Deeper than primes

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Let us use an analogy taken from a role film.

1 is a one second, and it is used as a domain that determines the number of frames per second.

This 1 second domain is equivalent to [0,1] interval, where each different frame is a member of this interval.

We run the film for exactly one second, where 0 is the initial frame (the initial episode) and 1 id the last episode.

If there is no episode after 0 episode, then after 1 second we are still at episode 0 (nothing was changed).

If we have at lest two episodes, then after 1 second we will be in another episode, which is not episode 0.

The normal frequency of "continuous" episodes is 24 frames per second (each episode is changed after 1/24 second).


If the frames were recorded in less than 24 episodes per second, then we get a Charily Chaplin movie, because there are bigger changes between episodes, that still are projected in a frequency of 24 frames per second.

If the frames that were recorded in less than 24 episodes per second, are projected in less than 24 episodes per second, we get a jumpy (discontinuous) movie.


If the frames were recorded in more than 24 episodes per second, then we get a Slow Motion movie, because there are smaller changes between episodes, that still are projected in the a frequency of 24 frames per second.

If the frames that were recorded in more than 24 episodes per second, are projected in more than 24 episodes per second, we get a High Definition movie.


If the frames were recorded in infinitely many episodes per second, then we get a No Motion movie, because there are 0 changes between episodes, that still are projected in the a frequency of 24 frames per second .

If the frames that were recorded in infinitely many episodes per second, are projected in infinitely many episodes per second, we still get a No Motion movie, because there are 0 changes between episodes.
 
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And you have to deal with the non-finite in X < Y case, exactly as you do in the case of the sum of infinitely many things.

If you do not do that then your reasoning does not deal with the non-finite, and this is exactly my argument about your proof by contradiction.

It simply does not deal with the non-finite, and therefore cannot be used in order to conculde anything about the non-finite X < Y case.

There is no requirement to deal with infinitely many cases. Just one case is sufficient to identify a contraction. Any additional cases would be redundant.
 
Jsfisher, your proof by contradiction is based on these trivial facts:

Any ordered collection of non-finite Q or R members does not have the largest member, if you take some object along the real line and exclude it from the collection.

In that case, this excluded object is > any member of the ordered collection.

Words mean things, doron, and what you have said here is trivially and colossally wrong. Consider the set {M : 1 <= M <= 2}. The number 0.5 is excluded from the set, yet the set has a largest member.

What did you really mean to say?
 
Words mean things, doron, and what you have said here is trivially and colossally wrong. Consider the set {M : 1 <= M <= 2}. The number 0.5 is excluded from the set, yet the set has a largest member.

What did you really mean to say?

doronshadmi said:
In that case, this excluded object is > any member of the ordered collection.

You know exactly that we are talking about "<Y" of [X,Y), so save your words for this case, exactly because words mean things.
 
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There is no requirement to deal with infinitely many cases. Just one case is sufficient to identify a contraction. Any additional cases would be redundant.

Your single case (where h has a room between Z and Y) is possible exactly because you do not deal with the non-finite.

There is no redunndancy here because your proof by contradiction is not related at all to [X<Y) non-finite interval.
 
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Your single case (where h has a room between Z and Y) is possible exactly because you do not deal with the non-finite.

That's a tautology.

ETA: Well, no, not really a tautology. Yes, a single case is not an infinite one - that part is tautological - but the former isn't "possible exactly because" of the latter. That part is just nonsense.​

There is no redunndancy here because your proof by contradiction is not related at all to [X<Y) non-finite interval.

You keep repeating the same misinformation as if it were true and significant. It is neither.

You also continue to abuse the language and notation.
 
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You know exactly that we are talking about "<Y" of [X,Y), so save your words for this case, exactly because words mean things.

Well, if you meant "the set of real numbers less than some number, Y, does not have a largest element" than just say that. You were the one that concocted the rather contorted "any ordered collection of non-finite Q or R members does not have the largest member, if you take some object along the real line and exclude it from the collection."

Your statement was convoluted and, more importantly, blatantly wrong. Since a correct, clear, and even shorter wording was available, why did you go for the twisty maze option?
 
That part is just nonsense.[/indent]
No, your first part, where you claim that the finite single case of X<h<Y can be used in order to conclude something about, [X,Y), this is nonsense.
 
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Your statement was convoluted and, more importantly, blatantly wrong. Since a correct, clear, and even shorter wording was available, why did you go for the twisty maze option?
In that case, this excluded object is > any member of the ordered collection.
It is simple and clear enough, but as usual you try to take the focus away from the failure of your "proof by contradiction" that does not hold water ( as clearly shown in http://www.internationalskeptics.com/forums/showpost.php?p=4783540&postcount=3380 ), because your Z<h<Y construction has nothing to do with the non-finite [X,Y) interval.
 
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No, your first part, where you claim that the finite single case of X<h<Y can be used in order to conclude something about, [X,y), this is nonsense.


I didn't try to conclude anything about [X,Y). That's another of your misrepresentations.

However, this still comes back to your baseless assertion about whether conclusions from finite examples can be drawn about infinite sets. Are you ever going to support your assertion with, you know, facts?

Also, if, as you insist, 0.999... < 1, then what is 1 - 0.999...? And if we let d = 1-0.999..., what would 0.999... + d/2 be?
 
I didn't try to conclude anything about [X,Y). That's another of your misrepresentations.
Really??

jsfisher said:
http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864

Assuming X < Y < Z, and any reasonable definition for "immediate successor", then, like it or not, doron, Y is in fact an immediate successor to [X,Y)

The last dialog actually started because you claimed that Y is an immediate successor of [X,Y). Go on jsfisher and continue to air your lies.
 
It is simple and clear enough, but as usual you try to take the focus away from the failure of your "proof by contradiction" that does not hold water ( as clearly shown in http://www.internationalskeptics.com/forums/showpost.php?p=4783540&postcount=3380 ), because your Z<h<Y construction has nothing to do with the non-finite [X,Y) interval.

Again [X,Y) is a finite interval as is (X,Y) as well as (Z,Y) with h as some member of that interval, which would be your Z<h<Y. Your whole notion seems to be you just simply and continually denying the fact that a finite interval (in the reals) would have a infinite amount of real numbers as possible values for h. Along with just mixing and matching your variables (Z instead of X) to make yourself believe you are talking about something different.
 
jsfisher said:
Also, if, as you insist, 0.999... < 1, then what is 1 - 0.999...? And if we let d = 1-0.999..., what would 0.999... + d/2 be?
This is another twisted way to get away from your failure, becuae you clearly know that we are talking here only about Standard Math framework where 0.999...=1.
 
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Again [X,Y) is a finite interval

Ok, I used the wrong name, but not names are important here but the fact that [X,Y) has infinitely many elements that we have to deal with.


EDIT: ( actually the name non-finite interval is ok for [X,Y) and not ok for [X,Y] )
 
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The Man said:
Your whole notion seems to be you just simply and continually denying the fact that a finite interval (in the reals) would have a infinite amount of real numbers as possible values for h.
Nonesene.

You simply do not get the difference between infinitely many finite cases (where each h is such a case that is < Y) and the real non-finite case, where "<Y" does not hold (exactly as can be found in the case of 0.9+0.09+0.009+...=Y=1).
 
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Ok, I used the wrong name, but not names are important here…

Names are always important here, specifically because that is how you seem to continue to confuse yourself.

…but the fact that [X,Y) has infinitely many elements that we have to deal with.

Right and they can all be represented by the variable h where X £ h < Y.

Nonesene.

You simply do not get the difference between infinitely many finite cases (where each h is such a case that is < Y) and the real non-finite case, where "<Y" does not hold (exactly as can be found in the case of 0.9+0.09+0.009+...=Y=1).

What the heck do you think you’re talking about? Each of your numbers “0.9”, “0.09”, “0.009” to “…” are each a finite number and thus a finite case that is < 1 (or Y) . Your sum “0.9+0.09+0.009+...=Y=1” as an infinite series is “infinitely many finite cases” (each a finite number and thus a finite case that is < 1 (or Y)) added together. You simple confirm my above assertion that you use different names (“infinitely many finite cases” or “the real non-finite case”) to confuse yourself about what you are talking about.
 
I didn't try to conclude anything about [X,Y). That's another of your misrepresentations.
Really??

Yes, really. The proof showed that the set {X : X < Y} has no largest member. There was no half-open finite interval [X, Y) anywhere in the proof.

jsfisher said:
http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864

Assuming X < Y < Z, and any reasonable definition for "immediate successor", then, like it or not, doron, Y is in fact an immediate successor to [X,Y)

The last dialog actually started because you claimed that Y is an immediate successor of [X,Y). Go on jsfisher and continue to air your lies.

Sure. Note the qualifier, 'any reasonable definition for "immediate successor"'. Further clarity was provided in a later post as to how the successor (predecessor) concept would be extended to include intervals along with the reals.

Are you unclear as to how that would work? It is really very simple.
 
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Yes, really. The proof showed that the set {X : X < Y} has no largest member. There was no half-open finite interval [X, Y) anywhere in the proof.
Is Y is a mamber of set {X : X < Y}?
Please answer by yes or no.


Sure. Note the qualifier, 'any reasonable definition for "immediate successor"'. Further clarity was provided in a later post as to how the successor (predecessor) concept would be extended to include intervals along with the reals.

Are you unclear as to how that would work? It is really very simple.
No, you are unclear of how that would works, so please direct us to some professional source that clearly talks about Y as am immediate successor of [X,Y) (as you claim).
 
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