A Revised Postulate for Peer Review

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Jun 2015
59
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America
Axiom

Let every number be arbitrarily composed of two numbers.


Let the number table exist as such…


0=(0,1)
1=(1,1)
2=(2,2)
3=(3,3)
4=(4,4)…and so on

Let the first number of the number chosen be labeled as z1


Let the second number of the number chosen be labeled as z2


Let multiplication exist as follows…


(A x B) = ( z1forA x z2forB ) = ( z2forA x z1forB ) = ( z1forB x z2forA ) = ( z2forB x z1forA )


Let division exist as follows…


(A/B) = ( z1forA/z2forB )

(B/A) = ( z1forB/z2forA )
 

topsquark

Forum Staff
Jan 2006
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3,458
Wellsville, NY
Axiom

Let every number be arbitrarily composed of two numbers.


Let the number table exist as such…


0=(0,1)
1=(1,1)
2=(2,2)
3=(3,3)
4=(4,4)…and so on

Let the first number of the number chosen be labeled as z1


Let the second number of the number chosen be labeled as z2


Let multiplication exist as follows…


(A x B) = ( z1forA x z2forB ) = ( z2forA x z1forB ) = ( z1forB x z2forA ) = ( z2forB x z1forA )


Let division exist as follows…


(A/B) = ( z1forA/z2forB )

(B/A) = ( z1forB/z2forA )
I don't see anything that tells me what you are doing! What's the point you are trying to make/derive?

So just to be clear:
(1 x 3) = (1 x 3) = (3)? Or some such?

(1/3) = (1/3) = 1/3?

Suggestion: Rewrite your table as
a = (0, 1)
b = (1, 1)
etc.

Then (a x c) = (1 x 3) = 3 as a real number?

-Dan
 
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Jun 2015
59
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America
Dan

I can not rewrite it as you suggest. If you do not use 0 in your equations... then you will find this to be of no use. Ill start off with the math here...


0 = (0,1)
1 = (1,1)
2 = (2,2)

Let the expression be...

( 0 * X )

(z1for0) * x = 0
(z2for0) * x = x
 

Plato

MHF Helper
Aug 2006
22,508
8,664
Axiom
Let every number be arbitrarily composed of two numbers.

Let the number table exist as such…
0=(0,1)
1=(1,1)
2=(2,2)
3=(3,3)
4=(4,4)…and so on

Let the first number of the number chosen be labeled as z1

Let the second number of the number chosen be labeled as z2


Let multiplication exist as follows…

(A x B) = ( z1forA x z2forB ) = ( z2forA x z1forB ) = ( z1forB x z2forA ) = ( z2forB x z1forA )


Let division exist as follows…

(A/B) = ( z1forA/z2forB )

(B/A) = ( z1forB/z2forA )
Having done pre-publication reviews, my comment is that what you posted is nonsense.
1) You have not said what the heck this is all about. Axioms for what?
2) What are the symbols used in the axioms. You used numbers 1,2,3 etc, without telling the reviewer what they are.
3) Then you use appears to be ordered pairs: what are they?
4) Because all of foundation papers in mathematics are set-theory based, are we to assume that the answers to the above are commonly understood. If so, what is the point?

Now the notation z1forA is nonsense if you do not give a coherent definition.
 
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Jun 2015
59
1
America
Having done pre-publication reviews, my comment is that what you posted is nonsense.
1) You have not said what the heck this is all about. Axioms for what?
2) What are the symbols used in the axioms. You used numbers 1,2,3 etc, without telling the reviewer what they are.
3) Then you use appears to be ordered pairs: what are they?
4) Because all of foundation papers in mathematics are set-theory based, are we to assume that the answers to the above are commonly understood. If so, what is the point?

Now the notation z1forA is nonsense if you do not give a coherent definition.


Use zero in binary muliplication and divsion.

again...


0=(0,1)
1=(1,1)
2=(2,2)
3=(3,3)
4=(4,4)…and so on

Let the first number of the number chosen be labeled as z1
Let the second number of the number chosen be labeled as z2


Let the expression be...

( x * 0 )

x * (z1for0) = 0
x * (z2for0) = x

we can then extrapolate for division by zero...in such a way as to NOT contradict any current field axiom...

Field Axioms -- from Wolfram MathWorld

I defined z1 and z2 respective to a NUMBER corresponding in the given table.
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
Use zero in binary muliplication and divsion.

again...


0=(0,1)
1=(1,1)
2=(2,2)
3=(3,3)
4=(4,4)…and so on

Let the first number of the number chosen be labeled as z1
Let the second number of the number chosen be labeled as z2


Let the expression be...

( x * 0 )

x * (z1for0) = 0
x * (z2for0) = x

we can then extrapolate for division by zero...in such a way as to NOT contradict any current field axiom...

Field Axioms -- from Wolfram MathWorld

I defined z1 and z2 respective to a NUMBER corresponding in the given table.
The pairs and operations you have chosen will not generate a field. You just showed that every element is zero. X*0 = 0 and x*0=x implies x=0.
 
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Plato

MHF Helper
Aug 2006
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What a monumental waste of time.
Why are any of us suckered into such by trolls?
 
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topsquark

Forum Staff
Jan 2006
11,602
3,458
Wellsville, NY
Dan

I can not rewrite it as you suggest. If you do not use 0 in your equations... then you will find this to be of no use. Ill start off with the math here...


0 = (0,1)
1 = (1,1)
2 = (2,2)

Let the expression be...

( 0 * X )

(z1for0) * x = 0
(z2for0) * x = x
You are using the symbol "1" (and 0, 2, 3, etc) to mean two different things. You are going to have to relabel something to make this work right.

-Dan
 
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Jun 2015
59
1
America
The pairs and operations you have chosen will not generate a field. You just showed that every element is zero. X*0 = 0 and x*0=x implies x=0.
Incorrect...I did not say

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

I said SPECEFICALLY

(z1for0) * x = 0
(z2for0) * x = x
 
Last edited:
Jun 2015
59
1
America
You are using the symbol "1" (and 0, 2, 3, etc) to mean two different things. You are going to have to relabel something to make this work right.

-Dan
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
 
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