Deeper than primes

[zooterkin]

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

Should we take that as a "No"?

 

[zooterkin]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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

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]

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]

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]

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.