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 October 12th, 2017, 02:40 PM #21 Senior Member   Joined: Sep 2016 From: USA Posts: 167 Thanks: 79 Math Focus: Dynamical systems, analytic function theory, numerics 1. I haven't raised 0 to any negative power. This is simply notation for zero's inverse. You are assuming that it is possible to satisfy the equation $x\cdot 0 = u$ where $u$ is a unit. Note that EVERY element in a field other than zero is a unit. I have shown this is impossible in a field, hence why we don't allow 0 to have an inverse. 2. Please read this: https://en.wikipedia.org/wiki/Well-defined 3. I still don't understand what $z_1,z_2$ are. You can't define the product of 2 elements to depend on a 3rd. This makes multiplication no longer a binary operation. Thanks from Conway51
October 12th, 2017, 03:58 PM   #22
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 Originally Posted by SDK 1. I haven't raised 0 to any negative power. This is simply notation for zero's inverse. You are assuming that it is possible to satisfy the equation $x\cdot 0 = u$ where $u$ is a unit. Note that EVERY element in a field other than zero is a unit. I have shown this is impossible in a field, hence why we don't allow 0 to have an inverse. 2. Please read this: https://en.wikipedia.org/wiki/Well-defined 3. I still don't understand what $z_1,z_2$ are. You can't define the product of 2 elements to depend on a 3rd. This makes multiplication no longer a binary operation.
If you where using (x = 0^-1) as the inverse of zero. And then you say there is not an inverse of zero...you leave me confused.

But in any case I am not claiming (x = 0^-1) is the inverse of 0.

I have stated to you very clearly on a previous post that 1 is the inverse of 0 and 0 in the inverse of 1.

I can define the product of two elements to depend on a third. This is called relativity. It specifically makes "binary" multiplication relative....

If you have trouble understanding z1 and z2 consider the following then..

Let z1 be the multiplicative property of 0
Let z2 be the multiplicative identity property of 1

Let zero posses both properties
Let only one property be used in any binary expression
If both numbers are 0 then z1 must be used as default.

A x 0(z1) = 0
0 x A(z1) = 0
A x 0(z2) = 1
0 x A(z2) = 1

no tables necessary....any better for you?

we then extrapolate for division
and then for multiplicative inverses....

 October 12th, 2017, 05:38 PM #23 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,595 Thanks: 620 Math Focus: Wibbly wobbly timey-wimey stuff. Conway, I think you are misunderstanding what SDK is driving at. He is mainly trying to tell you that your construction is not a field. Your take on this system is perfectly allowed so long as you realize that. -Dan
October 12th, 2017, 05:47 PM   #24
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 Originally Posted by topsquark Conway, I think you are misunderstanding what SDK is driving at. He is mainly trying to tell you that your construction is not a field. Your take on this system is perfectly allowed so long as you realize that. -Dan
Dan

Perhaps. Perhaps not. What is the point of your reply? Would you care to regulate this idea to some "new" field like meadows....fine....as you said...

"Your take on this system is perfectly allowed"....

I have no other way to interpret that other than to say you therefore see some form of "validity" in this idea....

October 12th, 2017, 07:26 PM   #25
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Quote:
 Originally Posted by Conway51 Dan Perhaps. Perhaps not. What is the point of your reply? Would you care to regulate this idea to some "new" field like meadows....fine....as you said... "Your take on this system is perfectly allowed".... I have no other way to interpret that other than to say you therefore see some form of "validity" in this idea....
Applicability is nice, too. However yes, there is some validity here I suppose. I don't know much about the mathematics of a system that includes an inverse for 0. The point of my reply is that I believe that SDK is trying to tell you that your system is not a field, and that you are trying to tell him that you believe that it is a field. (I still think your notation is ugly as sin, though. You need to work on that.)

-Dan

October 12th, 2017, 07:51 PM   #26
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 Originally Posted by topsquark Applicability is nice, too. However yes, there is some validity here I suppose. I don't know much about the mathematics of a system that includes an inverse for 0. The point of my reply is that I believe that SDK is trying to tell you that your system is not a field, and that you are trying to tell him that you believe that it is a field. (I still think your notation is ugly as sin, though. You need to work on that.) -Dan
Dan

Well now....my my....this is a far cry from how you have treated me on other forums. As well as your original post on this thread....

"I notice that, though you certainly may ask questions on any forum, you haven't changed anything since that last long conversation on MHB. What new thing do you expect here?"

-Dan

That said....THANK YOU....seriously.

To other matters...

I have never really intended to debate what sort of "construction" this is. I have only intended to validate the mathematical conclusions of the given axioms. This you have agreed has been done.

Now...I do consider it a field...(maybe falsely so)

1. "The multiplicative property of 0" is a valid property in a field
2. "The multiplicative identity property of 1" is a valid property in a field
3. If a unique solution can be shown for the "use" of both of these properties regarding zero....then this "construct" is valid in a field.

This doesn't really carry us all the way though does it. We must do the math....as SDK showed earlier with his equations. However his equations relied on the assumption that (A * 0 = A ) in all cases.

My reply here being that I can at "will" change the properties of zero to fit the desired product in the equation. IF and ONLY IF the solutions are unique.

So can you pose another equation showing contradiction....

The key here is to remember the expression (A * 0 ) is relative in product.

Thank you

P.S.

Should you wish to discuss "applicability" I would love to. But perhaps that is better done in another thread. Also I remind you of a "list" of goals I posted on another forum. These goals being your "applicability".

Last edited by Conway51; October 12th, 2017 at 07:58 PM.

October 12th, 2017, 08:37 PM   #27
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 Originally Posted by Conway51 Well now....my my....this is a far cry from how you have treated me on other forums. As well as your original post on this thread....
Actually, no. Your system is ambiguous as to how you are defining your symbols. You can't say that 1 = (1, 1) without saying 1 = (1, 1) = ( (1, 1), ( 1, 1) ). You can't simply say that you aren't because that doesn't fix the problem. Either the "1" on the left is an integer (as you seem to be saying) or the two 1's on the RHS are integers. You can't use the same symbols for both. This the same thing I've been saying to you all along. Since you haven't changed even that little bit I stand by what I've said in this thread and others.

Quote:
 Originally Posted by Conway51 I have never really intended to debate what sort of "construction" this is. I have only intended to validate the mathematical conclusions of the given axioms. This you have agreed has been done.
Quote:
 Originally Posted by Conway51 Now...I do consider it a field...(maybe falsely so)
Yes, falsely so.

The existence of a multiplicative inverse in the field means that, for any given element $\displaystyle a \neq 0$ of the field we have an element $\displaystyle a^{-1}$ belonging to the field.

You say that you have a multiplicative inverse for 0. According to the above axiom that can't happen in a field. What you have is a field plus one more condition...I have no idea what to call it. But no matter what the nomenclature that means you don't have a field. End of story.

-Dan

Last edited by topsquark; October 12th, 2017 at 08:41 PM.

October 12th, 2017, 08:54 PM   #28
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 Originally Posted by topsquark Actually, no. Your system is ambiguous as to how you are defining your symbols. You can't say that 1 = (1, 1) without saying 1 = (1, 1) = ( (1, 1), ( 1, 1) ). You can't simply say that you aren't because that doesn't fix the problem. Either the "1" on the left is an integer (as you seem to be saying) or the two 1's on the RHS are integers. You can't use the same symbols for both. This the same thing I've been saying to you all along. Since you haven't changed even that little bit I stand by what I've said in this thread and others. Yes, falsely so. The existence of a multiplicative inverse in the field means that, for any given element $\displaystyle a \neq 0$ of the field we have an element $\displaystyle a^{-1}$ belonging to the field. You say that you have a multiplicative inverse for 0. According to the above axiom that can't happen in a field. What you have is a field plus one more condition...I have no idea what to call it. But no matter what the nomenclature that means you don't have a field. End of story. -Dan
Dan on your first point you are wrong. If you observe the original post please...

"Let no "ordered pair" be represented by another further "ordered pair"

See Dan I do try to learn. I DID change something. You clearly missed it because you didn't read it. You were only interested in trolling me.

The existence of a multiplicative inverse in the field means that, for any given element
a≠0
of the field we have an element
a^-1
belonging to the field.

I agree...there however is not problem in the mathematics if you use zero as z1 instead of z2. I have shown this repeatedly. So we keep the stament and only add

the element 0 has an inverse of 1

Set this aside for a moment. Then consider with me Dan...if the idea is valid "except" as a field...then reapply the ideas of space and value as z2 and z1.

Then "field" or otherwise I could care less...

What it does mean is that division by zero is then defined.
Multiplication by zero is relative
Multiplicative inverses for 0 and 1
Varying amounts of 0

As well as a new way in which to consider mathematics.....

Field or otherwise......

Last edited by Conway51; October 12th, 2017 at 08:58 PM.

October 13th, 2017, 12:32 AM   #29
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 Originally Posted by Conway51 Dan on your first point you are wrong. If you observe the original post please... "Let no "ordered pair" be represented by another further "ordered pair"
You can't do that and still use the symbols you are using! As I mentioned on the other site you need to define things in some way where f(1) = (1 , 1) or perhaps 1 = (f(1), f(1)) or whatever. In both cases you can simply define f(x) to be an inductive set. Then there is no ambiguity to your work. Sloppy notation makes for sloppy explanations.

And it's good that you aren't calling it a field. I am good with that. SDK and others can worry about how your axioms work. As I've said before I don't know much about constructions with a multiplicative inverse for 0.

-Dan

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