Dear Dessi,
I certainly do not reject algebraic equivalence, I simply do not ignore the different levels of values that are used in some algebra, for example:
X = 0.999...
10X = 9.999...
10X - X = 9.999... - 0.999 = 9 (this is the critical operation, where we get rid of X (which is some value at the level of fractions) and what is left is the (positive, in this case) value at the level of whole numbers).
From now on X can't be but some positive whole number, which has nothing to do with the original value of X, as was given by X = 0.999...
Yes, it has everything to do with the original value of X = 0.999... . The value of X is
constant in your equation; 10X is the constant 9.999..., 10X - X is also constant, and our original X = 9/9 = 1 is also constant.
Every value in the equation is a constant. So why does X = 0.999... become X = 1? Because 0.999... is an algebraically identical representation of 1.
Try plugging a few different start values of X into your equation, and perform the same operations.
Code:
X = 0.999... X = 1
10X = 9.999... 10X = 10
10X - X = 9.999... - 0.999... 10X - X = 10 - 1
9X = 9 9X = 9
X = 1 X = 1
X = √2/2 X = 12
10X = 10√2/2 10X = 120
10X - X = 10√2/2 - √2/2 10X - X = 120 - 12
9X = 9√2/2 9X = 108
X = √2/2 X = 12
X = 0.11111... X = 0.12345 12345 12345...
10X = 1.1111... 10X = 1.2345 123451 23451...
10X - X = 1.11111... - 0.11111... 10X - X = 1.23451 23451 23451... - 0.12345 12345 12345...
9X = 1 9X = 1.11106 11106 11106...
X = 1/9 X = 0.12345 12345 12345...
X = 0.99999999 (finite) X = n
10X = 9.9999999 10X = 10n
10X - X = 9.9999999 - 0.99999999 10X - X = 10n = n
9X = 8.9999999 9X = 9n
X = 0.99999999 X = n
Every X = (10X - X)/9 = X. It starts and ends with the same value. Why wouldn't it? Why is X = 0.999... the singular exception?
As near as I can tell, you object to the X = 0.999... case because 9.999... - 0.9999... removes all of the trailing digits. Indeed, it does, 10X - X = 9, it has no other value. I can prove the algebra works out by showing that if X = 0.999..., then 10X - X = 9, therefore 9X = 9*(0.999...) = 9:
9 * .999...
= 8.999...
= Σ(n = 0, n -> ∞) (81/10)(1/10
n)
= (81/10) / (1 - 1/10) [see
this identity]
= (81/10) / (9/10)
= 9
The math works beautifully. From this, one can conclude that 0.999... is an equivalent representation of 1, for the same reason that 0.1111.... is an equivalent representation of 1/9.
Actually there is no problem to use serial_only observation of some finite framework in case of summation, but this is not the case if we deal with infinite summation in terms of process, because from this point of view the process can't be stopped and no exact result can be provided.
This statement is incorrect. The example above, showing 9*(0.999...) = 9, demonstrates one way of the many ways people can analyze infinite series. Let me explain:
Let's say we have a sequence {a
1, a
2, a
2, a
2, . . . }, the nth partial sum S
n is the sum of the first n terms of the sequence:
S
n = Σ(k = 1, k -> n) a
k
This series converges if the sequence of its partial sums, { S
1, S
2, S
3, . . . } converges. In other words, the series converges if there exists an L such that for any arbitrarily small positive number x > 0, there exists a large N so that for all n >= N,
| S
n - L | <= x
where |
expr| is the absolute value function. If a series converges (and there are many tests for convergence), then as x -> 0, S
n - L -> 0 and S
n -> L. A formal proof of this property
closely resembles this explanation.
I know it doesn't seem "intuitive" that an infinite series converges to anything, but that intuition is wrong. Your insistence that we need a different way of analyzing infinite series isn't based on anything, and in fact many ot of the same techniques for analyzing finite series hold for infinite series. I don't think you can articulate any counterargument to this point.
So, in this case we are using the brilliant notion of Cantor's transfinite cardinality that is definitely based also on parallel thinking, as explained in
http://www.internationalskeptics.com/forums/showpost.php?p=10330277&postcount=102.
The use of finite cardinality, countably infinite or uncountably infinite cardinality is essential to my theorem in
http://www.internationalskeptics.com/forums/showpost.php?p=10328657&postcount=73, where parallel thinking can't be avoided if we don't wish to find our framework stack in some endless process, if we deal with countably infinite or uncountably infinite cardinality.
Moreover, the whole idea of, for example, the accurate value of |
N| is possible only if we transcend some endless process, and this is done exactly by using parallel thinking that captures a given collection by using one step.
Doron, again, I view you as a very polite person with a serious passion for mathematics, but whatever you mean by "parallel thinking" is undefined, whatever operation you describe to compute an entire series in a single step without intermediate calculations is undefined. You aren't able to explain your to undefined concepts to anyone, so your proofs can't be analyzed for correctness. As near as I can tell, you don't have a proof nor anything novel to say about cardinality.
Since you mentioned Cantor, I encourage you to study the Cantor set and its analysis, as it has some fascinating and unexpected properties. The
Wiki article on the Cantor set is actually very accessible to readers like you who have at least some familiarity with series, sets, and numbers in other bases.
Lastly, I note that you must be a computer programmer or have some programming background from the statement "we find our framework stack in some endless process". Thinking in terms of call stacks, frameworks, and serial processes immediately tells me that you imagine operations on infinite series as a computer program that can never halt. I infer that "parallel thinking" is a very informal description of a hypothetical Turing machine which, by some miracle, instantaneously reads an infinite tape of inputs and halts with an output. This process happens inside a "black box"; we don't know how the black box works, we just know that it does. Let's call this model a
Super Duper Turing Machine, its what you call "parallel-computation". While the expression (2/1 + 2/3 + 2/9 + 2/27 + ...) - (2/3 + 2/9 + 2/27 + ....) will never halt on an ordinary Turing machine, the Super Duper Turing Machine, by some black box miracle, always halts with the answer 2. There are some inputs which even our powerful Super Duper Turing Machine cannot compute, jsfisher provided
one such example,
Busy Beaver numbers are another example.
An interesting question would be whether we can determine the halting behavior and output of our Super Duper Turing Machine, if such a machine existed?
Yes, in fact we can; the entire study of calculus, infinite series, and differential geometry does exactly that. Conventional analysis techniques like the ones in this thread are Turing equivalent to computations on a Super Duper Turing Machine, both give the same answers to the same questions,
there's no reason to think we would get a different result.
