Cont: Deeper than primes - Continuation 2

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Nay_Sayer said:
Sure; First understand that 1 really equals 3.104. This is key because it factors into directional degrees being assigned a shade of blue.
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Some suggested improvements: First understand that x/1 = x, where one of the possible cases is x=3.104
 
More about Daniel Dennett.

3.1 Skepticism about Unity

...

Dennett is interesting in this regard. He is thoroughly sceptical about the traditional picture of unified consciousness. Yet even he can invoke unity. He says, “What is it like to be an ant colony? Nothing, I submit … What is it like to be a brace of oxen? Nothing (even if it is like something to be a single ox)” (Dennett 2005). Why is the answer nothing? In such cases, “there is no functional unity … — no unity to distinguish an I from a we” (Dennett 2005).
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( http://plato.stanford.edu/entries/consciousness-unity/#SkeAboUni )


The keyword here is "about".

An awareness about Unity is not direct awareness of Unity.

Moreover, Unity is nor nothing neither yesthing.

It simply thing without complements, and this is exactly why it is the cause of multiplicity/diversity, but not vise versa.
 
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Nay_Sayer said:
Sayernetics: 1=3.104 and nothing else. Before you ask I won't define anything to stay within the par set for the course.
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Sayernetics: "My framework is always false (1=3.104) and nothing else."

So after all, Nay_Sayer, you can't stay within your course, since you have logically defined it.

Please move on.
 
doronshadmi said:
Sayernetics: "My framework is always false (1=3.104) and nothing else."

So after all, Nay_Sayer, you can't stay within your course, since you have logically defined it.

Please move on.
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You have not seen the rest of the rules yet,

Rule numero 2: Sometimes 1 = 2.87 when 3.104 is feeling fluish ; So 1=x and 3.104 when it has a headache is Y

X=3.104 or 2.87¬3/014=Y

Stay tuned for rules 3 through 1,000
 
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Nay_Sayer said:
You have not seen the rest of the rules yet,

Rule numero 2: Sometimes 1 = 2.87 when 3.104 is feeling fluish ; So 1=x and 3.104 when it has a headache is Y

X=3.104 or 2.87¬3/014=Y

Stay tuned for rules 3 through 1,000
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Before we continue, please write down all your rules such that they are based on logic.

Without doing it, I have no further discussion with you.

I am waiting :popcorn1

(Please try no to hurt your foot again (as you did in
15th July 2014 11:18 AM)
 
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doronshadmi said:
Before we continue, please write down all your rules such that they are based on logic.

Without doing it, I have no further discussion with you.

I am waiting :popcorn1

(Please try no to hurt your foot again (as you did in
15th July 2014 11:18 AM)
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Why should he, you don't write down yours. Not even straightforward definitions.
 
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abaddon said:
Why should he, you don't write down yours. Not even straightforward definitions.
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You win the cookie.

I appreciate him wishing me well on my foot and only 2 years late.

I'll give away the next rule

Rule: trio 2=5.9999999999999999 on leapyears there are exceptions like if 3.104 is feeling fluish AND 2.87 is unavailable on a leap year then 1=1 if it's a normal year 1=2.
 
turingtest said:
In kind, you mean? I can't match your deepity, but how about this? I give you Life, the universe, and everything, in one easy-to-grasp diagram-
[qimg]http://www.internationalskeptics.com/forums/picture.php?albumid=1257&pictureid=10847[/qimg]
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Patterns. Definitely patterns...
 
Now, let's look at Fuzzy Logic unbounded logical tree.

Also in this case we can define pairs of complements, for example:

Code: 
         COMPLEMENTS
┌───────────────────────────┐
│                           │
│        COMPLEMENTS        │
│      ┌─────────────┐      │
│      │             │      │
│      │             │      │
0______.      .      .______1

       A      C      B

In this example 0 (contradiction) and 1 (tautology) are pair of complements, and so is pair A,B (A,B is the notation of any arbitrary pair of complements between pair 0,1, which is not pair 0,1 AND not "pair" C,C , where C is the logical connective between 0 and 1 that is its own complement).

The number of arbitrary A,B pairs is indeterminable since they are logically not pair 0,1 AND not "pair" C,C.


In terms of the notion of set, 1 (tautology) is equivalent to the outer braces "{" and "}", 0 (contradiction) is equivalent to the void between the outer braces, C notates the notion of the identity of a given member to itself, and A,B is the notation of the notion of the difference between identified members (the void and the outer braces are not members, yet they determine the boundaries of finitely or infinitely many identified members).
 
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doronshadmi said:
Thank you, after all time is not a gap between your two legs, so you must be careful all the time.
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irrefutabllenumberillustration.jpg


This should clear up the confusion you may be having.

Rule number 3⅖: 1 will never equal 4 or any multiple of if the total sum is divisible by 0
 
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jsfisher said:
So can the triadic tri-state functions.
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By Standard Math one uses base 3 without digit 1, for example, in Cantor's set

Cantor_base_3.svg


(in this particular example 1.000...3 is replaced by 0.222...3 etc.) in order to prove that Cantor's set is uncountable ( https://en.wikipedia.org/wiki/Cantor_set#Cardinality ).

By using 3-valude logic as the logical basis of base 3, we get the following unbounded distinguished numbers:
Code: 
Unity                                                            
|\                                                               
| \                                                              
|  \                                                             
|   \                                                            
|    \                                                           
|     \                                                          
|      \                                                         
|       \                                                        
|        \                                                       
|         \                                                      
|          \                                                     
|           \                                                    
|            \                                                   
|             \                                                  
|              \                                                 
|               \                                                
|                \                                               
|                 \                                              
|\                 \                                             
| \                 \                                            
|  \                 \                                           
|   \                 \                                          
|    \                 \                                         
|     \                 \                                        
|      \                 \                                       
|       \                 \                                      
|        \                 \                                     
|         \                 \                                    
|          \                 \                                   
|           \                 \                                  
|            \                 \                                 
|             \                 \                                
|              \                 \                               
|               \                 \                              
|                \                 \                             
|                 \                 \                            
[B]0[/B] ----------------[B]1[/B]-----------------2-----------------Integers   
|[B]\                |[/B]\                |\                           
|[B] \               |[/B] \               | \                          
|  [B]\              |[/B]  \              |  \                         
|  [B] \             |[/B]   \             |   \                        
|   [B] \            |[/B]    \            |    \                       
|     [B]\           |[/B]     \           |     \                      
|\     [B]\          |[/B]\     \          |\     \          Fractions  
| \     [B]\         |[/B] \     \         | \     \                    
|  \     [B]\        |[/B]  \     \        |  \     \                   
|   \     [B]\       |[/B]   \     \       |   \     \                  
|    \     [B]\      |[/B]    \     \      |    \     \                 
|     \     [B]\     |[/B]     \     \     |     \     \                
0     1     [B]2     0[/B]     1     2     0     1     2                
|\    |\    |[B]\    |[/B]\    |\    |\    |\    |\    |\               
| \   | \   | [B]\   |[/B] \   | \   | \   | \   | \   | \              
|\ \  |\ \  |\ [B]\  |[/B]\ \  |\ \  |\ \  |\ \  |\ \  |\ \             
| \ \ | \ \ | \ [B]\ |[/B] \ \ | \ \ | \ \ | \ \ | \ \ | \ \            
0 1 2 0 1 2 0 1 [B]2 0[/B] 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2            
                                                                 
                        . . .

0.222... is logically not 1.000... , so by my direct (without free variables) logical framework 0.222...3 ≠ 1.000...3 , 0.0222...3 ≠ 0.1000...3 etc. (moreover, any irrational number between 0 and 1 that is represented by base 3, is included in this 3-valued logic unbounded tree) so I have more numbers than the standard framework, which means that even by Standard Math, the number of the unbounded distinguished branches of the tree of 3-valued logic, is uncountable.
 
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doronshadmi said:
0.222... is logically not 1.000...
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In base three, sure it is. Your continued denial doesn't change that basic fact of Mathematics. Even your fascination with abusing the term, free variable, has no impact on the identity you reject without basis.

...
moreover, any irrational number between 0 and 1 that is represented by base 3, is included in this 3-valued logic unbounded tree) so I have more numbers than the standard framework, which means that even by Standard Math, the number of the unbounded distinguished branches of the tree of 3-valued logic, is uncountable.
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More numbers than the standard framework? You continue to just make stuff up.
 
jsfisher said:
In base three, sure it is. Your continued denial doesn't change that basic fact of Mathematics. Even your fascination with abusing the term, free variable, has no impact on the identity you reject without basis.



More numbers than the standard framework? You continue to just make stuff up.
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jsfisher, by using Cantor's diagonal argument along the 2-valued logic unbounded tree

Code: 
Unity
|\
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
|         \
|          \
|           \
|            \
|             \
|              \
0---------------1---------------Integers
|\              |\
| \             | \
|  \            |  \
|   \           |   \
|    \          |    \
|     \         |     \         Fractions
|      \        |      \
0       1       0       1
|\      |\      |\      |\
| \     | \     | \     | \
|  \    |  \    |  \    |  \
0   1   0   1   0   1   0   1
|\  |\  |\  |\  |\  |\  |\  |\
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

              ...

please show some path (which is also some number between 0.000...2 and 1.111...2 that is not in that tree.

If you can't do it, it means that the cardinality of the distinguished paths is at least (according to the notion of Standard Math) of an uncountable collection of distinguished paths.
 
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doronshadmi said:
jsfisher, by using Cantor's diagonal argument along the 2-valued logic unbounded tree...please show some path (which is also some number between 0.000...2 and 1.111...2 that is not in that tree.
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Why am I restricted to a diagonal argument? Be that as it may, you were the individual who made a claim; you are the individual responsible for its proof. "Prove me wrong" might apply after you've made your claim credible, but not before.

And be that as it may...
If you can't do it, it means that the cardinality of the distinguished paths is at least (according to the notion of Standard Math) of an uncountable collection of distinguished paths.
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...isn't a valid prove method ("I'm right 'cause you didn't prove me wrong"), nor does it address the claim at issue right now. It was this:
I have more numbers than the standard framework
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Start there. We await your proof.
 
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jsfisher said:
Why am I restricted to a diagonal argument?
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Because you are the one who claimed (in http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965) about 2-valued unbounded logical tree that "all dyadic boolean functions can be enumerated" (where being enumerated means that a given collection has at most countably infinite distinct paths, as seen here:

Some authors use countable set to mean countably infinite alone.[1] To avoid this ambiguity, the term at most countable may be used when finite sets are included and countably infinite, enumerable,[2] or denumerable[3] otherwise.
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( https://en.wikipedia.org/wiki/Countable_set )

).

In that case please use the diagonal argument in order to provide a distinct unbounded path that is not in that tree, in order to prove that there are only countably infinite (or enumerable) distinct paths in the 2-valued unbounded logical tree.
 
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jsfisher said:
"Prove me wrong" might apply after you've made your claim credible, but not before.
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The 2-valued unbounded logical tree has only paths that are distinguished of each other (it is a self evident logical fact).

Start there. We await your proof that the 2-valued unbounded logical tree is at most enumerable.
 
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