A question on a<b

  • Thread starter Organic
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  • #26
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supppose we have p < q real numbers.

then p < { (a, 0.5(p+q) ) } < q for all a. this gives an infinite collection of connectors between any two real numbers.

do the set of connectors plus reals form a field? if not, since R is a field, i don't think you could call them numbers. http://mathworld.wolfram.com/Field.html

cheers,
phoenix
 
  • #27
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phoenixthoth,

Please read my last post to you, beacuse I updated it, sorry.

I hope my updates helps.


Organic
 
  • #28
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still wondering if the set of connectors plus reals is a field... that will decide whether or not connectors could be considered numbers, i think.
 
  • #29
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Hi Hurkyl,

Thank you for reply.

Are the "connectors" you talking about simply intervals?
The connectors, which are C elements, close the interval between any two different "normal" real numbers, which are D elements (singletons).

Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.

By the way, I hope you have the time to help me in:

https://www.physicsforums.com/showthread.php?s=&threadid=6427

Thank you.


Organic
 
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  • #30
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D + C is a field.

You wrote:
supppose we have p < q real numbers.

then p < { (a, 0.5(p+q) ) } < q for all a. this gives an infinite collection of connectors between any two real numbers.
Let us say that b=0.5(p+q), then between a and b there is an infinite collection of D and C elements, which is a field.


Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.
 
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  • #31
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Hi Anton A. Ermolenko,


By using the empty set (with the Von Neumann Heirarchy), we can construct the set of all positive integers {0,1,2,3,...}:
Code:
0 = { } 

1 = {{ }} = {0}
               
2 = {{ },{{ }}} = {0,1}
  
3 = {{ },{{ }},{{ },{{ }}}} = {0,1,2}

4 = {{ },{{ }},{{ },{{ }}},{{ },{{ }},{{ },{{ }}}}} = {0,1,2,3}

and so on.
So, as you see we can use nothing but the {} to construct numbers, which are not empty.

The common building block -the simplest and invariant(=symmetric) element- is {}.


Organic
 
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  • #32
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that there are infinitely many of them isn't enough.

how is the arithmetic (addition/subtraction, multiplication/division), of connectors defined?

since these are functions with one element...

for example, what are { (a, b) } + { (c, d) }
and { (a, b) } * { (c, d) }?

if this is to be a field extension of R, then it would have to be the case that
{ (a, a) } + { (b, b) } = { (a+b, a+b) }
and
{ (a, a) } * { (b, b) } = { (ab, ab) }
since the connector { (a, a) } is identified with the real number a.

if it can be shown that if this has ring structure (again, see http://www.mathworld.com for def. of ring), can you show it is an integral domain (ie, no zero divisors)? what is the formula for the multiplicative inverse of a connector { (a, b) }?

a question for the long term is that if the connectors plus reals forms a field, then is it algebraically closed?

cheers,
phoenix
 
  • #33
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phoenixthoth ,

You wrote:
...since the connector { (a, a) } is identified with the real number a.
Well {(a,a)} can't be but a D element, so there is no connactor (a C element, which exists only between two different Ds) in {(a,a)}.


Please look at the pdf that I sent to your personal email.

Then, please open and read my informal overview in:

http://www.geocities.com/complementarytheory/CATpage.html

I have a new point of view on the natural numbers, which are the simplest numbers in Math language.

In general I show that any natural number > 1 is several structural variations of the same quantity, where each structure is some tree-like element, constructed by an AND connective between C and D elements.

So any arithmetic operation must be found among those structures,
and there is no meaning to Cs XOR/OR Ds operations.


Please see if I am understood, and reply for any problem.


Thank you.

(By the way, I opned a thread in your forum at:

http://207.70.190.98/scgi-bin/ikonboard.cgi?;act=ST;f=2;t=48 )




Yours,


Organic
 
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  • #34
"However, if I get you right, your precisely empty set is not (also can’t be) an element of any definite superset."

sure it can.
Hi Phoenixthoth.
I still insist, it can’t.

let 0 denote the empty set. {0} is a superset containing 0. in fact, that is the definition of the number 1.
Let’s see. At first you’ve denoted an empty set as zero (but this denotation is not a number), then you’ve substituted this concept by zero (a number!).
The axiomatic theory of sets needs the concept of an empty set so that the result of any operation with sets is to be a set, too, but the results of all operations with sets form a class; this class is neither a set, nor a superset. In general, a class defines only the properties of some objects (which may be sets), not the objects themselves.
Zero (number) in fact can be an element of a numerical set of an algebraic system. However, it is a number – thus, an element of a numerical set.
But, if we keep on going this way, we’ll change the topic.
The idea I’d like to get across with in my message above is that an empty set can’t possess symmetry as it has no elements.

To Organic:

In general, your ideas aren’t new extension of mathematics, but may have application in the computer science.
 
  • #35
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I still insist, it can’t.


Let’s see. At first you’ve denoted an empty set as zero (but this denotation is not a number), then you’ve substituted this concept by zero (a number!).
The axiomatic theory of sets needs the concept of an empty set so that the result of any operation with sets is to be a set, too, but the results of all operations with sets form a class; this class is neither a set, nor a superset. In general, a class defines only the properties of some objects (which may be sets), not the objects themselves.
if you prefer, denote the empty set by e. then {e} is a superset of e containing e.

http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html

here, i'm using the unordered pair axiom (axiom 2) where i'm considering the pair e and e and saying {e, e} is a set. one can then use axiom 1 to show that {e, e} = {e}, so it is a set.




Zero (number) in fact can be an element of a numerical set of an algebraic system. However, it is a number – thus, an element of a numerical set.
in set theory, one defines zero to be the empty set. then it can be shown that all number systems can be built upon this.
But, if we keep on going this way, we’ll change the topic.
The idea I’d like to get across with in my message above is that an empty set can’t possess symmetry as it has no elements.
what is the definition of symmetry, because whatever it is may be vacuously true of the empty set?

cheers,
phoenix
 
  • #36
By using the empty set (with the Von Neumann Heirarchy), we can construct the set of all positive integers {0,1,2,3,...}:
code:

0 = { }
Huh... really? Let’s see. How do you define an empty set? The axiomatic theory of sets (ATS) defines it as:
("õ"="direct product", "!="="not equal", "Å"="direct addition", "Ú"="or")
A=Æ defined as "B(B !=Æ & AõB=Æ & A+B=B & "C(C=A equivalent to CõA=Æ)).
However,
"B(B !=Æ & AõB=Æ & A+B=B & "C(C=A equivalent to CõA=Æ)) Þ A=Æ Ú A="zero divisor", i.e. it is non-empty set.

i.e. without a non-empty set no one from mathematicians can’t defines an empty set. And this is nature of an empty set (or ATS) – it is necessary to definition of operations with sets.
1 = {{ }} = {0}
What it precisely means? Either an empty set is a subset of «1» (but «1» is not a natural number) or it is an element of «1». The ATS doesn’t let us define 1 (natural number) by non-number. Only a map correspond the elements of different nature to each other. And so on...
 
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  • #37
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Huh... really? Let’s see. How do you define an empty set? The axiomatic theory of sets (ATS) defines it as:
("õ"="direct product", "<>"="not equal", "+"="direct addition")
AõA=0, B<>0, AõB=0, A+B=0: A=0.
However,
AõA=0, B<>0, AõB=0, A+B=0: A=0 or A="zero divisor", i.e. it is non-empty set.
using axiom 5 in

http://mathworld.wolfram.com/Zermel...nkelAxioms.html [Broken]

one can define an empty set to be such an x. then you can prove that all empty sets are equal, so it makes sense to give them all one notation.

the word set is undefined.

i.e. without a non-empty set no one from mathematicians can’t defines an empty set. And this is nature of an empty set (or ATS) – it is necessary to definition of operations with sets.

What it precisely means? Either an empty set is a subset of «1» (but «1» is not a natural number) or it is an element of «1». The ATS doesn’t let us define 1 (natural number) by non-number. Only a map correspond the elements of different nature to each other. And so on...
let's look at the first statement and replace "empty set" by the phrase "concept X." i'll also rule out your double negative. it becomes this: without concept X mathematicans can't define concept X. i disagree with this statement. you can certainly define any concept X you like. whether concept X "exists" is another story...

what do you mean when you add (direct sum) and multiply (direct product) sets? is that the same as union and intersection?

the empty set is an element of 1 in set theory.
0 = { }
1 = {0}

n = {n-1}

n+1 = n U {n}
 
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  • #38
if you prefer, denote the empty set by e. then {e} is a superset of e containing e.
Excuse me, is e a subset of {e}, or e an element of {e}?
If e is a subset of {e}, then you’ll not define either {e} or e, because {e}=e, hence, {e, e}={e} imply e either empty set, or zero divisor. Without a non-empty set neither you, nor anybody else will not prove that e is empty set
If e is an element of {e}, then you’ll not prove that e is empty set, because in this case {e, e} always {e, e}.
 
  • #39
Originally posted by phoenixthoth
using axiom 5 in

one can define an empty set to be such an x. then you can prove that all empty sets are equal, so it makes sense to give them all one notation.

the word set is undefined.
Well, then how do you define zero divisor?
 
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  • #40
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Originally posted by Anton A. Ermolenko
Excuse me, is e a subset of {e}, or e an element of {e}?
If e is a subset of {e}, then you’ll not define either {e} or e, because {e}=e, hence, {e, e}={e} imply e either empty set, or zero divisor. Without a non-empty set neither you, nor anybody else will not prove that e is empty set
If e is an element of {e}, then you’ll not prove that e is empty set, because in this case {e, e} always {e, e}.
element.

it's not about proving the set is empty. its emptiness is a postulation.

for any x, {x, x} = {x}
 
  • #42
Originally posted by phoenixthoth
element.

it's not about proving the set is empty. its emptiness is a postulation.

for any x, {x, x} = {x}

0 = { }
1 = {0}

n = {n-1}

n+1 = n U {n}
Huh... well, well, well... I like it. Then following your logic:
0=1=2=3=4=5=6=7=8=9=...="infinity".
You really think so?
 
  • #43
Originally posted by phoenixthoth
using axiom 5 in

0 = { }
1 = {0}

n = {n-1}

n+1 = n U {n}
In the real
f: f({0})=1
The NBG (not ZF, because ZF hasn't the classes, hence, hasn't a the hierarchies) system of axioms.
 
  • #44
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Originally posted by Anton A. Ermolenko
Huh... well, well, well... I like it. Then following your logic:
0=1=2=3=4=5=6=7=8=9=...="infinity".
You really think so?
i don't see how you get 0 = 1 = 2 = ...

0 = { }
1 = {0}

two sets x and y are equal iff (a is an element of x iff a is an element in y).

0 is an element of 1 but 0 is not an element of 0 = { }.

therefore, it is not the case that a is an element of 1 iff a is an element of 0.

therefore, 0 != 1.
 
  • #45
Originally posted by phoenixthoth
i don't see how you get 0 = 1 = 2 = ...

0 = { }
1 = {0}

two sets x and y are equal iff (a is an element of x iff a is an element in y).

0 is an element of 1 but 0 is not an element of 0 = { }.

therefore, it is not the case that a is an element of 1 iff a is an element of 0.

therefore, 0 != 1.
Exactly 0!=1, because 0! is number of permutations (there is an order!) {01}=1
but if {01}=1, then neither {01,02} nor {02,01} is not equal to {01}, because either a set has the order relations between the elements (even if there is only one element), or hasn't the order relations...
In the real your {0} (0 is an element of {0}) is equal to 0õ{1}, otherwise this {0} is imply that 0 is a subset of {0}, hence, {0}=0...
 
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  • #46
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!= means "does not equal."

i said, "therefore, 0 != 1." if ! is factorial, then it would be directly adjacent to the zero, which it is not.

if i meant ! as factorial, the statement 0! = 1 is a non-sequitor from my argument.

the statements above the conclusion demonstrated how the pair 0 and 1 don't fit the definition of set equality.

the conclusion was that the two sets are not equal.

i urge you to check out "elements of set theory" by enderton, "set theory" by stoll, or "axiomatic set theory" by suppes for all the details.
 
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  • #47
Originally posted by phoenixthoth
!= means "does not equal."

i said, "therefore, 0 != 1." if ! is factorial, then it would be directly adjacent to the zero, which it is not.

if i meant ! as factorial, the statement 0! = 1 is a non-sequitor from my argument.

the statements above the conclusion demonstrated how the pair 0 and 1 don't fit the definition of set equality.

the conclusion was that the two sets are not equal.

i urge you to check out "elements of set theory" by enderton, "set theory" by stoll, or "axiomatic set theory" by suppes for all the details.
From the ZF system of axioms may be proved (almost) all mathematics. In spite of the fact that the actual number of axiom is equal to infinity (Z5 and ZF9 are not the axioms, only schemes), but there is no these: {0,0}={0}. Empty set (or zero element) defined as the result of these operations:
B != A, A õ A = A, A õ B= A, A + B = B: A=0.
Otherwise, how do you define a zero divisor or a nilpotent device?
 
  • #48
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Originally posted by Anton A. Ermolenko
From the ZF system of axioms may be proved (almost) all mathematics. In spite of the fact that the actual number of axiom is equal to infinity (Z5 and ZF9 are not the axioms, only schemes), but there is no these: {0,0}={0}. Empty set (or zero element) defined as the result of these operations:
B != A, A õ A = A, A õ B= A, A + B = B: A=0.
Otherwise, how do you define a zero divisor or a nilpotent device?
{0,0} = {0} is not an axiom but it can be proven from the axiom of extensionality.

what are the definitions of A and B?

x is a zero divisor if it is nonzero and there is a nonzero y such that x õ y = 0. (source: http://mathworld.wolfram.com/ZeroDivisor.html )

in your equations above, neither B nor A is conclusively a zero divisor because A = 0. in every ring, B õ 0 = 0; that doesn't make B a zero divisor since then, every element would be a zero divisor.

here's how i would define nilponency:
this is pseudo-code:
let N be given. let m be a natural number and x = 1.
N^1 := N. (by, :=, i mean, "is defined to equal")
1. N^(x+1) := N õ N^x. update x so that x = x+1.
2. repeat step 1 until x = m.
when finished, N^m is defined. intuitively, N^m is
N õ N õ ... õ N, where there are m copies of N.

definition: N is nilpotent if N^m = 0 for some m.
 
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  • #49
Originally posted by phoenixthoth
{0,0} = {0} is not an axiom but it can be proven from the axiom of extensionality.

what are the definitions of A and B?

sets

x is a zero divisor if it is nonzero and there is a nonzero y such that x õ y = 0. [/qoute]
I want to see your definition of a zero divisor by ZF system axioms.

in your equations above, neither B nor A is conclusively a zero divisor because A = 0. in every ring, B õ 0 = 0; that doesn't make B a zero divisor since then, every element would be a zero divisor.
You've misunderstood. I define an empty set (or zero element). In other words, iif B != A, A õ A = A, A õ B= A, A + B = B, then A=0. If B != A, A õ A = A, A õ B= A, A + B != B, then A is a zero divisor. Your postulate {0,0}={0} doesn't allow defining a zero divisor.
 
  • #50
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Originally posted by Anton A. Ermolenko
sets


You've misunderstood. I define an empty set (or zero element). In other words, iif B != A, A õ A = A, A õ B= A, A + B = B, then A=0. If B != A, A õ A = A, A õ B= A, A + B != B, then A is a zero divisor. Your postulate {0,0}={0} doesn't allow defining a zero divisor.
zero divisor isn't a term used in set theory. therefore, there's no need to relate it to ZF axioms. zero divisors occur in rings; investigate how rings develop out of set theory. to write down the ZF axioms and then a sequence of statements leading to the definition of zero divisor would take a while.

are A and B allowed to be any two different sets?

is A õ A = A an assumption or a theorem?

is A õ B = A an assumption or a theorem?

is A + B = B an assumption or a theorem?

from the last equation, that B can be "cancelled" is an assumption. cacellation presumes both that there is a zero element and that all elements have inverses. therefore, this is a circular argument.
 

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