conway

Oh I take it in strides...ever trying to improve....How do you take the fact your Bigjerk?

13 minutes ago, Sensei said:

It gave 1 on my calculator..

 

https://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml

https://brilliant.org/wiki/what-is-00/

The point being that "whatever" it is....

All representations of (-0) as exponents and logarithms work like 0 and without change from 0

thank you for your time and reply

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.

2 hours ago, MigL said:

Two expressions...

0*1=(0.z1)*1=0
0*1=(0.z2)*1=1

Divide through by 1...

[0*1]/1=[(0.z1)*1]/1=0/1
[0*1]/1=[(0.z2)*1]/1=1/1

And cancelling 1s, gives...

0=0.z1=0
0=0.z2=1

So unless  *, =, ( ), /, and 0 mean something else in your 'new' math, it is garbage, not just different.

Well nice...but maybe you should have read all the pertinent information before assaulting me...

0.z1 * 1  =/=  0.z1

0.z2 * 1  =/=  0.z2

yes you may divide through by 1 but...

0.z1 * 1 = 0

0.z2 * 1 = 1

therefore after your cancelation of 1 you will have

0 = 0 = 0

1 = 1 = 1

It was unfair of you to have read so little before making such a negative reply.

6 hours ago, conway said:

but...

0.z1 * 1 =/= 0.z2 * 0

Pzkpfw...

please note I made a mistake here

I meant to say

0.z1 * 1 =/= 0.z2 * 1

0.z1 * 1 = 0.z2 * 0

Apologies...

1 hour ago, John Cuthber said:

"A Brand New Approach"
The old approach works; yours is different.

Think about that...

No one is debating that the old approach works John.

Just because something works doesn't make it right John.

Just because something is different doesn't make it wrong John.

You have a history with me of passive aggressiveness and thread de-railing.  I have NO desire to communicate with you any further.  Further replies from you on this thread will not be responded to by me... and will receive a -1

39 minutes ago, pzkpfw said:

If you're going to completely redefine all of math, you'll never get anywhere.

You need to be consistent, too.

You wrote 

0 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z1) * 1 = 0
1 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z2) * 1 = 1

And then go to:

0 = 0 * 1 = 0
1 = 1 * 1 = 1

The bit I highlighted in red changes. That's nuts. One moment you have 1 = 0 * 1, the next you have 1 = 1 * 1.

There is no consistency to your "math".

0 * 1 = ( 0.z1 ) * 1 = 0

0 * 1 = ( 0.z2 ) * 1 = 1

0.z1 = 0

0.z2 = 1

No inconsistences....however the binary expression ( A * 0 ) is RELATIVE to what projection operator for zero was used.

If I use the projection operator 0.z1 in a binary operation for zero the product is 0

If I use the projection operator 0.z2 in a binary operation for zero the product is X

You MUST CONVERT 0 to a projection operator BEFORE you can solve for the binary equations involving zero.

If you do not understand this...or agree...or I am a lousy at explaining myself.... that is fine.  I thank you for your time.  Perhaps in a year or so I will find a better way to communicate with you. Thank you.

1 hour ago, pzkpfw said:

No.

Unless you have re-defined what "=" means, you show that 0 = 1.

Say we have: 1 + 1 + 1 = 2 + 1 = 3

Each of the parts is equal to all the others. We can remove the middle part and still be accurate: 1 + 1 + 1 = 3

Taking your equations:

0 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z1) * 1 = 0
1 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z2) * 1 = 1

We should be able to remove the wiggildybop bits and the other parts that are equal should still be equal:

0 = 0 * 1 = 0
1 = 0 * 1 = 1

So, you have both 0 and 1 equal to 0 * 1. If 0 and 1 both equal 0 * 1, then you imply 0 equals 1.

 

You're not explaining this well.

NO....

I never stated 0 = 1.....why don't you post a quote of this

I said 0.z2 =1.....

clearly your mistake was to replace my "wiggildybop" (0.z2) with 0...therefore 0 = 1

you MUST replace my "wigglildybop"  0.z1 "with" 0

you MUST replace my "wiggligdybop" 0.z2 "with" 1

If you replace it in the way in which I suggest what you end up with is

0 = 0 * 1 = 0
1 = 1 * 1 = 1

AS I STATED

0.z1 = 0

0.z2 = 1

therefore when you replace them correctly you have

0 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z1) * 1 = 0
1 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z2) * 1 = 1

0 = 0 * 1 = 0
1 = 1 * 1 = 1

notice 0 * 1 = (0.z1) * 1

notice 0 * 1 = (0.z2) * 1

but...

0.z1 * 1 =/= 0.z2 * 0

again please note the projection operators from the original post....

0.z1 is a projection operator FOR 0 but = 0

0.z2 is a projection operator FOR 0 but = 1

On ‎11‎/‎5‎/‎2017 at 4:27 PM, conway said:

 

 

0=(0.z 1 0.z 2  ) 

 

0.z1 = 0

0.z2 = 1

 

P 1 0=(1,0) (0.z 1 0.z 2  )=1⋅0.z 1 +0⋅0.z 2 =0.z 1  

 

P 2 0=(0,1) (0.z 1 0.z 2  )=0⋅0.z 1 +1⋅0z 2 =0.z 2  

 

 

 

 

 

 

3 hours ago, studiot said:

A word about your technique.

 

Please don't react the wrong way to this, it is meant to help you progress whatever your goal actually is.

 

The most important thing to learn here would be to stop using terms and notation that already have very well and tightly defined meanings for something entirely different.

I'm sure the moderators do not expect this when they said

 

Such presentation makes your posts impenetrable to even the most intelligent members and leads to comments like this

 

So start with your basic object and give it a name.

It is not a number so how about duplet or couplet since it appears to be formed of two simpler objects joined together in some way.

Then you can explain how the joining process works and what you can do with your object.

If you must use the term 'binary expression' (which establishes a connection between two objects) you (and others) will then be in a position to distinguish between your object (see why it needs a name of its own?) and objects already named and defined by you or anybody else.

 

I really hope this helps you progress.

 

 

Clearly you did NOT read my op in this thread did you Studiot.  This is more of the same typical behavior from you and frankly I would rather not hear from you again.

FACT

1.  NO new terms where introduced (or used unorthodoxly) in the newest version of this. (only new notations referring to CURRENT real numbers)

while I had a 0.z1 and a 0.z2 it was only a facilitator for 0 and for 1.....the thing is studiot I went OUT of my way here to do EXACTLY as you asked.  THERE IS NO NEW terminology here.

please reread the op in this thread.  copy/quote and past ANYTING that is NEW to mathematics.

Even the projection operators were typical.

Until you can ACTUALLY post a quote from THIS thread showing ANYTHING NEW .....-1....should you do so....I will withdraw this.

2 hours ago, pzkpfw said:

This seems to imply 0 = 1

Is that useful?

pzkpfw

I seem to think so.  Most do not.  0 and 1 are not equal. No where is that equivalency in my equations.  It does however imply that there is  a similarity between 0 and 1 yes.  But that would be philosophy and I would rather stick to the mathematics.

thank you

2 hours ago, Strange said:

I have occasionally done this in software to prevent divide by zero errors (yes, it's a bit of a hack). But I can't see it being useful in mathematics.

Perhaps Conway could provide an example where it could be useful?

Strange ...I have done so already.

12 hours ago, conway said:

1. Relative binary multiplication by zero.
2. Defined division by zero.
3. Create varying amounts of zero.
4. Unify semantics, and physics with theoretical mathematics.
5. Offer a new approach on the continuum theory.
6. Suggest solutions for the physics regarding the unification of quantum and classical mathematics.

2 hours ago, mathematic said:

What is the purpose of this?

mathematic

1. Relative binary multiplication by zero.
2. Defined division by zero.
3. Create varying amounts of zero.
4. Unify semantics, and physics with theoretical mathematics.
5. Offer a new approach on the continuum theory.
6. Suggest solutions for the physics regarding the unification of quantum and classical mathematics.

Also this might help you.  This is from a previous closed thread.  As I was asked by a moderator to come up with something new here...in order to continue.(As I have done)

*NOTE this is edited from the previous post.. because I have learned a great deal form the members of this community...and would like to reflect an ability to learn and grow.

(0.z1) = in a binary expression of multiplication yields the product 0 : in a binary expression of division is the numerator and yields the quotient 0 : if both numbers are 0 in an expression of binary multiplication the binary product is 0

(0.z2) = in a binary expression of multiplication yields the product x : in a binary expression of division is the denominator and yields the quotient x : if both numbers are 0 in an expression of binary division the binary quotient is 0


0 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z1) * 1 = 0
1 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z2) * 1 = 1



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



0 = x * ( 0 + 0 ) = x * (0z1) = (0z1) * x = ((0z1)/1) * (1/(0z2)) = (0z1) * x = 0

x = x * ( 0 + 0 ) = x * (0z2) = (0z2) * x = ((0z1)/1) * (1/(0z2)) = (0z2) * x = x

thank you for your time.

No number tables...no properties.

No axioms change (except) when involving zero.

The following projection operators allow for no further axioms......

[math]0 = \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) [/math]

0.z1 = 0

0.z2 = 1

[math] P_1 0 = (1, 0) ~ \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) = 1 \cdot 0.z_1 + 0 \cdot 0.z_2 = 0.z_1[/math]

2 hours ago, pzkpfw said:

 

Where did you get the 4 and 7 from? (You picked 5 arbitrary numbers, not three).

That should be:

1 * (2 + 3) = 1 * 2 + 1 * 3

From:

a * (b + c) = a * b + a * c

 

You start with: 1*(2+3), so (your three numbers) a=1, b=2 and c=3.

So on the right is: a*b + a*c = 1*2 + 1*3 (not picking another two random numbers)

 

e.g. today is October 3rd, 2017 for me, so I'll say a=3, b=10, c=17

left = a * (b + c) = 3 * (10 + 17) = 3 * 27 = 81

right = a * b + a * c = 3 * 10 + 3 * 17 = 30 + 51 = 81

pzkpfw

Well said...+1...I did finally after much exhaustion see why uncool chose this equation.  I chose not to reply on the matter as uncool has said some very rude things.  I also stated in my reply with uncool that ALL this shows is that a simple axiom of the nature that the distributive property remains without change...except zero....and then provided equations and expressions showing what I mean by this.  

You put much time into this reply....put a little more into it.

Use the original post....apply any number other than zero to the equation brought forth by uncool...and the equations remains equivalent.

As I was already changing other axioms regarding the nature of 0....and as I was frustrated I rushed my last replies to him and poorly chose my expressions and equations.

please note this has only happened with him.

please note I gave him a +1 for pointing this out

please note he then attempted a system of circular logic regarding the "order of operations" in an attempt to "trip" me up.

The math on the matter...

a * (b + c) = a * b + a * c 

1 * (2 + 3) = 1 * 2 + 1 * 3

uncool gave...

1 ( 0 + 0 ) = 1 * 0 + 1 * 0

uncool made a = 1, b = 0 , c = 0

then he said some mean things...successfully frustrating me...and tripping me up...

so here you are then....with HIS equalities for a, b and c

1 * ( 0 + 0 ) = 1 * 0 + 1 * 0

1 * (0 + 0) = 1 * (0.z1) = 1 * (0.z1) + 1 * (0.z2)

I hope you can understand the nature of my mistakes with him.

I even told him this was all that was necessary.  He would have seen it if he hadn't been so focused on being a troll.  Which is why no one else brought it up. Or supported him.

again thank you...and a well earned +1

10 minutes ago, studiot said:

The idea is sound.

But I also said that it is not pure maths.

So I have been seeking a place in applied maths place for it, where many similar ideas already operate.

However none of these run counter to the underlying pure maths - they all conform to it as the master plan. They also all have extra restrictions peculiar to their own application.

That is also probably why uncool has spent so much time trying to work it out with you. You should thank him for that.

 

The problem is that you want your idea to be more basic than the underlying maths rather than a restricted application like all the others.

I'm sorry but this it can never be.

 

 

So then....this "idea" is sound....in whatever application you wish

So then...I did for a FACT do my best to answer your questions.....you denied this

Uncool gave an equation that is NOT a valid representation of the distributive property (this can be proven).(I thanked him many times, verbally and with +1)

I believe the "idea" to be AS basic as the underlying math....I understand that you do not agree with this.

I thank you for your time and peer review

You did put and extraordinary amount of effort into it....towards the end....I mean you finally looked up my quote of you....to bad you didn't read it back when we where both happy and getting along.  But in all seriousness thank you.

9 minutes ago, studiot said:

What do you expect stamping your foot and showing off your temper to achieve?

I neither said that I did or that I did not like your explanations I said several times that I did not understand it.

That is why I repeatedly asked for further explanation.

I am still waiting for this.

It certainly appears to me that you are stamping your feet and showing  your temper.  I don't think it will get you anything.

I have given a link with a quote of YOU....agree with me... on the explanations of the definitions your continuously asking for.

I am waiting for you to click on it and read for yourself....yourself...answering these questions.

Please leave this alone.  This is dangerously close to getting closed.  If your done just leave me be.....please.

I apologize for BOTH our tempers flaring.....but please just leave me be.

+1 to get us back on the peace trail

1 hour ago, studiot said:

I would like to make it quite plain before this thread is closed as unproductive that I have only argued with you once in my seven posts in this thread. You quickly agreed that I was right and that you cannot place all the real numbers in a table of any sort. 

I congratulated you on this.

 

Apart from that all my post have been questions as I have tried to understand what your proposal actually is.

Each time you have failed to answer and finally stated that you cannot answer.

Each of my questions have been straightforward technical questions.

As the author of a hypothesis, how do you expect it to be accepted if you cannot answer questions about it?

 

Studiot

Do you pick and chose the post you want to read?  Why do you blatantly make stuff up?

challenge....

Post a quote of me saying that you where right about my number table and I was wrong. I NEVER said this.  I did say that I could not list all of them on a table.  That is obvious and has nothing to do with the table it's self.

case in point: YOUR PEER STRANGE does NOT agree with me on this idea.  But he does agree your point here is "unfair".....exact words.

Facts are...I did answer your questions...you didn't like the answers

I DEMAND you address my link and quote of you from two years ago.....AGREEING with me about space and value as definitions for z1 and z2...the "question" you insist I didn't answer.  Why did you not address this?  Mad that I caught you in a PROVEABLE lie?

If you did not like my latest rendition, why did you bother to reply? Your negative posts are what make this thread "unproductive".....-1 for replying without merit.

Admit you where wrong 2 years ago...or admit your wrong now....otherwise I will not continue to address your replies as they are (as you point out) unproductive

1 hour ago, uncool said:

I did not "force" them to be equal. I literally picked 3 arbitrary numbers. That's the entire point of the distributive property.

Also: you still have your equation wrong. It's not "a( b + c) = a*b + b*c"; it should be a*(b + c) = a*b + a * c.

Further: the fact that you have to say "excluding zero" is a huge problem. It means that for literally everything you do in your system, you may have to say "Unless we ran into zero somewhere along the way". And since operations (namely, multiplication) with zero are the only difference between your system and the usual system of multiplication, that takes away any possible advantage your system could possibly have.

If you don't want to talk distribution any more, that's fine; I'm done. I've given my "peer review" (although to be honest, no, I am not your peer), which is: your system does not add anything useful to the usual system of multiplication, and rather strongly takes something away. 

Uncool

Excluding zero is not a problem if you provide solutions.

1*(2 + 3) = 1*4 + 1*7

mmh that's funny I just picked three arbitrary numbers and the equation failed to be equivalent.....lol give this one up buddy it's embarrassing.

you are human that makes you my peer....even if you don't like it.

yes it is clear you are done' as you have nothing further to add....please note...not one other person complained about the distributive property.  I have posted links to other websites....many different one's in fact

NO ONE HAS ever complained that it fails the distributive property....so yes Im done with your peer review...thank you.

-1 for not claiming another human being is your peer.  ALL HUMAN BEINGS are PEERS....ALL HUMAN BEINGS are equal...including me and you.

Yes yes I still had it wrong...I was in a hurry....but clearly there was NO NEED for precision as I provided links for you......

1 hour ago, uncool said:

Change the 1 to 5, the first 0 to 2, the second 0 to 3. 

5*(2 + 3) = 25 = 10 + 15 = 5*2 + 5*3. Yes, it does work with numbers that aren't 0.

"a ( b + c ) = (a + b) * c" is not the distributive property whatsoever. 

LOL....LOL...no really....LOL

sure I'll give you that if you work hard enough you can find the right combination of numbers to "force" the equation to be equal

But then it would be NO good as an example of the distributive property.....

Yeah I messed up here...

a ( b + c ) = (a + b) * c

yeah I mixed up the associate and the distributive....I meant

a( b + c) = a*b + b*c

sorry bout that....

As I said chose any number other than zero.....apply to equation...distributive property holds

Any number other than zero see op.

Here you go...

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

Let there be NO FURTHER argument about the distributive property.

The given link gives our definition of the distributive property.  The op follows the linked definition excluding zero. The  op gives rules for 0.

48 minutes ago, uncool said:

You still haven't answered how distribution can possibly work, except by breaking the usual way distribution works. Which would make this useless to any normal use of multiplication.

I have actually.  I don't understand you at all on this matter.  For example...

This is the equation you presented me with...

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

This equation does NOT work for the distributive property with ANY number, 0 or otherwise. "change 0 to any other number"  "do the math" "equation fails"

In NO cases is the equation EVER equivalent (our debate about the adding of the 0's or otherwise).  

Therefore it is NOT an argument or a representation of the distributive property.

This is the OFFICAL equation showing the distributive property

NOTE any number OTHER than zero may be chosen and the equation holds true. (with-in this idea)(outside this idea)

In the cases involving a,b,c being zero....see op.

a ( b + c ) = (a + b) * c

Thank you for your consideration.  I understand our differences on this matter.

Strange, Studiot, Uncool

I have tried to take into account the points brought forth by all of you.  Perhaps a quick glance....would not be to much to ask?

   It is the inherent nature of all things that they are a compilation of two different and distinct things. It is axiomatic that these two things are space and value. The value of any given thing being what it is, while the space is what it occupies.

   It is true that, abstract or otherwise numbers are a thing, therefore they must also contain a compilation of space and value. It is an axiomatic truth that space is the labeling of quantities of dimensions. It is an axiomatic truth that value is the labeling of quantities of existence, other than dimensions.

  It is an axiomatic truth that space and value exist in one of two forms. So that any given quantity of space or value is first labeled as defined or undefined. It is reasonable to say that any given number, that has had both its quantities of space and value labeled as undefined, requires no further question as to its nature. If however a given number, has had both its quantities of space and value labeled as defined, it is then necessary to further define the given quantities. That is to say what is the nature of the space and value's that are defined.

  There are four axiomatic steps in the further defining of a defined quantity of space and value. First it is that, after a given quantity of space and value is labeled as defined, a symbol is given to identify the amount of quantities given. Second it is that the given amounts of defined space and value are labeled as finite or infinite. Third it is that the given amounts of defined space and value, that are finite or infinite, are labeled as fractional or whole. Fourth it is that the given amounts of defined space and value, that are finite or infinite, fractional or whole, are labeled as positive or negative.

   It is the case that all forms of the defining of quantities of space and value, are from the perspective of our humanity. This then shows that there is a collection of only four kinds of numbers. That is there are numbers that possess an undefined space and an undefined value. Otherwise represented as a ( Uv + Us ). Such a number not requiring further defining. There are numbers that possess a defined value and a defined space. Otherwise represented as a ( Dv + Ds ). Such a number requiring further defining. There are numbers that possess a defined value and an undefined space. Otherwise represented as a ( Dv + Us ). There are numbers that possess an undefined value and a defined space. Otherwise represented as a ( Uv + Ds ).

   It is reasonable to say that natural numbers have both their quantities of space and value labeled as defined. That is that a natural number is a ( Dv + Ds ). It is then through the process of further defining, that a natural number such as 2 is labeled as having ( 2Dv + 2Ds ). The symbol 2 then is the symbol identifying the amounts of quantities contained. It is then that the given quantities are labeled as finite. Otherwise represented as a ( 2DvF + 2DsF ). It is then that the given quantities are labeled as whole. Otherwise represented as a ( 2DvFW + 2DsFW ). It is then that a positive is assigned to the compilation of space and value, and it is so on for any natural number.

   It is also the case that fractions are labeled as a ( Dv + Ds ). That is any given fraction has both its quantities of space and value labeled as defined. So that such a number as .2 is labeled as ( 2DvFF + 2DsFF ). Then a positive is assigned to the compilation of space and value. Additionally, a fractional symbol may replace the decimal symbol.

   It is also the case that infinite numbers are labeled as a ( Dv + Ds ). So that such a number as 2infinite is defined as a ( 2DvIW + 2DsIW ). As well as fractional infinites, such as .2infinite. Which is labeled as ( 2DvIF + 2DsIF ). Then a positive is assigned to both compilations of space and value, and it is so on for any infinite or fractionally infinite number.

   Remaining are numbers that are a ( Uv + Ds ) and numbers that are a ( Dv + Us ). Such numbers do not necessarily require further defining. As an undefined quantity of space or value composites the given number. So then such numbers can only be limitedly defined relative to the given defined quantity. If then a number possess a defined value and an undefined space, the sum is then relative to the defined value. So that such a number as ( Dv + Us ) is then a 1 relative. Otherwise represented as a 1r.

   If then a number possess an undefined value and a defined space, the sum is then relative to the defined space. So that such a number as a ( Uv + Ds ) is then a zero. As no quantity of value is defined, and as one quantity of space is defined. The space of zero is clearly defined on any number line. The equation ( 1 + (-1) = 0 ) proves this in that, if zero did not occupy a defined space on the number line, then the equation would equal ( -1 ), and not zero.

   It is the case in multiplication and division, that neither number given is an actual number. Not in the fashion that each symbol contains both space and value. It is that one symbol is representing a value, and that one symbol is representing a space. It is the case that in multiplication the labeling of the given symbols as space or value in a specific order is not necessary. The sum yielded is always the same.

   It is the case that in division the labeling of the given symbols as space or value in a specific order changes the sum that is yielded. So that as an axiom the first given symbol is labeled as value, while the second given symbol is labeled as space.

   It is then that in multiplication the given value is placed additionally into the given spaces. Then all values are added in all spaces. It is then that in division the given value is placed divisionally into all given spaces. Then all values are subtracted except one.

   So that in the equation ( 2 x 0 = X ), there is a given defined value of ( 2DvFW ), that is placed additionally into the given defined space of ( Ds ). Then all values are added in all spaces. This process then yields the number 2.

   Whereas the equation ( 0 x 2 = X ), there is a given undefined value of ( Uv ), that is placed additionally into the defined space of ( 2DsFW ). Then all values are added in all spaces. This process then yields the number zero.

   So then in the equation ( 2 / 0 = X ), there is a defined value of ( 2DvFW ), that is placed divisionally into the defined space of ( Ds ). Then all values are subtracted except one. This process then yields the number 2.

   Whereas the equation ( 0 / 2 = X ), there is an undefined value of ( Uv ), that is placed divisionally into the defined space of ( 2DsFW ). Then all values are subtracted except one. This process then yields the number zero.

   As an addition to all current field axioms these ideas are expressed as stated.
   " For every A in S there exists a Z1 and Z2, constituting A, such that any A in operation of multiplication or division is only representing Z1 or Z2 in any given equation.  Such that Z1 for all A's other than zero equal A.  Such that Z2 for all A's other than zero equal A.  Such that Z1 for zero equals zero.  Such that Z2 for zero equals 1. "

   It is possible that further defining of the given defined value of a relative number, and the given defined space of a zero, is applicable and necessary.  It is possible to either leave the same, or adapt exponents and logarithms.  Naturally further axioms will be needed for adaption. Such as exponents of zero existing as a space representation of zero (z2).  Logarithms of zero existing as a value representation of zero (z1).

   It is possible to here-in re-address the idea of the continuum theory.  If the definitions for numbers and their groups, are adapted as stated, and with further exploration into the defining of ( Dv + Us ) relative numbers, ( Uv + Ds ) zero numbers, ( Uv + Us ) undefined numbers, and their placement onto the number line.  The idea here being to show all numbers originating from and returning to ( Uv + Us ) on any given number line.

Please forgive the members of this forum.  They can be very rude.

Cell phones emit waves not particles.  Your thought process was to consider the phones messages like particles (little bullets).  But they are waves.  While waves can interfere the rules are much different for how waves behave as opposed to particles.

Consider if you dropped a stone "your phone message" into a pool.  The "wave" or "ripple" would extend in "all" directions.

Further here is some info on the behavior of waves v.s. particles.

Your post showed a large amount of creative thinking I liked that.  +1

http://physics.ucr.edu/~wudka/Physics7/Notes_www/node64.html

3 hours ago, Bobby17 said:

When you’re outside and you push the send button on the screen of your smartphone to send a message to someone, the smartphone sends this message in every direction from your smartphone in order for your smartphone to reach the nearest cell-tower. 

 

This message in every direction must have a weight bigger than zero in every direction. Because what weighs zero or nothing is nothing and nothing can only remain itself. 

 

And when this message in every direction has a weight bigger than zero in every direction, for not being nothing, one can see this message in every direction as very tiny bullets that leave your smartphone in every direction at high speed while each tiny bullet has a weight bigger than zero. 

 

And that’s when the logic of a smartphone starts to fail because all these very tiny bullets in every direction crash at high speed with the other very tiny bullets in every direction from other smartphone users so that the message doesn’t reach the nearest cell-tower in the original form. 

 

That means that smartphones use a manifestation of the impossible to send their message in original form to the receiver and that leads to the study of manifestations of the impossible. 

 

Studiot

It took some time to find this.....

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3 hours ago, studiot said:

That is not an answer.

 

The difficulty I face is that what appears on one side of an equation must evaluate to whatever appears on the other side.

[ Remember you wrote  x = (x,x) ]

Your assertions do not explain, back up or develop how this can happen with what you wrote.

In particular using your discussion with uncool, whatever appears in parenthesis must evaluate to x by some process.

What is the process? You have not defined it, even for uncool.

I have told you this several times and have been left to guess what sort of process you might have in mind.

You introduced the mention of ''tables', so when I tried to discuss tables with you , you backed away and said that I should forget tales.

If they are so forgettable that they have no bearing on your assertion why did you introduce them?

 

So what the hell is going on in those parentheses that something  can evaluate to "itself, itself" ?

What does it mean to say

A number, x = (a number, x ; a number , x)

Which is what you have continued to assert.

 

Introducing further symbols like z1 and z2 or anything else that has not been defined only serves to further confuse matters.

 

Here is a way in which splitting the number makes some sort of sense (to me anyway) 

Consider the number 25  (let's avoid 1s and 0s for now)

We cannot directly combine the 2 and the 5 to make 7, because neither 2 nor 5 nor 7 equals 25.

Twentyfive means two tens and five units and could be written, (2,5) although that would be rather clumsy notation.

Considered this way the two tens and the five ones obey all the usual rules of arithmetic you have been arguing with uncool about.

I doubt this is what you mean, because I am still  forced to guess.

Studiot

First let me show you that I am fully listening to you.

You demand an explanation for HOW (x equates to (x,x)).

I have told you as an answer that I do not have to answer.  This is an axiom.

However I would love to answer......(philosophically)

All numbers are composed of a given value inside of a given space.

25 = 25 values each inside of 25 spaces

how we arrive at ( x = (x,x))

the value is "placed" into the space

observe...

value of three = (1,1,1)....NOT the number 3 or the number 1....any symbol here will actually do.

space of three= (_,_,_)....Not the number three.....see the "three" empty spaces....this is all abstract of course.

each value is "placed" into each "space"

so then the number 3 is

3 = (1,1,1)

so then the number 1 is

1 = (1)

0 = (0)

so then

(_) = the space of 1 and the space of 0

if a single value is "placed" into this space... it becomes the number 1.  If 0 value is "placed" into this space it becomes the number 0

in binary multiplication "one" symbol represents ONLY value.  "one" symbol represents only space.

But all this is unnecessary....the table is an AXIOM...I understand you do not agree or like it.  This is why I have asked you repeatedly to address the "properties" version.

That is why there is two version...to help with communication and axiomatic belief issues.

Strange

Thank you for going out of your way on this point.  I understand you do not agree with me....but I appreciate your help in trying to resolve this particular issue with Studiot...I hope that the two of us might still make progress.

1 hour ago, studiot said:

 

Above was my question and below was your non-answer.

 

 

Question What is the equals sign doing there? (Correctly followed by a question mark to tell you that it was a question)

Answer I am not using the equal sign exactly as intended.

 

No further explanation of how you have redefined a basic and standard mathematics symbol.

Assertion x = (x, x)

Question what is x?

No answer whatsoever and as far as I can see you have not mentioned 'x' again in this thread, except when quoting another.

Again I did answer..... you just didn't like the answer.

x is a number

(x,x) are two numbers that arbitrarily compose x

This was all in the op.

Or (x,x) is a value, and a space.(respectively)

Or you can perform this idea without using the table at all....in case you don't like the table....which is your case.....so then how about addressing the properties version of this idea since you don't agree with my answers for x and (x,x)....and the equal sign that is between them.

Look....You know very well that I can not define x = (x,x) without using philosophy.  Which would get me in greater trouble.  Further the table I gave in the op is a axiomatic table.  I don't have to define ANYTING inside the axioms.  You either take the axiomatic table as self inherently true or not.  

(I do have to define the math that follows...but NOT the axiom.)(define for my why anything multiplied by 1 is 1, or 0 by 0....they are self inherently true...there is no mathematical answer for these question. Only philosophical answers.)

Your post had nothing to do with me and everything to do with the idea thank you. ....+1

1 hour ago, studiot said:

 

If you look at page 1 of this thread, you can see that  we had an adult converstation spread over several posts where I was trying out a new idea in the hope of understanding your proposal..

Unfortunately you eventually baulked at answering my direct question, offering me all sorts of other things instead, but no answer to my question.

 

I reserve the right to call that half listening, not trolling and most certainly not a personal insult which I have not proffered to you throughout this thread.

Rather, and having withdrawn as above, upon seeing you making some sort of progress with another, I thought to offer some encouragement in the way of a +1 point.

 

I call that biting the hand that feeds.

When the hand that feeds you also slaps you...you may expect a bite.

I answered your question...for a fact....you just chose not to accept my answer as valid.  Which is fair...but claiming I did NOT answer it is not fair.  Return to topic...stay on topic...or cease your reply's...or expect a -1 for a continued tirade against me.

3 hours ago, studiot said:

I'll second that +1 to uncool

and also +1 to conway for (half) listening.

If I half listen or  not... is not anyone's point.  You make this commit to be a troll.  I feel I have always listened to others.  Just because I don't agree with you on the validity of my table does NOT mean I did not listen to you.  In fact it is my judgment that you only half listened to me.  While uncool gave it his/her best.....yet had no constructive critique... as far as I was concerned.  After all the equation we were discussing was equivalent.  Thereby validating the distributive property under my axioms.

-1 for making negative passive aggressive needless remarks about me.

Further more take the equation uncool gave me...

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

Now use any number other than 0 in place of 0

Fact....it fails the distributive property

Not sure where uncool even got this equation as an argument for the distributive property...but it clearly fails.

1 minute ago, uncool said:

And once again: it seems that you are choosing z1 or z2 in every case to simply match what happens for the original multiplication in order to check the axioms. Which means that your system doesn't add anything useful - in every case that matters, we'd just use basic multiplication. 

If the expression...

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

must be equivalent...

and if I had it right the "first" time and I must add the two zero's....

thereby making my equation NOT equivalent....and all I have to do is state something to the affect of ...

"Let any equality possessing multiple zero's consider the following application of z1 and z2"

then it is extremely useful....

I will ponder what you have shared with me....thank you.