Deeper than primes

doronshadmi · May 28, 2009

jsfisher said:
Don't move the goal posts. There is nothing in your declaration that restricts it to the reals. Here is your statement again. Notice all it mentions are finite collections and infinite collections; there are no restrictions or exclusions other than that:

I have also now given you several examples which contradict your statement. The statement is clearly bogus, so either try again to say what you really mean, clearly and succinctly, or just retract the whole thing.

There is a simple difference between you and me in this case jsfisher, which is:

I did answer to this question.

You did not answer to this question.

doronshadmi · May 28, 2009

The Man said:
Again you simply do not understand the invalid assertion that opens your argument. As has been repeatedly explained the interval [1,2] in the reals is not “a finite amount of R members”. Again if you think it is then list all those members or simply answer jsfishers direct question.

Instead of answering this simple and direct question you obfuscate with a dissertation about the “will to understand the fundamental notions that stand at the basis of any formal definition” while apparently lacking the will to clearly explain what “finite amount of R members” you are talking about.

So to put it in the format you are requesting above; “what is this” “finite amount of R members” you are referring to, “why it is” “a finite amount of R members” and not the infinite amount of members in any interval of the reals and "how it is” being used “in order to say any meaningful thing about a collection of all R members”? Only you can explain yourself and what you mean by “a finite amount of R members”, no amount of philosophical rambling, finite or infinite, can be substituted for a simple and direct explanation by you as to what “finite amount of R members” you are referring to.

The Man, the beautiful thing here is that a real understanding is not what other people understand, but what you, The Man, understand.

You cannot really understand X if you do not ask yourself questions about X. This is the right way to ensure that you did not make any short cut that prevents from you the understanding of X.

In this case, we are talking about the understanding of a real-line interval.

So here is the question again:

The Man, do you understand that you can show that "there is always another real number between any two real numbers" exactly because you are always based on a finite case?

integral · May 28, 2009

doronshadmi said:
The Man, do you understand that you can show that "there is always another real number between any two real numbers" exactly because you are always based on a finite case?

Actually, it's based off of the density property of real numbers - something than can be proven using basic deductive principles. Give it a try!

doronshadmi · May 28, 2009

integral said:
Actually, it's based off of the density property of real numbers - something than can be proven using basic deductive principles. Give it a try!

Please formally show us the density property of real numbers.

The Man · May 28, 2009

doronshadmi said:
The Man, the beautiful thing here is that a real understanding is not what other people understand, but what you, The Man, understand.

You cannot really understand X if you do not ask yourself questions about X. This is the right way to ensure that you did not make any short cut that prevents from you the understanding of X.

In this case, we are talking about the understanding of a real-line interval.

We asked you questions to try and find out what it is you think you are ’understanding’.

So here are those questions again.

jsfisher said:
Exactly what "finite amount of R members" are you referring to? Surely not the two numbers that establish the bounds for an interval. After all, the interval includes an infinite supply of numbers regardless of which two distinct bounds are chosen.

The Man said:
So to put it in the format you are requesting above; “what is this” “finite amount of R members” you are referring to, “why it is” “a finite amount of R members” and not the infinite amount of members in any interval of the reals and "how it is” being used “in order to say any meaningful thing about a collection of all R members”? Only you can explain yourself and what you mean by “a finite amount of R members”, no amount of philosophical rambling, finite or infinite, can be substituted for a simple and direct explanation by you as to what “finite amount of R members” you are referring to.

See Doron the “beautiful thing here is that” you can demonstrate “a real understanding” by answering these simple and direct questions, not about what we understand, “but what you”, Doron, “understand” about what your “finite amount of R members" refers to.

doronshadmi said:
So here is the question again:

The Man, do you understand that you can show that "there is always another real number between any two real numbers" exactly because you are always based on a finite case?

As I have asked you to clarify, I have no idea what you mean by “a finite case” or specifically “you are always based on a finite case”. As such I simply do not understand your question and have already asked you to clarify what you mean by “finite amount of R members” or as you now refer to it “a finite case”.

ETA:

So again you are professing a “step by step” or “serial” only method of understanding. In fact you even warn against taking “any short cut” thus bypassing one or more steps to that understanding. Have you now completely abandoned your notion of ’parallel thinking’? Perhaps it is just like most aspects of your notions that you simply invoke when you think it will help you obfuscate some point or issue raised by others yet completely ignore when you think you are raising some point or issue to others.

integral · May 28, 2009

doronshadmi said:
Please formally show us the density property of real numbers.

This board doesn't have LaTeX capability, and I can't post URLs yet. Just do a google search if you want to see it. Seriously though, a sophomore or junior level math major could reason through it.

If you've studied any introductory course or book in analysis or abstract algebra should cover this kind of deduction. "A course of pure mathematics" by Hardy is a good start to general modern mathematics.

doronshadmi · May 28, 2009

The Man said:
As I have asked you to clarify, I have no idea what you mean by “a finite case” or specifically “you are always based on a finite case”.

My question is a response to jsfisher's phrase. Here it is again (with some addition):

jsfisher said:
There is no immediate predecessor because there is always another real number between any two real numbers.

doronshadmi said:

Do you understand that you can show that "there is always another real number between any two real numbers" exactly because you are always based on a finite case (in order to derive to such a conclusion)?

Click to expand...

Please look both on jsfisher's explanation AND on my question to his explanation.

What do you understand when you do that?

The Man said:
ETA:

So again you are professing a “step by step” or “serial” only method of understanding.

No, in the last posts (and I clearly wrote it) I use only Standard Math framework.

dlorde · May 28, 2009

integral said:
Just do a google search if you want to see it. Seriously though, a sophomore or junior level math major could reason through it.

Wikipedia has this article on Dense Sets, where it mentions "The real numbers with the usual topology have the rational numbers and the irrational numbers as dense subsets". Seems straight-forward enough to me.

zooterkin · May 28, 2009

integral said:
This board doesn't have LaTeX capability

It does, actually.

jsfisher · May 28, 2009

doronshadmi said:
The Man said:

As I have asked you to clarify, I have no idea what you mean by “a finite case” or specifically “you are always based on a finite case”.

Click to expand...

My question is a response to jsfisher's phrase.

The Man was really asking you two important questions. You didn't answer them. You've been asked the same questions several times by the The Man and myself because they are important questions. You didn't answer them.

I'll give it one more try:

What do you mean by a finite case?
What do you mean by you are always based on a finite case?

Without your clarification for those two phrases, the following cannot be answered:

doronshadmi said:
Do you understand that you can show that "there is always another real number between any two real numbers" exactly because you are always based on a finite case

(Emphasis altered from the original.)

doronshadmi · May 28, 2009

x < y iff x ≤ y and x ≠ y , and this definition enables the existence of different elements along the real-line.

It means that we do not need more than two elements for the existence of difference along the real-line.

wiki said:
A partial order is a binary relation "≤" over a set P which is reflexive, antisymmetric, and transitive, i.e., for all a, b, and c in P, we have that:
• a ≤ a (reflexivity);
• if a ≤ b and b ≤ a then a = b (antisymmetry);
• if a ≤ b and b ≤ c then a ≤ c (transitivity).

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

In mathematics, a partial order ≤ on a set X is said to be dense (or dense-in-itself) if, for all x and y in X for which x < y, there is a z in X such that x < z < y.

Can set X be a set of integers (according to the above)?

jsfisher · May 29, 2009

doronshadmi said:
Can set X be a set of integers (according to the above)?

If you read the very next two sentences of the wiki article you posted, you'd know the answer already.

But why yet another tangent? Please answer the questions, now repeated many times:

What do you mean by a finite case?
What do you mean by you are always based on a finite case?

sympathic · May 29, 2009

doronshadmi said:
x < y iff x ≤ y and x ≠ y , and this definition enables the existence of different elements along the real-line.

It means that we do not need more than two elements for the existence of difference along the real-line.

Can set X be a set of integers (according to the above)?

And you ask us to show examples for your lack of knowledge in Math?

zooterkin · May 29, 2009

doronshadmi said:
x < y iff x ≤ y and x ≠ y

Why do you feel the need to add 'iff x ≤ y and x ≠ y'? In what other circumstances can x be less than y?

doronshadmi · May 29, 2009

jsfisher, you did not think about it, so here is it again:

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

In mathematics, a partial order ≤ on a set X is said to be dense (or dense-in-itself) if, for all x and y in X for which x < y, there is a z in X such that x < z < y.

I think (and maybe I am wrong) that there is no problem to show that X is a set of integers (according to the definition above) as follows:

If, for all x and y in X for which x < y (and x and y are integers), then there is (integer) z in X such that x < z < y.

Here are some examles:

x=1 < z=2 < y=3
x=2 < z=3 < y=4
x=4 < z=5 < y=6
x=6 < z=7 < y=8
...

zooterkin · May 29, 2009

doronshadmi said:
jsfisher, you did not think about it, so here is it again:

I think (and maybe I am wrong) that there is no problem to show that X is a set of integers (according to the definition above) as follows:

If, for all x and y in X for which x < y (and x and y are integers), then there is (integer) z in X such that x < z < y.

Here are some examles:

x=1 < z=2 < y=3
x=2 < z=3 < y=4
x=4 < z=5 < y=6
x=6 < z=7 < y=8
...

What are you trying to show? What about x=1, y=2?

How about answering the questions you've been asked above?

doronshadmi · May 29, 2009

zooterkin said:
What are you trying to show? What about x=1, y=2?

How about answering the questions you've been asked above?

Very good zooterkin, you start to get the point.

It is easy to show that my previous post is wrong, exactly because the integers' construction is based on a finite case of x < y (because there is nothing between x and y).

But this is not the case about R construction that is based on all non-finite elements that exist between x and y (where x < y).

Furthermore, it is possible to show that there is a room for z between x and y (where x < y) exactly because we are using only a finite case of 3 different elements (which are notated as x,z and y) such that x < z < y.

Since x < z < y (which is a finite case) is not the case of x,y AND the all non-finite elements between x and y, it cannot be used in order to conclude anything about x,y and the all non-finite elements between x and y.

In other words, we can use a finite construction, in order to prove something about the integers.

We cannot use a finite construction (like x < z < y) in order to prove something about the reals.

To make it more clear, z (which is a single arbitrary R member) is not all the members between x and y, and as a result we cannot use the finite case x < z < y in order to conclude what happens between z and y.

jsfisher · May 29, 2009

doronshadmi said:
I think (and maybe I am wrong) that there is no problem to show that X is a set of integers (according to the definition above) as follows:

You may think that, but you are wrong.

If, for all x and y in X for which x < y (and x and y are integers), then there is (integer) z in X such that x < z < y.

Here are some examles:

x=1 < z=2 < y=3
x=2 < z=3 < y=4
x=4 < z=5 < y=6
x=6 < z=7 < y=8
...

This is what is known as confirmation bias. You are considering only examples that work. Here's one that doesn't: Let X be 1 and Y be 2.

The Man · May 29, 2009

doronshadmi said:
Very good zooterkin, you start to get the point.

It is easy to show that my previous post is wrong, exactly because the integers' construction is based on a finite case of x < y (because there is nothing between x and y).

Only if Y = X+1, other then that particular circumstance in the integers there will be at least one integer between X and Y.

doronshadmi said:
But this is not the case about R construction that is based on all non-finite elements that exist between x and y (where x < y).

Furthermore, it is possible to show that there is a room for z between x and y (where x < y) exactly because we are using only a finite case of 3 different elements (which are notated as x,z and y) such that x < z < y.

Now was that so hard, all you had to do was specify what you were referring to by your “finite case”.

doronshadmi said:
Since x < z < y (which is a finite case) is not the case of x,y AND the all non-finite elements between x and y, it cannot be used in order to conclude anything about x,y and the all non-finite elements between x and y.

You do understand what a variable is don’t you?

doronshadmi said:
In other words, we can use a finite construction, in order to prove something about the integers.

We cannot use a finite construction (like x < z < y) in order to prove something about the reals.

To make it more clear, z (which is a single arbitrary R member) is not all the members between x and y, and as a result we cannot use the finite case x < z < y in order to conclude what happens between z and y.

Well only each instance of Z is a “a single arbitrary R member” for example Z₁ < Z₂ < Z₃… <Z_n if we were to specify the indices of Z in that fashion. Remember a variable is, well, variable, we are not limited to just one particular instance of that variable hence the use of index notation. The set of all possible values for Z (Z₁ ….Z_n where n has the interval [1,¥) in the integers) is in fact all real numbers between X and Y. So your finite case of only one value for Z is only finite if you choose to make it so limited of a consideration, that limitation is not inherent in the math or the notations. Index notation allows for considerable variability as we are not limited to a singular instance of Z, X or Y.

For example given X < Z < Y just in the integers and let X₁ = 1 with Y₁ =5 then we can have the following values for Z₁ in index notation.

Z_1,1 = 2
Z_1,2 = 3
Z_1,3 = 4

If we then let let X₂ = 3 with Y₂ =8 then we can have the following values for Z₂ in index notation.

Z_2,1 = 4
Z_2,2 = 5
Z_2,3 = 6
Z_2,4 = 7

The set for all values of Z₁ would have an intersection with the set for all values of Z₂ as the value 4 which would represent Z_1,3 and Z_2,1 in this case.

The point being again that variables are variable, how one chooses to limit those variables depends on the application being considered. That you choose to limit Z to just “a single arbitrary” value is just your chosen limitation and thus so is your ‘finite case’.

jsfisher · May 29, 2009

doronshadmi said:
It is easy to show that my previous post is wrong

Yes.

...exactly because the integers' construction is based on a finite case of x < y (because there is nothing between x and y).

No. The construction of the integers is not based on the less-than operator.

The Man's post deals with other doronisms in the post, so I'll not repeat them here.

doronshadmi · May 29, 2009

jsfisher said:
You may think that, but you are wrong.

This is what is known as confirmation bias. You are considering only examples that work. Here's one that doesn't: Let X be 1 and Y be 2.

I made this mistake for perpuse.

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=4759564&postcount=3217 .

doronshadmi · May 29, 2009

doronshadmi said:
Very good zooterkin, you start to get the point.

It is easy to show that my previous post is wrong, exactly because the integers' construction is based on a finite case of X < Y (because there is nothing between X and Y).

The Man said:

Only if Y = X+1, other then that particular circumstance in the integers there will be at least one integer between X and Y.

Click to expand...

You have missed my point here.

The finite construction that shows a difference between integers, is based on X < X (such that Y=X+1), where there is no other integer between X and Y.

This finite construction has a meaning according to Standard Math (for example, in the case of a non-finite set of all integers ≤ X) exactly because this finite construction is the essential building-block of what integer is.

The Man said:
Well only each instance of Z is a “a single arbitrary R member” for example Z1 < Z2 < Z3… <Zn if we were to specify the indices of Z in that fashion. Remember a variable is, well, variable, we are not limited to just one particular instance of that variable hence the use of index notation. The set of all possible values for Z (Z1 ….Zn where n has the interval [1,[1,¥) ) in the integers) is in fact all real numbers between X and Y.

The Man, look what you are doing:

1) you are using an index that is based on all natural numbers, in order to define an uncountable set of all real numbers between X and Y. As a result, the claim "all real numbers between X and Y" is false.

2) In order to conclude that X < Y, there must d, such that d > 0 AND Y=X+d

In the case of integers d=1, and there is no problem to define a collection (finite or not) of all different elements (according to Standard Math).

In the case of the reals d can be any arbitrary value > 0, and as a result there is a non-finite regression of d values, such that X < X + (non-finite regression of d value).

Because of this non-finite regression of d value, the claim "all real numbers between X and Y" is false.

But if we claim that "all real numbers between X and Y exist in that interval, without a single exaction", then Y must have an immediate predecessor.

Since Standard Math claims that "all"="non-finite regression" its reasoning about the reals is based on a contradiction.

jsfisher · May 29, 2009

doronshadmi said:
if we claim that "all real numbers between X and Y exist in that interval, without a single exaction", then Y must have an immediate successor.

No. And repeating this false claim isn't going to change its truth.

Since Standard Math claims that "all"="non-finite regression" its reasoning about the reals is based on a contradiction.

No. Standard Math makes no such claim, and so your conclusion is without merit.

jsfisher · May 29, 2009

doronshadmi said:
I made this mistake for perpuse.

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=4759564&postcount=3217 .

I did look at and have already responded to that post. It begins with a false premise, so nothing after that is of any merit.

doronshadmi · May 29, 2009

jsfisher said:
No. And repeating this false claim isn't going to change its truth.

Prove that this is a false claim:

"there is always another real number between any two real numbers" = "a non-finite regression of d value"

jsfisher said:
No. Standard Math makes no such claim, and so your conclusion is without merit.

Standard Math makes exactly this claim, which is:

"all" = "there is always another real number between any two real numbers" = "a non-finite regression of d value"

doronshadmi · May 29, 2009

jsfisher said:
I did look at and have already responded to that post. It begins with a false premise, so nothing after that is of any merit.

Respond but not understand.

zooterkin · May 29, 2009

doronshadmi said:
"there is always another real number between any two real numbers" = "a non-finite regression of d value"

Are you saying that there isn't always another real number between any two real numbers?

Standard Math makes exactly this claim, which is:

"all" = "there is always another real number between any two real numbers" = "a non-finite regression of d value"

I don't think "Standard Math(s)" defines 'all' like that anywhere.

jsfisher · May 29, 2009

doronshadmi said:
Prove that this is a false claim

Since the initial claim was yours, the burden of proof falls to you.

"there is always another real number between any two real numbers" = "a non-finite regression of d value"

Again, you attempt to move the goal posts. This isn't the claim currently at issue.

Standard Math makes exactly this claim, which is:

"all" = "there is always another real number between any two real numbers" = "a non-finite regression of d value"

Again, this is your claim, so the obligation to provide evidence is yours.

Permit me to provide a simple example: Please prove that in Standard Math, the "all" in

All Greeks are liars.

is equivalent to "there is always another real number between any two real numbers" or "a non-finite regression of d value". If nothing else, the following doesn't quite seem right:

A non-finite regression of d value Greeks are liars.

...but that's just me.

The Man · May 30, 2009

doronshadmi said:
You have missed my point here.

The finite construction that shows a difference between integers, is based on X < X (such that Y=X+1), where there is no other integer between X and Y.

So I missed your point because you are confirming what I said that Y=X+1 is just a particular case of X < Y in the intervals where no integer is between X and Y just in that particular case?

doronshadmi said:
This finite construction has a meaning according to Standard Math (for example, in the case of a non-finite set of all integers ≤ X) exactly because this finite construction is the essential building block of what integer is.

Um in the case of the set of all integers that would be an infinite construction as that would not be a finite interval, but of course that has been explained to your before.

doronshadmi said:
The Man, look what you are doing:

1) you are using an index that is based on all natural numbers, in order to define an uncountable set of all real numbers between X and Y. As a result, the claim "all real numbers between X and Y" is false.

Sorry but that is how indices work, mostly for simplicity. An index of X_1.3 or Y_0.46756 would just get a very confusing. That is the flexibility of indices you have an infinite number of instances with each index and you can use an infinite number of indices, basically giving infinite infinities. However I should have been more specific about that in the previous post and that was my mistake. It still does not change the fact that your “finite case” for “a single arbitrary” value of Z is just a limitation you have chosen to impose.

doronshadmi said:
2) In order to conclude that X < Y, there must d, such that d > 0. AND Y=X+d

See here you’ve got it all backasswords again X < Y is not the conclusion but the stipulation which of course requires some difference ‘d’ between X and Y that is simply trivial.

doronshadmi said:
In the case of integers d=1, and there is no problem to define a collection (finite or not) of all different elements (according to Standard Math).

Just as there is no problem defining a set (finite or not) in the case of real numbers, you do realize that the set of all integers is just a subset of the set of all real numbers.

doronshadmi said:
In the case of the reals d can be any arbitrary value > 0, and as a result there is a non-finite regression of d values, such that X < X + (non-finite regression of d value).

Because of this non-finite regression of d values, the claim "all real numbers between X and Y" is false.

What the heck is this “non-finite regression of d value” ?

doronshadmi said:
But if we claim that "all real numbers between X and Y exist in that interval, without a single exaction", then Y must have an immediate successor.

That is your claim so again you must define how that real number ‘immediate successor’ is determined.

doronshadmi said:
Since Standard Math claims that "all"="non-finite regression" its reasoning about the reals is based on a contradiction.

Please show use where any standard math text claims “"all"="non-finite regression"“ whatever that is suppose to mean

doronshadmi · May 30, 2009

jsfisher said:
Since the initial claim was yours, the burden of proof falls to you.

Again, you attempt to move the goal posts. This isn't the claim currently at issue.

Again, this is your claim, so the obligation to provide evidence is yours.

Permit me to provide a simple example: Please prove that in Standard Math, the "all" in

All Greeks are liars.

is equivalent to "there is always another real number between any two real numbers" or "a non-finite regression of d value". If nothing else, the following doesn't quite seem right:

A non-finite regression of d value Greeks are liars.

...but that's just me.

Jsfisher you still don't get it.

Let us do it this way:

Y=X+d

X < Y only if d is explicitly defined.

In other words, the existence of X<Y expression depends of the explicit value of d.

Since d's value is at non-finite regression (and therefore not explicitly defined), then X<Y expression does not exist.

As a result X < Z < Y expression does not exist.

Furthermore, no interval of R members exists.

The Man · May 30, 2009

doronshadmi said:
Jsfisher you still don't get it.

Let us do it this way:

Y=X+d

X < Y only if d is explicitly defined.

No it does not need to be explicitly defined, it just needs to be a non-zero and non-negative value which are the limits placed on it by your “X < Y” requirement.

doronshadmi said:
In other words, the existence of X<Y expression depends of the explicit value of d.

Again no, the limits of the variable ‘d’ depend on and are the result of your “existence of X<Y expression”, which sets those limits for that variable.

doronshadmi said:
Since d's value is at non-finite regression (and therefore not explicitly defined), then X<Y expression does not exist.

As a result X < Z < Y expression does not exist.

Again what the heck does this “d's value is at non-finite regression” mean anyway. Again “d” is a variable specifically because its value is “not explicitly defined” although some of the limits on the possible values of d might be defined to varying degrees.

doronshadmi said:
Furthermore, no interval of R members exists.

Great so now intervals of the real numbers do not exist.

jsfisher · May 30, 2009

doronshadmi said:
Furthermore, no interval of R members exists.

Wow! That's quite the leap into the darkness, Doron.

The Man · May 30, 2009

Ok let’s use Doron’s ‘d’ variable and interpretations to check something.

Doron claims that without an explicit defined value for ‘d’ the expression X<Y does not hold where Y =X + d.

He also claims that any real number has an immediate predecessor and immediate successor in the real numbers.

That gives us four variables or just a ‘finite case’, by Doron’s assertions, as follows

X = A real number

X_P = Immediate real number predecessor to X

X_S = Immediate real number successor to X

d = X- X_P = X_S -X

Such that X_P < X and X < X_S with no other real numbers between those values (sounds like integers where d = 1 doesn‘t it?).

As noted before without an explicit defined value for d, X_P < X and X < X_S do not hold.

So Doron give us your explicit defined value for ‘d’ that is the result of a difference between real numbers such that X_P + d /2 is not in the interval [X_P, X] and X + d/2 is not in the interval [X, X_S]?

I predict that Doron will claim his d value as an atom (his indivisible and non-composite atom as opposed to his atoms composed as divisions) as one of his ‘non-local’ thingies. Claiming it is a superposition of the X_P and X or X and X_S ‘state‘ values. Not realizing of course that superposition is just a linier addition in this case resulting in his d value then being actually 2X-d or 2X+d. Making 2d = 2X thus d = X or d-d = 2X. Unfortunately this would also mean his d can never have an ‘explicitly defined value’ that meets the given requirements. Even in his usual misinterpretation of superposition meaning ‘having more then one value’. Thus Doron’s augment about the deficiencies of ‘standard math’ are just a projection of the very deficiencies within his own notions that he apparently can not bring himself to admit.

doronshadmi · May 30, 2009

The Man said:
Ok let’s use Doron’s ‘d’ variable and interpretations to check something.

Doron claims that without an explicit defined value for ‘d’ the expression X<Y does not hold where Y =X + d.

He also claims that any real number has an immediate predecessor and immediate successor in the real numbers.

You have missed it.

Y have no immediate predecessor in Standard Math framework, exactly because d value is a non-finite regression, such that this non-finite regression > 0.

If the term all is used, then there is no non-finite regression > 0, and d=0 (there is no other way, because if we use all on the non-finite, then d (which is an absolute value) must have the minimal absolute value, which is 0) .

In that case X<Y=X+0, is false.

Standard Math is based on a contradiction, in this case, because it claims that:

d > 0 (there is no immediate predecessor to Y) AND d=0 (the non-finite universe between X and Y, is completely filled).

EDIT:

"there is always another real number between any two real numbers" = "a non-finite regression of d value" = "the non-finite universe between X and Y is not completely filled"

jsfisher · May 30, 2009

doronshadmi said:
Y have no immediate predecessor in Standard Math framework, exactly because d value is a non-finite regression, such that this non-finite regression > 0.

Ok, well, I think you've answered a previous question of mine. Regression is now the latest word de jour of Doronetics.

And, no surprise, you aren't using the term correctly.

...Standard Math is based on a contradiction, in this case, because it claims that:

d > 0 (there is no immediate predecessor to Y) AND d=0 (the universe between X and Y, is completely filled).

Standard Mathematics makes no such claim about values for some variable, d, being both positive and zero. That's just something you keeping trying to twist because you really don't have a very good understanding of infinite sets, or of the infinities, in general.

However, Mathematics does establish the fact that between any two real numbers there is a real number, and from this conclude that no real number has an immediate predecessor. The latter is not a contradiction; it is a simple consequence.

jsfisher · May 30, 2009

By the way, doron, note that the following expresses the notion that the real numbers are dense, and it does it without the appearance of any variable, d:

[latex]$$$ \forall x \forall y, (x < y) \Rightarrow \exists z, (x < z) \wedge (z < y) $$$[/latex]

(Ok, ok, I admit it. I just wanted to play with Latex.)

doronshadmi · May 30, 2009

jsfisher said:
Ok, well, I think you've answered a previous question of mine. Regression is now the latest word de jour of Doronetics.

And, no surprise, you aren't using the term correctly.

Standard Mathematics makes no such claim about values for some variable, d, being both positive and zero. That's just something you keeping trying to twist because you really don't have a very good understanding of infinite sets, or of the infinities, in general.

However, Mathematics does establish the fact that between any two real numbers there is a real number, and from this conclude that no real number has an immediate predecessor. The latter is not a contradiction; it is a simple consequence.

Since Standard Math claims that the universe between X and Y is (completely filled) AND (there is no immediate predecessor to Y), than Standard Math actually claims that d is both > AND = 0 (which is a contradiction under Standard Math framework).

jsfisher · May 30, 2009

doronshadmi said:
Since Standard Math claims that the universe between X and Y is (completely filled) AND (there is no immediate predecessor to Y), that Standard Math actually claims that d is both > AND = 0 (which is a contradiction under Standard Math framework).

Repeating a false statement doesn't change its truth value. This "d dependency" is just your invention to cover your misunderstanding.

doronshadmi · May 30, 2009

jsfisher said:
By the way, doron, note that the following expresses the notion that the real numbers are dense, and it does it without the appearance of any variable, d:

[latex]$$$ \forall x \forall y, (x < y) \Rightarrow \exists z, (x < z) \wedge (z < y) $$$[/latex]

(Ok, ok, I admit it. I just wanted to play with Latex.)

It appears under the name z.

doronshadmi · May 30, 2009

jsfisher said:
Repeating a false statement doesn't change its truth value. This "d dependency" is just your invention to cover your misunderstanding.

It it ture jsfisher.

Actually I used here your phrases: (completely filled) AND (there is no immediate predecessor to Y).

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