Deeper than primes

doronshadmi · May 31, 2009

jsfisher said:
Learn to read, and learn what words mean. I've already told which word in particular you are misinterpreting, here.

jsfisher, you contradict yourself. Simple at that.

jsfisher said:
Do you really not get that all I asked was for you to clarify what you meant by d in this context.

Accrding to Standard Math d=0, in the case of 0.999...[base 10], and I clearly wrote it in http://www.internationalskeptics.com/forums/showpost.php?p=4763819&postcount=3266.

jsfisher said:
Oh, I see. You use it inconsistently. No wonder you couldn't clarify. It's not one thing; it's infinitely many things.

You see exactly nothing.

The fact is that you don't understand what you read, in this case, and you contradict yourself by your own words, in this case ( as clearly seen at http://www.internationalskeptics.com/forums/showpost.php?p=4765238&postcount=3279 ).

Since (1) and (1) are actually the same case, then Standard Math is based on contradiction, in this case, because
by Standard Math d>0 (according to (1)) AND d=0 (according to (2)).

Jsfisher does his best to not see that (1) and (2) are actually the same case (and again, we can use Q members instead of R members, because "dense" holds for both of them, along some given interval), but jsfisher you can't escape form the fact that in this case Standard Math actually claims that d>0 (according to (1)) AND d=0 (according to (2)).

You can use the word "disjoint" as much as you like, but it will not change the fact that you contradict yourself, because in one hand (and in your words) you claim that there is a connection between (1) and (2) AND on the same case you claim that there is no connection between (1) and (2) (they are disjoint and no term in one of them is also the term of the other).

EDIT:

a) d must = 0 , if 0.999…[base 10] = 1

b) d must > 0 , if Y of X<Y does not have an immediate predecessor.

Since (a) and (b) are actually the same case, then Standard Math is based on this contradiction, in this case:

d>0 AND d=0

jsfisher · May 31, 2009

doronshadmi said:
You can use the word "disjoint" as much as you like, but it will not change the fact that you contradict yourself, because in one hand (and in your words) you claim that there is a connection between (1) and (2) AND on the same case you claim that there is no connection between (1) and (2) (they are disjoint and no term in one of them is also the term of the other).

So tell me, Doron, are the set of even integers and the set of odd integers disjoint or are they unrelated to each other. Must be one or the other. Which is it?

jsfisher · May 31, 2009

doronshadmi said:
a) d must = 0 , if 0.999…[base 10] = 1

b) d must > 0 , if Y of X<Y does not have an immediate predecessor.

Since (a) and (b) are actually the same case, then Standard Math is based on this contradiction, in this case:

d>0 AND d=0

Setting aside for a moment your refusal to clarify what you mean by d in the first case, the assertion high-lighted above is just that - another assertion.

You have provided no proof of the equivalence of (a) and (b).

The Man · May 31, 2009

doronshadmi said:
jsfisher, you contradict yourself. Simple at that.

Accrding to Standard Math d=0, in the case of 0.999...[base 10], and I clearly wrote it in http://www.internationalskeptics.com/forums/showpost.php?p=4763819&postcount=3266.

You see exactly nothing.

The fact is that you don't understand what you read, in this case, and you contradict yourself by your own words, in this case ( as clearly seen at http://www.internationalskeptics.com/forums/showpost.php?p=4765238&postcount=3279 ).

Since (1) and (1) are actually the same case, then Standard Math is based on contradiction, in this case, because
by Standard Math d>0 (according to (1)) AND d=0 (according to (2)).

Jsfisher does his best to not see that (1) and (2) are actually the same case (and again, we can use Q members instead of R members, because "dense" holds for both of them, along some given interval), but jsfisher you can't escape form the fact that in this case Standard Math actually claims that d>0 (according to (1)) AND d=0 (according to (2)).

You can use the word "disjoint" as much as you like, but it will not change the fact that you contradict yourself, because in one hand (and in your words) you claim that there is a connection between (1) and (2) AND on the same case you claim that there is no connection between (1) and (2) (they are disjoint and no term in one of them is also the term of the other).

EDIT:

a) d must = 0 , if 0.999…[base 10] = 1

Only if you set d equal to 1-1 because 0.9999… is just an infinite decimal representation of 1 meaning it extends to the infinite decimal place (does not terminate or have recurring zeros from a certain point on).

doronshadmi said:
b) d must > 0 , if Y of X<Y does not have an immediate predecessor.

Again you must first define what constitutes an “immediate predecessor” before you can determine if anything does or does not fit that definition.

As jsfisher notes it is all a matter of what you are using your d variable to represent. In (a) it is the difference between a real number in finite decimal representation and the same number in infinite decimal representation. In (b) it is in fact the difference between two different numbers.

doronshadmi said:
Since (a) and (b) are actually the same case, then Standard Math is based on this contradiction, in this case:

d>0 AND d=0

Again the misunderstanding and contradiction is yours and yours alone as (a) and (b) are clearly not the same case but simply employ the same letter ‘d’ to represent a ‘difference’ in entirely dissimilar concepts. One where the only difference being considered is in how a single value is represented, finite decimal or infinite decimal, which results in no value for that difference and the other where different values with the same representation are being considered.

I would object that you were deliberately trying to deceive since the difference between your (a) and (b) examples is so prominent that anyone doing the slightest research could understand your fallacy. However you have consistently demonstrated both your inability and unwillingness to actually do research or understand the concepts you attempt to refute. As such I am certain that you do not actually see (or simply do not choose to see) the difference between your (a) and (b) examples.

For those actually interested.

http://en.wikipedia.org/wiki/0.999...

doronshadmi · May 31, 2009

The Man said:
Only if you set d equal to 1-1 because 0.9999… is just an infinite decimal representation of 1 meaning it extends to the infinite decimal place (does not terminate or have recurring zeros from a certain point on).

So, you have no clue of what d is, but it does not stop you from writing a bunch of nonsense about it.

The rest of your post is nothing but a waste of time.

doronshadmi · May 31, 2009

jsfisher said:
Setting aside for a moment your refusal to clarify what you mean by d in the first case, the assertion high-lighted above is just that - another assertion.

You have provided no proof of the equivalence of (a) and (b).

Just that Q interval "dense" case and 0.999... is actually a non-finite sequence of Q members between 0.9 and 1.

But standard Math takes this same thing and claims that d>0 (the "dense" aspect of it) AND d=0 (the 0.999...=1 aspect of it).

jsfisher · May 31, 2009

It may be worth noting, too, that Doron has "proven" in other of this writings that 0.999... does not equal 1 at all. He has even pointed out that the difference between the two (i.e. 1 - 0.999...) is 0.000...1, where the ellipses represent an infinite number of 0's.

I bring this up because Doron is pre-disposed to believe the standard interpretation of 0.999... is already a contradiction all by itself.

jsfisher · May 31, 2009

doronshadmi said:
Just that Q inverval "dense" case and 0.999... is actually a non-finite sequence of Q members that have Y as their limit.

No, 0.999... is not an infinite sequence of rational numbers. It is a single number. The value of 0.999... happens to be the limit of an infinite sum, but so what?

And what does any of this have to do with immediate predecessors?

The Man · May 31, 2009

doronshadmi said:
So, you have no clue of what d is, but it does not stop you from writing a bunch of nonsense about it.

The rest of your post is is nothing but a waste of time.

Oh the irony of it all.

jsfisher said:
It may be worth noting, too, that Doron has "proven" in other of this writings that 0.999... does not equal 1 at all. He has even pointed out that the difference between the two (i.e. 1 - 0.999...) is 0.000...1, where the ellipses represent an infinite number of 0's.

I bring this up because Doron is pre-disposed to believe the standard interpretation of 0.999... is already a contradiction all by itself.

I must have missed those particular writings. So his ‘1’ is in the ‘infinite + 1’ decimal place? Kind of makes his interpretation of infinite rather meaningless, but then of course that would make it consistent with the rest of his interpretations.

jsfisher · May 31, 2009

The Man said:
I must have missed those particular writings. So his ‘1’ is in the ‘infinite + 1’ decimal place? Kind of makes his interpretation of infinite rather meaningless, but then of course that would make it consistent with the rest of his interpretations.

Just some examples:

http://www.physicsforums.com/archive/index.php/t-5267.html
http://www.geocities.com/complementarytheory/9999.pdf
http://www.internationalskeptics.com/forums/showthread.php?t=112582
http://forum.wolframscience.com/archive/topic/747-1.html

doronshadmi · May 31, 2009

jsfisher said:
No, 0.999... is not an infinite sequence of rational numbers. It is a single number.

0.999... = 0.9 + 0.09 + 0.009 + ... = 1 exactly because d=0 and 0.999... is a sum of a non-finite sequence of rationl numbers, whether you like it or not.

You can call it a single number exactly because d=0.

But when you are talking about the interval of Q members between X<Y, such that X=0.9 and Y=1 then 1 has no immediate predecessor iff d>0 for the non-finite Q sequence 0.9,0.99,0.999, ... ,1.

Both cases are actually the same mathematical object (0.999... and 0.9,0.99,0.999,... ,1 is two representations of a one mathematical object), and as a result Standard Math is based on a contradiction, because it claims that:

d>0 ( 1 does not have an immediate predecessor in 0.9,0.99,0.999,... ,1 representation ) AND d=0 (in 0.999...=1 representation)

jsfisher · May 31, 2009

doronshadmi said:
But when you are talking about the interval of Q members between X<Y, such that X=0.9 and Y=1 then 1 has no immediate predecessor iff d>0 for the non-finite Q sequence 0.9,0.99,0.999, ... ,1.

What do infinite sequences have to do with immediate predecessors? For that matter, what do intervals have to do with either one? And how about instead of the interval (0.9, 1), we us the interval (-17.8, 1). What now?

ETA: The bolded part is not an infinite sequence. See below.

Both cases are actually the same mathematical object, and as a result Standard Math is based on a contradiction, because it claims that:

d>0 ( 1 does not have an immediate predecessor in the case of the particular Q interval [X=0.9, Z=0.99,Z=0.999,Z=... ,Y=1), such that X < Z < Y ) AND d=0 (0.999...=1)[/QUOTE]

Ummm, [X=0.9, Z=0.99,Z=0.999,Z=... ,Y=1) is not an interval. What did you intend for this to mean?

Do you also realize that 0.999... does not ever appear in the infinite sequence 0.9, 0.99, 0.999, 0.9999, ..., right? You do know that, right? So, you can refer to the sequence all you like and to your ubiquitous [I]d[/I], but you won't be discussing 0.999....

doronshadmi · May 31, 2009

jsfisher said:
Ummm, [X=0.9, Z=0.99,Z=0.999,Z=... ,Y=1) is not an interval.[/quote]

[X,Y) interval such that [X=0.9, Z=0.99, Z=0.999, Z=... ,Y=1) is exactly the same mathematical object that is represented as 0.999...
[QUOTE="jsfisher, post: 4766738, member: 7169"]
Do you also realize that 0.999... does not ever appear in the infinite sequence 0.9, 0.99, 0.999, 0.9999, ..., right?

A = 0.999... = [X=0.9, Z=0.99, Z=0.999, Z=... ,Y=1)

How can A be one of its parts?

[QUOTE="jsfisher, post: 4765830, member: 7169"]Just some examples:

http://www.physicsforums.com/archive/index.php/t-5267.html
http://www.geocities.com/complementarytheory/9999.pdf
http://www.internationalskeptics.com/forums/showthread.php?t=112582
http://forum.wolframscience.com/archive/topic/747-1.html[/QUOTE]

This is not relevelt to this case, because here I am talking only about the framework of Stantard Math, by using claims of Standard Math.

doronshadmi · Jun 1, 2009

jsfisher said:
I will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

(In this case X = {0.9,0.99,0.999, ...} where any member of X < Y=1 , but it does not matter because this case can be used as a particular example Without loss of generality)

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

jsfisher · Jun 1, 2009

doronshadmi said:
Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

(In this case X = {0.9,0.99,0.999, ...,1})

This parenthetic part is multiply wrong. For one, X is not the name of the set. For another, you made a specific example of it in which, apparently, Y is taken as 1. You've also selected some example members that from a sequence implying something other than what is required to be in the set. E.g. -42 is in the set, but your representation would suggest otherwise.

Most important of the errors, though, is that you have 1 as an element of the set. It should not be since it is required that all members be strictly less than 1.

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

Except for where you got lost with that parenthetical aside you added on your own, you copy/pasted the proof just fine.

I'm guessing the original post is undergoing / will undergo an epic Doron edit.

doronshadmi · Jun 1, 2009

jsfisher said:
This parenthetic part is multiply wrong. For one, X is not the name of the set. For another, you made a specific example of it in which, apparently, Y is taken as 1. You've also selected some example members that from a sequence implying something other than what is required to be in the set. E.g. -42 is in the set, but your representation would suggest otherwise.

Most important of the errors, though, is that you have 1 as an element of the set. It should not be since it is required that all members be strictly less than 1.

Except for where you got lost with that parenthetical aside you added on your own, you copy/pasted the proof just fine.

I'm guessing the original post is undergoing / will undergo an epic Doron edit.

You are right I did not write it correctly, so let us try again:

Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set A={X : X<Y} does have a largest element, Z.

(In this case A = {0.9,0.99,0.999, ...} where any member of A < Y=1 , but it does not matter because this case can be used as a particular example Without loss of generality)

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set A={X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set A={X : X<Y} does not have a largest element.

jsfisher · Jun 1, 2009

doronshadmi said:
You are right I did not wirte it correctly, so let us try again:

Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set A={X : X<Y} does have a largest element, Z.

(In this case A = {0.9,0.99,0.999, ...} where any member of A < Y=1 , but it does not matter because this case can be used as a particular example Without loss of generality)

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set A={X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set A={X : X<Y} does not have a largest element.

You are still misrepresenting the membership of the set in your parenthetical aside. What's the point of this, anyway?

doronshadmi · Jun 1, 2009

jsfisher said:
You are still misrepresenting the membership of the set in your parenthetical aside.

Please explain in details.

jsfisher · Jun 1, 2009

doronshadmi said:
Please explain in details.

I did, but here goes again on one of the points:

By expressing it as A = {0.9,0.99,0.999, ...} you are implying the membership consists of only those values that fit the sequence. This is misleading. As written, you have implied that -42 is not it the set. It is. As written, you have implied that 0.99988888888888... isn't in the set. It is.

Also important, you haven't shown any value to the discussion by introducing this specific instance of the set {X : X<Y}.

doronshadmi · Jun 1, 2009

jsfisher said:
I did, but here goes again on one of the points:

By expressing it as A = {0.9,0.99,0.999, ...} you are implying the membership consists of only those values that fit the sequence. This is misleading. As written, you have implied that -42 is not it the set. It is. As written, you have implied that 0.99988888888888... isn't in the set. It is.

Also important, you haven't shown any value to the discussion by introducing this specific instance of the set {X : X<Y}.

Jsfisher, set A is exactly only the particular numbers of the form 0.9,0.99,0.999, ... etc.

Yet this limited case can be used Without loss of generality in your proof by contradiction.

jsfisher · Jun 1, 2009

doronshadmi said:
Jsfisher, set A is exactly only the particular numbers of the form 0.9,0.99,0.999, ... etc.

Yet this limited case can be used Without loss of generality in your proof by contradiction.

No, it cannot.

The thing to be proven is that the set {X : X<Y} has no largest value. However, since you keep mis-referring to it as a proof no real number has an immediate predecessor, and since that is substantially the same claim, I'll refer to it as such.

The proof introduces Y to represent an arbitrary real number.

By then fixing Y to be 1, as you did, you have now lost generality. If you continue the proof under this assumption, then all you end up proving is that 1 has no immediate predecessor.

Ok, so that's one fault. There is another.

The original proof uses the set {X : X<Y}, which is the set of all real numbers less then Y. More precisely, these are all the predecessors of Y. If there is an immediate predecessor, it must be in this set.

You, however, have restricted the set further, even though your notation is inadequate for the task. Instead of all predecessors, you've chosen just those that fit a tidy sequence (0.9, 0.99, 0.999, 0.9999, ... for the case at hand).

So, instead of all predecessors of Y, you've changed it to some predecessors of Y. Again, you have lost generality. All you will be able to prove with this restricted set is that the immediate predecessor of Y (if it exists) isn't in your restricted set.

doronshadmi · Jun 1, 2009

jsfisher said:
No, it cannot.

The thing to be proven is that the set {X : X<Y} has no largest value. However, since you keep mis-referring to it as a proof no real number has an immediate predecessor, and since that is substantially the same claim, I'll refer to it as such.

The proof introduces Y to represent an arbitrary real number.

By then fixing Y to be 1, as you did, you have now lost generality. If you continue the proof under this assumption, then all you end up proving is that 1 has no immediate predecessor.

Ok, so that's one fault. There is another.

The original proof uses the set {X : X<Y}, which is the set of all real numbers less then Y. More precisely, these are all the predecessors of Y. If there is an immediate predecessor, it must be in this set.

You, however, have restricted the set further, even though your notation is inadequate for the task. Instead of all predecessors, you've chosen just those that fit a tidy sequence (0.9, 0.99, 0.999, 0.9999, ... for the case at hand).

So, instead of all predecessors of Y, you've changed it to some predecessors of Y. Again, you have lost generality. All you will be able to prove with this restricted set is that the immediate predecessor of Y (if it exists) isn't in your restricted set.

I was not clear enough also in this case.

What I did here is to limit by purpose the universe only to non-finite sets of finite Q members in order to ask the following question:

Let us take the non-finite set of the finite Q members of the form {0.9,0.09,0.009, …}

Why the sum of the non-finite set of the finite Q members of the form {0.9,0.09,0.009, …}, = 1 , where any finite Q member of the non-finite set {0.9, 0.99, 0.999, …}, < 1?

After all in both cases we deal with a non-finite set of finite Q members.

EDIT:

Can we conclude that the sum is greater than its parts in the case of 0.999... = 1?

jsfisher · Jun 1, 2009

doronshadmi said:
I was not clear enough also in this case.

What I did here is to limit by purpose the universe only to non-finite sets of finite Q members in order to ask the following question:

Let us take the non-finite set of the finite Q members of the form {0.9,0.09,0.009, …}

Why the sum of the non-finite set of the finite Q members of the form {0.9,0.09,0.009, …}, = 1 , where any finite Q member of the non-finite set {0.9, 0.99, 0.999, …}, < 1?

After all in both cases we deal with a non-finite set of finite Q members.

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.

By the way, you may notice that even though 1 does not appear as an element of the second set (in either 1 or 0.999... form), it is the limit of the sequence represented by the set. This exactly parallels the summations. Even though 1 is not the sum of any finite sequence from the first set, it is the limit of those sums.

doronshadmi · Jun 1, 2009

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.

By the way, you may notice that even though 1 does not appear as an element of the second set (in either 1 or 0.999... form), it is the limit of the sequence represented by the set. This exactly parallels the summations. Even though 1 is not the sum of any finite sequence from the first set, it is the limit of those sums.

I think that you have missed the point.

1 is the result of infinitely many finite values (each one of the members of {0.9, 0.09, 0.009, ...} is a finite case).

Also, can we conclude that the sum is greater than its parts in the case of 0.999... = 1?

<the last part of this post was deleted>

jsfisher · Jun 1, 2009

doronshadmi said:
I think thay you have missed the point.

1 is the result of infinitely many finite values (each one of the members of {0.9, 0.09, 0.009, ...} is a finite case).

It is not a finite case. There are infinitely many addends involved in the sum.

Also, can we conclude that the sum is greater than its parts in the case of 0.999... = 1?

We can conclude 0.999... = 1 because the whole is exactly equal to the sum of its parts.

doronshadmi · Jun 1, 2009

jsfisher said:
It is not a finite case. There are infinitely many addends involved in the sum.

The sum is the result of infinitely many addends where each added value is finite.

So why do you think that inifintely many finite values can reach their limit?

jsfisher · Jun 1, 2009

doronshadmi said:
The sum is the result of infinitely many addends where each added value is finite.

So why do you think that inifintely many finite values can reach their limit?

I think that because I have quite the understanding of Mathematics, including how limits work. By Standard Mathematics (which you have already conceded as the domain of discourse, here), it is a geometric series of the form:

[latex]$$ S = a + ar + ar^2 + ar^3 + ... $$[/latex]

The series converges to a value given by:

[latex]$$ S = \sum_{n \ge 0} ar^n = \frac{a}{1 - r} $$[/latex]

Simple substitution with a=0.9 and r = 0.1 gives S = 0.9/(1-0.1) = 0.9/0.9 = 1.

doronshadmi · Jun 1, 2009

AGAIN:

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, ...}.

jsfisher I going to use your argument avobe in this post:

A = {0.9, 0.99, 0.999, ...}

B = {0.9, 0.09, 0.009, ...}

Both cases are based on infinitely many finite values (where each value < 1).

Each member of A set is the result of the sum of a finite amount of members of B set.

As a result it is obvious that any member of A set < 1 (your argument is used here).

Now let us check B set.

C = {0.1, 0.01, 0.001, …}

Each member of B set is the result of a the sum of a finite amount of members of C set.

As a result it is obvious that any member of B set < 1 (your argument is used here).

Please ignore the fact that the sum of A members > 1 and the fact that the sum of C members < 1, and focused on the common fact that each member of A or B sets < 1 and each member of A or B set is the sum of a finite amount of members of another set.

By understanding the following, why do you think that the sum of the members of B set reaches the limit 1?

jsfisher · Jun 1, 2009

doronshadmi said:
AGAIN:

A = {0.9, 0.99, 0.999, ...}

B = {0.9, 0.09, 0.009, ...}

Both cases are based on infinitely many finite values.

Both are sets of infinitely many rational numbers. If you feel it is necessary to point out that every rational number is of finite value, well, ok.

Each member of A set is the result of the sum of finite amount of members of B set.

As a result it is obvious that any member of A set < 1.

Yes, and with only a little bit of notation it could be made mathematically precise what you mean the members of A to be.

Now let us check B set.

C = {0.1, 0.01, 0.001, …}

Each member of B set is the result of a the sum of finite amount of members of C set.

No. What did you really mean to write?

As a result it is obvious that any member of B set < 1

Please ignore the fact that the sum of A members > 1

No, that is not a correct statement. The series does not converge, so it has no defined sum.

...and focused on the common fact that each member of A or B sets < 1 and each member of A or B set is the sum of a finite amount of members of another set.

Well, except you failed to show B as derived as the sum of members from some other set.

By understanding the following, why do you think that the sum of the members of B set reaches the limit 1?

Already asked and answered. By the way, "reaches the limit", although picturesque, isn't quite an accurate description. The limit is 1. It doesn't get there by a sequence of events; it simply is.

doronshadmi · Jun 1, 2009

jsfisher said:
Well, except you failed to show B as derived as the sum of members from some other set.

Not a all.

0.9 of B set = 9*0.1 of C.

0.09 of B set = 9*0.01 of C.

...

Please ignore the fact that the "sum" of A members > 1 and the fact that the sum of C members < 1, and focused on the common fact that each member of A or B sets < 1 and each member of A or B set is the sum of a finite amount of members of another set.

By understanding the following, why do you think that the sum of the members of B set is the limit 1?

(More general, why do you think that the sum of the members of a non-finite set that is based on finte Q numbers (where each one of tham < 1) is the limit 1?

jsfisher · Jun 1, 2009

doronshadmi said:
By understanding the following, why do you think that the sum of the members of B set is the limit 1?

Already asked and answered.

(More general, why do you think that the sum of the members of a non-finite set that is based on finte Q numbers (where each one of tham < 1) is the limit 1?

Now your are just being silly. Consider the infinite series:

S = ( 0.5, 0.5, 0, 0, 0, 0, 0, 0, ...)

It meets all your high-lighted characteristics. It is an infinite sequence. It is based on rational numbers (and only the finite rational numbers at that), and each member of the sequence is < 1.

And even with all that, the sum of the entire sequence is obviously 1.

doronshadmi · Jun 1, 2009

jsfisher said:
Already asked and answered.

Now your are just being silly. Consider the infinite series:

S = ( 0.5, 0.5, 0, 0, 0, 0, 0, 0, ...)

It meets all your high-lighted characteristics. It is an infinite sequence. It is based on rational numbers (and only the finite rational numbers at that), and each member of the sequence is < 1.

And even with all that, the sum of the entire sequence is obviously 1.

This is a silly example, because non-finite sum of 0 = 0

As a result we deal with the finite case of 0.5+0.5+0=1

doronshadmi · Jun 1, 2009

jsfisher said:
Already asked and answered.

Now your are just being silly. Consider the infinite series:

S = ( 0.5, 0.5, 0, 0, 0, 0, 0, 0, ...)

It meets all your high-lighted characteristics. It is an infinite sequence. It is based on rational numbers (and only the finite rational numbers at that), and each member of the sequence is < 1.

And even with all that, the sum of the entire sequence is obviously 1.

This is a silly example, because non-finite sum of 0 = 0

As a result we deal with the finite case of 0.5+0.5+0=1

But you are right about < 1 term.

So let us change it to "a finite Q number that is > 0 AND < 1"

doronshadmi · Jun 1, 2009

jsfisher said:
Already asked and answered.

Now your are just being silly. Consider the infinite series:

S = ( 0.5, 0.5, 0, 0, 0, 0, 0, 0, ...)

It meets all your high-lighted characteristics. It is an infinite sequence. It is based on rational numbers (and only the finite rational numbers at that), and each member of the sequence is < 1.

And even with all that, the sum of the entire sequence is obviously 1.

This is a silly example, because non-finite sum of 0 = 0

As a result we deal with the finite case of 0.5+0.5+0=1

But you are right about < 1 term.

So let us change it to "a finite Q number that is > 0 AND < 1"

More general, why do you think that the sum of the members of a non-finite set that is based on finite Q numbers
(where each one of tham is > 0 AND < 1) is the limit 1?

jsfisher · Jun 1, 2009

doronshadmi said:
This is a silly example, because non-finite sum of 0 = 0

As a result we deal with the finite case of 0.5+0.5+0=1

The case meets every criteria you suggested. It was meant to point out that your criteria were not particularly meaningful.

If you can't accept that 0.999... is equal to 1, then please tell us what the difference between the two is. That is, what is 1 - 0.999..., and please restrict you response to standard mathematics constructs.

(When I do the subtraction, I get 0.000..., which for all practical purposes is just 0.)

doronshadmi · Jun 1, 2009

jsfisher said:
The case meets every criteria you suggested. It was meant to point out that your criteria were not particularly meaningful.

If you can't accept that 0.999... is equal to 1, then please tell us what the difference between the two is. That is, what is 1 - 0.999..., and please restrict you response to standard mathematics constructs.

(When I do the subtraction, I get 0.000..., which for all practical purposes is just 0.)

jsfisher, thank you for your corrections, but you still did not answer to this (corrected) verstion:

AGAIN:

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, ...}.

jsfisher I going to use your argument avobe in this post:

A = {0.9, 0.99, 0.999, ...}

B = {0.9, 0.09, 0.009, ...}

Both cases are based on infinitely many finite values (where each value < 1).

Each member of A set is the result of the sum of a finite amount of members of B set.

As a result it is obvious that any member of A set < 1 (your argument is used here).

Now let us check B set.

C = {0.1, 0.01, 0.001, …}

Each member of B set is the result of a the sum of a finite amount of members of C set:

0.9 of B set = 9*0.1 of C.

0.09 of B set = 9*0.01 of C.

...

As a result it is obvious that any member of B set < 1 (your argument is used here).

Please ignore the fact that the "sum" of A members > 1 and the fact that the sum of C members < 1, and focused on the common fact that each member of A or B sets < 1 and each member of A or B set is the sum of a finite amount of members of another set.

By understanding the following, why do you think that the sum of the members of B set is the limit 1?

More general, why do you think that the sum of the members of a non-finite set that is based on finite Q numbers
(where each one of tham is > 0 and < 1) is the limit 1?

jsfisher · Jun 1, 2009

doronshadmi said:
...
By understanding the following, why do you think that the sum of the members of B set is the limit 1?

More general, why do you think that the sum of the members of a non-finite set that is based on finite Q numbers
(where each one of tham is > 0 and < 1) is the limit 1?

Already asked. Already answered.

All this hand-waving with sequences derived from other sequences is irrelevant. The original answer stands.

doronshadmi · Jun 1, 2009

jsfisher said:
(When I do the subtraction, I get 0.000..., which for all practical purposes is just 0.)

It is possible because you already say that 1=0.999... so your "practical purposes" is based on circular reasoning.

jsfisher · Jun 1, 2009

doronshadmi said:
It is possible because you already say that 1=0.999... so your "practical purposes" is based on circular reasoning.

You are using this (incorrect) assessment to avoid answering the question for yourself. What do you get for the difference?

doronshadmi · Jun 1, 2009

jsfisher said:
Already asked. Already answered.

All this hand-waving with sequences derived from other sequences is irrelevant. The original answer stands.

In other words, I am right.

You simply do not wish to understand things that are not derived from "How to define and\or use"? question.

As a result you do not get that your reasoning is nothing but an arbitrary deremination of some snapshot of an "hand-waving".

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