (-2 (1 + sqrt(5)))² - 8 (-2 (1 + sqrt(5))) + 32

= (1 + sqrt(5))² - 2 (1 + sqrt(5)) - 4 = 0

Which shows us that

R(1 + sqrt(5)) = 0

where

R(y) = y² - 2 y - 4

So, to recap, we started out with a number, namely 1 + sqrt(5), with the property that its products with 3 and -2 are both algebraic integers. Because of this, we were able to find monic polynomials with integer coefficients P and Q that have 3 (1 + sqrt(5)) and -2 (1 + sqrt(5)) as roots, respectively. Then, using those two polynomials, we constructed a new monic polynomial with integer coefficients R that has 1 + sqrt(5) - that is, the number we started with - as a root. Thus we have shown that 1 + sqrt(5) is also an algebraic integer.

There is nothing particularly special about 1 + sqrt(5) that made this possible; given any other complex number whose products with 3 and -2 are algebraic integers, if I have monic polynomials with integer coefficients that have those products as roots then I can construct another monic polynomial with integer coefficients that has the original number as a root. This shows that every complex number that yields an algebraic integer when multiplied by 3 and -2 is itself an algebraic integer.

You picked monic polynomials with an excess factor. So with first 3 divides across, and gives a monic, and with second 2 divides across.

No, once again the polynomials I picked are

P(y) := y² - 6 y - 36

Q(y) := y² + 4 y - 16

Both of them are monic, and therefore not divisible by any non-unit integer.

 

THEN you still solve as if quadratics, when equations now only allow one solution.

 Which equations? The equations I labelled (3), (4) and (5) don't involve any variables and so I don't know what it could mean for them to have "solutions" - they're just true.

[JSH]

On Sunday, November 26, 2017 at 5:34:05 PM UTC-5, Rotwang wrote:

On Sunday, November 26, 2017 at 4:38:41 PM UTC, JSH wrote:

On Saturday, November 25, 2017 at 5:21:46 PM UTC-5, Rotwang wrote:

On Sunday, November 19, 2017 at 12:49:47 PM UTC, JSH wrote:

[...]

x is not a specific number until you give a value, which you DO below.

Am literally explaining algebra to you.

Just because you say, x is a number, is no different from any other algebra using x. Like x+y = 5. x and y are some numbers.

But x = 1, is set.

Algebraically these distinctions matter. As saying you have x as set in your mind, you go through an argument where it is a variable where you engage in direct contradiction, which have repeatedly explained to you.

No, the "contradiction" you keep explaining is a consequence of things you're adding to my argument, not what's actually there.

Here it is simply for you. If P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+5) = 0. No contradiction because x+1 = 0 can be true for BOTH cases.

But if:  P(x) = (x+1)(x+2) = 0, and Q(x) = (x+3)(x+5) = 0. Contradiction. As no value of x can satisfy both.

But those aren't examples of the P and Q I use in my argument. Remember, this is how I introduced P and Q:

    Since 3x is an algebraic integer, there exist monic polynomials with

    integer coefficients that have 3x as a root. We pick one such

    polynomial and call it P. And since -2x is an algebraic integer, there

    exist monic polynomials with integer coefficients that have -2x as a

    root. We pick one such polynomial and call it Q.

As you can see, the P and Q I use in my argument aren't just any old monic polynomials with integer coefficients - they are polynomials that are chosen so as to have particular roots. Obviously, if you ignore the part of my argument where I say what P and Q are, and instead just make up arbitrary polynomials and pretend I'm talking about those, the things I write about the P and Q I use in my argument will not in general be true of the P and Q you made up. That doesn't mean there's anything wrong with my argument, though.

Not going to go beyond this statement, so will just admit did not read anything else. And barely read it.

You have P(3x) = 0, and Q(-2x) = 0, for your arbitrary polynomials. Where I've simply noted that determines solutions, where is only valid for an intersection. Gave a simple example of the concept: P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+3) = 0 is valid, as x = -1 works! For both.

While P(x) = (x+1)(x+2) = 0 and Q(x) = (x+3)(x+4) = 0, cannot be valid. As no x satisfies both.

THAT is my point entirely. If you do not at any point in your argument set polynomials in a way as I describe, then you can continue.

If you DO set polynomials by setting anything to zero then you can explain further how my point in your opinion does not apply.

If you DO accept does apply, as is basic algebra, then you can talk the intersection of solutions for your polynomials.

___JSH 

[Rotwang]

On Monday, November 27, 2017 at 4:54:25 PM UTC, JSH wrote:

[...]

But those aren't examples of the P and Q I use in my argument. Remember, this is how I introduced P and Q:

    Since 3x is an algebraic integer, there exist monic polynomials with

    integer coefficients that have 3x as a root. We pick one such

    polynomial and call it P. And since -2x is an algebraic integer, there

    exist monic polynomials with integer coefficients that have -2x as a

    root. We pick one such polynomial and call it Q.

As you can see, the P and Q I use in my argument aren't just any old monic polynomials with integer coefficients - they are polynomials that are chosen so as to have particular roots. Obviously, if you ignore the part of my argument where I say what P and Q are, and instead just make up arbitrary polynomials and pretend I'm talking about those, the things I write about the P and Q I use in my argument will not in general be true of the P and Q you made up. That doesn't mean there's anything wrong with my argument, though.

Not going to go beyond this statement, so will just admit did not read anything else. And barely read it.

You have P(3x) = 0, and Q(-2x) = 0, for your arbitrary polynomials.

No, my polynomials are not arbitrary. In fact, in the material you quoted above, I explicitly wrote (emphasis added)

    if you ignore the part of my argument where I say what P and Q are,

    and instead just make up arbitrary polynomials and pretend I'm talking

    about those, the things I write about the P and Q I use in my argument

    will not in general be true of the P and Q you made up.

As you can see from the quoted material, I explicitly stated that the polynomials I call P and Q in my argument are not arbitrary; if you want to follow my argument, you need to look at the part where I state what they are.

 

Where I've simply noted that determines solutions, where is only valid for an intersection. Gave a simple example of the concept: P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+3) = 0 is valid, as x = -1 works! For both.

While P(x) = (x+1)(x+2) = 0 and Q(x) = (x+3)(x+4) = 0, cannot be valid. As no x satisfies both.

THAT is my point entirely. If you do not at any point in your argument set polynomials in a way as I describe, then you can continue.

I don't, see above. 

[JSH]

On Monday, November 27, 2017 at 1:05:20 PM UTC-5, Rotwang wrote:

On Monday, November 27, 2017 at 4:54:25 PM UTC, JSH wrote:

[...]

But those aren't examples of the P and Q I use in my argument. Remember, this is how I introduced P and Q:

    Since 3x is an algebraic integer, there exist monic polynomials with

    integer coefficients that have 3x as a root. We pick one such

    polynomial and call it P. And since -2x is an algebraic integer, there

    exist monic polynomials with integer coefficients that have -2x as a

    root. We pick one such polynomial and call it Q.

Yeah. You CAN do that, and you either have multi-variables, which I noted WAY back, with something like P(x,y), or you can TRY to keep a single variable, which is what you do.

If you have multi-variable polynomials, then you need to show. Like again: P(x,y) 

As you can see, the P and Q I use in my argument aren't just any old monic polynomials with integer coefficients - they are polynomials that are chosen so as to have particular roots. Obviously, if you ignore the part of my argument where I say what P and Q are, and instead just make up arbitrary polynomials and pretend I'm talking about those, the things I write about the P and Q I use in my argument will not in general be true of the P and Q you made up. That doesn't mean there's anything wrong with my argument, though.

I do no such thing. And wow, you got me to actually read through what I already knew, from before.

If you want multi-variable polynomials, then so declare. But if you have P(x), and Q(x), then you have single variable.

Not complicated.

 

Not going to go beyond this statement, so will just admit did not read anything else. And barely read it.

You have P(3x) = 0, and Q(-2x) = 0, for your arbitrary polynomials.

No, my polynomials are not arbitrary. In fact, in the material you quoted above, I explicitly wrote (emphasis added)

    if you ignore the part of my argument where I say what P and Q are,

    and instead just make up arbitrary polynomials and pretend I'm talking

    about those, the things I write about the P and Q I use in my argument

    will not in general be true of the P and Q you made up.

As you can see from the quoted material, I explicitly stated that the polynomials I call P and Q in my argument are not arbitrary; if you want to follow my argument, you need to look at the part where I state what they are.

 

Where I've simply noted that determines solutions, where is only valid for an intersection. Gave a simple example of the concept: P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+3) = 0 is valid, as x = -1 works! For both.

While P(x) = (x+1)(x+2) = 0 and Q(x) = (x+3)(x+4) = 0, cannot be valid. As no x satisfies both.

THAT is my point entirely. If you do not at any point in your argument set polynomials in a way as I describe, then you can continue.

I don't, see above. 

Now you seem confused about variables in algebra with regard to polynomials. Yes, you can have a monic polynomial which has, say 3x as a root! That is trivial. But generally that would mean you have TWO variables. like P(x,y).

For example: P(x,y) = (y - 3x)(y + x) = 0

There 3x is a root of the polynomial. And you have: y = 3x as a solution.

___JSH 

[Rotwang]

On Tuesday, November 28, 2017 at 4:05:57 PM UTC, JSH wrote:

On Monday, November 27, 2017 at 1:05:20 PM UTC-5, Rotwang wrote:

On Monday, November 27, 2017 at 4:54:25 PM UTC, JSH wrote:

[...]

But those aren't examples of the P and Q I use in my argument. Remember, this is how I introduced P and Q:

    Since 3x is an algebraic integer, there exist monic polynomials with

    integer coefficients that have 3x as a root. We pick one such

    polynomial and call it P. And since -2x is an algebraic integer, there

    exist monic polynomials with integer coefficients that have -2x as a

    root. We pick one such polynomial and call it Q.

Yeah. You CAN do that, and you either have multi-variables, which I noted WAY back, with something like P(x,y), or you can TRY to keep a single variable, which is what you do.

If you have multi-variable polynomials, then you need to show. Like again: P(x,y) 

I don't. The polynomials P and Q I use in my argument are single-variable. For example, in the numerical example I posted earlier, in which x was 1 + sqrt(5), P(y) was y² - 6 y - 36. Note that P(y) only has a single variable, y.

 

As you can see, the P and Q I use in my argument aren't just any old monic polynomials with integer coefficients - they are polynomials that are chosen so as to have particular roots. Obviously, if you ignore the part of my argument where I say what P and Q are, and instead just make up arbitrary polynomials and pretend I'm talking about those, the things I write about the P and Q I use in my argument will not in general be true of the P and Q you made up. That doesn't mean there's anything wrong with my argument, though.

I do no such thing.

Well, you say that, and yet...

 

And wow, you got me to actually read through what I already knew, from before.

If you want multi-variable polynomials,

(I don't)

 

then so declare. But if you have P(x), and Q(x), then you have single variable.

...I don't have P(x) and Q(x). Seriously, if you don't believe me then search through every post I've made in this thread for either of the strings "P(x)" or "Q(x)"; apart from where they appear in quoted material that was originally written by you, the only time either one appears in one of my posts is when I explicitly pointed out that I wasn't using them:

    Note, however, that my equations (1) and (2) state that P(3x) = 0 and

    Q(-2x) = 0, not P(x) = 0 and Q(x) = 0

So as you can see, my argument does not have P(x) and Q(x). However, it does have single-variable polynomials P and Q - though I don't ever use x to denote the variable in those polynomials, because I am already using x for something else.

 

Not complicated.

 

Not going to go beyond this statement, so will just admit did not read anything else. And barely read it.

You have P(3x) = 0, and Q(-2x) = 0, for your arbitrary polynomials.

No, my polynomials are not arbitrary. In fact, in the material you quoted above, I explicitly wrote (emphasis added)

    if you ignore the part of my argument where I say what P and Q are,

    and instead just make up arbitrary polynomials and pretend I'm talking

    about those, the things I write about the P and Q I use in my argument

    will not in general be true of the P and Q you made up.

As you can see from the quoted material, I explicitly stated that the polynomials I call P and Q in my argument are not arbitrary; if you want to follow my argument, you need to look at the part where I state what they are.

 

Where I've simply noted that determines solutions, where is only valid for an intersection. Gave a simple example of the concept: P(x) = (x+1)(x+2) = 0, and Q(x) = (x+1)(x+3) = 0 is valid, as x = -1 works! For both.

While P(x) = (x+1)(x+2) = 0 and Q(x) = (x+3)(x+4) = 0, cannot be valid. As no x satisfies both.

THAT is my point entirely. If you do not at any point in your argument set polynomials in a way as I describe, then you can continue.

I don't, see above. 

Now you seem confused about variables in algebra with regard to polynomials. Yes, you can have a monic polynomial which has, say 3x as a root! That is trivial. But generally that would mean you have TWO variables. like P(x,y).

No. x is not a variable. For example, in the numerical example I posted earlier, in which x was 1 + sqrt(5) and P(y) was y² - 6 y - 36, P was a single-variable polynomial. 1 + sqrt(5) is not a variable. If I had chosen a different value for x then I would also have chosen a different polynomial for P; in that sense, P indeed depends on x. But obviously the coefficients of P cannot be chosen to be polynomial functions of x.

Remember what my argument is about. It is an argument that purports to show that, whenever a number has the property that its products with 3 and -2 are both algebraic integers, then that number is also an algebraic integer. Obviously, there are infinitely many such numbers. So I can't write all of them down and write a proof for each one. What I do instead is, I use a letter - I chose the letter x - to denote a non-specific number, and I proceed with steps that are valid independent of which number x denotes. This is a very standard technique for proving things about multiple numbers (or other mathematical objects) at once. But apparently there's no way for me to use this standard technique without you inserting a bunch of false assumptions into my argument. So let's try a different approach. You claim that there exists a number x with the property that 3x and -2x are both algebraic integers, but x is not. Please give an example of such an x. That way I can go through my argument using the number you give me. I will show that the number does not have all the properties you claim of it.

[JSH]

I've answered you repeatedly and been very patient. This discussion is over.

___JSH