Deeper than primes

[doronshadmi]

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

 

[doronshadmi]

1 = 0.999...

You have yet to show otherwise, Doron.

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).

 

[epix]

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.

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.

 

[The Man]

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.

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.

 

[jsfisher]

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

It works if only addition is used

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.)

 

[jsfisher]

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).

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.

 

[doronshadmi]

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.

The Man you simply lack of any understanding of what is written in the links of my previous replay to you.

 

[doronshadmi]

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.)

Now can't understand what is "if only addition is used", which is not the case in your http://www.internationalskeptics.com/forums/showpost.php?p=7374476&postcount=15948 post.

 

[doronshadmi]

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.

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 .

 

[doronshadmi]

but you wrongly credited the association property of a redundant transformation with affecting the difference in the results.

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).

 

[epix]

Crap! Lol.

 

[epix]

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).

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

 

[RandFan]

Still going!

 

[epix]

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.

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.

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]

Still going!

Yes, not that fast as the others

Location: Southern California
Posts: 31,968

but going.

 

[RandFan]

Yes, not that fast as the others

Location: Southern California
Posts: 31,968

but going.

Like molasses running down Mt Everest.

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.

 

[RandFan]

If one travels some distance but ends up at the beginning is that progress? I guess in some sense.

2008!

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

 

[epix]

2008!

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

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.

 

[jsfisher]

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.

"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 .

Again, you accuse others of failings that are uniquely your own.

 

[jsfisher]

Still going!

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?