Deeper than primes

[doronshadmi]

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]

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]

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]

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]

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.

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.

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.

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]

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]

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):

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

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)?

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

What do you understand when you do that?

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]

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]

This board doesn't have LaTeX capability

It does, actually.

 

[jsfisher]

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.

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:

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]

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.

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]

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]

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]

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]

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]

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]

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]

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]

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.

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”.

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?

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]

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.