Deeper than primes

[jsfisher]

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]

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]

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]

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]

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]

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]

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]

AGAIN:

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]

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]

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]

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]

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]

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]

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]

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]

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:

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]

...
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]

(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]

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]

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".