# Thread: A Revised Postulate for Peer Review

1. ## A Revised Postulate for Peer Review

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 )

2. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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?

a = (0, 1)
b = (1, 1)
etc.

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

-Dan

3. ## Re: A Revised Postulate for Peer Review

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

4. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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.

5. ## Re: A Revised Postulate for Peer Review

Originally Posted by Plato
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.

6. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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.

7. ## Re: A Revised Postulate for Peer Review

What a monumental waste of time.
Why are any of us suckered into such by trolls?

8. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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

9. ## Re: A Revised Postulate for Peer Review

Originally Posted by SlipEternal
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

10. ## Re: A Revised Postulate for Peer Review

Originally Posted by topsquark
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

11. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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?

12. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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

13. ## Re: A Revised Postulate for Peer Review

Originally Posted by SlipEternal
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

14. ## Re: A Revised Postulate for Peer Review

Originally Posted by topsquark
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.......

15. ## Re: A Revised Postulate for Peer Review

Originally Posted by Conway
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.

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