conway

If I had it right the first time then all I have to do is change the application of z2 to z1 on the second half of the equation......for the equation to be equivalent...purely a matter of pinning down a finalized set of axioms.... once again....

Yes they are equal....if I do NOT add the zero's as I thought I was supposed to. If I do it according to the fashion you suggest then the statements are equivalent. Thanks for your hard work...you deserve the +1's....

I was AGREEING with you! OK it is a matter of equality....therefore I must multiple 1 by both 0's in the first set of parenthesis..therefore 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) these two expressions are equivalent......because of the distributive property....

Look I was under the assumption that you added the two zero's first that is.... 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 * 0 (as z2) =/= 1 * 0 (as z2) + 1* 0 (as z2) 1 =/= 2 but if I USE the distributive property before adding the two zero's.... 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 so if I do it as you suggest.....the operations are equivalent....that is I "carry then multiple" the one to both zero's before adding them.

actually I think we had a misunderstanding regarding the axioms....consider if I use the distributive property first as you suggested. 1*(0 + 0)(as z2) =/= 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 2 = 2 If there is a unique solution to each product where is the problem?

Not so... While binary multiplication by zero is RELATIVE...that is it yields two products...each product has a unique solution. Yes further "clarity" is needed as to these unique solutions. I appreciate your help.

Let's assume your right. It does NOT mean it fails. IT means that in the cases involving zero...for the distributive property to hold a specific application of z1 and z2 must occur remember when I said you must acknowledge that 1 * 0 (as z1) = 0 Do you see how I can rewrite the equations following the idea that the distributive property is a matter of equality and not multiplication. Look lets say i do the distributive property first.... 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) 1 + 1 = 1 + 1 2 + 2 Perhaps this was a drastic misunderstanding on my part...for that I'm sorry.

distributive is an act of multiplication.....the parenthesis must be solved before you can multiple in ANY fashion distributive or not. https://www.khanacademy.org/math/pre-algebra/pre-algebra-arith-prop/pre-algebra-ditributive-property/a/distributive-property-explained

Im sorry but your not. ANYTHING inside a set of PARENTHESIS must operate first. The addition of ( 0 + 0 ) MUST be performed BEFORE the distributive property. Because it is in the parenthesis. The "distributive" property which is multiplication.... can occur only after what is in the parenthesis is solved.

Ok....now explain how this fails the distributive property https://www.khanacademy.org/math/pre-algebra/pre-algebra-arith-prop/pre-algebra-ditributive-property/a/distributive-property-explained lol I'm sorry I see your caring it over to both zeros....................you must add both zeros first PEMDOS my friend....lol

Well you will have to show me in an equation how I have broken the distributive property. I do not see it. Besides.....I can manipulate z1 and z2 at will. Any attempt to show the distributive property failing out of a placement of z1 and z2...simply requires that I edit z1 and z2 accordingly. Then.....re apply to the axiom. Only regarding 0 of course. I wait and see....

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... Please Note 1 * 1 = 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

"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.

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.

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

Yes...you are correct...i repeated myself...and then introduced new terms. This made it clear that I had no other way to answer your question I also offered an alternative route which you clearly ignored....im fine with this thank you for your time.

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. Studiot Thank you for your time. I had hoped that I had answered your questions. It appears I have not. I apologize. I will try here again. 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.

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? 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. 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.

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.

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.

Very interesting article. Perhaps I chose my examples poorly. And yes I did mean epicycles...my apologies....I concede to you here as you are clearly correct....+1 for being nice about it....however I stand by my point that facts change. Including mathematics. The very fact that math has evolved over time shows a "changing" of the "facts". I will not take up further time from the op in this matter...my apologies.

Thank you for your sincere, polite reply ....+1 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 )

Incorrect... It is a FACT that it was ONCE considered a FACT the world was flat It is a FACT that it was ONCE considered a FACT that Greek Ellipses where a model for the solar system Aristotle thought the world was flat...for a FACT. Socrates thought the ellipses proves the movement of the solar system...for a FACT. Facts are nothing more than a objective decision by the majority that a particular subjective truth holds true. FACTS change...even in science.... That said I like your response to the op...as such I will give you a +1....despite your trollishness.....

What do you think Greek ellipses are? They ARE mathematics. It WAS considered FACT. Yet it was still wrong wasn't it. It's not your committing to a reply that bothers me. Its your rudeness and megalomania.