Deeper than primes

zooterkin · Jun 3, 2009

doronshadmi said:
What do you think about this book http://francis.williams.edu/record=b2104607 ?

Should we take that as a "No"?

zooterkin · Jun 3, 2009

doronshadmi said:
Jsfisher by your reply it is clearly understood that there is a common reasoning to both X<Y non-immediate predecessor case and the sum of the members of set {0.9, 0.09, 0.009, …}, exactly because you say that "If 0.999... were not identical to 1, then you'd have an inconsistency between statements (1) and (2)."

Why do you persist in linking two unrelated ideas? Is it that you haven't quite accepted the idea of 0.999... being equal to 1, and are trying to prove "standard maths" wrong by trying to find some unforeseen consequence of it?

jsfisher · Jun 4, 2009

doronshadmi said:
Jsfisher by your reply it is clearly understood that there is a common reasoning to both X<Y non-immediate predecessor case and the sum of the members of set {0.9, 0.09, 0.009, …}, exactly because you say that "If 0.999... were not identical to 1, then you'd have an inconsistency between statements (1) and (2)."

You seem to have taken "immediate predecessor of Y" and replaced it with the gibberish "X<Y non-immediate predecessor case".

Why?

...
Please do not reply to any part of it before you read all of it and then think about it, thank you.

How very instructive of you.

doronshadmi · Jun 4, 2009

jsfisher said:
You seem to have taken "immediate predecessor of Y" and replaced it with the gibberish "X<Y non-immediate predecessor case".

Why?

It seems that you have missed the common reasoning of the difference between:

a) "<Y" (which is the result of infinitely many comparisons with finite cases).

b) "=Y" (which is the result of all non-finite values that are converge to a given limit).

Since your proof by contradiction (Z < h < Y) and "dense" definition (X < Z < Y)
are both equivalent to (a) case, they cannot be used in order conclude anything about (b) case.

My claim is that:

If we use the term all on a non-finite interval, then we deal with a problem that is equivalent to (b) case ("<Y" expression does not hold).

jsfisher · Jun 4, 2009

doronshadmi said:
It seems that you have missed...

No, but you are ignoring the responses you receive. Instead, you just cycle back to one of your baseless assertions, still without any proof.

doronshadmi · Jun 4, 2009

jsfisher said:
No, but you are ignoring the responses you receive. Instead, you just cycle back to one of your baseless assertions, still without any proof.

Here is what we find at wikipadia:

http://en.wikipedia.org/wiki/Continuum_(mathematics)

The term the continuum sometimes denotes the real line. Somewhat more generally a continuum is a linearly ordered set of more than one element that is "densely ordered", i.e., between any two members there is another, and it lacks gaps in the sense that every non-empty subset with an upper bound has a least upper bound.

So, by using an example (based on a finite case) that between a finite amount of two distinct members, there is another member, Standard Math concludes (and I would say guesses) that this is also the case about an interval of the all non-finite elements.

"Any two members" is an (a) reasoning, and so is your proof by contradiction that is based on infinitely many finite Z < h < Y cases.

jsfisher · Jun 4, 2009

doronshadmi said:
Here is what we find at wikipadia:

So, by using an example (based on a finite case) that between a finite amount of two distinct members, there is another member

Ignoring your butchery of terminology, yes, between any two real numbers exists a third real number.

...Standard Math concludes (and I would say guesses)

And when saying that, you would be wrong.

...that this is also the case about an interval of the all non-finite elements.

Here, the butchery has exceeded my parser. "Of the all" is peculiar construct, indeed.

"Any two members" is an (a) reasoning, and so is your proof by contradiction that is based on infinitely many finite Z < h < Y cases.

Nope, just one case.

Be that as it may, you are still working from a baseless assertion. Are you going to support it any way?

doronshadmi · Jun 4, 2009

jsfisher said:
Nope, just one case.

Be that as it may, you are still working from a baseless assertion. Are you going to support it any way?

One finite case, jsfisher, exactly as each member of the set {0.9, 0.99, 0.999, ...} is < 1.

Z < h < Y holds in the same manner as set {0.9, 0.99, 0.999, ...} does not have the largest member, and any member of
set {0.9, 0.99, 0.999, ...} is < 1.

Since we deal with the non-finite, then you have to show that your proof by contradiction holds also in the case that is equivalent to the sum of the non-finite set {0.9, 0.09, 0.009, ...}, exactly because you claim that there is a big difference between sums over finite and infinite sequences (and your proof by contradiction is equivalent to what you call "sums over finite", where any given Z or h w.r.t Y is equivalent to any arbitrary member of the set {0.9, 0.99, 0.999, ...} w.r.t 1).

Jsfisher your proof does not hold exactly because it does not deal with the non-finite in the terms that were provided by you.

jsfisher · Jun 4, 2009

doronshadmi said:
One finite case, jsfisher, exactly as each member of the set {0.9, 0.99, 0.999, ...} is < 1.

Z < h < Y holds in the same manner as set {0.9, 0.99, 0.999, ...} does not have the largest member, and any member of
set {0.9, 0.99, 0.999, ...} is < 1.

No, the contradiction is deduced from one case. Not an infinite number of cases (as is the claim all members of {0.9, 0.99, 0.999, ...} are < 1). Just one.

Since we deal with the non-finite, then you have to show that your proof by contradiction holds also in the case that is equivalent to the sum of the non-finite set {0.9, 0.09, 0.009, ...}

In what way are the two equivalent?

...exactly because you claim that there is a big difference between sums over finite and infinite sequences (and your proof by contradiction is equivalent to what you call "sums over finite", where any given Z or h w.r.t Y is equivalent to any arbitrary member of the set {0.9, 0.99, 0.999, ...} w.r.t 1).

You continue to make baseless assertions. Are you ever going to substantiate any of them?

Jsfisher your proof does not hold exactly because it does not deal with the non-finite in the terms that were provided by you.

I didn't provide an infinite number of terms. The proof doesn't need to deal with an infinite number of terms. Just one example is sufficient.

It is no different than disproving "all apples are red" by exhibiting a single green apple. Just one example is sufficient.

The Man · Jun 4, 2009

doronshadmi said:
One finite case, jsfisher, exactly as each member of the set {0.9, 0.99, 0.999, ...} is < 1.

Z < h < Y holds in the same manner as set {0.9, 0.99, 0.999, ...} does not have the largest member, and any member of
set {0.9, 0.99, 0.999, ...} is < 1.

Since we deal with the non-finite, then you have to show that your proof by contradiction holds also in the case that is equivalent to the sum of the non-finite set {0.9, 0.09, 0.009, ...}, exactly because you claim that there is a big difference between sums over finite and infinite sequences (and your proof by contradiction is equivalent to what you call "sums over finite", where any given Z or h w.r.t Y is equivalent to any arbitrary member of the set {0.9, 0.99, 0.999, ...} w.r.t 1).

Jsfisher your proof does not hold exactly because it does not deal with the non-finite in the terms that were provided by you.

No, Doron the largest member of your first set would be 0.9999.... (given the sequence provided and that it is infinite), which is 1. Unless you are claiming 3 times 1/3 does not equal 1? The rest of your post is just your usual nonsense based on your usual and easily falsifiable misunderstanding of some simple mathematical concept.

jsfisher · Jun 4, 2009

The Man said:
No, Doron the largest member of your first set would be 0.9999....

Well, that's the least upper bound on the membership, but just like +inf doesn't appear in the set of positive integers, 0.999... doesn't appear in the set {0.9, 0.99, 0.999, ...}.

The Man · Jun 5, 2009

jsfisher said:
Well, that's the least upper bound on the membership, but just like +inf doesn't appear in the set of positive integers, 0.999... doesn't appear in the set {0.9, 0.99, 0.999, ...}.

I thought that was what the "..." signified that it was an infinite progression and that 0.999... would be a member. Oh well, least upper bound and not a member, I can buy that. Still it doesn’t change my point that Doron can not argue 0.999... does not equal 1 without arguing that 3 times 1/3 does not equal 1.

doronshadmi · Jun 5, 2009

jsfisher said:
Just one example is sufficient.

This is the whole point.

Your Z < h < Y one example is based on some finite case (of Z and h) w.r.t Y , exactly as 0.99 or 0.999 are some finite cases w.r.t to 1.

In both cases we get "< 1" or "< Y" exactly because both cases do not deal with the non-finite (in the sense of 0.999...).

You have a room for h between Z and Y exactly because you do not deal with the non-finite (in the sense of 0.999...).

In other words, your proof by contradiction does not deal with the non-finite (in the sense of 0.999...).

jsfisher · Jun 5, 2009

The Man said:
I thought that was what the "..." signified that it was an infinite progression and that 0.999... would be a member. Oh well, least upper bound and not a member, I can buy that. Still it doesn’t change my point that Doron can not argue 0.999... does not equal 1 without arguing that 3 times 1/3 does not equal 1.

Exactly right.

jsfisher · Jun 5, 2009

doronshadmi said:
This is the whole point.

Your Z < h < Y one example is based on some finite case (of Z and h) w.r.t Y , exactly as 0.99 or 0.999 are some finite cases w.r.t to 1.

Please stop torturing the word finite. That there is a value h between Z and Y is exactly one case. That all the members of {0.9, 0.99, 0.999, ...} are less then 1 is infinitely many cases.

In both cases we get "< 1" or "< Y" exactly because both cases do not deal with the non-finite (in the sense of 0.999...).

0.999... is not "non-finite." It is just one number and therefore one case.

You have a room for h between Z and Y exactly because you do not deal with the non-finite (in the sense of 0.999...).

One has nothing to do with the other (nor is 0.999... "non-finite").

In other words, your proof by contradiction does not deal with the non-finite (in the sense of 0.999...).

It doesn't deal with strawberry preserves either. What's your point?

The Man · Jun 5, 2009

doronshadmi said:
This is the whole point.

Your Z < h < Y one example is based on some finite case (of Z and h) w.r.t Y , exactly as 0.99 or 0.999 are some finite cases w.r.t to 1.

In both cases we get "< 1" or "< Y" exactly because both cases do not deal with the non-finite (in the sense of 0.999...).

You have a room for h between Z and Y exactly because you do not deal with the non-finite (in the sense of 0.999...).

In other words, your proof by contradiction does not deal with the non-finite (in the sense of 0.999...).

How about "in the sense of" 0.33333... or 0.2222.... or 0.1111... or 0.44444... , all possible values of 'h' if Z £ 0.1 and Y ³ 0.44445.

doronshadmi · Jun 5, 2009

jsfisher said:
0.999... is not "non-finite." It is just one number and therefore one case.

jsfisher said:
Every element of your second set {0.9, 0.99, 0.999, ...} is the sum of a finite sequence of elements from your first set {0.9, 0.09, 0.009, ...}.

The value 1 is the sum of an infinite sequence of elements. Big difference between sums over finite and infinite sequences.

So you contradict yourself again jsfisher.

jsfisher · Jun 5, 2009

doronshadmi said:
So you contradict yourself again jsfisher.

Nope. You just took two things, stripped them of context, and forced them together in an unnatural way.

0.999... is a single number. Just one thing.

0.9 + 0.09 + 0.009 + ... is also a single number. Just one thing. However, this second "just one thing" is an infinite series. An infinite series is the sum of infinitely many things.

doronshadmi · Jun 5, 2009

jsfisher said:
An infinite series is the sum of infinitely many things.

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.

doronshadmi · Jun 5, 2009

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.

But then it is not an immediate successor (as you claim in http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864) to any member < Y , of the non-finite Q or R collection.

Since you claim that Y is an immediate successor of the open interval [X,Y) you actually take anything that starts with X and < Y as a one mathematical object, exactly as the non-finite sequence 0.9+0.09+0.009+ … is considered as a one mathematical object.

But in the case of the one mathematical object that is based on 0.9+0.09+0.009+ … you really deal (according to Standard Math) with the non-finite (0.9+0.09+0.009+ … =1), where in the case of [X,Y) you do not deal with the non-finite, and this is exactly the reason of why "<Y" expression holds in the first place in X<Y, Z<Y or Z<h<Y cases.

doronshadmi · Jun 5, 2009

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.

jsfisher · Jun 5, 2009

doronshadmi said:
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 · Jun 5, 2009

doronshadmi said:
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?

doronshadmi · Jun 5, 2009

jsfisher said:
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.

doronshadmi · Jun 5, 2009

jsfisher said:
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.

jsfisher · Jun 5, 2009

doronshadmi said:
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.

jsfisher · Jun 5, 2009

doronshadmi said:
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?

doronshadmi · Jun 5, 2009

jsfisher said:
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.

doronshadmi · Jun 5, 2009

jsfisher said:
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?

jsfisher said:
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.

jsfisher · Jun 5, 2009

doronshadmi said:
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?

doronshadmi · Jun 5, 2009

jsfisher said:
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.

The Man · Jun 5, 2009

doronshadmi said:
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.

doronshadmi · Jun 5, 2009

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.

doronshadmi · Jun 5, 2009

The Man said:
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] )

doronshadmi · Jun 5, 2009

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).

The Man · Jun 6, 2009

doronshadmi said:
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.

doronshadmi said:
…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.

doronshadmi said:
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.

jsfisher · Jun 6, 2009

doronshadmi said:
jsfisher said:

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

Click to expand...

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:
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)

Click to expand...

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.

doronshadmi · Jun 6, 2009

The Man said:
What the heck do you think you’re talking about?

Look at this:

http://www.internationalskeptics.com/forums/showpost.php?p=4767201&postcount=3303
http://www.internationalskeptics.com/forums/showpost.php?p=4781744&postcount=3370
http://www.internationalskeptics.com/forums/showpost.php?p=4781767&postcount=3371
http://www.internationalskeptics.com/forums/showpost.php?p=4781867&postcount=3372

and please stop your zigzag.

zooterkin · Jun 6, 2009

doronshadmi said:
Look at this:

http://www.internationalskeptics.com/forums/showpost.php?p=4767201&postcount=3303
http://www.internationalskeptics.com/forums/showpost.php?p=4781744&postcount=3370
http://www.internationalskeptics.com/forums/showpost.php?p=4781767&postcount=3371
http://www.internationalskeptics.com/forums/showpost.php?p=4781867&postcount=3372

and please stop your zigzag.

Are you trying to say that 0.999... is the immediate predecessor of 1?

doronshadmi · Jun 6, 2009

jsfisher said:
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.

jsfisher said:
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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