Deeper than primes

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Another thing about associative property ( a(bc) = (ab)c ).

It works if only addition is used, in the case of infinite interpretation, for example:

1 + (0.999...[base 10] + 0.000...1[base 10]) = (1 + 0.999...[base 10]) + 0.000...1[base 10]) = 2
 
jsfisher said:
1 = 0.999...

You have yet to show otherwise, Doron.
Click to expand...

Wrong.

You still get 0.999...[base 10] as some numeral of number 1, exactly because you do not understand non-local numbers (where 0.999...[base 10] is an example of such a number).
 
doronshadmi said:
As you can see, my previous post is wrong (I wrote - instead of + in some places), let's correct it.

Let us use once again associative property ( a(bc) = (ab)c ).

Case a:
1 - 0.99 =
1 - (0.9 + 0.09) =
(1 - 0.9) - 0.09 =
0.1 - 0.09 = 0.01

Case b:
10 - 9.9 =
10 - (9 + 0.9) =
(10 - 9) - 0.9 =
1 - 0.9 = 0.1

Case a ≠ Case b if only finite interpolation is used.

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

Case c:
1 - 0.999... =
1 - (0.9 + 0.0999...) =
(1 - 0.9) - 0.0999... =
0.1 - 0.0999... = 0.000...1

Case d:
10 - 9.999... =
10 - (9 + 0.999...) =
(10 - 9) - 0.999... =
1 - 0.999... = 0.000...1

Case c = Case d if also infinite interpolation is used.

As we see by the provided example, associative property is too weak in order to distinguish between 10 - 9.999..., 1 - 0.999... and 0.1 - 0.0999... exactly because all these expressions are resulted by 0.000...1 Conclusion: Associative property is too weak in order to deal with infinite interpolation.
Click to expand...
The associative property (algebra term) doesn't affect the difference in results in Case a and b; it is one of the properties of equality (algebra term) that does, namely if a = b then a*c = b*c. It means that if a - b = c then d*(a - b) = d*a - d*b = d*c The quotient in Case a and b is 0.1/0.01 = 10. This quotient multiplies the initial left side in Case a: 10*(1 - 0.99) = 10 - 9.9 The result equals the initial left side in Case b, and so you did the arithmetic right, but you wrongly credited the association property of a redundant transformation with affecting the difference in the results. The same apply to the Case c and d.
 
doronshadmi said:
You still can try to grasp what is written in:

http://www.internationalskeptics.com/forums/showpost.php?p=7325618&postcount=15816

http://www.internationalskeptics.com/forums/showpost.php?p=7318785&postcount=15807


You still can try to grasp 1-dimensional space as a non-composed thing.
Click to expand...

So are you saying you will not even try to define your "superposition" without superposition? That’s understandable as you have already put yourself at quite a disadvantage by just calling it what you claim it is not and not just calling it what you do claim it is, "indeterminate". So not only does the simple lack of any definition seem to be your intent but also the apparently deliberate deception.
 
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doronshadmi said:
Another thing about associative property ( a(bc) = (ab)c ).

It works if only addition is used
Click to expand...


So, you have changed your mind yet again. First, addition is associative, then it was not, but now it is again.

Please make up your mind. (You can and will always change it later, it seems.)
 
doronshadmi said:
Wrong.

You still get 0.999...[base 10] as some numeral of number 1, exactly because you do not understand non-local numbers (where 0.999...[base 10] is an example of such a number).
Click to expand...


Just because you made something up is no reason for me to believe you.

1 = 0.999... and you have provided nothing to contradict that.
 
The Man said:
So are you saying you will not even try to define your "superposition" without superposition? That’s understandable as you have already put yourself at quite a disadvantage by just calling it what you claim it is not and not just calling it what you do claim it is, "indeterminate". So not only does the simple lack of any definition seem to be your intent but also the apparently deliberate deception.
Click to expand...
The Man you simply lack of any understanding of what is written in the links of my previous replay to you.
 
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jsfisher said:
Just because you made something up is no reason for me to believe you.

1 = 0.999... and you have provided nothing to contradict that.
Click to expand...
You simply can't grasp that there is no homeomorphism between 0-dimensional space and 1-dimensional space, and as a result a 1-dimensional element is irreducible into 0-dimensional element.

The non-local number 0.999...[base 10] is an expression of this irreducibility.

Furthermore, you can't get http://www.internationalskeptics.com/forums/showpost.php?p=6514886&postcount=12204 .
 
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epix said:
but you wrongly credited the association property of a redundant transformation with affecting the difference in the results.
Click to expand...
What I show is that if associative property is used only among finite interpolation and not only addition is used, then there can be a difference between case a and base b results:

Case a (1 - (0.9 + 0.09) = (1 - 0.9) - 0.09) ≠ Case b (10 - (9 + 0.9) = (10 - 9) - 0.9)

By using associative property among finite interpolation and infinite interpolation, where not only addition is used,
we get the same result in c and d cases:

Case c (1 - (0.9 + 0.0999...) = (1 - 0.9) - 0.0999...) = Case d (10 - (9 + 0.999...) = (10 - 9) - 0.999...)

In other words, the associative property is too weak if also infinite interpolation is used (the results are the same).
 
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doronshadmi said:
What I show is that if associative property is used only among finite interpolation and not only addition is used, then there can be a difference between case a and base b results:

Case a (1 - (0.9 + 0.09) = (1 - 0.9) - 0.09) ≠ Case b (10 - (9 + 0.9) = (10 - 9) - 0.9)

By using associative property among finite interpolation and infinite interpolation, where not only addition is used,
we get the same result in c and d cases:

Case c (1 - (0.9 + 0.0999...) = (1 - 0.9) - 0.0999...) = Case d (10 - (9 + 0.999...) = (10 - 9) - 0.999...)

In other words, the associative property is too weak if also infinite interpolation is used (the results are the same).
Click to expand...
Just let a = (1 - (0.9 + 0.09) = (1 - 0.9) - 0.09) and b = (10 - (9 + 0.9) = (10 - 9) - 0.9).

What is responsible for the inequality between both cases is simply the multiplication by 10.

a ≠ (10a = b).

All members in Case b were mutiplied by 10 and that's the only cause of the inequality. The type of association between the members doesn't play any role in the relationship.

You should consider an alternative:

10 - 9.999... = 0.000...1

1 - 0.999... = 0.000...1

The expresion 0.000...1 is simply too weak to detect the change caused by (10 - 9.999...)/10 = 1 - 0.999
 
doronshadmi said:
You simply can't grasp that there is no homeomorphism between 0-dimensional space and 1-dimensional space, and as a result a 1-dimensional element is irreducible into 0-dimensional element.

The non-local number 0.999...[base 10] is an expression of this irreducibility.
Click to expand...
Doron, homeomorphism means the reshaping of objects and it got little to do with what you try to use it for. Reshaping preserves mass and the process doesn't reduce it. Just make a ball out of chewing gum, weigh it, stretch it, weigh it again and you'll see. Algebra is sort of similar to homeomorfism: if you reduce the left side of the equation by certain amount then the same amount has to be added to the right side. The equation would look different but the balance (mass) between both sides would be preserved. But there is a curious exemption, which regards a magnitude expressed as 0.999... which is not assignable to x in order to perform algebraic manipulation, otherwise a contradiction may show up. When the set theory was young, it brought with it paradoxes, which were taken care of, and the same goes for the 0.999... = 1.000... contradiction.

Since the question of 0.999... does not affect the formal development of mathematics, it can be postponed until one proves the standard theorems of real analysis.
Click to expand...

In order to preserve the "shape" 0.999... so it wouldn't look like 1, you use the reduction of distance by division -- which is an infinite process -- as an argument. At the same time, you claim that real line segment cannot be "fully covered by points." Give someone immortal and illiterate a word processor. How many years would go by before a string of letters appears identical to the text in The Old Testament?

That's right. Many years.
 
epix said:
Yes, not that fast as the others

Location: Southern California
Posts: 31,968

but going.
:D
Click to expand...
Like molasses running down Mt Everest. :D

It does amaze me though. I've been gone from the forum since January of 2010. I wondered if this thread were still going and I'll be damn. It's here. FTR, I posted the pic of the bunny awhile back.
 
circle-children.jpg


RandFan said:
2008!

Wow, we've come some distance and we are still at the beginning. Progress, of a kind.
Click to expand...

Yes, but we take into the account the analytic rigor of things and so the progress isn't that fast. But it's safe, defined and scholarly.

HartCircle_700.gif
 
doronshadmi said:
You simply can't grasp that there is no homeomorphism between 0-dimensional space and 1-dimensional space, and as a result a 1-dimensional element is irreducible into 0-dimensional element.

The non-local number 0.999...[base 10] is an expression of this irreducibility.
Click to expand...

"Non-local" is just something you invented to cover your ignorance. The real numbers are everywhere dense. Your protests don't change that.

As for 0.999..., on a technical note, it is not a number, but a representation of a number. 0.999... relies on the common positional notation, and it literally means the limit of a particular infinite summation of 9 times successively smaller powers of 10. The limit is exactly 1.

If you don't like the fact 0.999... = 1 in standard Arithmetic, that's fine. Go invent your own, but that will have no impact on the validity of Arithmetic whatsoever. In fact it will have no impact on Arithmetic whatsoever.

Furthermore, you can't get http://www.internationalskeptics.com/forums/showpost.php?p=6514886&postcount=12204 .
Click to expand...

Again, you accuse others of failings that are uniquely your own.
 
RandFan said:
Still going!
Click to expand...

Think of it as a courtesy we provide. We know you occasionally are distant from the JREF forums for extended periods, and we want you to feel comfortable upon returning. We therefore sacrifice ourselves to maintain a level of constant in this one thread as an ever-present welcome-back to you.

I suppose, too, we are the lighthouse, vigilant in guiding would-be sailors from the rocky shores of Doronetics.

Then again, we are mostly bored, and this thread is a good outlet for battling never-ending nonsense. If only the back story were better, it would rival the immortal Realistice thread.

Care to join?
:D
 
doronshadmi said:
The Man you simply lack of any understanding of what is written in the links of my previous replay to you.
Click to expand...

What is written in those links lacks any definition of your “superposition” without superposition and simply claims it is “indeterminate”, “intermediate” as well as you asserting that you just call such a “superposition”. Those links just confirm your intent not to define your “superposition” without superposition and your deliberate attempt to deceive by continuing to refer to it as what you specifically claim it is not. As noted before that was already perceived to be your intent, so it comes as no big surprise.

At least your assertion “AB means that "A=bottom and B=top" is indeterminate so we can't determine if it is "bottom", "top" or some intermediate state between them.” Would place your “AB” “between” "A=bottom and B=top" inclusively. So it would seem your “non-strict” and “indeterminate” “AB” does have some restrictions and some determination making it “intermediate” and “between”.

Care to expand on that Doron or simply contradict yourself? I project that even if the former, soon to come the latter.
 
jsfisher said:
"Non-local" is just something you invented to cover your ignorance. The real numbers are everywhere dense. Your protests don't change that.
Click to expand...
If a person believes in "death by entropy" paradigm I am the last person that is going to argue with him in terms of belief.

If a person enables to re-consider his agreed paradigm, I am the first person that is going to argue with him in terms of paradigm's-shift.

jsfisher, you are not a person that enables to re-consider his agreed paradigm.
 
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The Man said:
So it would seem your “non-strict” and “indeterminate” “AB” does have some restrictions and some determination making it “intermediate” and “between”.
Click to expand...
The right phrase is this:

"So it would seem my “non-strict” and “indeterminate” AB is different than determined strict A or B"
 
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The Man said:
Care to expand on that Doron or simply contradict yourself? I project that even if the former, soon to come the latter.
Click to expand...
Contradiction, in this case, is the result of forcing determination on the indeterminate or the strict on the non-strict, and this is exactly what you are doing, The Man.
 
The Man said:
At least your assertion “AB means that "A=bottom and B=top" is indeterminate so we can't determine if it is "bottom", "top" or some intermediate state between them.” Would place your “AB” “between” "A=bottom and B=top" inclusively.
Click to expand...
You are still forcing strictness and determination on AB, and as a result you can't get it.
 
jsfisher said:
I suppose, too, we are the lighthouse, vigilant in guiding would-be sailors from the rocky shores of Doronetics.
:D
Click to expand...
All you can get is found under the spot of your lighthouse, where from this spot Doronetics can't be seen.

Bible%20with%20spotlight%202_0.jpg
 
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doronshadmi said:
All you can get is found under the spot of your lighthouse, where from this spot Doronetics can't be seen.

[qimg]http://www.christianfaith.com.au/system/files/Bible%20with%20spotlight%202_0.jpg[/qimg]
Click to expand...

D.D. means Doctor of Divinity, but also Doctor of Doronetics. If the curriculum is basically the same, then there doesn't have to be an adjustment to the acronym of both titles, right?

Your pic is a tough one. It says, "I'm thinking a passage in the Bible. Which one is it?"

Can that be done?
 
doronshadmi said:
[qimg]http://farm7.static.flickr.com/6008/5961689020_957e098143_b.jpg[/qimg]
Click to expand...


No.

(1) That isn't how 0.999... is shown to be 1 in any formal way.
(2) Getting 0.999... from 1/1 is a trivial exercise left for the interested reader.
 
jsfisher said:
No.

(1) That isn't how 0.999... is shown to be 1 in any formal way.
(2) Getting 0.999... from 1/1 is a trivial exercise left for the interested reader.
Click to expand...

By Traditional Math (which is actually a verbal_sequential-only reasoning) 0.111...[base 2] = 0.999...[base 10] = 1

By using verbal_sequential\visual_spatial reasoning as follows:

5962015728_d2fe37cc5f_z.jpg


it is a simple exercise for the interested mind to immediately get that 0.111...[base 2] < 1
 
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doronshadmi said:
By Traditional Math (which is actually a verbal_sequential-only reasoning) 0.111...[base 2] = 0.999...[base 10] = 1
Click to expand...

No, it is not.

By using verbal_sequential\visual_spatial reasoning as follows:

[qimg]http://farm7.static.flickr.com/6142/5962015728_d2fe37cc5f_z.jpg[/qimg]

it is a simple exercise for the interested mind to immediately get that 0.111...[base 2] < 1
Click to expand...

(1) No, it doesn't show that. It only helps demonstrate (which is far short of proving) that any finite such "decimal" is less that 1. It says nothing about the the infinite version, 0.111...2.

(2) Since it gets the wrong result, the only real conclusion here is that this personal invention of yours, verbal_sequential\visual_spatial reasoning, isn't very useful.
 
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doronshadmi said:
By Traditional Math (which is actually a verbal_sequential-only reasoning) 0.111...[base 2] = 0.999...[base 10] = 1

By using verbal_sequential\visual_spatial reasoning as follows:

[qimg]http://farm7.static.flickr.com/6142/5962015728_d2fe37cc5f_z.jpg[/qimg]

it is a simple exercise for the interested mind to immediately get that 0.111...[base 2] < 1
Click to expand...
I can only see three vertically organized values:

1/10 (dec)
2 (bin)
10 (bin)
 
doronshadmi said:
[qimg]http://farm7.static.flickr.com/6008/5961689020_957e098143_b.jpg[/qimg]
Click to expand...
You have a penchant for turning something simple into something needlessly complicated. You don't need to convert to another base to uncover a problem with this particular proof that only works when the left side of the equation is strictly kept in the exact format and the right side in the approximate format all throughout. Here it is once again:

nines.jpg


In this type of progress, the members of the left side are not evaluated in the approximate format, which allows the flow to reach the last line where the contradiction emerges. But in the all-format flow, the trouble start already in line 4 where

9x = 9

If you let the left side go to the approximate format and execute 9*x with x previously defined as x = 0.999..., then

9*0.999... = 8.999...1

and therefore it cannot equal 9.

The choice of preferring 0.999... = 1, baring the limit, instead of invoking a contradiction is simply wrong. It can be shown why: A contradiction of this kind is almost always preceded by a contradiction in some assumption, in this case that the expression "0.999..." is a real number. Here is the problem:

All manipulation that enter the various proofs, including the one that involves Dedekind cut, ask sooner or later and through different expressions for 0.999... = p/q. In other words, they ask for 0.999... to be a rational number. But there is no solution to p/q = 0.999... and therefore 0.999... cannot be a rational number. If not, then the only option left to place it is that 0.999... is an irrational number. That's can't be true, coz 0.999... has periodic fractional part and that fact places it into Q -- the rational numbers. So by one definition, the expression 0.999... is a rational number, but by the other it is not. This contradiction must be resolved. There is an ongoing effort to resolve the issue. (I already posted a quote addressing that.) The only equation that sets correctly 0.999... = 1 is the one which shows that 1 is the limit of 0.999...

Hey, do you feel like do some base converting here?

nines1.jpg


That's the place where "0.999..." is asked to become a rational number and therefore a real number. It refuses, but the math folks would go ahead anyway, coz the God of Number Crunching would get upset, and you don't want that to happen, coz He is defined mean. Yep.
 
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jsfisher said:
No, it is not.



(1) No, it doesn't show that.
Click to expand...

It shows that if the mind understands that no branch of that tree actually reaches any another branch of that tree, even if there are infinity many laves of that tree (where 1 is some branch of that tree), it immediately understands that this tree is actually some case of an infinite interpolation.

Infinite interpolation is a direct result of the fact that there is no homeomorphism between 1-dimensional space and 0-dimensional space or between between 2-dimensional space and 1-dimensional space etc. ad infinitum ... (or more general,
between any k-dimensional space (where k = 1 to ∞) and any n-dimensional space (where n = 0 to ∞ AND n < k)), where ∞ is an infinite extrapolation, which can't reach ∞(that has no successor) (Non-locality/Locality co-existence is an actual fact)).

Verbal_sequential-only minds, like jsfisher,
jsfisher said:
It says nothing about the the infinite version, 0.111...2.
Click to expand...
simply can't get infinite interpolation (where some case of it is shown in http://www.internationalskeptics.com/forums/showpost.php?p=7392225&postcount=16031) or infinite extrapolation (where ∞ is an infinite extrapolation, which can't reach ∞(that has no successor)).
 
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epix said:
So by one definition, the expression 0.999... is a rational number, but by the other it is not.
Click to expand...
epix, any given number that has infinity many symbols is a non-local number, for example:

0.333...[base 10] < 1/3
 
Some correction at the beginning of http://www.internationalskeptics.com/forums/showpost.php?p=7394481&postcount=16035.

Instead of
doronshadmi said:
It shows that if the mind understands that no branch of that tree actually reaches any another branch of that tree, even if there are infinity many laves of that tree (where 1 is some branch of that tree), it immediately understands that this tree is actually some case of an infinite interpolation.
Click to expand...
it has to be:

"It shows that if the mind understands that no branch of that tree actually reaches any another branch of that tree, even if there are infinitely many levels of that tree (where 1 is some branch of that tree), it immediately understands that this tree is actually some case of an infinite interpolation."

(Also in http://www.internationalskeptics.com/forums/showpost.php?p=7394523&postcount=16036 , please replace "infinity" by "infinitely".)

Anyway, jsfisher's "Death by entropy" reasoning can't get the openness of infinite interpolation, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=7392225&postcount=16031 and http://www.internationalskeptics.com/forums/showpost.php?p=7394481&postcount=16035 (including the correction in this post).
 
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Jsfisher and The Man are two minds that can't comprehend that no collection has the power of the continuum, where the power of the continuum is a property of Non-locality.
 
Originally Posted by epix
So by one definition, the expression 0.999... is a rational number, but by the other it is not.

doronshadmi said:
epix, any given number that has infinity many symbols is a non-local number, for example:

0.333...[base 10] < 1/3
Click to expand...

What's that for? 0.333... is a rational number, coz there exists p/q such that when p and q are entered into the Long Division algorithm, the result is 0.333... which satisfies the other definition of a rational number -- there is no contradiction that my quote refers to.

Your inequality lacks a definition of "1/3" and therefore you can't prove that inequality.
 
epix said:
Originally Posted by epix
So by one definition, the expression 0.999... is a rational number, but by the other it is not.



What's that for? 0.333... is a rational number, coz there exists p/q such that when p and q are entered into the Long Division algorithm, the result is 0.333... which satisfies the other definition of a rational number -- there is no contradiction that my quote refers to.

Your inequality lacks a definition of "1/3" and therefore you can't prove that inequality.
Click to expand...
epix, you still do not get non-local numbers, which are not rational or irrational numbers.

because Traditional Math does not get Non-locality, it has no choose but to define 0.333...[base 10] as 1/3 or 3/14...[base 10] as Pi , etc. ...

You simply ignore verbal_sequential\visual_spatial reasoning as used in http://www.internationalskeptics.com/forums/showpost.php?p=6465716&postcount=12075 and http://www.internationalskeptics.com/forums/showpost.php?p=6470162&postcount=12091 , and continue to use verbal_sequential-only reasoning, exactly like jsfisher and The Man.
 
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