Deeper than primes

[jsfisher]

I believe you.

Excellent! Now everyone accepts that what you say about Mathematics isn't true.

 

[doronshadmi]

The fact that none of what you wrote about appears in what I have written suggests the failure is solidly yours.

Really?

Z < Y or h < Y are two finite cases, where each finite case is between two distinct cases.

It is easily understood that no infinitely many cases of finite cases (where each finite case has cardinality 2) can conclude something about the non-finite case of [X,Y].

Jsfisher, you play with words and symbols without notion.

 

[doronshadmi]

Excellent! Now everyone accepts that what you say about Mathematics isn't true.

Here is another example of a step-by-step serial thinker.

He does not read all of the post, in order to get any part of the post.

 

[doronshadmi]

Interesting aside:

The work function for cesium is 1.95 eV. That divided Planck's constant is 471 THz. This can also be expressed as 636 nm, not that that matters, of course.

Does anyone else find this interesting?

These are all finite cases, so?

 

[jsfisher]

Really?

Yes, really.

Z < Y or h < Y are two finite cases, where each finite case is between two distinct cases.

It is easily understood that no infinitely many cases of finite cases (where each finite case has cardinality 2) can conclude something about the non-finite case of [X,Y].

And just what does that have to do with what I wrote? The interval [X,Y] nowhere appears in it.

Seriously, find some remedial reading course. It can only help.

 

[jsfisher]

These are all finite cases, so?

It doesn't surprise me at all that the meaning of my post completely escaped you. Perhaps someone else will be kind enough to explain it to you.

(Hint: What does the work function divided by Planck's constant give you?)

 

[doronshadmi]

It doesn't surprise me at all that the meaning of my post completely escaped you. Perhaps someone else will be kind enough to explain it to you.

(Hint: What does the work function divided by Planck's constant give you?)

The same result as you give, if the number system is the particular case of distinct elements.

Again, Planck's constant is used exactly because Physics can't deal with the non-finite (Planck's constant is used or Renormalization is used in Physics in order to avoid the non-finite).

 

[doronshadmi]

Yes, really.



And just what does that have to do with what I wrote? The interval [X,Y] nowhere appears in it.

Seriously, find some remedial reading course. It can only help.

Are Z , Y or h have distinct values?

If they have, then [Z,Y] or [h,Y] are twe finite cases of distinct values, that can't be used in order to conclude anything about [X,Y] non-finite interval.

 

[zooterkin]

Still waiting for the definition of 'crisp' and some example crisp ids.

 

[doronshadmi]

1. If h is an element of the interval (Z,Y), can we not conclude h < Y?
2. If h is an element of the interval (Z,Y), can we not conclude Z < h?

Since Z or h or Y must have distinct values in your argumant, you deal with the finite case of Z < h < Y, that cannot be used on order to conculde anything about the non-finite [X,Y] interval (you still do not get the red photon analogy).

 

[jsfisher]

Are Z , Y or h have distinct values?

If they have, then [Z,Y] or [h,Y] are twe finite cases of distinct values, that can't be used in order to conclude anything about [X,Y] non-finite interval.

(a) [X,Y] is not a part of anything I wrote, so your continual referral to it is so far irrelevant.

(b) [X,Y] is a finite interval.


Seriously, learn to read. Don't leave out the part about what words actually mean, either.

 

[doronshadmi]

(b) [X,Y] is a finite interval.

Wrong.

[X,Y] interval is exactly the all infinitely many R members between distinct X and distinct Y.

 

[doronshadmi]

Here is some part of your "proof":

Let h be any element of the interval (Z,Y)

In other words, you deal with distinct Z value and distinct Y value.

X=Z and we still have to deal with the non-finite [X,Y] (a,b constriction) interval.

X is obviously not the immediate predecessor of Y, simply because X < Y is nothing but a relation between finite amount of two distinct cases, which can't be used in order to conclude anything about the non-finite [X,Y] interval.

 

[jsfisher]

Wrong.

[X,Y] interval is exactly the all infinitely many R members between distinct X and distinct Y.

Perhaps you should consult a good reference for mathematical terms before you embarrass yourself like this.

 

[doronshadmi]

Perhaps you should consult a good reference for mathematical terms before you embarrass yourself like this.

[X,Y] interval is exactly the all infinitely many R members between (the included) distinct X and distinct Y.

This is an exact mathematical term.


You embarrass yourself by not get it.

 

[jsfisher]

This is an exact mathematical term.


You embarrass yourself by not get it.

Here, perhaps this will help. Scroll down to the Terminology section:

An interval is said to be left-bounded or right-bounded if there is some real number that is, respectively, smaller than or larger than all its elements. An interval is said to be bounded if it is both left- and right-bounded; and is said to be unbounded otherwise. Intervals that are bounded at only one end are said to be half-bounded. The empty set is bounded, and the set of all reals is the only interval that is unbounded at both ends. Bounded intervals are also commonly known as finite intervals.

 

[doronshadmi]

Bounded intervals are also commonly known as finite intervals.

So what?

This is another example of how mathematicians play with words (in this case, "finite interval"), which do not change the fact that there are
the all infinitely many R members between the included distinct X and distinct Y of [X,Y] interval.

Do you really think that by exclude or include two distinct values, you change a non-finite collection (ordered or not) to a finite collection (ordered or not), or vice versa?

You embarrass yourself by not get it.

 

[jsfisher]

This is another example of how mathematicians play with words (in this case, "finite interval")...

No, this is another example of where Doron Shadmi goes off on a tangent not knowing what he's talking about. Then, when it is pointed out how misinformed Doron Shadmi is about things, he goes off on another tangent to claim he's right all along and the rest of the world is completely wrong.

Doesn't work that way, my ignorant Internet pen pal.

You don't understand so many, many things, but instead of using it as an opportunity to learn, you run off to create an imaginary world of contradiction and inconsistency.

Do you really think that by exclude or include two distinct values, you change a non-finite collection (ordered or not) to a finite collection (ordered or not), or vice versa?

I never thought that, nor would I. You really, really must do something about your reading comprehension. Did you really think I said [X,Y] was a finite interval and (X,Y) was not? They are both finite intervals.

 

[jsfisher]

Let's try this again, but only one question at a time:

If B is in the open interval (A,C), then is B < C ?​

Simple enough question, isn't it?

 

[The Man]

No.

Actual infinity existence is beyond the collection's existence.

Since you force actual infinity on the level of the collection's existence, you naturally get a contradiction.

But this is your invalid framework, not mine.

No Doron again and as usual the contradiction is simply in your own assertions

This inability to conclude something about the non-finite, by using infinitely many finite cases, is a fundamental problem of the mathematical science if the term all is used on a non-finite collection, because no finite case alone has the quality of a non-finite collection of all distinct members.

“infinitely many finite cases” has “the quality of a non-finite collection” because you specifically gave it that ‘quality’ by declaring that you are concerned with “using infinitely many finite cases”. Then you say "no finite case alone has the quality of a non-finite collection", but your assertion was about "infinitely many finite cases” not any one "finite case alone".