Deeper than primes

[The Man]

By exactly how much?

What “By exactly how much?”. Are the simple relational and conditional statements in that quote just too complex for you to actually apply?

EDIT: (you still do not get your own question and my answer in http://www.internationalskeptics.com/forums/showpost.php?p=6349487&postcount=11598).

You gave no answer, just a question indicating that you did not read or simply did not understand the reference you later cited in support of a false statement related to your question.

So Doron show us that you actually read and understood the reference you cited by showing where it claimed anything about “no definable result to aleph1 - aleph0” as you claimed in your post that cited that reference?

 

[doronshadmi]

Are the simple relational and conditional statements in that quote just too complex for you to actually apply?

Is the simple x < ∞ just too complex for you to actually apply?

You gave no answer, just a question indicating that you did not read or simply did not understand the reference you later cited in support of a false statement related to your question.

This question is exactly the answer, which you don't get.

 

[doronshadmi]

It will be unique (and equal to σ) if and only if μ < σ.

The Man, since μ can be also infinite, then if, for example, σ=aleph1 and μ=aleph0, then what is the exact result of aleph1 - aleph0 ?

Another question: what is the exact result of aleph0-aleph0 ?

 

[epix]

You have missed my claim.

My claim is that "=" and "A" are not the same thing ("=" represents the observer and "A" represents the observed) and because of this difference observation is possible.

See what happens when claims take over definitions?

Originally Posted by doronshadmi
If the observer and the observed are identical, then there is no observation, and the expression "under observation" is false.

If there is no observation that builds premises, then there is no conclusion. What decides that the expression under no observation is false? What is the "concluder?"

The standard math allows to look at the mirror and conclude. That's why organic mathematicians must grow beards and they do it the healthy way -- no chemical fertilizers. BUT...!

The Barber paradox is a puzzle derived from Russell's paradox. It was used by Bertrand Russell himself as an illustration of the paradox, though he attributes it to an unnamed person who suggested it to him. It shows that an apparently plausible scenario is logically impossible.

Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men in town who do not shave themselves.

Under this scenario, we can ask the following question: Does the barber shave himself?

Asking this, however, we discover that the situation presented is in fact impossible:

If the barber does not shave himself, he must abide by the rule and shave himself.
If he does shave himself, according to the rule he will not shave himself.

Hence the organic beards?

 

[doronshadmi]

Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men in town who do not shave themselves.

Under this scenario, we can ask the following question: Does the barber shave himself?

Asking this, however, we discover that the situation presented is in fact impossible:

If the barber does not shave himself, he must abide by the rule and shave himself.
If he does shave himself, according to the rule he will not shave himself.

This scenario is based on the assumption that the shaved (the beard) is identical to the shaver, which is a false assumption.

Since any member (where one of them is the beard) of "that has no successor" is not identical to "that has no successor" then that has ∞ "magnitude of existence" (the shaver), is not identical to that has x "magnitude of existence" (the beard).

Again, the beard belongs to the shaver but it is not identical to the shaver.

By understanding x < ∞, there is no paradox.

 

[doronshadmi]

If the axiom of choice holds and given an infinite cardinal σ and a cardinal μ, there will be a cardinal κ such that μ + κ = σ if and only if μ ≤ σ. It will be unique (and equal to σ) if and only if μ < σ.

( http://en.wikipedia.org/wiki/Cardinal_arithmetic#Subtraction )

So κ is unique only if κ = σ and μ < σ. In that case k does not change the fact that μ < σ.

The Man, the result of κ = σ has nothing to do with result of σ - μ, so please tell us what is the exact result of σ - μ?

 

[doronshadmi]

Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men in town who do not shave themselves.

Under this scenario, we can ask the following question: Does the barber shave himself?

Asking this, however, we discover that the situation presented is in fact impossible:

If the barber does not shave himself, he must abide by the rule and shave himself.
If he does shave himself, according to the rule he will not shave himself.

There is another solution of the barber's paradox that usually is understood as a contradiction (The barber shaves AND does not shaves himself, according the rule).

According to QM two opposite states can be in a superposition. In that case "The barber shaves AND does not shave himself" describes the barber in terms of superposition, and by this term there is no paradox.

Russel's paradox is a direct result of distinct identities that share the same level of existence, and this kind of existence (called "flat-land" by me) is definitely not a general model for any possible universe.

In other words, Russel's paradox is limited to "flat-land" universe.

 

[The Man]

The self-contradictory, nonsense-only, deliberate ignorance-only, is a direct result of your relative-only framework, that can't get totalities like "that has no successor or "that has no predecessor".

Another edict of Dorons fiat-land, No Doron it is just a result of your self-contradictory and nonsense-only assertions that evidently result only from your deliberate ignorance.

In that case what you call trunks is at the level of the branches, where what you call roots is the level beyond them.

Doron, roots branch as well so they are certainly not “beyond” branching. Face it, your tree analogy just stinks and your ‘levels’ are just your own subjective fiat-land edicts.

Is the simple x < ∞ just too complex for you to actually apply?

Doron it is simply your fiat-land edict, no one need apply it and it is only you that chooses to simply insist upon it.

Is
This question is exactly the answer, which you don't get.

Doron a question is not an answer, which evidently you just don’t want to “get”. If you can not answer the question just say so.

The Man, since μ can be also infinite, then if, for example, σ=aleph1 and μ=aleph0, then what is the exact result of aleph1 - aleph0 ?

Doron you just quoted the part of the reference that tells you the result, apply it.

Another question: what is the exact result of aleph0-aleph0 ?

The reference you cited before gives that information, but unfortunately you still evidently just don’t understand it.

( http://en.wikipedia.org/wiki/Cardinal_arithmetic#Subtraction )

So κ is unique only if κ = σ and μ < σ.

No

It will be unique (and equal to σ) if and only if μ < σ

Read what you quoted

In that case k does not change the fact that μ < σ.

What do you mean “does not change the fact that μ < σ”?

if and only if μ < σ

It is specifically dependent on that fact.

The Man, the result of κ = σ has nothing to do with result of σ - μ, so please tell us what is the exact result of σ - μ?

Doron “κ = σ” is the “result of σ – μ” “if and only if μ < σ”

You have conclusively demonstrated that reading and applying very simple statements is just beyond your abilities. So it is no wonder that you create your own fiat-land as anything else is evidently simply beyond your comprehension.

 

[doronshadmi]

Doron “κ = σ” is the “result of σ – μ” “if and only if μ < σ”

Look what you actually say:

EDIT:

If infinite cardinal σ > cardinal μ , then κ=σ, σ – μ = σ + μ = σ, or in other words, "+" and "-" are indistinguishable operations.

What is the meaning of the word "subtraction" if a "result" is defined by indistinguishable operations, where σ > μ?

In other words, in flat-land one using opposite AND indistinguishable operations between different sizes, in order to get "results".

The reason is very simple, flat-land is a relative-only framework (no totalities) so it extremely sensitive to incompleteness (it can't deal with real uncertainty), so it forces strict results even if "opposite AND indistinguishable" operations are used.

 

[doronshadmi]

Doron, roots branch as well so they are certainly not “beyond” branching.

You certainly do not get Y beyond flat-land. Face it.

 

[doronshadmi]

Doron it is simply your fiat-land edict, no one need apply it and it is only you that chooses to simply insist upon it.

Let's clarify something about http://www.internationalskeptics.com/forums/showpost.php?p=6353058&postcount=11629.

0-0=0+0=0 where "+" and "-" are distinguishable (opposites) because the operations are used on cardinality that is not a successor of any cardinality.

This is not the case with σ, which is a successor of smaller cardinals (μ < σ) and in this case "+" and "-"
are both distinguishable AND indistinguishable, which is a contradiction.

In other words The Man, you insist about contradiction, in your flat-land.

--------------------------------

EDIT:

As for cardinals 0 and ∞ :

That has cardinality ∞ is non-local w.r.t anything that has smaller cardinality.

That has cardinality 0 is local w.r.t anything that has greater cardinality.

By understanding 0 < x < ∞, we get:

{
{},{{}}
}

where the outer "{" "}" is "that has no successor" (the totally strong space).

{} represents 0 dimensional space, and {{}} represents 1 dimensional space.

Accordinng to the given example, it is clearly seen that the 1 dimensional space is not less than included AND excluded w.r.t 0 dimensional space (it is non-local w.r.t 0 dimensional space, and it is valid because {} is not nothing), where 0 dimensional space is at most included w.r.t 1 dimensional space (it is local w.r.t 1 dimensional space).

 

[doronshadmi]

Doron, roots branch as well so they are certainly not “beyond” branching. Face it, your tree analogy just stinks and your ‘levels’ are just your own subjective fiat-land edicts.

In flat-land one can't distinguish between analogy of X and X, as clearly seen in The Man's case.

 

[The Man]

Look what you actually say:

EDIT:

If infinite cardinal σ > cardinal μ , then κ=σ, σ – μ = σ + μ = σ, or in other words, "+" and "-" are indistinguishable operations.

What is the meaning of the word "subtraction" if a "result" is defined by indistinguishable operations, where σ > μ?

In other words, in flat-land one using opposite AND indistinguishable operations between different sizes, in order to get "results".

The reason is very simple, flat-land is a relative-only framework (no totalities) so it extremely sensitive to incompleteness (it can't deal with real uncertainty), so it forces strict results even if "opposite AND indistinguishable" operations are used.

No Doron the operations remain “distinguishable”, even if they can give the same results.

http://en.wikipedia.org/wiki/Cardinal_arithmetic#Cardinal_addition

Tell us Doron what is the result of ∞ + 5?

How about ∞ - 5?

You certainly do not get Y beyond flat-land. Face it.

You certainly don’t get anything outside of your fiat-land, and you just can’t face that.

Let's clarify something about http://www.internationalskeptics.com/forums/showpost.php?p=6353058&postcount=11629.

0-0=0+0=0 where "+" and "-" are distinguishable (opposites) because the operations are used on cardinality that is not a successor of any cardinality.

This is not the case with σ, which is a successor of smaller cardinals (μ < σ) and in this case "+" and "-"
are both distinguishable AND indistinguishable, which is a contradiction.

One typical of your OM nonsense, again that the results can be the same does not make the operations “indistinguishable”.

Tell us Doron what is the result of 5+0?
How about 5-0?

In other words The Man, you insist about contradiction, in your flat-land.

Nope Doron, again that is simply your OM that “insist about contradiction”.

--------------------------------

EDIT:

As for cardinals 0 and ∞ :

That has cardinality ∞ is non-local w.r.t anything that has smaller cardinality.

That has cardinality 0 is local w.r.t anything that has greater cardinality.

By understanding 0 < x < ∞, we get:

{
{},{{}}
}

where the outer "{" "}" is "that has no successor" (the totally strong space).

{} represents 0 dimensional space, and {{}} represents 1 dimensional space.

Accordinng to the given example, it is clearly seen that the 1 dimensional space is not less than included AND excluded w.r.t 0 dimensional space (it is non-local w.r.t 0 dimensional space, and it is valid because {} is not nothing), where 0 dimensional space is at most included w.r.t 1 dimensional space (it is local w.r.t 1 dimensional space).

“1 dimensional space is not less than included AND excluded w.r.t 0 dimensional space”

See your OM nonsense specifically requires contradictions to be valid. Again it is quite humorous that you only seem to dislike contradictions when you are trying to ascribe them to someone else.

In flat-land one can't distinguish between analogy of X and X, as clearly seen in The Man's case.

In your fiat-land you can’t (or simply do not want to) distinguish between a good analogy and a bad analogy or between simply making analogies and actually doing some research or learning something.

 

[doronshadmi]

Tell us Doron what is the result of 5+0?
How about 5-0?

In both cases you have used cardinality that is not used as a successor of any cardinality.

Are you going to tell us that in flat-land μ is not a successor of any cardinality?

You still do not get that by your ridicules reasoning "-" and "+" are distinct operations only if μ = 0 (the cardinality that is not used as a successor of any cardinality).

Since this is not the case (μ can be also any cardinal > 0), your example does not hold water.

You are still using distinct AND non-distinct operations, which is a contradiction.

Flat-land is indeed an "interesting" framework, isn't it The Man?

Tell us Doron what is the result of ∞ + 5?

How about ∞ - 5?

Since when you are using ∞ as a unique number?

If you mean ∞ + 5, then since ∞ does not have a successor then this expresssion is invalid.

Also 0 - 5 is an invalid expresssion if 0 is a cardinal.

See your OM nonsense specifically requires contradictions to be valid. Again it is quite humorous that you only seem to dislike contradictions when you are trying to ascribe them to someone else.

Since Non-locality is different than Locality, there is no contradiction.

But in your case "+","-" operations (σ-0 is excluded) are distinct AND non-distincet, which is a contradiction.

 

[jsfisher]

This part of your reply is ridicules.

I was just citing your response to my comment. How does quoting your posts make my post ridiculous?

Perhaps you should stop wasting so much effort to claim that what you don't get is a lie , by simply get out of your box.

Perhaps you should worry less about where you think others must be (which you have been fabricating) and focus on finding something that makes Doronetics worth all this attention you've given it.

Blaming everyone else doesn't build your case.

Surely, you can define something, no?
Surely, you can demonstrate Doronetics superiority, no?

So far, all we have seen from you is vagaries, contradictions, and bold assertions without backing. Surely, you can do better, no?

 

[epix]

Quote:
Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men in town who do not shave themselves.

Under this scenario, we can ask the following question: Does the barber shave himself?

Asking this, however, we discover that the situation presented is in fact impossible:

If the barber does not shave himself, he must abide by the rule and shave himself.
If he does shave himself, according to the rule he will not shave himself.

This scenario is based on the assumption that the shaved (the beard) is identical to the shaver, which is a false assumption.

The word "beard" never appears in the description of the paradox, so any assumption based on that word doesn't exist. The paradox is based on the existence of a particular set, which you cannot render, coz if you could comprehend the logic of the paradox, you would be able to do it. I bet that you won't be able to come up with set A that describes the scenario.

So your "axiomatic framework" holds '=' as an observer, but you still haven't come up with the "concluder."

A guy looks at the mirror whether he needs a shave or not. Can OM come up with the conclusion?
Under no circumstances it can, coz the axiomatic foundation of OM holds the observer and the observed identical and therefore there is no observation with the implication that the guy is a vampire that can't see itself and be seen by others in the mirror. And that's the limitation OM has. Actually it is the virtue of OM: it is dead but at the same time it is alive, like a vampire. That's the paradox that supports the axiomatic framework of OM. Proof:

Q: Let {O M} be a subset of a set. What is the set?

A: The set is {{Z {O M} B I E}} of the famous dead/alive paradox.

 

[Robin]

This scenario is based on the assumption that the shaved (the beard) is identical to the shaver,

It is clearly not based on this assumption.

 

[Robin]

There is another solution of the barber's paradox that usually is understood as a contradiction (The barber shaves AND does not shaves himself, according the rule).

According to QM two opposite states can be in a superposition. In that case "The barber shaves AND does not shave himself" describes the barber in terms of superposition, and by this term there is no paradox.

Russel's paradox is a direct result of distinct identities that share the same level of existence, and this kind of existence (called "flat-land" by me) is definitely not a general model for any possible universe.

In other words, Russel's paradox is limited to "flat-land" universe.

So you can solve the paradox by removing the tautology "( x implies not x ) implies not x".

Fine. That was the easy part.

Now rewrite the 2,300 years (at least) of mathematics you have just wiped out.

 

[doronshadmi]

The word "beard" never appears in the description of the paradox, so any assumption based on that word doesn't exist.

It does not matter, you can replace "beard" by "shaves", and in this case "shaves" is not identical to "barber".

 

[doronshadmi]

More about http://www.internationalskeptics.com/forums/showpost.php?p=6354803&postcount=11634.

0 < x < ∞

X is a placeholder for 0, x or ∞



Definition C: "That has no predecessor is not a successor".

Example: X+0=X

Details:

5+0=5, where 0 is not a successor.

|infinite collection|+0=|infinite collection|, where 0 is not a successor.

∞+0=∞, where 0 is not a successor.



Definition D: "That has no successor is not a predecessor".

Example: ∞-X=∞

Details:

∞-5=∞, where ∞ is not a predecessor.

∞-|infinite collection|=∞, where ∞ is not a predecessor.

∞-∞=∞, where ∞ is not a predecessor.