Deeper than primes

[doronshadmi]

You are just making this nonsense up on the fly, aren't you? Ok, so you decided you needed to readjust your fantasy because of reality encroachment. According to this latest goal post placement, we have:

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

We also have:

10 - 9.999... = (9 + 1) - (9 + 0.999..) = (9 - 9) + (1 - 0.999...) = 0 + 0.000...1.


Therefore, 0.000...10 = 0.000...1, and from there it immediately follows that they both equal 0.

Let me help you:

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

10*(1 - 0.999...) = 10*(0.000...1) = 0.000...10

By your (9 - 9) + (1 - 0.999...) nonsense you simply eliminated the fact that both equation sides were multiplied by 10.

Let us demonstrate your nonsense by local numbers:

1 - 0.99 = 0.01

10*(1 - 0.99) = 10*(0.01) = 0.10

((9 - 9) + (1 - 0.99) = 0.01) ≠ (10*(1 - 0.99) = 10*(0.01) = 0.10)

You are just making this nonsense up on the fly, aren't you?

Care to readjust things in the hopes of not repeating your nonsense again?

 

[jsfisher]

Let me help you:

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

10*(1 - 0.999...) = 10*(0.000...1) = 0.000...10

Yes, that's your latest notational nonsense.

By your (9 - 9) + (1 - 0.999...) nonsense you simply eliminated the fact that both equation sides were multiplied by 10.

It is not nonsense at all. It's called arithmetic. It provides several equivalent ways to evaluate an expression, all providing identical answers. Arithmetic is nicely consistent that way.

By one sequence of operations, I can transform 10-9.999... into 10*(1-0.999...) and then into 0.000...10. By another sequence, I can transform 10-9.999... into (10-9)-0.999... and then into 0.000...1.

Nothing was eliminated and no facts were lost in the process. However, it does show that 0.000...10 and 0.000...1 have exactly the same value, and that value must be 0.


Heck, given Doronetics accepts that 10*0.999... = 9.999..., it can be directly shown 0.999... is identical to 1, despite your inconsistent assertion to the contrary.

 

[doronshadmi]

By one sequence of operations, I can transform 10-9.999... into 10*(1-0.999...) and then into 0.000...10. By another sequence, I can transform 10-9.999... into (10-9)-0.999... and then into 0.000...1.

You provide two different results by two different sequences.

Transforming results do not change the fact that there is more than one result.

However, it does show that 0.000...10 and 0.000...1 have exactly the same value, and that value must be 0.

Only in your nonsense dreams.

Transforming results do not change the fact that there is more than one result.

Heck, given Doronetics accepts that 10*0.999... = 9.999..., it can be directly shown 0.999... is identical to 1, despite your inconsistent assertion to the contrary.

It can't be shown since 10 - 9.999... = 0.000...10

jsfisher, once again your local-only reasoning forces its view on the co-existence of non-locality with locality.

 

[epix]

Let us demonstrate your nonsense by local numbers:

1 - 0.99 = 0.01

10*(1 - 0.99) = 10*(0.01) = 0.10

((9 - 9) + (1 - 0.99) = 0.01) ≠ (10*(1 - 0.99) = 10*(0.01) = 0.10)
You are just making this nonsense up on the fly, aren't you?

Care to readjust things in the hopes of not repeating your nonsense again?

But that argument isn't a part of your "correction" which says

10*(1.0-0.999...) = 10*(0.000...1) = 0.000...10 , which is 10 times greater than 0.000...1

In other words spoken by induction, your corrective measure holds 0.10 > 0.1; it doesn't mention any difference between 0.10 and 0.01 that you are trying to sneak in and which jsfisher never referred to.

I think it's the Straw man fallacy that you loaded your musket with, but there is the Red herring showing up too, I think. The Man knows all of them fallacies, so maybe he can place your latest activity right.

 

[jsfisher]

You provide two different results by two different sequences.

Transforming results do not change the fact that there is more than one result.

No. Arithmetic guarantees the results are identical. Didn't you ever learn any arithmetic, Doron?

Whether you admit it or not,

10 - 9.999... = (10*1) - (10*0.999...) = 10 * (1 - 0.999...) and
10 - 9.999... = 10 - (9 + 0.999...) = (10 - 9) - 0.999... = 1 - 0.999...

Arithmetic is obsessively consistent that way, unlike Doronetics.

 

[doronshadmi]

No. Arithmetic guarantees the results are identical. Didn't you ever learn any arithmetic, Doron?

Whether you admit it or not,

10 - 9.999... = (10*1) - (10*0.999...) = 10 * (1 - 0.999...) and
10 - 9.999... = 10 - (9 + 0.999...) = (10 - 9) - 0.999... = 1 - 0.999...

Arithmetic is obsessively consistent that way, unlike Doronetics.

Let us demonstrate your nonsense (whether you admit it or not) by using only local numbers:

1 - 0.99 = 0.01

10*(1 - 0.99) = 10*(0.01) = 0.10

0.01 ≠ 0.1

Let us demonstrate your nonsense (whether you admit it or not) by using also non-local numbers:

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

10*(1 - 0.999...) = 10*(0.000...1) = 0.000...10

0.000...1 ≠ 0.000...10

-----------

Didn't you ever learn the fundamental principle of Equation (changes are equally done at both sides)?

 

[doronshadmi]

But that argument isn't a part of your "correction" which says

In other words spoken by induction, your corrective measure holds 0.10 > 0.1; it doesn't mention any difference between 0.10 and 0.01 that you are trying to sneak in and which jsfisher never referred to.

I think it's the Straw man fallacy that you loaded your musket with, but there is the Red herring showing up too, I think. The Man knows all of them fallacies, so maybe he can place your latest activity right.

http://www.internationalskeptics.com/forums/showpost.php?p=7374411&postcount=15946

 

[jsfisher]

Let us demonstrate your nonsense (whether you admit it or not) by using only local numbers:

1 - 0.99 = 0.01

10*(1 - 0.99) = 10*(0.01) = 0.10

0.01 ≠ 0.1

Nobody has disputed this. Why do you insist on irrelevant arguments made of straw?

Let's stick to the basic derivation, as presented. You claim 0.000...1 and 0.000...10 are different numbers. So, where is the mistake in the following?

(1a) 10 - 9.999... = (10*1) - (10*0.999...)
(1b) (10*1) - (10*0.999...) = 10 * (1 - 0.999...)
(1c) 10 * (1 - 0.999...) = 10 * (0.000...1)
(1d) 10 * (0.000...1) = 0.000...10

(2a) 10 - 9.999... = 10 - (9 + 0.999...)
(2b) 10 - (9 + 0.999...) = (10 - 9) - 0.999...
(2c) (10 - 9) - 0.999... = 1 - 0.999...
(2d) 1 - 0.999... = 0.000...1

(3) Therefore, 0.000...10 = 0.000...1

Surely, Doron, there must be a mistake in the arithmetic, right? Otherwise, Doronetics would, once again, be fundamentally wrong.

 

[epix]

Let us demonstrate your nonsense (whether you admit it or not) by using only local numbers:

1 - 0.99 = 0.01

10*(1 - 0.99) = 10*(0.01) = 0.10

0.01 ≠ 0.1

Let us demonstrate your nonsense (whether you admit it or not) by using also non-local numbers:

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

10*(1 - 0.999...) = 10*(0.000...1) = 0.000...10

0.000...1 ≠ 0.000...10

-----------

Didn't you ever learn the fundamental principle of Equation (changes are equally done at both sides)?

Why do you use these local and non-local numbers to verify correctness of a statement when algebraic substitution does the job better?

10 - 9.999... = (10*1) - (10*0.999...) = 10 * (1 - 0.999...) and
10 - 9.999... = 10 - (9 + 0.999...) = (10 - 9) - 0.999... = 1 - 0.999...

Just set 10 = a, 9.999...= b and run jsfisher's statement that way:

1. a - b = (a*1) - (a*b/10) = a*(1 - b/10) = a - ab/10

2. a - b = a - (a - 1 + b/10) = (a - (a - 1)) - b/10 = 1 - b/10

Now you can clearly see that the results don't equal and therefore don't follow the premise (a - b) = (a - b). And so what? Inequalities are a part of mathematics, aren't they?

 

[doronshadmi]

Nobody has disputed this. Why do you insist on irrelevant arguments made of straw?

Let's stick to the basic derivation, as presented. You claim 0.000...1 and 0.000...10 are different numbers. So, where is the mistake in the following?

(1a) 10 - 9.999... = (10*1) - (10*0.999...)
(1b) (10*1) - (10*0.999...) = 10 * (1 - 0.999...)
(1c) 10 * (1 - 0.999...) = 10 * (0.000...1)
(1d) 10 * (0.000...1) = 0.000...10

(2a) 10 - 9.999... = 10 - (9 + 0.999...)
(2b) 10 - (9 + 0.999...) = (10 - 9) - 0.999...
(2c) (10 - 9) - 0.999... = 1 - 0.999...
(2d) 1 - 0.999... = 0.000...1

(3) Therefore, 0.000...10 = 0.000...1

Surely, Doron, there must be a mistake in the arithmetic, right? Otherwise, Doronetics would, once again, be fundamentally wrong.

You still do not get all of what is written in http://www.internationalskeptics.com/forums/showpost.php?p=7374411&postcount=15946, isn't it jsfisher?

 

[jsfisher]

You still do not get all of what is written in http://www.internationalskeptics.com/forums/showpost.php?p=7374411&postcount=15946, isn't it jsfisher?

No, you are again projecting onto others your own failings. Your cited post doesn't do anything to dispute my proof 0.000...1 = 0.000...10.

All you need do is point out which step is wrong and state why its wrong. What could be easier?

 

[doronshadmi]

No, you are again projecting onto others your own failings. Your cited post doesn't do anything to dispute my proof 0.000...1 = 0.000...10.

All you need do is point out which step is wrong and state why its wrong. What could be easier?

In

(2b) 10 - (9 + 0.999...) = (10 - 9) - 0.999...
(2c) (10 - 9) - 0.999... = 1 - 0.999...
(2d) 1 - 0.999... = 0.000...1

You isolate 9 from 0.999... in order to get your 0.000...1 result.

 

[epix]

Didn't you ever learn the fundamental principle of Equation (changes are equally done at both sides)?

So why don't you use the fundamental principle to prove that 0.10 > 0.1? See, inequalities are solved by the same laws that govern over the solution of equations. Or are they?

Well let's see. We multiply each side of the Doronian inequality by 10, and then we compare:

a) 0.10 * 10 = 01.0

b) 0.1 * 10 = 01

Since the result of (a) is a collection of 4 ASCII characters and the result of (b) is a collection of only 2 characters, 01.0 > 01, which is consistent with the initial inequality 0.10 > 0.1 (4 characters > 3 characters) and we're done. Well, not yet.
OM to the classrooms, OM to the classrooms...!!!
Now we are.

 

[jsfisher]

In

You isolate 9 from 0.999... in order to get your 0.000...1 result.

Which step? Be specific.

Are you claiming 9.999... does not equal 9 + 0.999... ? That would be a most bizarre claim, even by Doronetics standards. That rates right up there with your insistence 2 may not be a member of set, {2}.

 

[The Man]

Wrong.

AND connective provides a strict result only if both inputs are strict.

Since AB is non-strict, the output is non-strict and the commutativity of AND connective no influence on it.

and agian

Wrong again, as shown before any binary variable AND FALSE is strictly FALSE, while any binary variable AND TRUE is strictly just that binary variable.



Again as stated before that commutative property’s “influence” on the output is that the output can not change due to changes in the ordering of the variables.

You can go around this circle as many times as you like Doron, it won't change anything but it will continue to show how nonsensical your assertions are.

 

[The Man]

The Man, you are unable to get the generalization of my non-local number/ local number example,
which is equivalent to non-strict AB\(strict A;strict B) case.

You are still closed under your fragmented context-dependent reasoning.

So you’re still just going to deliberately ignore the actual question....

So are you now claiming your "AB" is something other than just being strictly your "AB"? That would be typically self contradictory of you. The simple fact remains Doron that you simply do not like limitation least of all your own.

….you are still just deliberately being ignorant.

 

[doronshadmi]

Which step? Be specific.

EDIT:

Let us demonstrate your nonsense (whether you admit it or not) by using only local numbers:

1 - 0.99 = 0.01

10*(1 - 0.99) = 10*1 – 10*0.99 = 10 - 9.9 = 10*(0.01) = 0.10

(1a) 10 - 9.9 = (10*1) - (10*0.99)
(1b) (10*1) - (10*0.99) = 10 * (1 - 0.99)
(1c) 10 * (1 - 0.99) = 10 * (0.01)
(1d) 10 * (0.01) = 0.10

(2a) 10 - 9.9 = 10 - (9 + 0.9)
(2b) 10 - (9 + 0.9) = (10 - 9) - 0.9
(2c) (10 - 9) - 0.9 = 1 - 0.9
(2d) 1 - 0.9 = 0.10

As you see, in both cases you have got the same 0.10 result, which is different than 1 - 0.99 = 0.01

The same reasoning holds in the case of 10-9.999… = 0.000…10 (which is different than 1 - 0.999… = 0.000…1) as long as no nonsense (2b) trick is used on infinite interpolation.

Are you claiming 9.999... does not equal 9 + 0.999... ?

No, I claim that your arithmetic changed 9.999... to 0.999... by your (10-9)+0.999... nonsense (2b) trick.

In order to get it clearer, your arithmetic eliminates the "shift to the left" of 1 and 0.999..., which are actually 10 and 9.999..., and as a result you get 0.000...1 instead of 0.000...10

Your arithmetic does not have any influence on the result, in the case of finite interpolation, as can be seen above.

Your local-only arithmetic can't show the non-locality of 0.000...1, which is 10 times closer to a given limit (1, in this case) than 0.000...10 (10, in this case) along an infinite interpolation, such that both non-local numbers are < AND = to a given limit, which is a property that no locality has.

 

[doronshadmi]

You can go around this circle as many times as you like Doron, it won't change anything but it will continue to show how nonsensical your assertions are.

The Man, you are the one who running in circles in your context-dependent-only closed box, exactly because you are not using the co-existence of Cross-contexts with Context-depended reasoning.

 

[doronshadmi]

….you are still just deliberately being ignorant.

So, do you still can't get that AB is a non-strict value?

 

[The Man]

The Man, you are the one who running in circles in your context-dependent-only closed box, exactly because you are not using the co-existence of Cross-contexts with Context-depended reasoning.

Again stop simply trying to posit aspects of your own failed reasoning onto others.