# Thread: A New Relative Mathematics

1. ## A New Relative Mathematics

In every R there exists an integer zero element ( -0 )
( -0 ) =/= 0
|0| = |(-0)|
( -0 ) : possesses the additive identity property
( -0 ) : does not possess the multiplication property of 0
( -0 ) : possesses the multiplicative identity property of 1
The zero elements ( 0 ) and ( -0 ) in an expression of division can only exist as: (0)/( -0 )

0 + ( -0 ) = 0 = ( -0 ) + 0
( -0 ) + ( -0 ) = 0
1 + ( -0 ) = 1 = ( -0 ) + 1

0 * ( -0 ) = 0 = ( -0 ) * 0
1 * ( -0 ) = 1 = ( -0 ) * 1
n * ( -0 ) = n = ( -0 ) * n

Therefore, the zero element ( -0 ) is by definition also the multiplicative inverse of 1 .

And as division by the zero elements requires ( - 0 ) as the divisor ( x / ( -0 )) is defined as the quotient ( x ) .

0 / n = 0
0 / ( -0 ) = 0
n / ( -0 ) = n

0 / 1 = 0
1 / ( -0 ) = 1
1 / 1 = 1

( 1/( -0 ) = 1 )

The reciprocal of ( -0 ) is defined as 1/( -0 )

1/(-0) * ( -0 ) = 1

(-0)^(-1) = ( 1/( -0 ) = 1

(-0)(-0)^(-1) = 1 = ( -0 )^(-1)

Any element raised to ( -1 ) equals that elements inverse.

0^0 = undefined
0^(-0) = undefined
1^0 = 1
1^(-0) = 1

Therefore, all expressions of ( -0 ) or ( 0 ) as exponents or as logarithms are required to exist without change.
Therefore, division by zero is defined.
Therefore, the product of multiplication by zero is relative to which integer zero is used in the binary expression of multiplication.

3. ## Re: A New Relative Mathematics

Originally Posted by skeeter
Its on far more web sites than just that skeeter. Nor am I ashamed. Perhaps you have something to offer...? NO.....? Didn't think so.

4. ## Re: A New Relative Mathematics

Originally Posted by Conway
Its on far more web sites than just that skeeter. Nor am I ashamed. Perhaps you have something to offer...? NO.....? Didn't think so.
I have something to add. You keep posting what you say is new mathematics, but do not provide any basis to allow anyone other than you to understand what you are doing. Others have tried to parse it both here and on other forums, but your system is full of inconsistencies. You have redefined every mathematical symbol, but not bothered to inform anyone what the new symbols mean in a way they could comprehend. You simply use them and assume their new meaning will be understood through context (or in other posts through philisophical interpretation, which is typically different for different people). You change definitions at your whim, and again, refuse to tell anyone the new definitions. This makes any type of peer review utterly impossible. At the moment, unfortunately, your work is designed to only be understandable to you. No one else should be capable of understanding it because the definitions for everything you are attempting are known only by you. To give you a better idea of what I am saying, consider the gibberish statement, "Ever are condors eating tulips." You see the statement, but it does not hold any meaning to you. Until I tell you that I decided to redefine all of the words of the sentence. Now, "Ever" means "This post" and "are condors eating tulips" means "is intractable". So, the evaluation of the statement in my mind is "This post is intractable". It is perfectly understandable to me, but no one else, because no one else defines the word or phrases that way.

Does this make sense? It would be awesome to see everything the way you do, but you have not provided a translation from what we know to what you are attempting to demonstrate to make it meaningful in any way. It causes an absolute breakdown of communication. It is entirely possible that what you are describing is a tremendous breakthrough in mathematics, but if you are unable to convey even the simplest concepts, then it is all for naught.

5. ## Re: A New Relative Mathematics

Originally Posted by SlipEternal
I have something to add. You keep posting what you say is new mathematics, but do not provide any basis to allow anyone other than you to understand what you are doing. Others have tried to parse it both here and on other forums, but your system is full of inconsistencies. You have redefined every mathematical symbol, but not bothered to inform anyone what the new symbols mean in a way they could comprehend. You simply use them and assume their new meaning will be understood through context (or in other posts through philisophical interpretation, which is typically different for different people). You change definitions at your whim, and again, refuse to tell anyone the new definitions. This makes any type of peer review utterly impossible. At the moment, unfortunately, your work is designed to only be understandable to you. No one else should be capable of understanding it because the definitions for everything you are attempting are known only by you. To give you a better idea of what I am saying, consider the gibberish statement, "Ever are condors eating tulips." You see the statement, but it does not hold any meaning to you. Until I tell you that I decided to redefine all of the words of the sentence. Now, "Ever" means "This post" and "are condors eating tulips" means "is intractable". So, the evaluation of the statement in my mind is "This post is intractable". It is perfectly understandable to me, but no one else, because no one else defines the word or phrases that way.

Does this make sense? It would be awesome to see everything the way you do, but you have not provided a translation from what we know to what you are attempting to demonstrate to make it meaningful in any way. It causes an absolute breakdown of communication. It is entirely possible that what you are describing is a tremendous breakthrough in mathematics, but if you are unable to convey even the simplest concepts, then it is all for naught.

While in the past, this idea has changed many times. That is the nature of incomplete works.

Other than (-0)

find a single definition of an element that has changed form current mathematics.....

THIS time it does not exist. The ONLY thing new...The ONLY thing defined even is (-0)
ALL definitions of (-0)...USE CURRENT definitions

I thank you for you time.

Go ahead....copy and past something that is defined differently then it CURRENTLY is....other than (-0)
other than 2 additive identities and 2 multiplicative identities (even these are defined the same -except now there are two of each)

6. ## Re: A New Relative Mathematics

Originally Posted by Conway
While in the past, this idea has changed many times. That is the nature of incomplete works.

Other than (-0)

find a single definition of an element that has changed form current mathematics.....

THIS time it does not exist. The ONLY thing new...The ONLY thing defined even is (-0)
ALL definitions of (-0)...USE CURRENT definitions

I thank you for you time.

Go ahead....copy and past something that is defined differently then it CURRENTLY is....other than (-0)
other than 2 additive identities and 2 multiplicative identities (even these are defined the same -except now there are two of each)
This is false. You have defined a system R for which you have not provided the definition. This system R has binary operators addition, multiplication, inversion, and exponentiation. You have provided definitions by example for them and left it to us to assume the rest. What other elements does R have? You said for any R. You alluded to the fact that R contains 0 and 1 as well as (-0). What if R is the set containing the elements 0, (-0), 1, and 2. How is addition defined on 2? Multiplication? Exponentiation and inversion? If that is not a valid R, what are valid R's? You have not told us. You just said for any R. But, I don't know what R is because you never told me. Finite math is a robust and growing discipline of mathematics.

If R is the set of real numbers, and you are adding an element (-0), then you are still extending addition, multiplication, exponentiation, and inversion to accommodate (-0), which implies a different operator than the one we are used to. You seem to assume that everyone has your thoughts in their head. We do not. What you have stated may make perfect sense to you, but if you do not communicate it well, no one else will be able to understand what you are saying.

7. ## Re: A New Relative Mathematics

Originally Posted by SlipEternal
This is false. You have defined a system R for which you have not provided the definition. This system R has binary operators addition, multiplication, inversion, and exponentiation. You have provided definitions by example for them and left it to us to assume the rest. What other elements does R have? You said for any R. You alluded to the fact that R contains 0 and 1 as well as (-0). What if R is the set containing the elements 0, (-0), 1, and 2. How is addition defined on 2? Multiplication? Exponentiation and inversion? If that is not a valid R, what are valid R's? You have not told us. You just said for any R. But, I don't know what R is because you never told me. Finite math is a robust and growing discipline of mathematics.

If R is the set of real numbers, and you are adding an element (-0), then you are still extending addition, multiplication, exponentiation, and inversion to accommodate (-0), which implies a different operator than the one we are used to. You seem to assume that everyone has your thoughts in their head. We do not. What you have stated may make perfect sense to you, but if you do not communicate it well, no one else will be able to understand what you are saying.

It is understood (by all mathematicians) R works as it always does...unless otherwise specified.

I gave specific examples of how all operators work...see here is an example

0 + ( -0 ) = 0 = ( -0 ) + 0
( -0 ) + ( -0 ) = 0
1 + ( -0 ) = 1 = ( -0 ) + 1

0 * ( -0 ) = 0 = ( -0 ) * 0
1 * ( -0 ) = 1 = ( -0 ) * 1
n * ( -0 ) = n = ( -0 ) * n

If you chose to not really observe what it is I post. That is on you...but then...yes...you will have no idea what I am thinking.

Perhaps you care to offer something about the op...as opposed to your opinions on how I do or do not communicate?

What? No mathematical opinions relative to the op....I thought so.

8. ## Re: A New Relative Mathematics

Originally Posted by Conway
It is understood (by all mathematicians) R works as it always does...unless otherwise specified.
@Conway, Have you studied model theory to any level of proficiency?
I must tell you that I see no evidence that you have.
I agree with slip that $(-0)$ seems to be an enlargement of $\mathcal{R}$ that is not consistent with other operations the way that $\bf{i}$ does to give the complex numbers or that the enlargement gained by the infinitesimals gives the hypereals, $\mathbb{H}$.

How does $(-0)$ enteract with all others in $\mathcal{R}~?$

9. ## Re: A New Relative Mathematics

With all due respect...I gave that information in the original post...look...here is some of it again.....

0 / n = 0
0 / ( -0 ) = 0
n / ( -0 ) = n

0 / 1 = 0
1 / ( -0 ) = 1
1 / 1 = 1

( 1/( -0 ) = 1 )

The reciprocal of ( -0 ) is defined as 1/( -0 )

1/(-0) * ( -0 ) = 1

(-0)^(-1) = ( 1/( -0 ) = 1

(-0)(-0)^(-1) = 1 = ( -0 )^(-1)

Any element raised to ( -1 ) equals that elements inverse.

0^0 = undefined
0^(-0) = undefined
1^0 = 1
1^(-0) = 1

But in any case I no longer care. I despise mathematicians. The likes of these have an utter disrespect for philosophical and spiritual truths. That is why they deserve the ever so slow crawl of advancement that they are currently mired in. Good luck.

10. ## Re: A New Relative Mathematics

Originally Posted by Conway
With all due respect I despise mathematicians. The likes of these have an utter disrespect for philosophical and spiritual truths. That is why they deserve the ever so slow crawl of advancement that they are currently mired in. Good luck.
Oh my goodness. Oskar Levant said that “a critic is like a eunuch who knows great wonderful things are happening all about him but he does not have the equipment necessary to take part”. That describes you my friend.

11. ## Re: A New Relative Mathematics

You requested a mathematical response. Here goes:

1 = 1 + (-0) (additive identity)
(-0)*1 = 1 (multiplicative identity)
So, if we multiply both sides of the top equation by 1, we get:
1*1 = 1*[1 + (-0)]
1 = 1*1 + 1*(-0) (by distribution)
1 = 1 + 1 = 2 (because you said (-0) is a multiplicative identity)
In fact, it can be shown that any rational number n where n is not 0 or (-0) in fact equals 1.

I have not thought enough about how this extends to the reals. Still, you have a rather meaningless system. So much for your "philosophical and spiritual truths". We mathematicians would respect them if you provided at least one.

12. ## Re: A New Relative Mathematics

Oh my goodness. I once said that "a troll is a person who continues to make replies when they are no longer relevant and necessary."

13. ## Re: A New Relative Mathematics

Originally Posted by Conway
Oh my goodness. I once said that "a troll is a person who continues to make replies when they are no longer relevant and necessary."
So you admit you are a troll. Good job!

14. ## Re: A New Relative Mathematics

Originally Posted by SlipEternal
You requested a mathematical response. Here goes:

1 = 1 + (-0) (additive identity)
(-0)*1 = 1 (multiplicative identity)
So, if we multiply both sides of the top equation by 1, we get:
1*1 = 1*[1 + (-0)]
1 = 1*1 + 1*(-0) (by distribution)
1 = 1 + 1 = 2 (because you said (-0) is a multiplicative identity)
In fact, it can be shown that any rational number n where n is not 0 or (-0) in fact equals 1.

I have not thought enough about how this extends to the reals. Still, you have a rather meaningless system. So much for your "philosophical and spiritual truths". We mathematicians would respect them if you provided at least one.
regardless of the distributive property PEMDOS requires you add what is in the parenthesis first.....observe...i don't care about your perspective of my truths. I made a statement that I no longer wish to discuss this. SHut up....! LEave! LOCK this F_ing thread....there did that do it?

15. ## Re: A New Relative Mathematics

Originally Posted by Conway
regardless of the distributive property PEMDOS requires you add what is in the parenthesis first.....observe...i don't care about you perspective of my truths. I made a statement that I no longer wish to discuss this. SHut up....! LEave! LOCK this F_ing thread....there did that do it?
Nope. Rejecting a refutation of your crappy math does nothing. It remains refuted.

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