I’ve done it … the 283 was sucking oil so we quit … just go buy it, much easier.
Your thought on my statement about taking liberties with establish principles?
I applied axioms, if you see that as “taking liberties” so be it.
They were applied correctly to arrive at an answer, some of the algorithms for multiplying and dividing infinite sequence real numbers are not well defined…so we hash them out here on this board…thats the way I see it.
I am not discarding the remainder. I am up to the 718th decimal place though.
What I don’t get is how you did it so fast and what you have done with the remainder. Are you using the same long division algorithm as me?
I don’t know, but I can demonstrate my result up to an ad nauseum level of repitition…that is, I see the workings of the process that is generating the result as one sees a wheel turning…so I believe I could demonstrate repitition to any number of digits. I could design a program to do this for the doubtful.
I believe I can put an ellipsis down for this reason…there is a threshold of repitition for each of us…and mine is when I can see the process repeating…not the digits recurring.
The problem is that you begin with 1 ≠ 0.999…
Yet later in your proof you rely on 1 = 0.999…
It has to be one or the other, or you violate an establish axiom.
I end with 1 ≠ 0.999…
It is stated as an objective
And the objective is reached when the contradiction in the system is revealed
Lets put it succinctly this way.
Great! That is exactly what I was hoping. I would love to know the decimal place where I stop getting these darned threes.
Nope. Really happy with just pen and paper. There is something – I don’t know – empirical and scientific about it. By all means set up a camera and make a video though. Come back when you are done.
Th contradiction is that 1 must equal 0.999… in that one step … if this is false, then this step is false and must be changed.
Every step in a proof MUST be true … or the conclusion is FALSE
The only way to invalidate my proof that the Real number system contains a logical contradiction.
Is to go back and attempt to do it adhering to 1 ≠ 0.999…
If it cannot be performed due to the lack of the ability to perform the substitutions…
you would have your answer. The system would no longer be invalidated by the proof
FINITE!!!
I have made this clear all throughout this thread
So EXPLICITLY stated…repitition of any FINITE number of digits…then at some point you have to make a leap of “pattern recognition faith”
Nope, sorry…you don’t get it…we can continue to drill down on the logic but you don’t seem to understand that a set must be internally consistent and operations must yield consistent results. 2 + 2 must always be 4, not just sometimes.
Not much more to say here…you guys are coming up weak.
I suggest you go back and provide some rigorous arguments to refute my proof.
Over and out for now.
Right, the proof is wrong, therefore 1 = 0.999…
As the system of a vector space is validated by the proof that has been falsified.
I do have a finite number of digits. 737 so far. So far I don’t have any difference between what I am doing and the number we represent by 0.33333… I am beginning to think that 0.33333… might be exactly the same as what I have on my piece of paper. But of course I have this remainder that I still have to deal with. It ain’t just a sprig of parsley you know.
The iteration begins.
Proof for .999… ≠ 1
Assume .999… = 1
Use this assumption via substitution to arrive at a clearly
erroneous result by applying the axioms of the system within the system
3 x .111… = .333…
=> 1/3 = .111…/.333…
1/3 = .1 + .0111…
.333…
The key step, square both sides
1/9 = [.1 + .0111…]^2
[333…]^2
1/9 = [.1 + .0111…][.1 + .0111…]
[.333…]^2
Cross Multiply
[.333…][.333…] = 9[.01 + .00111… + .00111… + [.0111…]^2]
[.333…][.333…] = .09 + 9x .00222… + 9x [.0111…]^2
Multiply through by 3
[.333…] = .27 + 27 x .002222… + 27 x [.0111…]^2
Multiply through by 3 again
1 = .81 + 81 x .00222… + 81 x [.0111…]^2
α = 81[.0111…]^2
1 = .81 + .18 + α
Show that α is less than .01
81[.0111…]^2 < .01
From this point on it is necessary to consider BOTH what is going on conceptually
AND practically via the use of multiplication and division algorithms.
The division of .01/81 requires the use of a division algorithm as it is not immediately apparent
what the answer could be…that is, it is not known a priori to have a finite decimal representation
such as 1/2 = .5. I assert that any algorithm used to arrive at a decimal representation for .01/81 will have
a remainder which can not be seen as clearly converging to zero…thus it can not be discarded.
The multiplication of .0111… by itself can be approximated and expanded iteratively via various algorithms
but it is clear that any form of multiplication will NOT result in any remainder.
The iterative process of multiplying .0111… by itself and dividing .01 by 81, if run in parallel, and if stopped
at ANY point in the iterative process, will always show the emergence of a pattern which clearly shows that
the left side is ALWAYS less than the right side due to the exist of a remainder from the division process which
CANNOT be seen as converging to zero due to its periodic nature.
[.0111…][.0111…] < .000123456789 + .000000000091/81
Halting the parallel iteration at the beginning of repetition of the integer counting sequence gives.
.000123456789 < .000123456789 + .000000000091/81
So ANY iterative real world process shows that
81[.0111…]^2 < .01
and therefore
.999… ≠ 1
I thought you were telling us that 0.333… is not exactly equal to 1/3.
How is it that you multiply by 3 and just get rid of one of your factors on the left hand side?
Then of course, if 0.3333… is exactly equal to 1/3 I can put away my 18 pieces of paper, multiply both sides of that equation by three and quickly conclude that 0.99999…=1
ninja’ed
Total nonsense.
I notice it was posted just yesterday. It was you who posted that, isn’t Cognitive Tide?
Just because you post it on youtube doesn’t make it true.
ETA: The post I was responding to seems to have disappeared 