Cont: Deeper than primes - Continuation 2

Status
Not open for further replies.
Dessi said:
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.
Click to expand...
There can be finitely or infinitely many operators of all kinds on finitely or infinitely many arranged levels that simply are calculated in one step, called parallel-summation.

Please think about parallel-summation more in terms of synthesis (more like parallel thinking) and less in terms of analysis (less like serial thinking).
 
Last edited:
Doron,

I'm seriously interested in your opinion: does your parallel-summation operator (however it is defined) give a different result than the ordinary summation operator Σ?

Part of the beauty of mathematics is that many different approaches to a problem can yield the same solution, many different expressions are in fact equivalent. What difference should it make if we add terms together one at a time, or add them up in one operation?

As near as I can tell:

0.999...
= 0.9 + 0.09 + 0.009 + 0.0009 + . . .
= Σ(n = 0, n -> infty) of (9/10)(1/10n)
= Σ(n = 0, n -> infty) of akn for a = 9/10, k = 1/10

Using the generalized form, note that the sum S is:

S = Σ(n = 0, n -> infty) of aknS = ak0 + ak1 + ak2 + . . .

Note that Sk = k(ak0 + ak1 + ak2 + . . .) = ak1 + ak2 + ak3 + . . .

We can infer

S - Sk = (ak0 + ak1 + ak2 + . . .) - (ak1 + ak2 + ak3 + . . .)
S - Sk = ak0 + (ak1 - ak1) + (ak2 - ak2) + (ak3 - ak3) + . . . .
S - Sk = ak0 + 0 + 0 + 0 + . . .
S - Sk = ak0S - Sk = a
S(1 - k) = a
S = a / (1 - k)

Substituting our original values, a = 9/10, k = 1/10:

S = (9/10) / (1 - (1/10))
S = (9/10) / (9/10)
S = 1

Why would you get a different result computing S = Σ(akn) vs S = parallel-summation-operator(akn)?
 
Last edited:
Dessi said:
What difference should it make if we add terms together one at a time, or add them up in one operation?
Click to expand...
When you deal with infinity, using step-by-step (serial) thinking style gives the illusion that some kind of "process" is involved.

In order to avoid such an illusion, it is better to use parallel thinking style.
 
Last edited:
doronshadmi said:
When you deal with infinity, using step-by-step (serial) thinking style gives the illusion that some kind of "process" is involved.

In order to avoid such an illusion, it is better to use parallel thinking-style.
Click to expand...

Let P = the parallel-summation operator
Let S = P(0.9, 0.09, 0.009, . . .) = 0.999...

(0.1)S = 0.0999...

S - (0.1)S = 0.999... - 0.0999...
S - (0.1)S = 0.9
S(1 - 0.1) = 0.9
S(0.9) = 0.9
S = (0.9) / (0.9)
S = 1
0.999... = 1

It's the exact same algebra whether we use parallel or serial thinking. Why do you think you get a different result?
 
Last edited:
Dessi said:
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/p412/julietrosenthal/orderhistory_zps8b63eee9.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.
Click to expand...

Yes, I certainly bow to your 1337 math skills (and your taste in comics and cartoons), but, as someone who does security evangelism and who teaches hacking academically, I do frown somewhat on your privacy skills.

Juliet, I would not recommend using a mixed private/public repo for images on fora where some, I do not say I count Doron to these, decidedly disturbed people hang out.

I'd recommend imgur or somesuch.

Other than that: <insert bowing smiley which I am too lazy too look for>
 
Dessi said:
Let P = the parallel-summation operator
Let S = P(0.9, 0.09, 0.009, . . .) = 0.999...

(0.1)S = 0.0999...

S - (0.1)S = 0.999... - 0.0999...
S - (0.1)S = 0.9
S(1 - 0.1) = 0.9
S(0.9) = 0.9
S = (0.9) / (0.9)
S = 1
0.999... = 1

It's the exact same algebra whether we use parallel or serial thinking. Why do you think you get a different result?
Click to expand...
What you wrote above is not considered as a rigorous proof of the discussed subject.

But since you have used it, I wish to show you how it looks by using parallel thinking:

5961689020_957e098143_z.jpg
 
Last edited:
doronshadmi said:
The difference between get X at once (parallel thinking) instead of step-by-step (serial thinking).
Click to expand...

I think this is visualized by Doron as 'just measure the height' on a 2D graph; instead of putting all values on the X-axis and then following the line, just put them on one point on the X-axis and the values on the Y-Axis. Then the number of your steps on the X-Axis is just 1.

Or to put into an analogy (which is not correct, I know, but I try to mediate in communications here):

- Imagine you have a really big amount of pebbles and you want to know their total weight.
- You can measure each pebble, one by one and add each weight to the total sum.
- Doron says: I just put all of the pebbles on my scale and know the total sum immediately.


At least, this is what I imagine he thinks. But who's to say?

EDIT: And then again...Seeing the post just above this one I am lost again.
 
Last edited:
doronshadmi said:
What you wrote above is not considered a rigorous proof of the discussed subject.
Click to expand...
Note that 0.22222...3
= 2*3^-1 + 2*3^-2 + 2*3^-3 + 2*3^-4 + 2*3^-5
= 2/3 + 2/9 + 2/27 + ....

And 2.2222...3 =
= 2*3^0 + 2*3^-1 + 2*3^-2 + 2*3^-3 + 2*3^-4 + 2*3^-5
= 2/1 + 2/3 + 2/9 + 2/27 + ...

2.2222...3 - 0.2222...3
= (2/1 + 2/3 + 2/9 + 2/27 + ...) - (2/3 + 2/9 + 2/27 + ....)
= 2/1 + (2/3 - 2/3) + (2/9 - 2/9) + (2/27 - 2/27) + ...
= 2 + 0 + 0 + 0 ...
= 2

Where's the missing fraction? Have I misunderstood you?
 
Last edited:
Dessi said:
Note that 0.22222...3
= 2*3^-1 + 2*3^-2 + 2*3^-3 + 2*3^-4 + 2*3^-5
= 2/3 + 2/9 + 2/27 + ....

And 2.2222...3 =
= 2*3^0 + 2*3^-1 + 2*3^-2 + 2*3^-3 + 2*3^-4 + 2*3^-5
= 2/1 + 2/3 + 2/9 + 2/27 + ...

2.2222...3 - 0.2222...3
= (2/1 + 2/3 + 2/9 + 2/27 + ...) - (2/3 + 2/9 + 2/27 + ....)
= 2/1 + (2/3 - 2/3) + (2/9 - 2/9) + (2/27 - 2/27) + ...
= 2 + 0 + 0 + 0 ...
= 2

Where's the missing fraction?
Click to expand...

Dear Dessi,

You are using 2.2222...3 - 0.2222...3 and get 2, and then you ask me where is the missing fraction?

I like your humor ;)

Please look very carefully at my page in http://www.internationalskeptics.com/forums/showpost.php?p=10329336&postcount=88.

Thank you.

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

Also please look at http://www.internationalskeptics.com/forums/showpost.php?p=10329220&postcount=79 for some parallel thinking training.

I think that after some training you can get http://www.internationalskeptics.com/forums/showpost.php?p=10328657&postcount=73 without any problems.

Also please do not ignore that fact that 0.999...10 = 1 if the real-line is observed from |N| level (as explained in that link), so understanding the different levels of cardinality is essential to my theorem, where parallel thinking is the best way to thing about infinite cardinality, if one wishes to avoid the illusion of "process".
 
Last edited:
doronshadmi said:
Dessi said:
Note that 0.22222...3
= 2*3^-1 + 2*3^-2 + 2*3^-3 + 2*3^-4 + 2*3^-5
= 2/3 + 2/9 + 2/27 + ....

And 2.2222...3 =
= 2*3^0 + 2*3^-1 + 2*3^-2 + 2*3^-3 + 2*3^-4 + 2*3^-5
= 2/1 + 2/3 + 2/9 + 2/27 + ...

2.2222...3 - 0.2222...3
= (2/1 + 2/3 + 2/9 + 2/27 + ...) - (2/3 + 2/9 + 2/27 + ....)
= 2/1 + (2/3 - 2/3) + (2/9 - 2/9) + (2/27 - 2/27) + ...
= 2 + 0 + 0 + 0 ...
= 2

Where's the missing fraction?
Click to expand...

Dear Dessi,

You are using 2.2222...3 - 0.2222...3 and get 2, and then you ask me where is the missing fraction?

I like your humor ;)
Click to expand...
I'm sorry, I don't understand your comment. I could not infer an error from your diagram because, unfortunately, you don't show a proof for the statement we discover the fundamental mistake in this theorem, because the omitted purple part 0.222... is definitely not the result of 2x/2 (the green part).

If you say there is an error in my proof, tell me in clear, precise language where I've made an error.

Also please look at http://www.internationalskeptics.com/forums/showpost.php?p=10329220&postcount=79 for some parallel thinking training.
Click to expand...
Note that addition in that notation sums elements one by one, not in a single step.
 
doronshadmi said:
There is no such thing like parallel-summation operator, as I explicitly wrote in http://www.internationalskeptics.com/forums/showpost.php?p=10329270&postcount=81.
Click to expand...
I disagree, you did not make that statement explicitly in any manner.

That aside, your proof for 0.999... < 1 depends on the existence of some process or operation which sums up all the elements in a sequence in a single step. This operation is the parallel-summation operator.

If your parallel-summation operator doesn't exist, then parallel-summation is undefined.
 
Last edited:
doronshadmi said:
What you wrote above is not considered as a rigorous proof of the discussed subject.
Click to expand...

Actually, it is sufficient. One might demand an explicit reference to summations, but since that is implicit in the meanings of 0.999... and 0.0999..., the added detail is unnecessary.

As with many things, more than one, equally rigorous proof is possible. The most direct, I suppose, would rely on the meaning of 0.999..., itself, as an infinite series then show that the limit of the corresponding partial summation sequence must be 1.

Then again, limit proofs are more difficult to grasp for most, so Dessi's approach has the advantage.
 
jsfisher said:
Then again, limit proofs are more difficult to grasp for most, so Dessi's approach has the advantage.
Click to expand...

We already did that last year or the year before (my, how time flies), then we went on to the 'two islands' thought experiment debacle, skipped over to metaconsciousness, skipped back here and are now, in extreme fastforward, repeating 7 years of argumentation.

The prediction I make is that Dessi will find out that being right merely will disqualify her as someone using the wrong 'set of glasses', 'view, like |N| vs |R|' or 'just don't be getting it'.

Doron controls the horizontal and the vertical here, so it's no use trying to adjust your set.
 
realpaladin said:
We already did that last year or the year before (my, how time flies), then we went on to the 'two islands' thought experiment debacle, skipped over to metaconsciousness, skipped back here and are now, in extreme fastforward, repeating 7 years of argumentation.

The prediction I make is that Dessi will find out that being right merely will disqualify her as someone using the wrong 'set of glasses', 'view, like |N| vs |R|' or 'just don't be getting it'.

Doron controls the horizontal and the vertical here, so it's no use trying to adjust your set.
Click to expand...

She may never get to enjoy a close encounter with Doron's exclusive "direct perception", then. How sad for her.
 
Dessi said:
I disagree, you did not make that statement explicitly in any manner.
Click to expand...
Here is the explicit statement:
doronshadmi said:
There can be finitely or infinitely many operators of all kinds on finitely or infinitely many arranged levels that simply are calculated in one step, called parallel-summation.
Click to expand...
Once again, parallel-summation is not some particular operator, it is simply using parallel approach in order to get a result in one step by not being influenced by the possibly complex structure of mathematical operations and their related variables\constant values, upon finitely or infinitely many arranged levels.

Dessi said:
That aside, your proof for 0.999... < 1 depends on the existence of some process or operation which sums up all the elements in a sequence in a single step.
Click to expand...
Dear Dessi,

At this stage you are simply forcing serial (step-by-step) thinking style, which is involved with some kind of process.

This is explicitly not the case if one using parallel thinking style of the considered subject.

As long as this is the case, there can't be any useful communication between us about this fine subject.
 
Last edited:
Dessi said:
If you say there is an error in my proof, tell me in clear, precise language where I've made an error.
Click to expand...
By using parallel thinking style, you are able to know in one step that the green and purple prats in the diagram are (at least) two different levels.

By using 2.222...3 - 0.222...3 we actually get rid of all levels below the floating point (marked by purple rectangle), and from now on we work only with the values at the level above the floating point and get the result 2X/2 = 1 that has nothing to do with with 0.222...3, because we already got rid of 0.222...3 by using 2.222...3 - 0.222...3, and from now on it is not used anymore as a factor of our conclusions.
 
Last edited:
Status
Not open for further replies.

Back
Top Bottom