# A Request for Peer Review

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

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 )

Edited by conway

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On ‎10‎/‎8‎/‎2017 at 4:39 PM, conway said:

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[/quote]

So this is an extension of the integers? That is (a, b) includes those numbers but also "numbers" of the form, say (3, 2)?

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

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 )

so, for example ((3, 2) x (5, 4))= (3 x 4)= (2 x 5)= (4 x 2)= (5 x 3).  What do those "="s mean?  Also those are no longer pairs of integers.  Don't you want the product of two pairs to be a pair?  Perhaps you meant (A x B) = ((z1forA  x  z2forB ) ,(  z2forA  x  z1forB ) )  so that ((3, 2) x (5, 4))= (3 x 4, 2 x 5)= (12, 10). But then  (( z1forB  x  z2forA ) , (  z2forB  x  z1forA )) would be (10, 12).  Are you saying that (x, y) and (y, x) are the same?

Let division exist as follows…

(A/B) = ( z1forA/z2forB )

(B/A) = ( z1forB/z2forA )

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2 hours ago, HallsofIvy said:

The table given is an extension of all numbers.....integers....rational...real....irrational...imaginary...whole.....natural

You can ONLY have one number in a "ordered pair" in each equation, from each number.....

if our expression is ( 2 x 3 )

It is only z1 for A....which is 2

It is only z2 for B....which is 3

or it is

Only z2 for A...which is 2

and

Only z1 for B....which is 3

So then you can see.....at NO time is a number ever represented in ANY expression as it's "entire" ordered pair.

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

The purpose of this is to show that the "equal" sign is not "typical"  as you point out.  Another way to consider this is to say that z1 and z2 are value and space...not numbers.

The value "occupies" the space.....a bit like vectors....

Additionally the table is not necessary to have in order for this idea to function......consider...

Let z1 be the multiplicative property of 0

Let z2 be the multiplicative identity property of 1

Let 0 have both properties

Let only one property be used at a time in any binary expression

If both numbers given are 0 then z1 is always used as default

We then extrapolate for division and multiplicative inverses

if our expression is ( A * 0 )

(A * 0 (as z1) = 0 )

(A * 0 (as z2) = A )

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aSorry, I am still trying to be "polite' but this simply makes no sense to me.  When you say "Let z1 be the multiplicative property of 0 ", what does that mean?  z1 is a "number" in this system, not a "property".  The "multiplicative property" of 0, in any algebraic system that has both multiplication and addition with more than 1 number and "0" is the additive identity, is that 0 times anything is equal to 0.

When you say "Let z2 be the multiplicative identity property of 1" do you mean that z2 is the multiplicative identity?

But then you say "Let 0 have both properties".  What happened to z1 and z2?  If you mean that this "number", is both the additive identity and the multiplicative identity, then, as I said before, there must exist only one number in your system, 0.

Proof:  let "0" be the additive identity.  Then x+ 0= x.  So, by the "distributive law", for any a, a(x+ 0)= ax+ a(0)= ax.  Subtracting ax from both sides, a(0)= 0.

Does your "number system" not satisfy the distributive law?  If so that sharply restricts its usefulness!

Now, suppose it is true that "0" is also the multiplicative identity.  That is,  a(0)= a for any a in the system.  From above, a(-0)= 0= a.  That is, assuming that your system satisfies the distributive law, every member member of the system is equal 0.

Again, I did not check to see if your system satisfies the "distributive law".  If it does, then "Let 0 have both properties" is impossible.  If it does not, then it simply won't be very interesting!

Edited by HallsofIvy

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7 hours ago, HallsofIvy said:

aSorry, I am still trying to be "polite' but this simply makes no sense to me.  When you say "Let z1 be the multiplicative property of 0 ", what does that mean?  z1 is a "number" in this system, not a "property".  The "multiplicative property" of 0, in any algebraic system that has both multiplication and addition with more than 1 number and "0" is the additive identity, is that 0 times anything is equal to 0.

When you say "Let z2 be the multiplicative identity property of 1" do you mean that z2 is the multiplicative identity?

But then you say "Let 0 have both properties".  What happened to z1 and z2?  If you mean that this "number", is both the additive identity and the multiplicative identity, then, as I said before, there must exist only one number in your system, 0.

Proof:  let "0" be the additive identity.  Then x+ 0= x.  So, by the "distributive law", for any a, a(x+ 0)= ax+ a(0)= ax.  Subtracting ax from both sides, a(0)= 0.

Does your "number system" not satisfy the distributive law?  If so that sharply restricts its usefulness!

Now, suppose it is true that "0" is also the multiplicative identity.  That is,  a(0)= a for any a in the system.  From above, a(-0)= 0= a.  That is, assuming that your system satisfies the distributive law, every member member of the system is equal 0.

Again, I did not check to see if your system satisfies the "distributive law".  If it does, then "Let 0 have both properties" is impossible.  If it does not, then it simply won't be very interesting!

I understand that this makes no sense to you.  I came to that understanding in previous conversations with you.  As such I was extremely surprised you responded here.  But if "likes" is what your after just be nice and ill keep giving them to you.

First point.....the very first thing you should have done is checked to see if the distributive property still holds true.  Additionally I have posted links in other forums to a list of axioms.....I have shown how not a single axiom changes or is dissolved.  Including the additive identity property of 0 and the distributive property.  You must do a small amount of homework here if you really wish to offer a fair peer review.

steps...

1.  Take the number table given in the op....set it aside

2.  Take a fresh piece of paper....write down the expression ( A x B )

3.  Now make A and B whatever number you want....other than zero.

4.  Now take a z1 and a z2 from the table given according to the numbers you chose.

5.  Apply specifically as directed in the op...to a binary expression

Now do it all again for 0....

Lastly.....I have shown this work to 3 phd's.  All three agree that there is NOT an issue with the mathematics (that they could see).  If an issue exists it is in the "validity" of the axioms. Such as whether zero can or can not ALSO posses/have/compose/ "the multiplicative identity property of 1".  Axioms can not be proven or disproven.  The are subjective truths.

Most importantly I can quote from swansont ( a fine member of this community) and from strange ( a not so fine member) showing that to some degree validity is in this idea.  If a flaw is found it is deep therein.  Hence a lifetime of work....unless I have help.

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On 08/10/2017 at 10:39 PM, conway said:

Axiom

Let every number be arbitrarily composed of two numbers.

Let the number table exist as such…

One problem I see is that there exists no table large enough to hold all the real numbers.

So are you restricting this to integers?

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2 hours ago, studiot said:

One problem I see is that there exists no table large enough to hold all the real numbers.

So are you restricting this to integers?

The table must be extended in your imagination.  Chose a number not on the table.  Then create it's ordered pair exactly as shown with any other number than zero.

On ‎10‎/‎18‎/‎2017 at 2:30 PM, conway said:

The table given is an extension of all numbers.....integers....rational...real....irrational...imaginary...whole.....natural

On ‎10‎/‎18‎/‎2017 at 2:30 PM, conway said:

Additionally the table is not necessary to have in order for this idea to function......consider...

Let z1 be the multiplicative property of 0

Let z2 be the multiplicative identity property of 1

Let 0 have both properties

Let only one property be used at a time in any binary expression

If both numbers given are 0 then z1 is always used as default

We then extrapolate for division and multiplicative inverses

if our expression is ( A * 0 )

(A * 0 (as z1) = 0 )

(A * 0 (as z2) = A )

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57 minutes ago, conway said:

The table must be extended in your imagination.  Chose a number not on the table.  Then create it's ordered pair exactly as shown with any other number than zero.

Thank you for your reply, I see you did not catch my meaning.

To able to place all the real numbers in a table is equivalent to counting them as every place or pigeonhole in any table can be given a unique (serial) number, which is another way of saying that you cannot put all the real numbers in one-to-one correspondence with the integers.

The real numbers are not countable. The fact that there is at least one that doesn't fit into any table was Cantor's original proof that there are more reals than integers.

There are just too many real numbers so a table containing all of them is wishful thinking.

However is your idea equivalent to saying that every number fits into a unique pigeonhole which may be referenced by a column and row reference?

This is what database designers call a flat table.

Edited by studiot

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On 08/10/2017 at 10:39 PM, conway said:

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

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 )

By itself, I can't see that this would ever lead to anything but an integer n being represented by (n,n).

So for example, if we make A = 2 = (2, 2) and B = 3 = (3, 3) then, using your rule for multiplication:

A*B = (2x3) = (2x3) = (3x2) = (3x2)

And, presumably, based on the initial axiom, all of these are equivalent to (6,6).

Even division leads to the same thing:

A/B = (2/3)

Which can, presumably be represented as (0.666, 0.666)

B/A = (3/2)

Which can, presumably be represented as (1.5, 1.5)

So, a couple of questions:

1. Are all numbers of them form (n, n); i.e. where z1 = z2? If not, how do we get a number where z1 and z2 are not equal?

2. In your multiplication and division rules, why don't we end up with a pair of numbers as the result? For example, I would have expected something more like:

A x B = (  z1forA  x  z2forB ,  z2forA  x  z1forB) ... or something similar

Also, just as an aside, the "z1forA" notation isn't very easy to read. I would go for something like A[z1] or A.z1 (both notations used in programming, not sure about pure math). But not important.

27 minutes ago, studiot said:

There are just too many real numbers so a table containing all of them is wishful thinking.

I'm not sure this is a problem. If the table is just an aid to understanding the concept, it could probably be extended to reals without needing an explicit table.

27 minutes ago, studiot said:

However is your idea equivalent to saying that every number fits into a unique pigeonhole which may be referenced by a column and row reference?

Good question. That might be one way of understanding it. But then, wouldn't this be like complex numbers (except the multiplication rules are different).

On 18/10/2017 at 8:30 PM, conway said:

Let z1 be the multiplicative property of 0

Let z2 be the multiplicative identity property of 1

Let 0 have both properties

Does this mean that, unlike the other numbers, 0 is represented as (0, 1) ?

Ah, yes. Sorry just noticed that is explicitly stated in your table.

So if we have A = 3 = (3, 3) then A * 0 = (3 x 0) = (3 x 1). But that seems to imply that multiplying a number by 0 can give two possible results, 0 or A. Is there a rule for choosing which to use? When you say "z1 by default" do you mean always z1?

Edited by Strange

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5 hours ago, studiot said:

Thank you for your reply, I see you did not catch my meaning.

To able to place all the real numbers in a table is equivalent to counting them as every place or pigeonhole in any table can be given a unique (serial) number, which is another way of saying that you cannot put all the real numbers in one-to-one correspondence with the integers.

The real numbers are not countable. The fact that there is at least one that doesn't fit into any table was Cantor's original proof that there are more reals than integers.

There are just too many real numbers so a table containing all of them is wishful thinking.

However is your idea equivalent to saying that every number fits into a unique pigeonhole which may be referenced by a column and row reference?

This is what database designers call a flat table.

Studiot

Thank you.  I apologize for missing the point of your reply.  I understand quite clearly Cantor's "Diagonal Argument". It as you point out....shows we can not count the real numbers.  I understand this.  Nor...as you also point out....can I fit all numbers in the table.  I don't have to.  Perhaps you can see Strange's reply here.  The point of the table was to show that...

for every X in R =/= 0 : x = (x,x)....."with no FURTHER representations of the given "ordered pair"

So yes...and thank you....a "flat table".......

If then we have come to an understanding on the "table" what then is your next concern?

4 hours ago, Strange said:

By itself, I can't see that this would ever lead to anything but an integer n being represented by (n,n).

So for example, if we make A = 2 = (2, 2) and B = 3 = (3, 3) then, using your rule for multiplication:

A*B = (2x3) = (2x3) = (3x2) = (3x2)

And, presumably, based on the initial axiom, all of these are equivalent to (6,6).

Even division leads to the same thing:

A/B = (2/3)

Which can, presumably be represented as (0.666, 0.666)

B/A = (3/2)

Which can, presumably be represented as (1.5, 1.5)

So, a couple of questions:

1. Are all numbers of them form (n, n); i.e. where z1 = z2? If not, how do we get a number where z1 and z2 are not equal?

2. In your multiplication and division rules, why don't we end up with a pair of numbers as the result? For example, I would have expected something more like:

A x B = (  z1forA  x  z2forB ,  z2forA  x  z1forB) ... or something similar

Also, just as an aside, the "z1forA" notation isn't very easy to read. I would go for something like A[z1] or A.z1 (both notations used in programming, not sure about pure math). But not important.

I'm not sure this is a problem. If the table is just an aid to understanding the concept, it could probably be extended to reals without needing an explicit table.

Good question. That might be one way of understanding it. But then, wouldn't this be like complex numbers (except the multiplication rules are different).

Does this mean that, unlike the other numbers, 0 is represented as (0, 1) ?

Ah, yes. Sorry just noticed that is explicitly stated in your table.

So if we have A = 3 = (3, 3) then A * 0 = (3 x 0) = (3 x 1). But that seems to imply that multiplying a number by 0 can give two possible results, 0 or A. Is there a rule for choosing which to use? When you say "z1 by default" do you mean always z1?

Strange

Yes every number other than zero, N = (N,N)....integer or otherwise.  We can assume only integers for now.

In all case of N (z1 = z2)  where N =/= 0

Your second questions answer lies in the nature of (z1,z2).  I have represented them only as numbers.  As such I would then agree with your assumption on the extrapolation of z1 and z2 while performing a binary operation.  However....z1 and z2 are not exactly numbers.  z1 is a value and z2 is a space. ( a bit like vectors ).  Then we see that in a binary expression the "numbers" given are only pieces of a whole "number".  And the pieces go together very specially.  This is the nature of the equation...

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

I am more than willing to change any and all notation.  Either is great with me.  Specify which you would prefer and I will here after refer to z1 and z2 as such.

Yes the expression (A * 0 ) yields two sums.  There is nothing saying that it can not.  If and ONLY if each product has a unique solution.  The unique solution if found in the application of z1 and z2 in the binary expression.

On ‎10‎/‎18‎/‎2017 at 2:30 PM, conway said:

Additionally the table is not necessary to have in order for this idea to function......consider...

Let z1 be the multiplicative property of 0

Let z2 be the multiplicative identity property of 1

Let 0 have both properties

Let only one property be used at a time in any binary expression

If both numbers given are 0 then z1 is always used as default

We then extrapolate for division and multiplicative inverses

if our expression is ( A * 0 )

(A * 0 (as z1) = 0 )

(A * 0 (as z2) = A )

Yes when I say that if both numbers given in a binary expression are 0 then z1 must ALWAYS be used as default.  This is if you are using the "properties axioms" as opposed to the "table axioms".

Above and beyond in your politeness and consideration of the idea.  +1....as well as a public recant of a statement by me about you.  Cleary you can be a "top notch" community member.

Edited by conway

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According to your rules, 0x2 = (0x2) = (1x2) = (2x0) = (2x1)

So you get 0 = 2. Which is a problem, unless you add a special rule for multiplication by 0, just as you added a special rule that 0 was (0, 1).

You seem to be trying to get past the fact that you can't divide by 0 (or that multiplication by 0 always gives 0, if I remember your earlier threads correctly) by writing "special rules" for 0 that don't apply to any other integers. And that won't work well. You can attempt to force it to work, by breaking all of the other real rules, but it will be ugly, and it won't be useful.

Edited by uncool

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16 minutes ago, conway said:

Thank you.  I apologize for missing the point of your reply.  I understand quite clearly Cantor's "Diagonal Argument". It as you point out....shows we can not count the real numbers.  I understand this.  Nor...as you also point out....can I fit all numbers in the table.  I don't have to.  Perhaps you can see Strange's reply here.  The point of the table was to show that...

for every X in R =/= 0 : x = (x,x)....."with no FURTHER representations of the given "ordered pair"

So yes...and thank you....a "flat table".......

If then we have come to an understanding on the "table" what then is your next concern?

At this time I don't have any 'other concerns' because I'm sorry but this line is meaningless misuse of symbolism and I don't know what you are trying to say.

Quote

for every X in R =/= 0 : x = (x,x)

I understand the first part:

For every element, X, of the set of real numbers, R, and with the sole exception of zero,

It is the next part of the statement that is unintelligable.

What is x, what is (x,x) and what is the equals sign doing there?

Please state what you mean in English, then we can sort out what symbolism can be used and proceed.
I have given you an example above.
There is nothing you can state in mathematical symbols that can't be stated in English, indeed the first time any symbolism is introduced it must perforce be stated in English first.

Without this I am reduced to scratching around trying to guess what you might possibly mean by the statement.

Thank you for the + point but I would rather you had answered the question I asked in that post, which concerned my latest guess.

This is important, and also answers Strange's point.

An instruction in machine code might be to go to such and such an address in memory,
Take what you find at that address
Do something with it, say double it.

The point is the separation of the address and the data that is held in that address.

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2 hours ago, uncool said:

According to your rules, 0x2 = (0x2) = (1x2) = (2x0) = (2x1)

So you get 0 = 2. Which is a problem, unless you add a special rule for multiplication by 0, just as you added a special rule that 0 was (0, 1).

You seem to be trying to get past the fact that you can't divide by 0 (or that multiplication by 0 always gives 0, if I remember your earlier threads correctly) by writing "special rules" for 0 that don't apply to any other integers. And that won't work well. You can attempt to force it to work, by breaking all of the other real rules, but it will be ugly, and it won't be useful.

Uncool

Your point is excellent!  I intentionally left this out for fear of a long post.  Suffice to say that the equation...

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

is true for all A and B =/= 0

if A and B are both zero then the product is always 0

if A or B is 0.... but not both then

A * 0(as z1) = 0

A * 0(as z2) = A

My assumption is that for the most part this is self evident once a person applies the numbers....

Lastly.....as I have said it is already "mostly decided" that the "field axioms" are not BROKEN or ALTERED with this idea.  I can post a plethora of information here.  However if you still take issue with any specific axiom let me know.  I can provide equations proving my claims regarding any specific axiom and or property.  If I can provide for a way to divide by zero without changing or breaking any current axiom then the idea is extremely useful.

2 hours ago, studiot said:

At this time I don't have any 'other concerns' because I'm sorry but this line is meaningless misuse of symbolism and I don't know what you are trying to say.

I understand the first part:

For every element, X, of the set of real numbers, R, and with the sole exception of zero,

It is the next part of the statement that is unintelligable.

What is x, what is (x,x) and what is the equals sign doing there?

Please state what you mean in English, then we can sort out what symbolism can be used and proceed.
I have given you an example above.
There is nothing you can state in mathematical symbols that can't be stated in English, indeed the first time any symbolism is introduced it must perforce be stated in English first.

Without this I am reduced to scratching around trying to guess what you might possibly mean by the statement.

Thank you for the + point but I would rather you had answered the question I asked in that post, which concerned my latest guess.

This is important, and also answers Strange's point.

An instruction in machine code might be to go to such and such an address in memory,
Take what you find at that address
Do something with it, say double it.

The point is the separation of the address and the data that is held in that address.

Studiot

Perhaps it would be best then for the two of us to not consider the "number table".  I understand that I am not using the equal sign exactly as it is intended.  I have acknowledged this.  So then if we can not come to a consensus as to the validity of the axiomatic table presented...then we should address only the axiomatic properties as presented.  If here again you take issue fundamentally...then I have failed to change or improve this idea from the year plus that it was last presented.  In this case I thank you for your time....

the axiomatic properties "version"

Let z1 be the multiplicative property of 0

Let z2 be the multiplicative identity property of 1

Let only one property be used in any binary expression

If both numbers given are 0 than z1 must be used by default

so then if our expression is ( A * 0 )

A * (0(z1)) = 0

A * (0(z2)) = A

I also stated very specially what the "ordered pair" on the RHS of the table was.  One is a value.  One is a space.  An enormous amount of issues arrive from these declarations however.  As such we use them only as numbers.  This effects the equality sign.  Only in that I am forced to make the statement....

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

Other than this the equality sign functions exactly as it currently does.

Or I may say....

Never allow a number to be represented by it's entire ordered pair in any equation.

It is only in binary multiplication and division that you use these "ordered pair"  and it is only pieces at a time that you use them.

Edited by conway

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Fair enough, I wish you well in your quest.

I will leave you with the observation that in addition to not defining any of the terms I asked for, your last post introduced yet more undefined terms that for everyone else have different definitions and follow different rules .

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31 minutes ago, studiot said:

Fair enough, I wish you well in your quest.

I will leave you with the observation that in addition to not defining any of the terms I asked for, your last post introduced yet more undefined terms that for everyone else have different definitions and follow different rules .

Yes...you are correct...i repeated myself...and then introduced new terms.

I also offered an alternative route which you clearly ignored....im fine with this thank you for your time.

Edited by conway

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On 10/20/2017 at 4:54 PM, conway said:

Uncool

Your point is excellent!  I intentionally left this out for fear of a long post.  Suffice to say that the equation...

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

is true for all A and B =/= 0

if A and B are both zero then the product is always 0

if A or B is 0.... but not both then

A * 0(as z1) = 0

A * 0(as z2) = A

My assumption is that for the most part this is self evident once a person applies the numbers....

Lastly.....as I have said it is already "mostly decided" that the "field axioms" are not BROKEN or ALTERED with this idea.  I can post a plethora of information here.  However if you still take issue with any specific axiom let me know.  I can provide equations proving my claims regarding any specific axiom and or property.  If I can provide for a way to divide by zero without changing or breaking any current axiom then the idea is extremely useful.

Yes, the field axioms would have to be either broken or altered in order for the equation "A * 0(as z2) = A" to make sense. I'll walk you through exactly why by asking questions.

In your system, what is 1 * 0 (as z2)?

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44 minutes ago, uncool said:

Yes, the field axioms would have to be either broken or altered in order for the equation "A * 0(as z2) = A" to make sense. I'll walk you through exactly why by asking questions.

In your system, what is 1 * 0 (as z2)?

Uncool

No they are not.

Let the following be the set of "axioms" to search for contradiction with.

http://mathworld.wolfram.com/FieldAxioms.html

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

1 = (1(z1),1(z2))
2 = (2(z1),2(z2))
3 = (3(z1),3(z2))

"see op for details"

Now to the math...

No application of this idea is ever used in addition or subtraction.

PROOF the first five properties under addition exits without change or contradiction.

(A * B) = (B * A)

if A and B =/= 0

(A(z1) * B(z2)) = (B(z2) * A(z1)) = (B(z1) * A(z2)) = (A(z2) *B(z1)) = (A * B ) = (B * A)

PROOF the Commutativity property of multiplication exists without change or contradiction

( A * (B * C) ) = ( (A * B) C )

if A and B and C =/= 0

( A(z1 or z2) * ( B(z1) * C(z2) ) ) = ( (A(z1) * B(z2) ) C(z1 or z2) ) = ( A(z1 or z2) * ( B(z2) *C(z1) ) )= ( (A(z2) * B(z1) ) C(z1 or z2) )

PROOF the associative property of multiplication exists without change or contradiction

Induction could be fairly used at this point for the following "remaining" properties to also hold true

distributive
identity
inverses

as long as X =/= 0

if not I can continue to lay this out for anyone....

what then matters of course is 0.....

I would like to "quote" Thurston from his book "The Number System"

Chapter 3...page 13

"As a slightly harder example, let us prove from the laws of arithmetic only that (0 * X = 0) "

" y * x + 0 * x = ( y + 0 ) x by the distributive law "
" = y * x by the neutrality of zero for addition"
" = y * x + 0 by the neutrality of zero for addition"
"Therefore 0 * x = 0 by the cancelation law for addition"

In the previous page he states the cancelation law of addition is...

"If x + y = x + z then y = z"

therefore 0 * X =

but for this expression to yield 0 as the sum.......

(0(z1) * X ) = 0 ... zero must be z1 by the properties assigned to the table
(0(z2) * X ) = X ... zero must be used as z2 by the properties assigned to the table

So then per the axiom set forth in the original post...

0 * X = 0 is a valid equation....if and only if 0 is z1
0 * X = X is a valid equation....if and only if 0 is z2

PROOF all equations expressed by Thurston regarding 0 and multiplication remain unchanged and without contradiction

It is only that an additional "path" is available with 0 relative to the property used in its binary operation.

Multiplication by zero is relative to zero used as z1 or as z2. not.  Proof....

Perhaps you should have just asked for these proofs.  I told you I would supply them.

1 * 0(as z1) = 0

1 * 0(as z2) = 1

Edited by conway

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Your "proofs" all make the assumption that none of your numbers are 0. That is very explicitly not an assumption in basic math. So yes, you have altered the axioms.

So you say that 1 * 0(as z2) = 1.

In your system, do you have that 0 + 0 = 0?

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1 minute ago, uncool said:

Your "proofs" all make the assumption that none of your numbers are 0. That is very explicitly not an assumption in basic math. So yes, you have altered the axioms.

So you say that 1 * 0(as z2) = 1.

In your system, do you have that 0 + 0 = 0?

Ok.  Fair enough.  As long as no given number is zero....all axioms hold the same.  I have then supplied axioms for what to do with zero.  Play fair.  But your point is well taken.

Yes zero is still the additive identity.

In no cases is the table used regarding addition or subtraction.  Only in binary expressions involving multiplication and division.

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3 minutes ago, conway said:

Ok.  Fair enough.  As long as no given number is zero....all axioms hold the same.  I have then supplied axioms for what to do with zero.  Play fair.  But your point is well taken.

Yes zero is still the additive identity.

In no cases is the table used regarding addition or subtraction.  Only in binary expressions involving multiplication and division.

That's my point. You have "defined" division by 0, only by making an exception as to what multiplication is when you use 0. You've made division by 0 make sense, by making nonsense of division. I'm sorry, but this system isn't useful.

According to your system, then, do you agree that 1*(0 + 0)(as z2) = 1?

But a hugely important point of the field axioms is the interplay between addition and multiplication. And as such, if you try to modify one, you need to check whether that interplay still works.

Edited by uncool

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1 hour ago, uncool said:

That's my point. You have "defined" division by 0, only by making an exception as to what multiplication is when you use 0. You've made division by 0 make sense, by making nonsense of division. I'm sorry, but this system isn't useful.

According to your system, then, do you agree that 1*(0 + 0)(as z2) = 1?

But a hugely important point of the field axioms is the interplay between addition and multiplication. And as such, if you try to modify one, you need to check whether that interplay still works.

"you've made division by 0 make sense, by making nonsense of division."

I am confused here...do you many multiplication.  I don't think I have made non sense out of it.  Perhaps that is your opinion.  Thank you for it.

1 * ( 0 + 0)(as z2) = 1

However....

1 * (0 + 0) (as z1) = 0

Thurston's arguments require only that (A * 0 = 0)...NOT....that there is only ONE product for the expression (A * 0).  I have shown that (A * 0 = 0) is still a valid equation.  Therefore satisfying Thurston.

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2 minutes ago, conway said:

"you've made division by 0 make sense, by making nonsense of division."

I am confused here...do you many multiplication.  I don't think I have made non sense out of it.  Perhaps that is your opinion.  Thank you for it.

1 * ( 0 + 0)(as z2) = 1

However....

1 * (0 + 0) (as z1) = 0

Thurston's arguments require only that (A * 0 = 0)...NOT....that there is only ONE product for the expression (A * 0).  I have shown that (A * 0 = 0) is still a valid equation.  Therefore satisfying Thurston.

I think you have made nonsense of both multiplication and division, yes.

So you agree that 1 * (0 + 0)(as z2) = 1 in your system. In your system, what is 1*0(as z2) + 1*0(as z2)?

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16 minutes ago, uncool said:

I think you have made nonsense of both multiplication and division, yes.

So you agree that 1 * (0 + 0)(as z2) = 1 in your system. In your system, what is 1*0(as z2) + 1*0(as z2)?

Yes

1 * (0 + 0)(as z2) = 1

but you must acknowledge that...

1 * (0 + 0)(as z1) = 0

I can chose z1 and z2 at will.  Any expression or equation given by you requiring ( A * 0 = 0 ) will still be possible under the given solutions.

1 * 0 (as z2) + 1* 0 (as z2) = 2

because...

(1 * 1) + (1 * 1) = 2

Edited by conway

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So in your system, the following is NOT true:

1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2)

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18 minutes ago, uncool said:

So in your system, the following is NOT true:

1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2)

Correct...

1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2)

But...

1* (0 + 0)(as z1) = 1*0(as z1) + 1*0(as z1).....IS TRUE

as well as

1* (0 + 0 )(as z2) = 1 *0(asz1) + 1*0(as z2).....IS TRUE

and so on...