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Old 15th November 2014, 06:24 AM   #41
doronshadmi
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It is well known that mathematicians avoid diagrams as rigorous mathematical proofs because of optical illusions that may involved with them.

Nevertheless, proof without words is accepted if it
Quote:
... can be demonstrated as self-evident by a diagram without any accompanying explanatory text.
(http://en.wikipedia.org/wiki/Proof_without_words)

I claim that there are also symbolic illusions that can be discovered if both visual AND symbolic methods are combined in some mathematical work, for example:



In this diagram (where visual AND symbolic methods are involved) 0.111...2 = 1 if only |N| observation of the real-line is used.

By using the same diagram (where visual AND symbolic methods are involved) 0.111...2 < 1 if also |R| observation of the real-line is used.

In this case 0.111...2 < 1 by exactly 0.000...12, where the value 0.000...12 is rigorously analyzed as follows:

The value 0.000...12 acts differently than value 0, as follows:

The ".000..." is used as a |N| size place value keeper that is inaccessible to "...12" that is at |R| size.

----------------------
The following is an example of symbolic illusion (that is discovered by using also |R| observation of the real-line) simply because n = 0 to ∞ is a hands-waving that misses the fact that |N|<|R| (as rigorously given by the diagram above (where visual AND symbolic methods are involved)):
Quote:
As for n = 0 to ∞, 0.999...10 obviously does not involved with (9/10)*(1/10|N|) simply because |N| is not one of the values of any collection of |N| finite values (and this is exactly the case of the parallel-summation of finite |N| values 0.9+0.09+0.009+... that are based on the convergent sequence of <0.9,0.09,0.009,...> finite |N| values).
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Old 15th November 2014, 08:47 AM   #42
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Old 15th November 2014, 10:13 AM   #43
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Quote:
In this case 0.111...2 < 1 by exactly 0.000...12, where the value 0.000...12 is rigorously analyzed as follows:

The value 0.000...12 acts differently than value 0, as follows:

The ".000..." is used as a |N| size place value keeper that is inaccessible to "...12" that is at |R| size.
I have two functions, graphed for your convenience:
f(x) = 1 + 1/x
g(x) = 1 - 1/x
Therefore:
f > 1 > g for all x in (0, ∞)
In your notation, there must be a value L such that:
f - L = 1 = g + L as x -> ∞
But notice that:
f = 1 + L as x -> ∞
f - 1 = L as x -> ∞

and

g = 1 - L as x -> ∞
g - 1 = L as as x -> ∞

Therefore

f - 1 = g - 1 as x -> ∞
f - g = 1 - 1 as x -> ∞
f - g = 0 as x -> ∞
The difference separating between f and g is exactly 0 as x -> ∞. Therefore:
f = 1 = g as x -> ∞
QED

---

Let's carry this a little further. Note that we can derive f' and g' in terms of f and g as follows:
f'(x) = f(x) - L = 1 + 1/x - L
g'(x) = g(x) + L = 1 - 1/x + L
Observe that:
f'(1) > f'(2) > f'(3) > . . . > f'(x) > 1 for all x in (0, ∞)
and
f(x) > f'(x) for all x in (0, ∞)

Therefore f(x) > f'(x) > 1 for all x in (0, ∞)
Likewise:
g'(1) < g'(2) < g'(3) < . . . < g'(x) < 1 for all x in (0, ∞)
and
g(x) < g'(x) for all x in (0, ∞)

Therefore g(x) < g'(x) < 1 for all x in (0, ∞)
Because f' > 1 and 1 > g', the following inequality must hold:
f' > 1 > g' for all x in (0, ∞)
In your notation, there must be a value M such that:
f' - M = 1 = g' + M as x -> ∞
Everything looks good here ^_^ From this, we can infer:
f - L = f' - M
and
g + L = g' + M
And with a little algebra, we can write:
f - f' = L + M
and
g - g' = M - L
But remember, f' was defined as f - L, and g' was defined as g + L. Let's re-write the expressions above in terms of f and g:
f - (f - L) = L + M
and
g - (g + L) = M - L
Simplifying:
f - (f - L) = L + M
f - f + L = L + M
0 + L = L + M
L = L + M
(L - L) = M
0 = M

and

g - (g + L) = M - L
g - g - L = M - L
0 - L = M - L
-L = M - L
(-L + L) + M
0 = M
So, M = 0 and (L - L) = M implies L = 0. Because L = M = 0, then
f(x) = f'(x) and g(x) = g'(x) as x -> ∞
and
f - L = 1 = g + L as x -> ∞ implies that f - 0 = 1 = g + 0 as x -> ∞

Therefore f = 1 = g as x -> ∞
QED.

--

You can generalize this further for f''(x) = f'(x) - M, f'''(x) = f''(x) - N, . . . , and so on to show that L = M = N = ... = 0.

We assume that 0.000...1 must be some C in an expression fa(x) = fa - 1(x) - C for any a > 1, so C = L = M = N = .... = 0.

Therefore, there exists is no C = 0.000...1 such that C > 0.

QED.
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Old 15th November 2014, 12:08 PM   #44
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Awesome!

Edit: Dessi, please be persistent... I am nowhere near as rapierlike in mathematical argumentation, I am just muleheaded persistent when it comes to Doronetics.

I am so going to enjoy this thread (together with people like jsfisher we are in this thread like over 7 or 8 years now...)

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Old 15th November 2014, 12:12 PM   #45
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Originally Posted by Dessi View Post
Therefore f = 1 = g as x -> ∞
Dear Dessi,

All your poofs are based on x -> ∞.

We know that there are infinitely many levels of infinity, for example:

|N| < |P(N)| < |P(P(N))| < ... etc. ad infinitum.

So when you are using ∞, what level of infinity you are using?

Again, 0.999...10 = 1 if we observe the real-line from level |N|.

This is not the case if we observe the real-line from level |R|.

In this case 0.999...10 is the result of parallel-summation of countable |N| finite values, where this result is inaccessible to 1 exactly by the value 0.000...110, such that ".000..." is used as a countable |N| level place value keeper that is inaccessible to "...110" that is at uncountable |R| level.

As long as you are not rigorously define ∞, x -> ∞ has nothing to do with the fact that there are infinitely many levels of infinity, and my argument is based on this fact.

So if you really wish to write any meaningful reply about my argument, you have to do that by explicitly not ignore the fact that there are infinitely many levels of infinity.

Can you demonstrate how your proofs hold also if you do not ignore the fact that there are infinitely many levels of infinity?
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Old 15th November 2014, 12:18 PM   #46
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Originally Posted by doronshadmi View Post
As long as you are not rigorously define ∞, x -> ∞ has nothing to do with the fact that there are infinitely many levels of infinity, and my argument is based on this fact.

So if you really wish to write any meaningful reply about my argument, you have to do that by explicitly not ignore the fact that there are infinitely many levels of infinity.

Can you demonstrate how your proofs hold also if you do not ignore the fact that there are infinitely many levels of infinity?
Doron, when have you, in all these years, *ever* defined *anything* rigorously?

And stop playing 'reductio ad absurdum'.
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Old 15th November 2014, 12:20 PM   #47
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Originally Posted by doronshadmi View Post
All your poofs are based on x -> ∞.
Your proofs are just going *poof*
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Old 15th November 2014, 01:52 PM   #48
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Originally Posted by doronshadmi View Post
In this case 0.999...10 is the result of parallel-summation of countable |N| finite values, where this result is inaccessible to 1 exactly by the value 0.000...110, such that ".000..." is used as a countable |N| level place value keeper that is inaccessible to "...110" that is at uncountable |R| level.
I promise you, I am far from a math noob, but I'm afraid the statement above is unreadable. I don't understand what you are describing.

Its not clear to me how the expression 0.000...1 > 0 can be constructed at all. Could you provide a function which computes C = 0.000...1, then a provide a proof that C > 0.

Quote:
As long as you are not rigorously define ∞, x -> ∞
Let's go back to our equations:
f(x) = 1 + 1/x as x -> ∞
f(x) = 1 - 1/x as x -> ∞
Let's annotate this graph a little:



Note that x -> ∞ is the same as saying c -> 0, so:
f(x) = 1 + 1/x as c -> 0 = g(x) 1 - 1/x as c -> 0
So, for every M > 0, there exists a number c such that 2c = f(M) - g(M). A c = 0 must exist because:



If M > 0, then c = 0. Therefore, existence of a limit L at c -> 0 implies the existence of L at x -> ∞. The rigorousness of f(x) as x -> ∞ is proven.

QED

Quote:
Can you demonstrate how your proofs hold also if you do not ignore the fact that there are infinitely many levels of infinity?
Let R bet the set of real numbers
Let N be a superset of real numbers R
Let f be the injective function from R to N
Let P be a proof function on the set of real numbers R
Let P' be the proof P(f(R))
Therfore, for every P(R) there exists a P'(N)

Which is a very pretentious way of saying my proof in the set of real numbers implies existence of a proof in any superset of real numbers. In other words, the expression C = 0.000...1 where C > 0 cannot be constructed any superset of real numbers, for any |N| > |R|, where |s| refers to the cardinality of the set s.

QED.

Apologies if I've misunderstood what you are trying to communicate.
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Old 15th November 2014, 02:58 PM   #49
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Originally Posted by Dessi View Post
Show your proof.
Again, the following quote is true because |N|<|R| is true:
Quote:
In this case 0.999...10 is the result of parallel-summation of countable |N| finite values, where this result is inaccessible to 1 exactly by the value 0.000...110, such that ".000..." is used as a countable |N| level place value keeper that is inaccessible to "...110" that is at uncountable |R| level.
As long as you are missing it, you actually missing my theorem.



Originally Posted by Dessi View Post
In other words, the expression C = 0.000...1 where C > 0 cannot be constructed any superset of real numbers, for any |N| > |R|, where |s| refers to the cardinality of the set s.

Apologies if I've misunderstood what you are trying to communicate.
Dear Dessi,

For example:

C = 0.000...110 is constructable only if you observe the real-line from |R|uncountable level.

By doing this the place value example 0.999...10 (which is at most at |N| countable level of the real-line) is a value of its own along the real that is < 1 exactly by C.
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Old 15th November 2014, 03:16 PM   #50
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Originally Posted by doronshadmi View Post
C = 0.000...110 is constructable only if you observe the real-line from |R|uncountable level.

By doing this the place value example 0.999...10 (which is at most at |N| countable level of the real-line) is a value of its own along the real that is < 1 exactly by C.
Can you show a proof?

Could also clarify the expression "observe the real-line from |R|uncountable level"? I've never heard that expression before, I don't know what you are describing.
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Old 15th November 2014, 03:20 PM   #51
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Originally Posted by Dessi View Post
I promise you, I am far from a math noob, but I'm afraid the statement above is unreadable. I don't understand what you are describing.

Its not clear to me how the expression 0.000...1 > 0 can be constructed at all.
It is constructable only if you observe the real-line from |R| level, and as a result, for example, 0.999...10 (which is a value of its own at |N| level, if it is observed from |R| level) < value 1, exactly by 0.000...110

Please do not ignore anything of post http://www.internationalskeptics.com...5&postcount=41 if you really wish to know my theorem.
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Old 15th November 2014, 03:29 PM   #52
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Originally Posted by doronshadmi View Post
Originally Posted by Dessi
I promise you, I am far from a math noob, but I'm afraid the statement above is unreadable. I don't understand what you are describing.
It is constructable only if you observe the real-line from |R| level, and as a result, for example, 0.999...10 (which is a value of its own at |N| level, if it is observed from |R| level) < value 1, exactly by 0.000...110
The highlighted statement clarifies nothing. I have no idea what you are saying. You are obviously talking about some mathematical abstraction, but I'm afraid I can't guess that abstraction is or how its derived.

Could you tell me what you think the real line is, what |R| means, what the expression "observe the real line from |R| level" refers to? After that, can you show how these things allow you to write a function which computes 0.000...1 > 0?
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Old 15th November 2014, 03:35 PM   #53
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Originally Posted by Dessi View Post
Can you show a proof?

Could also clarify the expression "observe the real-line from |R|uncountable level"? I've never heard that expression before, I don't know what you are describing.
You never heard of that expression before because it is (probably) novel thing about the real-line, that was not used (yet) by traditional mathematicians, even if |N|<|R| is an established mathematical fact.

Did you read all of http://www.internationalskeptics.com...7&postcount=40 before you replied?
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Old 15th November 2014, 03:47 PM   #54
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Originally Posted by Dessi View Post
Could you tell me what you think the real line is, what |R| means,
The real line is the set of real numbers where |R| is the cardinality of that set.

So, as you see, I use well-defined mathematical knowledge.
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Old 15th November 2014, 03:52 PM   #55
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Originally Posted by doronshadmi View Post
The real line is the set of real numbers where |R| is the cardinality of that set.

So, as you see, I use well-defined mathematical knowledge.
Me: Could you tell me what you think the real line is
You: The real line is the set of real numbers
Me: I understand this.

Me: what does |R| mean?
You: |R| is the cardinality of that set of real numbers
Me: I understand this.

Me: what does the expression "observe the real line from |R| level" refer to?
You:
Me:

Me: how does one construct the expression 0.000...1 > 0 at all?
You: It is constructable only if you observe the real-line from |R| level
Me:

Whatever manner one "observes the real line from |R| level" is most certainly not well-defined mathematical knowledge.
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Old 15th November 2014, 03:52 PM   #56
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Originally Posted by Dessi View Post
Could you tell me what you think the real line is, what |R| means, what the expression "observe the real line from |R| level" refers to? After that, can you show how these things allow you to write a function which computes 0.000...1 > 0?
Please read very carefully the first part of http://www.internationalskeptics.com...0&postcount=49.
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Old 15th November 2014, 03:56 PM   #57
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Originally Posted by Dessi View Post
You:
Me:
Already given in http://www.internationalskeptics.com...postcount=4282.

Originally Posted by Dessi View Post
Whatever manner one "observes the real line from |R| level" is most certainly not well-defined mathematical knowledge.
Only if |N|<|R| is also not well-defined mathematical knowledge.
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Old 15th November 2014, 03:59 PM   #58
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Originally Posted by Dessi View Post
Can you show a proof?
Ooh, ooh, I know, I know. No, Doron cannot.

Quote:
Could also clarify the expression "observe the real-line from |R|uncountable level"? I've never heard that expression before, I don't know what you are describing.
Ooh, ooh, I know, I know. No, Doron cannot.


Curiously enough, Doron also rejects Cantor's Theorem, yet he relies on its results. Limits are an issue for him, too, but you probably noticed that by now.
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Old 15th November 2014, 04:04 PM   #59
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Originally Posted by doronshadmi View Post
I promise you, I am not being intentionally obtuse. I've read your post, can you clarify the what the following expression means: |N|<|R|

Does |N| here refer to the cardinality of natural numbers, and |R| refer to the cardinality of natural numbers?
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Old 15th November 2014, 04:07 PM   #60
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Originally Posted by Dessi View Post
Me: how does one construct the expression 0.000...1 > 0 at all?
Dear Dessi,


There are two levels of infinity in |N|<|R|.

There are |N| place value keepers in ".000...", which are symbols at |N| level that are inaccessible to "...1", because it is a symbol at |R| level.
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Old 15th November 2014, 04:09 PM   #61
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Originally Posted by doronshadmi View Post
There are two levels of infinity in |N|<|R|.

There are |N| place value keepers in ".000...", which are symbols at |N| level that are inaccessible to "...1", because it is a symbol at |R| level.
What does |N| mean, what is N? I understand its an infinite set of some kind, but what is in the set N? Is N the set of natural numbers?
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Old 15th November 2014, 04:13 PM   #62
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Originally Posted by Dessi View Post
I promise you, I am not being intentionally obtuse. I've read your post, can you clarify the what the following expression means: |N|<|R|

Does |N| here refer to the cardinality of natural numbers, and |R| refer to the cardinality of natural numbers?
|N| is the cardinality of natural numbers and |R| is the cardinality of the real numbers, and I use the fact that |N|<|R|, or in other words |N| and |R| are two different levels of infinity that can be used in order to observe the real-line, where |N| is a countable observation and |R| is an uncountable observation.
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Old 15th November 2014, 04:14 PM   #63
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Originally Posted by Dessi View Post
What does |N| mean,
|N| is the cardinality of any countable collection.
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Old 15th November 2014, 04:16 PM   #64
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Originally Posted by doronshadmi View Post
|N| is the cardinality of any countable collection.
{A, B, C} is a countable collection.
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Old 15th November 2014, 04:30 PM   #65
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Originally Posted by doronshadmi View Post
|N| is the cardinality of natural numbers and |R| is the cardinality of the real numbers, and I use the fact that |N|<|R|, or in other words |N| and |R| are two different levels of infinity that can be used in order to observe the real-line.
Ok, |N| is the cardinality of natural numbers, |R| is the cardinality of reals. That is exactly the information I wanted.

Yes, |N| < |R|, but the language "two levels of infinity" is very imprecise. It is more precise to say that N is enumerable, and R is non-enumerable, meaning:

Natural numbers are the set {1, 2, 3, 4, 5, . . . Nk-2, Nk-1, Nk . . . }, this set is enumerable. In other words, it is possible to name every number in this set, given an infinite amount of time.

Rational numbers are the set of all numbers naturals numbers a and b of the form a/b:
{
{ N1/N1, N1/N2, N1/N3, . . . } ∪
{ N2/N1, N2/N2, N2/N3, . . . } ∪
{ N3/N1, N3/N2, N3/N3, . . . } ∪
. . .
}
This set is also enumerable. It is also clearly the cartesian product of N, so the cardinality of rationals > cardinality of naturals.

Real numbers are not enumerable, which is proved by Cantor's diagonal argument.

To put this another way, N ⊆ R or ∀x{x∈N → x∈R}.

Formality out of the way, I want to comment on:

Quote:
There are |N| place value keepers in ".000...", which are symbols at |N| level that are inaccessible to "...1", because it is a symbol at |R| level.
Show me how to construct C = 0.000...1 where C > 0.
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Old 15th November 2014, 04:36 PM   #66
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Originally Posted by doronshadmi View Post
|N| is the cardinality of any countable collection.
Counter example: the cardinality of {1, 2, 3} is less than naturals.
Counter example: the cardinality of the countable set of even naturals < the cardinality of naturals.
Counter example: the cardinality of the countable set of primes < the cardinality of naturals.
Counter example: the cardinality of the countable set of rational numbers > the cardinality of naturals, see theorem Q.
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Old 15th November 2014, 05:20 PM   #67
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Originally Posted by doronshadmi View Post
It is well known that mathematicians avoid diagrams as rigorous mathematical proofs because of optical illusions that may involved with them.

Nevertheless, proof without words is accepted if it
(http://en.wikipedia.org/wiki/Proof_without_words)

I claim that there are also symbolic illusions that can be discovered if both visual AND symbolic methods are combined in some mathematical work, for example:

http://farm9.staticflickr.com/8416/1...dfa4cebf_z.jpg
doronshadmi, thank you for responding to my posts, they provide much needed background to make sense of your diagram.

I'm still struggling with this part:

Quote:
In this diagram (where visual AND symbolic methods are involved) 0.111...2 = 1 if only |N| observation of the real-line is used.

By using the same diagram (where visual AND symbolic methods are involved) 0.111...2 < 1 if also |R| observation of the real-line is used.
Your diagram mentions something about a green and a blue line, and some function of these lines, shows that they compute the binary number 0.111..., but its not clear to me what this function is.

I understand that "proof without words" does not need accompanying text, but could you humor me here and provide an explanation anyway? Can you explain in more detail how one derives 0.111... from the blue and green lines?
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Old 15th November 2014, 08:07 PM   #68
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Originally Posted by Dessi View Post
Counter example: the cardinality of {1, 2, 3} is less than naturals.
Yes, of course.

Quote:
Counter example: the cardinality of the countable set of even naturals < the cardinality of naturals.
Counter example: the cardinality of the countable set of primes < the cardinality of naturals.
Counter example: the cardinality of the countable set of rational numbers > the cardinality of naturals, see theorem Q.
No. In each case, these sets have the same cardinality.
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Old 15th November 2014, 08:57 PM   #69
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Originally Posted by jsfisher View Post
Yes, of course.



No. In each case, these sets have the same cardinality.
Much appreciated, will review my set theory before guessing about the properties of sets
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Old 15th November 2014, 09:44 PM   #70
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Originally Posted by Dessi View Post
Much appreciated, will review my set theory before guessing about the properties of sets
"Infinity" is one of those things were intuition betrays you. For example, the even integers and the odd integers are both half the size of the integers, right? Or so it would seem.

For cardinality, the test for equality is can you put the members of the two sets into one-to-one correspondence. For the evens vs. integers, there is a simple correspondence between any integer, n, and any even, e: e = 2n.

The rational numbers are a bit less obvious, but the link you cited and its Theorem Q provides a correspondence method.

The primes are even less so, but it is straight-forward to show there is no largest prime, and from that we can conclude the set of primes is infinitely large. The primes being a subset of the integers assures us the primes are merely countably infinite.
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Old 16th November 2014, 12:55 AM   #71
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Originally Posted by jsfisher View Post
{A, B, C} is a countable collection.
Thank you for the correction, the right one is:

Any countably infinite collection has the same cardinality as |N| (which is the cardinality of natural numbers).

For example, the convergent sequence <0.9,0.09,0.009,...> has |N| values.
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Old 16th November 2014, 04:02 AM   #72
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Originally Posted by Dessi View Post
I understand that "proof without words" does not need accompanying text, but could you humor me here and provide an explanation anyway? Can you explain in more detail how one derives 0.111... from the blue and green lines?
Think "Erich von Daniken" but mathematically.
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Old 17th November 2014, 02:18 AM   #73
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Ok, let's address the ability to get a given value by using parallel-summation of values.

parallel-summation is a given value, which is the result of at lest two added values, such that the addition is done in parallel (no "process" of one-after-one steps (no serial steps) are involved).

It means that only one step is used among more than one added values.

The investigated object is [0,1].

Let's investigate some finite example of two added values:

0+1 is some example of how to get value 1 by parallel-summation among finite number of values (countably finite cardinality |2|).

0.910+0.0910+0.00910+0.00110 is some example of how to get value 1 by parallel-summation among finite number of values (countably finite cardinality |4|).

etc. ...

-----------------------------

Now let's investigate [0,1] by using countably infinite cardinality |N|.

Let's use, for example, 0.910+0.0910+0.00910+... that is a parallel-summation among infinite number of values (countably infinite cardinality |N|).

It is obvious that no parallel-summation of countably finite cardinality |n| (where n is some natural number > 1) among 0.910+0.0910+0.00910+... gets value 1.

Actually, we are using here (at the level of countably infinite cardinality |N|) the fact that |N|+1=|N| in order to conclude that 0.910+0.0910+0.00910+...=1

-----------------------------

Now let's investigate [0,1] by using uncountably infinite cardinality |R|.

Again, let's use, for example, 0.910+0.0910+0.00910... that is a parallel-summation among infinite number of values (countably infinite cardinality |N|).

It is obvious that no parallel-summation of countably infinite cardinality |N| among 0.910+0.0910+0.00910... gets value 1 simply because of the fact that |N|+1=|N|<|R|=|P(N)|.

-----------------------------

So, the general conclusion is as follows:

We get value 1 by parallel-summation only if we observe the values by some particular cardinality.

If a greater cardinality is also used, value 1 is inaccessible to the parallel-summations that are based on smaller cardinalities.

-----------------------------

Let's transcend the real-line as follows:

BY using the tower of powers line |P(P(N))| < |P(P(P(N)))| < |P(P(P(P(N))))| < ... there are obviously the following values:

Between any two irrational values along the tower of powers line, there are |P(P(N))| values.

Between any two |P(P(N))| values along the tower of powers line, there are P(P(P(N)))| values.

Between any two P(P(P(N)))| values along the tower of powers line, there are |P(P(P(P(N))))| values.

...

etc. ... ad infinitum, where the inaccessible limit of the tower of powers line is simply the non-composed 1-dimesional space (which is not necessarily a metric space).

-----------------------------

My theorem is false only if |n>1| < |N| < |P(N)| < |P(P(N))| < |P(P(P(N)))| < |P(P(P(P(N))))| < ... is false.
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Old 17th November 2014, 09:13 AM   #74
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Thank you for replying to my request ^_^

Originally Posted by doronshadmi View Post
Ok, let's address the ability to get a given value by using parallel-summation of values.

parallel-summation is a given value, which is the result of at lest two added values, such that the addition is done in parallel (no "process" of one-after-one steps (no serial steps) are involved).
So, parallel-summation is... addition? But different because it adds an entire sequence in one operation, without intermediate steps?

Short of representing a sequence of additions using some kind of equivalent identity (for example, and identity using multiplication ×, summation Σ, product Π, integral, etc), I don't really think a "parallel-summation" operator exists, intermediate additions are unavoidable:

"a1 + a2 + a3 + . . . + an" involves at least n-1 intermediate steps: ((((a1 + a2) + a3) + a3) + . . . an)

Even if not shown explicitly in a proof, intermediate steps -- some kind of process which perform the additions -- is implied.

If the basis of your proof depends on the existence of a parallel-summation operator, how is it defined mathematically? I can't analyze your proof beyond this point without a meaningful description of a parallel-summation.

Quote:
Let's use, for example, 0.910+0.0910+0.00910+... that is a parallel-summation among infinite number of values (countably infinite cardinality |N|).

It is obvious that no parallel-summation of countably infinite cardinality |N| among 0.910+0.0910+0.00910... gets value 1 simply because of the fact that |N|+1=|N|<|R|=|P(N)|.
Ignoring the problem of that the "parallel-summation" operator may not exist, I want to stress again that you proof is very unclear to me.

For a start, the statement that you say is "obvious" is absolutely not "obvious" in any sense. Your statement requires justification, you must define your parallel-summation operator and prove that it has the property you say it has. I cannot infer your property from the diagram you posted earlier. Please be as explicit as you can by writing out your proof for the highlighted statement, step-by-step.
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Old 17th November 2014, 09:52 AM   #75
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Originally Posted by Dessi View Post
Thank you for replying to my request ^_^

So, parallel-summation is... addition? But different because it adds an entire sequence in one operation, without intermediate steps?

Short of representing a sequence of additions using some kind of equivalent identity (for example, and identity using multiplication ×, summation Σ, product Π, integral, etc), I don't really think a "parallel-summation" operator exists, intermediate additions are unavoidable:

"a1 + a2 + a3 + . . . + an" involves at least n-1 intermediate steps: ((((a1 + a2) + a3) + a3) + . . . an)

Even if not shown explicitly in a proof, intermediate steps -- some kind of process to compute perform the additions -- is implied.
Doron is trying to resolve a criticism that his concepts rely on some form of step-by-step process (impossible processes at that) rather than being simply a result. An infinite series is not determined by an infinite sequence of additions, although Doron has in the past expressed that view. He is now trying to reduce the infinitely many additions down to just one by hand-waving "parallel summation" into existence. It is still a process, albeit only one step, if he could ever formally define parallel summation, which he cannot.

Also, as you so correctly point out, the full set of intermediate additions is necessary, if for no other reason than the order of operations can be important. Consider the series:

1 - 1 + 1 - 1 + 1 - 1 + ...

This is a non-convergent series since the partial summations oscillate between 1 and 0. It does not have a limit. However, if I change the order of operations a bit to be:

(1 - 1) + (1 - 1) + (1 - 1) + ... = 0 + 0 + 0 ...

That would be 0, right? So the original series has a value of 0, right? Then again:

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) ... = 1 + 0 + 0 + 0 + ...

And that would be 1.

The point of all this being that order of operations matters. Doron's "parallel summation" is completely bogus (for this and other reasons, too).

Quote:
If the basis of your proof depends on the existence of a parallel-summation operator, how is it defined mathematically? I can't analyze your proof beyond this point without a meaningful description of a parallel-summation.
Good luck with that. Soon you will be part of the "you don't get it" club because you lack the visual/spatial thinking skills necessary to overcome the conceptual limitations of Mathematics. The problem is with you and not Doron's concepts.
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Old 17th November 2014, 10:45 AM   #76
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Doron, before you comment on my lack of visual/spacial thinking, I want to show you a subset of my Amazon order history:

http://i346.photobucket.com/albums/p...ps8b63eee9.png

I encourage you to enjoy some posts I've written on computer science, be amused by my implementation of the cons/car/cdr combinators in Ruby, be more amused by my AVL tree implementation using only lambda combinators, and enjoy my book on functional programming.

Trust me, I am a goddamn nerd of the highest order. I'm confident my mathematical, visual, and spatial reasoning are more than adequate to analyze a proof that 0.999... < 1 as long as you provide the details.
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Old 17th November 2014, 10:51 AM   #77
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Originally Posted by Dessi View Post
Doron, before you comment on my lack of visual/spacial thinking, I want to show you a subset of my Amazon order history:

http://i346.photobucket.com/albums/p...ps8b63eee9.png

I encourage you to enjoy some posts I've written on computer science, be amused by my implementation of the cons/car/cdr combinators in Ruby, be more amused by my AVL tree implementation using only lambda combinators, and enjoy my book on functional programming.

Trust me, I am a goddamn nerd of the highest order. I confident my mathematical, visual, and spatial reasoning are more than adequate to analyze a proof that 0.999... < 1 as long as you provide the details .
+1

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Old 17th November 2014, 11:35 AM   #78
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Originally Posted by jsfisher View Post

Also, as you so correctly point out, the full set of intermediate additions is necessary, if for no other reason than the order of operations can be important. Consider the series:

1 - 1 + 1 - 1 + 1 - 1 + ...

This is a non-convergent series since the partial summations oscillate between 1 and 0. It does not have a limit. However, if a change the order of operations a bit to be:

(1 - 1) + (1 - 1) + (1 - 1) + ... = 0 + 0 + 0 ...

That would be 0, right? So the original series has a value of 0, right? Then again:

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) ... = 1 + 0 + 0 + 0 + ...

And that would be 1.

The point of all this being that order of operations matters. Doron's "parallel summation" is completely bogus (for this and other reasons, too)
The number of arranged levels that are involved have no impact on the result of a given parallel-summation because parallel-summation is done in one step on all arranged levels (again, no step-by-step is involved with parallel-summation).

Actually, for example, a value like 0.009 is also a parallel-summation among several arranged levels, for example:

0 (at the level of whole numbers) + (level of .1)*0 + (level of .01)*0 + (level of .001)*9
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Old 17th November 2014, 12:07 PM   #79
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Originally Posted by Dessi View Post
So, parallel-summation is... addition? But different because it adds an entire sequence in one operation, without intermediate steps?
It at least an addition among two values (including negative values), and yes, parallel-summation is done on all arranged levels in one step.

If you wish to understand better the notion of parallel-summation that is not influenced by arranged levels, please think in terms of the cardinality of, for example:

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Old 17th November 2014, 12:20 PM   #80
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Originally Posted by doronshadmi View Post
It at least an addition among two values (including negative values), and yes, parallel-summation is done on all levels in one step.
Your proof relies on the existence of a "parallel-summation" operator which lacks a proper definition. I can't infer the correctness of your proof from undefined operators.

Would you mind replying to the rest of my post?
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