A Revised Postulate for Peer Review

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SlipEternal

MHF Helper
Nov 2010
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Incorrect...I did not say

x * 0 = 0
x * 0 = x
therefore
x = 0

I said SPECEFICALLY

(z1for0) * x = 0
(z2for0) * x = x
Incorrect. You SPECIFICALLY defined multiplication as $(z_{1_A,} z_{2_A}) \times (z_{1_B}, z_{2_B}) = z_{1_A} z_{2_B} = z_{2_A} z_{1_B} = z_{1_B} z_{2_A} = z_{2_B} z_{1_A}$

So if $x=(x,x)$ you have $x\times 0 = x 1 = x 0 = 0 x = 1x $. This was YOUR rule.

If you cannot understand your own rules, how do you expect anyone else to?
 
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topsquark

Forum Staff
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I am not redefineing the number 1 in ANY way whatsoever...

"Let all numbers be composed of two numbers"

0 = (0(z1),1(z2))
1 = (1(z1),1(z2))

(z1for0) * x = 0
(z2for0) * x = x
Last try.

So when you have 4 = (4, 4) you are calling the number 4 by what? The symbol 4 can't be the integer 4 on both sides! Your notation implies 4 = (4, 4) = ( (4, 4), (4, 4) ) = ( ( (4, 4), (4, 4) ), ( (4, 4), (4, 4) ) ) = ... which is just garbage without meaning.

-Dan
 
Jun 2015
59
1
America
Incorrect. You SPECIFICALLY defined multiplication as $(z_{1_A,} z_{2_A}) \times (z_{1_B}, z_{2_B}) = z_{1_A} z_{2_B} = z_{2_A} z_{1_B} = z_{1_B} z_{2_A} = z_{2_B} z_{1_A}$

So if $x=(x,x)$ you have $x\times 0 = x 1 = x 0 = 0 x = 1x $. This was YOUR rule.

If you cannot understand your own rules, how do you expect anyone else to?




This is all WELL said.....it should have been obvious however...my apologies it was not. I was trying to keep the original post short so ...

(z1forA * z2forB) = (z1forB * z2forA) = (z1forB * z2forA) = (z2forA * z2forB)

this equation is true as long as A and B =/=

if A = 0
(z1forA) * (z2forB) = 0
(z2forB) * (z1forA) = 0

(z2forA) * (z1forB) = B
(z1forB) * (z2forA) = B


if B = 0
(z1forA) * (z2forB) = A
(z2forB) * (z1forA) = A


(z1forB) * (z2forA) = 0
(z1forB) * (z1forB) = 0

If and A and B = 0
(z1forA) * (z2forB) = 0
(z2forB) * (z1forA) = 0
(z1forB) * (z2forA) = 0
(z2forA) * (z1forB) = 0



Perhaps "leaving" this out was not right. I was afraid of a lengthy post
 
Jun 2015
59
1
America
Last try.

So when you have 4 = (4, 4) you are calling the number 4 by what? The symbol 4 can't be the integer 4 on both sides! Your notation implies 4 = (4, 4) = ( (4, 4), (4, 4) ) = ( ( (4, 4), (4, 4) ), ( (4, 4), (4, 4) ) ) = ... which is just garbage without meaning.

-Dan
I disagree...I have implied

4 = (4,4)

NOT
4 = ((4,4)(4,4))

"Let every number be composed of TWO numbers"..... NOT any other amount.......
 

SlipEternal

MHF Helper
Nov 2010
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I disagree...I have implied

4 = (4,4)

NOT
4 = ((4,4)(4,4))

"Let every number be composed of TWO numbers"..... NOT any other amount.......
You do not seem to understand that equality is transitive. If a=b and b=c then a=c.

You have 4 = (4,4). Now, on the RHS, we can replace 4 with (4,4), so (4,4) becomes ((4,4),4) or we can write (4,(4,4)), or we can write ((4,4),(4,4)).

I repeat, if you are not able to understand the basic rules of mathematics, then we are not going to be able to have a meaningful dialog. You seem to be picking and choosing rules to follow if you like them and ignoring them if they do not match what you "thought" it was supposed to be.
 

SlipEternal

MHF Helper
Nov 2010
3,728
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This is all WELL said.....it should have been obvious however...my apologies it was not. I was trying to keep the original post short so ...

(z1forA * z2forB) = (z1forB * z2forA) = (z1forB * z2forA) = (z2forA * z2forB)

this equation is true as long as A and B =/=

if A = 0
(z1forA) * (z2forB) = 0
(z2forB) * (z1forA) = 0

(z2forA) * (z1forB) = B
(z1forB) * (z2forA) = B


if B = 0
(z1forA) * (z2forB) = A
(z2forB) * (z1forA) = A


(z1forB) * (z2forA) = 0
(z1forB) * (z1forB) = 0

If and A and B = 0
(z1forA) * (z2forB) = 0
(z2forB) * (z1forA) = 0
(z1forB) * (z2forA) = 0
(z2forA) * (z1forB) = 0



Perhaps "leaving" this out was not right. I was afraid of a lengthy post
As stated in my previous post, equality is transitive. If you want to redefine equality so that it is no longer a transitive operator, that is fine, but you should explain explicitly what you mean by that. At this point, you are redefining symbols on the fly to suit your purposes, and you are not letting anyone else know what you redefine them to. I highly recommend that you go back and figure out exactly what you mean before asking mathematicians to peer review your work. At the moment, we do not share a common language with which to have any meaningful dialog. Good luck, though!
 
Jun 2015
59
1
America
You do not seem to understand that equality is transitive. If a=b and b=c then a=c.

You have 4 = (4,4). Now, on the RHS, we can replace 4 with (4,4), so (4,4) becomes ((4,4),4) or we can write (4,(4,4)), or we can write ((4,4),(4,4)).

I repeat, if you are not able to understand the basic rules of mathematics, then we are not going to be able to have a meaningful dialog. You seem to be picking and choosing rules to follow if you like them and ignoring them if they do not match what you "thought" it was supposed to be.

Incorrect...It seems to be you who does not understand the nature of axioms.

If I state....let a number be composed of two numbers...and provide a table...then the table must be adhered to as AN AXIOM....no properties, that is.. "additive""distributive""equalities" or any property WHATSOEVER exists until AFTER the axiom is set. The math FOLLOWS the axiom.

I repeat in a more specific manner....if you do not understand the basic rules of mathematics "axioms" then you are not going to be able to have a meaningful dialog. Be nice.....Or don't reply....or continue to exist as the fine troll that you are.
 
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SlipEternal

MHF Helper
Nov 2010
3,728
1,571
Incorrect...It seems to be you who does not understand the nature of axioms.

If I state....let a number be composed of two numbers...and provide a table...then the table must be adhered to as AN AXIOM....no properties, that is.. "additive""distributive""equalities" or any property WHATSOEVER exists until AFTER the axiom is set. The math FOLLOWS the axiom.

I repeat in a more specific manner....if you do not understand the basic rules of mathematics "axioms" then you are not going to be able to have a meaningful dialog. Be nice.....Or don't reply....or continue to exist as the fine troll that you are.
I was being extremely nice. I was not trolling you. I am the ONLY one still trying to help you. You insulted the forum moderator when he questioned you. You insulted me when I tried to give you advice. Your mathematics is not clear to anyone who has commented (all of the forum experts). The only thing you have made abundantly clear is that you do not want to answer questions that people have for you. Instead you take them as "trolling". I struggle to understand what it is you hoped to accomplish by posting this.

I am done trying to help. I do, however, leave you with one thought. Since you are attempting to define division by zero, there is an easier way. One that is far more understandable than what you have written:

https://en.wikipedia.org/wiki/Wheel_theory

I would pay specific attention to the part labeled "Wheel of Fractions".
 
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Jun 2015
59
1
America
I was being extremely nice. I was not trolling you. I am the ONLY one still trying to help you. You insulted the forum moderator when he questioned you. You insulted me when I tried to give you advice. Your mathematics is not clear to anyone who has commented (all of the forum experts). The only thing you have made abundantly clear is that you do not want to answer questions that people have for you. Instead you take them as "trolling". I struggle to understand what it is you hoped to accomplish by posting this.

I am done trying to help. I do, however, leave you with one thought. Since you are attempting to define division by zero, there is an easier way. One that is far more understandable than what you have written:

https://en.wikipedia.org/wiki/Wheel_theory


I insulted no one. I only told the mod that I could not do as he asked.......partly for the same reason you and I disagree. I am aware of wheel theory and meadows.....in case your not aware of both...

https://arxiv.org/abs/1406.6878

I leave you with one thought. An axiom requires no definition. By definition it can NOT be defined. If I set a number table as an axiom.....then no properties apply to the number table....properties only extend from a mathematical "use" of the number table.

Thank you for your time and good luck.
 
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topsquark

Forum Staff
Jan 2006
11,602
3,458
Wellsville, NY
You are defining the symbol 4 only once. If 4 = (4, 4) then 4 = ( (4, 4) , (4, 4) ) by your own logic, which is flawed.

We've all been here before.

Thread closed.

-Dan
 
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