Deeper than primes

[doronshadmi]

Learn to read, and learn what words mean. I've already told which word in particular you are misinterpreting, here.

jsfisher, you contradict yourself. Simple at that.

Do you really not get that all I asked was for you to clarify what you meant by d in this context.

Accrding to Standard Math d=0, in the case of 0.999...[base 10], and I clearly wrote it in http://www.internationalskeptics.com/forums/showpost.php?p=4763819&postcount=3266.

Oh, I see. You use it inconsistently. No wonder you couldn't clarify. It's not one thing; it's infinitely many things.

You see exactly nothing.

The fact is that you don't understand what you read, in this case, and you contradict yourself by your own words, in this case ( as clearly seen at http://www.internationalskeptics.com/forums/showpost.php?p=4765238&postcount=3279 ).


Since (1) and (1) are actually the same case, then Standard Math is based on contradiction, in this case, because
by Standard Math d>0 (according to (1)) AND d=0 (according to (2)).

Jsfisher does his best to not see that (1) and (2) are actually the same case (and again, we can use Q members instead of R members, because "dense" holds for both of them, along some given interval), but jsfisher you can't escape form the fact that in this case Standard Math actually claims that d>0 (according to (1)) AND d=0 (according to (2)).

You can use the word "disjoint" as much as you like, but it will not change the fact that you contradict yourself, because in one hand (and in your words) you claim that there is a connection between (1) and (2) AND on the same case you claim that there is no connection between (1) and (2) (they are disjoint and no term in one of them is also the term of the other).


EDIT:

a) d must = 0 , if 0.999…[base 10] = 1

b) d must > 0 , if Y of X<Y does not have an immediate predecessor.

Since (a) and (b) are actually the same case, then Standard Math is based on this contradiction, in this case:

d>0 AND d=0

 

[jsfisher]

You can use the word "disjoint" as much as you like, but it will not change the fact that you contradict yourself, because in one hand (and in your words) you claim that there is a connection between (1) and (2) AND on the same case you claim that there is no connection between (1) and (2) (they are disjoint and no term in one of them is also the term of the other).

So tell me, Doron, are the set of even integers and the set of odd integers disjoint or are they unrelated to each other. Must be one or the other. Which is it?

 

[jsfisher]

a) d must = 0 , if 0.999…[base 10] = 1

b) d must > 0 , if Y of X<Y does not have an immediate predecessor.

Since (a) and (b) are actually the same case, then Standard Math is based on this contradiction, in this case:

d>0 AND d=0

Setting aside for a moment your refusal to clarify what you mean by d in the first case, the assertion high-lighted above is just that - another assertion.

You have provided no proof of the equivalence of (a) and (b).

 

[The Man]

jsfisher, you contradict yourself. Simple at that.



Accrding to Standard Math d=0, in the case of 0.999...[base 10], and I clearly wrote it in http://www.internationalskeptics.com/forums/showpost.php?p=4763819&postcount=3266.



You see exactly nothing.

The fact is that you don't understand what you read, in this case, and you contradict yourself by your own words, in this case ( as clearly seen at http://www.internationalskeptics.com/forums/showpost.php?p=4765238&postcount=3279 ).


Since (1) and (1) are actually the same case, then Standard Math is based on contradiction, in this case, because
by Standard Math d>0 (according to (1)) AND d=0 (according to (2)).

Jsfisher does his best to not see that (1) and (2) are actually the same case (and again, we can use Q members instead of R members, because "dense" holds for both of them, along some given interval), but jsfisher you can't escape form the fact that in this case Standard Math actually claims that d>0 (according to (1)) AND d=0 (according to (2)).

You can use the word "disjoint" as much as you like, but it will not change the fact that you contradict yourself, because in one hand (and in your words) you claim that there is a connection between (1) and (2) AND on the same case you claim that there is no connection between (1) and (2) (they are disjoint and no term in one of them is also the term of the other).


EDIT:

a) d must = 0 , if 0.999…[base 10] = 1

Only if you set d equal to 1-1 because 0.9999… is just an infinite decimal representation of 1 meaning it extends to the infinite decimal place (does not terminate or have recurring zeros from a certain point on).

b) d must > 0 , if Y of X<Y does not have an immediate predecessor.

Again you must first define what constitutes an “immediate predecessor” before you can determine if anything does or does not fit that definition.

As jsfisher notes it is all a matter of what you are using your d variable to represent. In (a) it is the difference between a real number in finite decimal representation and the same number in infinite decimal representation. In (b) it is in fact the difference between two different numbers.

Since (a) and (b) are actually the same case, then Standard Math is based on this contradiction, in this case:

d>0 AND d=0

Again the misunderstanding and contradiction is yours and yours alone as (a) and (b) are clearly not the same case but simply employ the same letter ‘d’ to represent a ‘difference’ in entirely dissimilar concepts. One where the only difference being considered is in how a single value is represented, finite decimal or infinite decimal, which results in no value for that difference and the other where different values with the same representation are being considered.

I would object that you were deliberately trying to deceive since the difference between your (a) and (b) examples is so prominent that anyone doing the slightest research could understand your fallacy. However you have consistently demonstrated both your inability and unwillingness to actually do research or understand the concepts you attempt to refute. As such I am certain that you do not actually see (or simply do not choose to see) the difference between your (a) and (b) examples.

For those actually interested.

http://en.wikipedia.org/wiki/0.999...

 

[doronshadmi]

Only if you set d equal to 1-1 because 0.9999… is just an infinite decimal representation of 1 meaning it extends to the infinite decimal place (does not terminate or have recurring zeros from a certain point on).

So, you have no clue of what d is, but it does not stop you from writing a bunch of nonsense about it.

The rest of your post is nothing but a waste of time.

 

[doronshadmi]

Setting aside for a moment your refusal to clarify what you mean by d in the first case, the assertion high-lighted above is just that - another assertion.

You have provided no proof of the equivalence of (a) and (b).

Just that Q interval "dense" case and 0.999... is actually a non-finite sequence of Q members between 0.9 and 1.

But standard Math takes this same thing and claims that d>0 (the "dense" aspect of it) AND d=0 (the 0.999...=1 aspect of it).

 

[jsfisher]

It may be worth noting, too, that Doron has "proven" in other of this writings that 0.999... does not equal 1 at all. He has even pointed out that the difference between the two (i.e. 1 - 0.999...) is 0.000...1, where the ellipses represent an infinite number of 0's.

I bring this up because Doron is pre-disposed to believe the standard interpretation of 0.999... is already a contradiction all by itself.

 

[jsfisher]

Just that Q inverval "dense" case and 0.999... is actually a non-finite sequence of Q members that have Y as their limit.

No, 0.999... is not an infinite sequence of rational numbers. It is a single number. The value of 0.999... happens to be the limit of an infinite sum, but so what?

And what does any of this have to do with immediate predecessors?

 

[The Man]

So, you have no clue of what d is, but it does not stop you from writing a bunch of nonsense about it.

The rest of your post is is nothing but a waste of time.

Oh the irony of it all.

It may be worth noting, too, that Doron has "proven" in other of this writings that 0.999... does not equal 1 at all. He has even pointed out that the difference between the two (i.e. 1 - 0.999...) is 0.000...1, where the ellipses represent an infinite number of 0's.

I bring this up because Doron is pre-disposed to believe the standard interpretation of 0.999... is already a contradiction all by itself.

I must have missed those particular writings. So his ‘1’ is in the ‘infinite + 1’ decimal place? Kind of makes his interpretation of infinite rather meaningless, but then of course that would make it consistent with the rest of his interpretations.

 

[jsfisher]

I must have missed those particular writings. So his ‘1’ is in the ‘infinite + 1’ decimal place? Kind of makes his interpretation of infinite rather meaningless, but then of course that would make it consistent with the rest of his interpretations.

Just some examples:

http://www.physicsforums.com/archive/index.php/t-5267.html
http://www.geocities.com/complementarytheory/9999.pdf
http://www.internationalskeptics.com/forums/showthread.php?t=112582
http://forum.wolframscience.com/archive/topic/747-1.html

 

[doronshadmi]

No, 0.999... is not an infinite sequence of rational numbers. It is a single number.

0.999... = 0.9 + 0.09 + 0.009 + ... = 1 exactly because d=0 and 0.999... is a sum of a non-finite sequence of rationl numbers, whether you like it or not.

You can call it a single number exactly because d=0.

But when you are talking about the interval of Q members between X<Y, such that X=0.9 and Y=1 then 1 has no immediate predecessor iff d>0 for the non-finite Q sequence 0.9,0.99,0.999, ... ,1.

Both cases are actually the same mathematical object (0.999... and 0.9,0.99,0.999,... ,1 is two representations of a one mathematical object), and as a result Standard Math is based on a contradiction, because it claims that:

d>0 ( 1 does not have an immediate predecessor in 0.9,0.99,0.999,... ,1 representation ) AND d=0 (in 0.999...=1 representation)

 

[jsfisher]

But when you are talking about the interval of Q members between X<Y, such that X=0.9 and Y=1 then 1 has no immediate predecessor iff d>0 for the non-finite Q sequence 0.9,0.99,0.999, ... ,1.

What do infinite sequences have to do with immediate predecessors? For that matter, what do intervals have to do with either one? And how about instead of the interval (0.9, 1), we us the interval (-17.8, 1). What now?

ETA: The bolded part is not an infinite sequence. See below.

Both cases are actually the same mathematical object, and as a result Standard Math is based on a contradiction, because it claims that:

d>0 ( 1 does not have an immediate predecessor in the case of the particular Q interval [X=0.9, Z=0.99,Z=0.999,Z=... ,Y=1), such that X < Z < Y ) AND d=0 (0.999...=1)[/QUOTE]

Ummm, [X=0.9, Z=0.99,Z=0.999,Z=... ,Y=1) is not an interval. What did you intend for this to mean?

Do you also realize that 0.999... does not ever appear in the infinite sequence 0.9, 0.99, 0.999, 0.9999, ..., right? You do know that, right? So, you can refer to the sequence all you like and to your ubiquitous [I]d[/I], but you won't be discussing 0.999....

 

[doronshadmi]

Ummm, [X=0.9, Z=0.99,Z=0.999,Z=... ,Y=1) is not an interval.[/quote]

[X,Y) interval such that [X=0.9, Z=0.99, Z=0.999, Z=... ,Y=1) is exactly the same mathematical object that is represented as 0.999...
[QUOTE="jsfisher, post: 4766738, member: 7169"]
Do you also realize that 0.999... does not ever appear in the infinite sequence 0.9, 0.99, 0.999, 0.9999, ..., right?

A = 0.999... = [X=0.9, Z=0.99, Z=0.999, Z=... ,Y=1)

How can A be one of its parts?

[QUOTE="jsfisher, post: 4765830, member: 7169"]Just some examples:

http://www.physicsforums.com/archive/index.php/t-5267.html
http://www.geocities.com/complementarytheory/9999.pdf
http://www.internationalskeptics.com/forums/showthread.php?t=112582
http://forum.wolframscience.com/archive/topic/747-1.html[/QUOTE]

This is not relevelt to this case, because here I am talking only about the framework of Stantard Math, by using claims of Standard Math.

 

[doronshadmi]

I will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

(In this case X = {0.9,0.99,0.999, ...} where any member of X < Y=1 , but it does not matter because this case can be used as a particular example Without loss of generality)

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

 

[jsfisher]

Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set {X : X<Y} does have a largest element, Z.

(In this case X = {0.9,0.99,0.999, ...,1})

This parenthetic part is multiply wrong. For one, X is not the name of the set. For another, you made a specific example of it in which, apparently, Y is taken as 1. You've also selected some example members that from a sequence implying something other than what is required to be in the set. E.g. -42 is in the set, but your representation would suggest otherwise.

Most important of the errors, though, is that you have 1 as an element of the set. It should not be since it is required that all members be strictly less than 1.

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set {X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set {X : X<Y} does not have a largest element.

Except for where you got lost with that parenthetical aside you added on your own, you copy/pasted the proof just fine.

I'm guessing the original post is undergoing / will undergo an epic Doron edit.

 

[doronshadmi]

This parenthetic part is multiply wrong. For one, X is not the name of the set. For another, you made a specific example of it in which, apparently, Y is taken as 1. You've also selected some example members that from a sequence implying something other than what is required to be in the set. E.g. -42 is in the set, but your representation would suggest otherwise.

Most important of the errors, though, is that you have 1 as an element of the set. It should not be since it is required that all members be strictly less than 1.



Except for where you got lost with that parenthetical aside you added on your own, you copy/pasted the proof just fine.

I'm guessing the original post is undergoing / will undergo an epic Doron edit.

You are right I did not write it correctly, so let us try again:


Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set A={X : X<Y} does have a largest element, Z.

(In this case A = {0.9,0.99,0.999, ...} where any member of A < Y=1 , but it does not matter because this case can be used as a particular example Without loss of generality)

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set A={X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set A={X : X<Y} does not have a largest element.

 

[jsfisher]

You are right I did not wirte it correctly, so let us try again:


Let us follow jsfisher's proof by contradiction:

We will use a simple proof by contradiction. As with all such proofs, it begins with an assumption then proceeds to construct a contradiction, thereby showing the assumption to be false.

Assume the set A={X : X<Y} does have a largest element, Z.

(In this case A = {0.9,0.99,0.999, ...} where any member of A < Y=1 , but it does not matter because this case can be used as a particular example Without loss of generality)

For Z to be an element of the set, Z < Y.
Let h be any element of the interval (Z,Y).
By the construction of h, Z < h < Y.
Since h < Y, h is an element of the set A={X : X<Y}.
Since Z < h, the assumption Z was the largest element of the set has been contradicted.

Therefore, the set A={X : X<Y} does not have a largest element.

You are still misrepresenting the membership of the set in your parenthetical aside. What's the point of this, anyway?

 

[doronshadmi]

You are still misrepresenting the membership of the set in your parenthetical aside.

Please explain in details.

 

[jsfisher]

Please explain in details.

I did, but here goes again on one of the points:

By expressing it as A = {0.9,0.99,0.999, ...} you are implying the membership consists of only those values that fit the sequence. This is misleading. As written, you have implied that -42 is not it the set. It is. As written, you have implied that 0.99988888888888... isn't in the set. It is.

Also important, you haven't shown any value to the discussion by introducing this specific instance of the set {X : X<Y}.

 

[doronshadmi]

I did, but here goes again on one of the points:

By expressing it as A = {0.9,0.99,0.999, ...} you are implying the membership consists of only those values that fit the sequence. This is misleading. As written, you have implied that -42 is not it the set. It is. As written, you have implied that 0.99988888888888... isn't in the set. It is.

Also important, you haven't shown any value to the discussion by introducing this specific instance of the set {X : X<Y}.

Jsfisher, set A is exactly only the particular numbers of the form 0.9,0.99,0.999, ... etc.

Yet this limited case can be used Without loss of generality in your proof by contradiction.