Deeper than primes

[sympathic]

Yes, that is a sound advice. However, Doron never got past step 1 and, if the last 340 or so pages are any indication, he never will.

Also: you are playing right into his hands, and he would actually take that as a compliment. He would say that Copernicus was also considered mad.

 

[laca]

Also: you are playing right into his hands, and he would actually take that as a compliment. He would say that Copernicus was also considered mad.

Yes, probably. Although there is a difference between being considered mad and actually being mad...

 

[sympathic]

Yes, probably. Although there is a difference between being considered mad and actually being mad...

I agree completely. I think Doron has considerable issues with the perception of reality. The question is: why do we bother with him? It is like trying to explain to someone that the skies are blue when he continues to insist they are pink.

 

[doronshadmi]

You have not explained how your diagonal element would be constructed.

Again, it is done in http://www.internationalskeptics.com/forums/showpost.php?p=6820757&postcount=14148.

Why can't you get it?

 

[doronshadmi]

Ironically, your <0,1> argument calls for an isolation of the square matrix from the whole |2^m| set

Not Ironically m² is a partial case (a part of) 2^m so it is definitely not isolated from 2^m.

Furthermore, the objects of m² and 2^m share the same form (if translatable to <0,1> form).

 

[zooterkin]

Again, it is done in http://www.internationalskeptics.com/forums/showpost.php?p=6820757&postcount=14148.

No, it's not. Do you really think someone could apply your diagonal method from reading that post?

 

[doronshadmi]

If a lot of independent people disagree with you

This is not the case here, because the people here share a common paradigm about the considered subjects, and therefore they are not independent.

The rest of Mirrorglass propositions are irrelevant in this case, because we are talking about not less than a paradigm-shift of this common paradigm.

 

[doronshadmi]

No, it's not. Do you really think someone could apply your diagonal method from reading that post?

Yes, by using also the link of this post.

 

[zooterkin]

Yes, by using also the link of this post.

No, because that post only has an example, no instructions. From that example it is clear you don't really know what you're doing. Why, for example, do you choose more bits than are necessary to represent the number of elements in the power set?

 

[doronshadmi]

No, because that post only has an example, no instructions.

You are wrong, the diagonal is explicitly constructed, and it is done such that P(N) members , P(P(N)) members (etc. ad infinitum) are translatable to <0,1> forms.

Why, for example, do you choose more bits than are necessary to represent the number of elements in the power set?

Exactly because it does not matter, after all each <0,1> form is infinitely long (in this case) AND incomplete.

The important fact is the existence of the inverse of the diagonal beyond the range of P(N) , P(P(N)) ... etc. ... ad infinitum ...

 

[jsfisher]

You are wrong, the diagonal is explicitly constructed

There is an easy way to prove us wrong, Doron. Simply recite the steps to explicitly construct a diagonal from some collection of these bit maps with which you have become so fond. I bet you cannot, though, since all your referenced post contains is an example with a diagonal marked in a graphic -- no steps, nothing general.

 

[doronshadmi]

There is an easy way to prove us wrong, Doron. Simply recite the steps to explicitly construct a diagonal from some collection of these bit maps with which you have become so fond. I bet you cannot, though, since all your referenced post contains is an example with a diagonal marked in a graphic -- no steps, nothing general.

On the contrary, {},{1,2,3,...} and any given P(N) member between {} and {1,2,3,...} is translatable to <0,1> form and so are P(P(N)) members, etc. ... ad infinitum ... , exactly as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6820757&postcount=14148 .

-- no understanding, nothing general.

 

[jsfisher]

On the contrary, {},{1,2,3,...} and any given P(N) member between {} and {1,2,3,...} is translatable to <0,1> form and so are P(P(N)) members, etc. ... ad infinitum ....

Your reading comprehension limitations are kicking in again. Nobody is asking about how to translate between conventional set notation and these bit maps. That is not the issue before us right now.

The issue is, very simply, given a collection of bitmaps, what are the precise steps one follows to construct the diagonal?

It seems like such a simple question, yet, you, Doron, so far have been incapable of understanding the question let alone answer it.

 

[epix]

Not Ironically m² is a partial case (a part of) 2^m so it is definitely not isolated from 2^m.

Furthermore, the objects of m² and 2^m share the same form (if translatable to <0,1> form).

No, it is not isolated until you cut it out and use it for your own purpose of infecting a finite set with it, thus passing the bug of incompleteness on it. Speaking of incompleteness . . . Some of the arguments that can show you the fallacy of your reasoning may be formulas. But most of them never make it alive to your attention. Look again at what I wrote:

m² < 2^m for any integer m

When things are genuinly incomplete, you don't intervene, such as in this case. Here is the complete version:

m² < 2^m for any positive integer m except 2, 3, and 4.

So the evidence of you not being able to consider and evaluate counter-arguments is aproaching its final stages. But it may take another two years before you run out material that builds the impression that there may be a tiny speck of credence in your jolly constructs. Apart from this, you made a significant contribution to the A/D argument -- quite unexpectedly.

 

[doronshadmi]

The issue is, very simply, given a collection of bitmaps, what are the precise steps one follows to construct the diagonal?

It seems like such a simple question, yet, you, Doron, so far have been incapable of understanding the question let alone answer it.

The construction of the inverse of any diagonal is very precise, and it is done along P(N) members, P(P(N)) members, etc. ... ad infinitum, which are translatable to <0,1> form.

Yet, you, jsfisher, so far have been incapable of understanding the answer exactly because your indices=natural numbers illusion prevents from you to get http://www.internationalskeptics.com/forums/showpost.php?p=6820757&postcount=14148.

 

[doronshadmi]

m² < 2^m for any positive integer m except 2, 3, and 4.

This is a false proposition.

Anyway:

|N| < |x|

x² < 2^x, where x² is not isolated from 2^x, otherwise you can't conclude that x² < 2^x.

Furthermore, x² is not the same as 2^x, and this conclusion can't be concluded if x² is isolated from 2^x, and vice versa.

 

[jsfisher]

The construction of the inverse of any diagonal is very precise, and it is done along P(N) members, P(P(N)) members, etc. ... ad infinitum, which are translatable to <0,1> form.

Yes, you said that before, and it still doesn't answering the question. Come on. Surely you can just string together a few words to describe step #1. Which element of the collection of bit maps do we consider first when constructing the diagonal?

It also seems that you now insist on having the number of elements in the collection being equal the number of bits in each element. Is that correct? Looks like another hidden assumption to me, Doron. You really aren't detail oriented, now, are you?

 

[doronshadmi]

Which element of the collection of bit maps do we consider first when constructing the diagonal?

It does not matter, only the Distinction among the <0,1> forms is significant.

It also seems that you now insist on having the number of elements in the collection being equal the number of bits in each element. Is that correct?

No, you are wrong again, look:

Why, for example, do you choose more bits than are necessary to represent the number of elements in the power set?

Exactly because it does not matter, after all each <0,1> form is infinitely long (in this case) AND incomplete.

The important fact is the existence of the inverse of the diagonal beyond the range of P(N) , P(P(N)) ... etc. ... ad infinitum ...

Your last, jsfisher, reply reinforces my claim about you in http://www.internationalskeptics.com/forums/showpost.php?p=6828391&postcount=14195 .

 

[The Man]

“X” and “beyond X” are simply not parts of each other.

Nor do they share any “parts” in common and in fact they specifically exclude each other, which again is why they are mutually exclusive.

But if observed from a higher level, then they are parts (or shell we say local) w.r.t to the higher level, where the higher level is the whole (or shell we say non-local) w.r.t “X” and “beyond X”.

The higher level is not a part of "X" or “beyond X”, yet it is used as a common environment for both of them.

Once again just because they can be included as parts of some “common environment for both of them” does not preclude or restrict them in any way from being mutually exclusive, as they in fact are.

Simultaneity means that no step-by-step observation is involved.

Actually it just means ‘at the same time’ it makes no assertions about “step-by-step observation” or what may or may not happen or what has or has not happened at other times.

 

[doronshadmi]

Once again just because they can be included as parts of some “common environment for both of them” does not preclude or restrict them in any way from being mutually exclusive, as they in fact are.

This is the beautiful thing of being mutually exclusive, they are disjoint at their level but also have a common environment for both of them on a higher level.

Actually it just means ‘at the same time’ it makes no assertions about “step-by-step observation” or what may or may not happen or what has or has not happened at other times.

Because there is only sameness about the concept of Time, and therefore no process. It is called parallel observation.