conway

A Request for Peer Review

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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.

Edited by conway

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On 20/10/2017 at 11:33 PM, studiot said:

So you do not wish to answer the question I asked.

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 .

 

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Rather, and having withdrawn as above, upon seeing you making some sort of progress with another, I thought to offer some encouragement in the way of a +1 point.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

Edited by studiot
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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.

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On 20/10/2017 at 9:04 PM, studiot said:
Quote

conway

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?

 

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

 

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

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.

 

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.

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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

Edited by conway

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1 hour ago, conway said:

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

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.

Edited by studiot
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49 minutes ago, studiot said:

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) ]

I think you are being a little unfair. The equals sign is commonly used to mean "defined as" as well as "equal to". While I'm sure the notation could be improved, I think the meaning is fairly clear, especially given the examples (the "table" you objected to in the opening posts and others later).

Maybe it would be better to say something like: every non-zero number, x, is represented by a tuple, X, such that X = (x, x). And so on. But this doesn't really make any difference to the argument presented.

(Whether the result is valid or useful is another question.)

To take your example, the number 25 would, in this system, be represented by the pair (25,25).

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19 minutes ago, Strange said:

I think you are being a little unfair. The equals sign is commonly used to mean "defined as" as well as "equal to". While I'm sure the notation could be improved, I think the meaning is fairly clear, especially given the examples (the "table" you objected to in the opening posts and others later).

Maybe it would be better to say something like: every non-zero number, x, is represented by a tuple, X, such that X = (x, x). And so on. But this doesn't really make any difference to the argument presented.

(Whether the result is valid or useful is another question.)

To take your example, the number 25 would, in this system, be represented by the pair (25,25).

 

Not at all unfair. I said "evaluate to" which is the most common use of the equals sign.

I agree is also is used to represent "defined as" although there is a perfectly good and proper symbol available for that meaning.

That is of course the identity symbol which is what some that is defined as is.

Unfortunately, x was also 'defined'  as a number, causing endless confusion.

 

My question is simple, and if the definitions provided in this thread would enable the proverbial Man on the Clapham Omnibus to determine what x is for any conceivable situation, I would be glad to have it pointed out to me.

(25,25), if I ask someone else does not cut it.

And I still don't know what process goes on inside that bracket.

 

 

Edited by studiot
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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.

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Studiot

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

 

 

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Quote

 

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.

 

 

First the good news to encourage you.

 

This idea is sound, but really in the province of applied maths or physics, rather than pure maths.

 

 

 

This is quoting you from two years ago.  (x,x) = the first x is value, the second x is space.....you agreed this particular idea was sound.  I have merely searched for a better way to communicate it.  Perhaps I have failed.  

Bare with me Studiot....I think in time we can come to some form of understanding and hopefully pin this thing down and make some real progress.

 

Edited by conway

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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?

 

0 = (0.z1) , (0.z2)
 
 
(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

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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.

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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.

 

 

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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. 

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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.

Edited by conway
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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. 

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44 minutes ago, conway said:

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.

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?

 

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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......

Edited by conway
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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.

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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

Edited by conway

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On ‎10‎/‎25‎/‎2017 at 2:07 AM, conway said:

Studiot

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

 

 

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First the good news to encourage you.

 

This idea is sound, but really in the province of applied maths or physics, rather than pure maths.

 

 

 

This is quoting you from two years ago.  (x,x) = the first x is value, the second x is space.....you agreed this particular idea was sound.  I have merely searched for a better way to communicate it.  Perhaps I have failed.  

Bare with me Studiot....I think in time we can come to some form of understanding and hopefully pin this thing down and make some real progress.

 

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.

 

 

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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.

Edited by conway

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23 hours ago, conway said:

...

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.

...

 

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 * (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

Edited by pzkpfw
  • Upvote 1

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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 * (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

 

 

Edited by conway

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Moderator Note

conway, you place far too much importance on people being agreeable with you, and not enough on explaining your ideas to a group of peers. This thread has been reported for your insistence in the face of multiple attempts to help. Hand-waving and foot-stomping aren't part of the peer review process. You need to support your ideas in the face of criticism, or show where that criticism fails. 

Next time. This subject stays closed until you have more to discuss.

 

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