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April 19th, 2018, 08:10 PM   #11
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Quote:
 Originally Posted by SDK So your argument that you can have multiple additive identities is to link to a proof that this is not possible? Amazing. Also, I'd like to point out (once again), that your construction isn't even a group under either addition or multiplication. You keep getting hung up on whether or not its a field which is missing the point by a mile. Lots of constructions which aren't fields are useful (e.g. the integers). Constructions which aren't a ring or even an additive group on the other hand....are useless nonsense.
In fact it is amazing!

If one wants to contradict a commonly held belief "like I am"...then one should offer an example of what it is they are trying to contradict. I was showing everyone...that the reason why "we" say there can only be "one" additive identity is......and then showing how I circumnavigated it.......this WAS the appropriate type of link...perhaps you don't often try to show others what current contradictions exist for your arguments...but I am honest SDK...

Address the idea....don't jabber....show some math if you disagree..and why.......not just why....that is how this is done!

April 19th, 2018, 08:15 PM   #12
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Quote:
 Originally Posted by Conway51 In fact it is amazing! If one wants to contradict a commonly held belief "like I am"...then one should offer an example of what it is they are trying to contradict. I was showing everyone...that the reason why "we" say there can only be "one" additive identity is......and then showing how I circumnavigated it.......this WAS the appropriate type of link...perhaps you don't often try to show others what current contradictions exist for your arguments...but I am honest SDK... Address the idea....don't jabber....show some math if you disagree..and why.......not just why....that is how this is done!
You linked to a proof that identity elements are unique for any group. Let me repeat that. It was a PROOF. Not an opinion, or another point of view. It is a mathematical fact and it doesn't require any arguing because this fact is proved in the link and its veracity can be verified by anyone who can read and understand words.

April 19th, 2018, 08:37 PM   #13
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Quote:
 Originally Posted by SDK You linked to a proof that identity elements are unique for any group. Let me repeat that. It was a PROOF. Not an opinion, or another point of view. It is a mathematical fact and it doesn't require any arguing because this fact is proved in the link and its veracity can be verified by anyone who can read and understand words.
lol....I certainly agree.....that very specfic reason is why I chose to link it....

let me repeat that....

bc it is a PROOF....contradicting my statement....I NEEDED IT.....in order to show you why....

That very PROOF.....relies on the definition of the equalities sign for its argument....

I did NOT dis prove the PROOF...I agreed with it.....I invented a new form of equalities in order to solve the issue....

again you jabber......show some math...copy and paste from the link I provided if you are unable to do so on your own.......

 April 19th, 2018, 09:02 PM #14 Senior Member   Joined: Oct 2009 Posts: 753 Thanks: 261 God, why are crackpots always so arrogant...
April 19th, 2018, 10:49 PM   #15
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Quote:
 Originally Posted by Conway51 bc it is a PROOF....contradicting my statement....I NEEDED IT.....in order to show you why.... That very PROOF.....relies on the definition of the equalities sign for its argument.... I did NOT dis prove the PROOF...I agreed with it.....I invented a new form of equalities in order to solve the issue....
The issue being what, exactly? I think others have proved any number of times that you don't have a field for the exact same reasons your link would agree...a field can't have two additive identities! How many more times does this need to be said? Including an extra equality relation simply makes the problem worse as there is only one form of equality relation for a field.

Hey, if you want to create a system that allows you to define division by 0 then you seem to have done something. But don't expect it to have any validity for constructions that don't allow it.

-Dan

April 19th, 2018, 11:15 PM   #16
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Quote:
 Originally Posted by topsquark The issue being what, exactly? I think others have proved any number of times that you don't have a field for the exact same reasons your link would agree...a field can't have two additive identities! How many more times does this need to be said? Including an extra equality relation simply makes the problem worse as there is only one form of equality relation for a field. Hey, if you want to create a system that allows you to define division by 0 then you seem to have done something. But don't expect it to have any validity for constructions that don't allow it. -Dan
You must be specific...

It is that you can not have more than one additive identity in a field...under the current definition of equality...

1. I did NOT break this rule.
2. I proposed an alternative equality.

as of yet I have heard nothing for why only one equality may exist in a field...please expand...very intriguing...

April 20th, 2018, 03:32 AM   #17
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Quote:
 Originally Posted by Conway51 It is that you can not have more than one additive identity in a field...under the current definition of equality...
This "current definition of equality" is precisely the one used in defining fields/rings/groups/etc. Whenever you see the symbol $=$ in the definition of one of these objects, it doesn't mean "an equality-like thing", it means "the bog standard usual equality". If you want to use a different equality-like thing, then you'll end up with a very, very different object.

That being said, to actually see if your idea leads to anything interesting, first you might actually want to define the symbol |=|. So far you've given a vague description of some properties you would like it to have, but you haven't actually told us exactly what it means to say a |=| b for elements a and b of your object.

In particular, how could you define |=| so that it's not the case that 0 |=| (-0), even though 0 = (-0) (which means 0 and (-0) are the same element)?

Last edited by cjem; April 20th, 2018 at 04:30 AM.

April 20th, 2018, 05:20 AM   #18
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Quote:
 Originally Posted by Conway51 Attention : Studiot As my posting privileges have been removed on the other forum... Your quote: "What I do not understand (because you haven't told me) is what are all the members of your field." You chose: It does NOT matter...... And yes it is a pity the two of us are not able to communicate. And yes it is a pity you misread what I wrote.....

What you say about the other forum has nothing to do with me.

There and now here better pure mathematicians than I am have attempted to help you in various ways.

There and here I have kept an open mind since I don't know what you are trying to achieve.

I do realise that before commenting on your statements it is necessary to have a clear understanding of them, which is why I asked you to identify the Field concerned.

The point about these algebraic sets is that Groups have at least one binary operation, Fields and Rings have at least two defined on them.
The minimum operations must conform to certain rules (axioms) of a Field, Group or Ring as appropriate.
Other operations may or may not also be defined with different rules.
The sets therefore have elements and operations involving these elements.

But there are two principles that stand out.

1) The principle that any application of one of the fundamental binary operations produces another member of the set.

2) The principle that one operation does not interfere with the axioms of another in any way whatsoever. This means you cannot 'get round' a rule you don't like.

I'm sure you really know all this.

But what you perhaps don't know

is that you can create (define) an Extension Field which contains extra elements and / or operations such as your elements |0| or your operation |=|

However, in line with principle2, you cannot use these to get around the basic ones that are required in any event.

Does this help?

April 20th, 2018, 06:49 AM   #19
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Quote:
 Originally Posted by cjem This "current definition of equality" is precisely the one used in defining fields/rings/groups/etc. Whenever you see the symbol $=$ in the definition of one of these objects, it doesn't mean "an equality-like thing", it means "the bog standard usual equality". If you want to use a different equality-like thing, then you'll end up with a very, very different object. That being said, to actually see if your idea leads to anything interesting, first you might actually want to define the symbol |=|. So far you've given a vague description of some properties you would like it to have, but you haven't actually told us exactly what it means to say a |=| b for elements a and b of your object. In particular, how could you define |=| so that it's not the case that 0 |=| (-0), even though 0 = (-0) (which means 0 and (-0) are the same element)?
I can not respond quickly here...I must think for a while....

Is this not the first real honest response to this thread?
Thank you....I will consider deeply......!

April 20th, 2018, 07:00 AM   #20
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Quote:
 Originally Posted by studiot What you say about the other forum has nothing to do with me. There and now here better pure mathematicians than I am have attempted to help you in various ways. There and here I have kept an open mind since I don't know what you are trying to achieve. I do realise that before commenting on your statements it is necessary to have a clear understanding of them, which is why I asked you to identify the Field concerned. The point about these algebraic sets is that Groups have at least one binary operation, Fields and Rings have at least two defined on them. The minimum operations must conform to certain rules (axioms) of a Field, Group or Ring as appropriate. Other operations may or may not also be defined with different rules. The sets therefore have elements and operations involving these elements. But there are two principles that stand out. 1) The principle that any application of one of the fundamental binary operations produces another member of the set. 2) The principle that one operation does not interfere with the axioms of another in any way whatsoever. This means you cannot 'get round' a rule you don't like. I'm sure you really know all this. But what you perhaps don't know is that you can create (define) an Extension Field which contains extra elements and / or operations such as your elements |0| or your operation |=| However, in line with principle2, you cannot use these to get around the basic ones that are required in any event. Does this help?
It does...as well as cjem's reply.....here I must think for a while....I have debated this with you going back to 2015 studiot. I have always shown you the ability to admit when I am wrong.....observe the equations uncool and I debated at scienceforums.net.........clearly I am right and he is wrong yet he doesn't admit it and I get banned. He was right about a separate issues...which i acknowledge.

I am the crank here...Thus I get treated like crap constantly....it is evidenced in every single post i have ever made......but look again....just like in this case....when something is presented I acknowledge it....what I don't understand is why you started treating me like crap.....observe our old conversations.....the mathematician's here are just as bad about their treatment of others then the other forum....observe romesk's replies to me.

I challenge you.....re read some of this crap going back to 2012....I have strived to be "nice" and "polite"......yet these "mathematician's" you speak of....treat all people like me.....like pure and utter shit.

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