conway

A Brand New Approach

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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]

 
[math] P_2 0 = (0, 1) ~ \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) = 0 \cdot 0.z_1 + 1 \cdot 0z_2 = 0.z_2[/math]
 
 
The distributive property (all combinations of a, b, and c as zero)
 
a * (b + c) = a * b + a * c 

 

a = 1, b = 0 , c = 0

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

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

 

a = 1, b = 1 , c = 0

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

 

a = 0, b = 0 , c = 0

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

 

a = 1, b = 0 , c = 1

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

 

1 = 0, b = 1, c = 0

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

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

 

Edited by conway

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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.

Edited by conway

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9 hours ago, conway said:

because I have learned a great deal form the members of this community.

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

9 hours ago, conway said:

As I was asked by a moderator to come up with something new here

 

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

11 hours ago, mathematic said:

What is the purpose of 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.

 

:)

 

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9 hours ago, conway said:

...


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

...

This seems to imply 0 = 1

Is that useful?

Edited by pzkpfw
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25 minutes ago, pzkpfw said:

This seems to imply 0 = 1

Is that useful?

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?

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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.

 

Edited by conway
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4 hours ago, conway said:

... 0 and 1 are not equal. No where is that equivalency in my equations.  ...

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.

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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  )=10.z 1 +00.z 2 =0.z 1  

 
P 2 0=(0,1) (0.z 1 0.z 2  )=00.z 1 +10z 2 =0.z 2  
 
 
 

 

 

 

 

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

You need to be consistent, too.

You wrote 

= ((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

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".

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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 

= ((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

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.

 

Edited by conway

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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

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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.

Edited by MigL

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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...

Edited by conway

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!

Moderator Note

I suggest you actually read the mod notes we post if you want to have any threads that remain open. For example:

 
On 04/11/2017 at 0:05 AM, Phi for All said:
!

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.

 

 

!

Moderator Note

I don't see where you have attempted to rectify this at all. Moreover, I see very little substantive difference between this thread and the thread the above note was quoted from. As such, this is closed, and you are not permitted to reopen it. 

 
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