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 October 10th, 2017, 07:27 PM #1 Banned Camp   Joined: Aug 2012 Posts: 153 Thanks: 3 Request for Peer Review Axioms 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 ) Last edited by Conway51; October 10th, 2017 at 07:31 PM.
October 10th, 2017, 07:39 PM   #2
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Quote:
 Originally Posted by Conway51 Axioms 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
0 doesn't fit the pattern. Is that a typo?

Is there more? There doesn't seem to be any mathematical argument. What kind of structure and properties does your set of ordered pairs have under those operations you defined?

October 10th, 2017, 07:52 PM   #3
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Quote:
 Originally Posted by Maschke 0 doesn't fit the pattern. Is that a typo? Is there more? There doesn't seem to be any mathematical argument. What kind of structure and properties does your set of ordered pairs have under those operations you defined?
Maschke

Thank you for your time. No (0=(0,1)) is not a typo. All properties and axioms as currently set forth continue to exist without change. However if you look closely at how multiplication is defined as pairs of "pieces" of two different numbers you will begin to see the point. That is regarding zero. This set of axioms allow "0" to use the "multiplicative identity property of 1". It generates two unique sums for (A * 0). (A * 0 = 0) is still a valid equation. Thereby solving for division by zero. Multiplicative inverses of zero and 1. As well as creating a structure in mathematics where binary multiplication is "relative".

 October 10th, 2017, 09:11 PM #4 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,855 Thanks: 750 Math Focus: Wibbly wobbly timey-wimey stuff. 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
October 10th, 2017, 09:40 PM   #5
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Quote:
 Originally Posted by Conway51 Axioms 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 ... (A x B) = ( z1forA x z2forB ) = ( z2forA x z1forB )
If I'm understanding your notation, 0 x 1 = (0,1) x (1,1) = 0 by the first equality, but 1 by the second. Isn't that a problem?

And why is the output of the product of two ordered pairs NOT another ordered pair, but a single number?

Last edited by Maschke; October 10th, 2017 at 09:43 PM.

October 10th, 2017, 10:26 PM   #6
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Quote:
 Originally Posted by Maschke If I'm understanding your notation, 0 x 1 = (0,1) x (1,1) = 0 by the first equality, but 1 by the second. Isn't that a problem? And why is the output of the product of two ordered pairs NOT another ordered pair, but a single number?
Maschke

The combinations of "ordered pairs" while operating a binary expression must be labeled as given in the op. That is...

only z1forA and only z2forB

or...

only z2forA and only z1forB

So then ( 0 x 1 )

must be only

z1for0 x z2for1

or only...

z2for0 x z1for1

*note we can "flip" either previous expression

So then we have

0 x 1 = z1for0 x z2for1 = 0

or

0 x 1 = z2for0 x z1for1 = 1 x 1 = 1

If a unique solution can be offered for both products in the previous expressions of ( 0 x A ) then there is no "mathematical" contradiction.

What remains is deciding when to use 0 as z1 or z2. I believe this question is "inherent" in the purpose of the mathematical "expression".

As such "theoretically"...this apparently has not purpose.

However...it is my opinion that it is a valid "work around" to allow for division by zero as well as the other things I have mentioned. As well as a few possible "theoretical" consequences much to distant to be relevant at the moment.

The inclusion of the axiom...

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

Is to show that a number is NEVER represented as its "entire" pair in ANY expression. That is 0 is never represented as (0,1). If we allowed this then we would have

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

It is only one number from each "ordered pair" in any given expression.

I hope I have addressed your questions sufficiently.

Thank you.

Last edited by skipjack; October 11th, 2017 at 10:21 AM.

 October 11th, 2017, 08:00 AM #7 Senior Member   Joined: Sep 2016 From: USA Posts: 417 Thanks: 231 Math Focus: Dynamical systems, analytic function theory, numerics I think you misunderstand why we don't divide by zero. It isn't because we can't figure out a way to do it meaningfully. It is a basic fact that in a field, every element is invertible except 0 (the label for the additive identity). Conversely, if you try to define an inverse for 0, you are no longer working in a field. The properties of a field are extremely nice. In particular, fields are closed under division since every element is invertible. Not allowing division by zero is a small price to pay for this utility. This is not a problem that needs solving. Furthermore, the operation you are defining here is not useful even for dividing by zero. Namely, as people have pointed out, you have a binary operation which is not consistent with itself. As you yourself have computed that $0 = 0 \times 1 = 1 \times 0 = 1$ which makes it pretty useless. Thanks from Conway51
 October 11th, 2017, 08:38 AM #8 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 Axioms for rational numbers: (BM for example) Q(Z) of Z consists of all ordered pairs (a,b) with a,b $\displaystyle \epsilon$ Z and b $\displaystyle \neq 0$, and std definitions of equality, multiplication, and division of rational numbers. What is the point of a sloppy, confusing, paraphrase? Thanks from Denis and topsquark
October 11th, 2017, 09:11 AM   #9
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Quote:
 Originally Posted by zylo What is the point of a sloppy, confusing, paraphrase?
C'mon Zylo...dontcha know that 4*0 = 0+0+0+0 ?

October 11th, 2017, 03:08 PM   #10
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Quote:
 Originally Posted by SDK I think you misunderstand why we don't divide by zero. It isn't because we can't figure out a way to do it meaningfully. It is a basic fact that in a field, every element is invertible except 0 (the label for the additive identity). Conversely, if you try to define an inverse for 0, you are no longer working in a field. The properties of a field are extremely nice. In particular, fields are closed under division since every element is invertible. Not allowing division by zero is a small price to pay for this utility. This is not a problem that needs solving. Furthermore, the operation you are defining here is not useful even for dividing by zero. Namely, as people have pointed out, you have a binary operation which is not consistent with itself. As you yourself have computed that $0 = 0 \times 1 = 1 \times 0 = 1$ which makes it pretty useless.

SDK

You must know that the reason there is no "inverse operation" for division by zero (the reason why we cant divide by zero) is because (A * 0 = 0). I have shown that not only is this equation true, but (A*0 = A) is also true. Therefore there is now and inverse operation for division by zero.

Another way that you may consider this is to ask me about "multiplicative inverses". I can show you many different equations using the "table" showing 0 and 1 as multiplicative inverses. Therefore solving for division by zero. 0 and 1 are both "invertible" in this way.

The additive identity still exists without change.

At no time is a number ever used as it's entire "ordered pair". Therefore you can never arrive at the equation 0 = 0 x 1 = 1 x 0 = 1
You may ONLY have ONE number from EACH ordered pair...what I very specifically said was

0 = z1for0 x z2for1 = z2for1 x z1for0 = 0 = 0 x 1

or you may have

1 = zzfor0 x z1for1 = z1for1 x z2for0 = 1 = 1 x 1

Thank you.

Last edited by Conway51; October 11th, 2017 at 03:46 PM.

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