Deeper than primes

[jsfisher]

It appears under the name z.

You are reading things into the predicate that simply are not there. You are going to have trouble understanding Mathematics if you continue to add see extra stuff over that is actually there.

 

[jsfisher]

It it ture jsfisher.

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

...and those weren't the phrases being challenged.

 

[doronshadmi]

You are reading things into the predicate that simply are not there. You are going to have trouble understanding Mathematics if you continue to add see extra stuff over that is actually there.

As I say jsfisher, you play with notations without notions.

By notation d is not there.

By notion d is there, under the name z.

Anyone who uses notions, can't ascape from http://www.internationalskeptics.com/forums/showpost.php?p=4762726&postcount=3234 result.

 

[doronshadmi]

...and those weren't the phrases being challenged.

These is exactly my argument about Standard Math (within the framework of Standard Math), in this case.

 

[jsfisher]

As I say jsfisher, you play with notations without notions.

By notation d is not there.

By notion d is there, under the name z.

Where would that be?

 

[jsfisher]

These is exactly my argument about Standard Math (within the framework of Standard Math), in this case.

Well, actually it is your assertion. You haven't presented enough substance to raise it to argument status.

 

[doronshadmi]

Well, actually it is your assertion. You haven't presented enough substance to raise it to argument status.

No jsfisher, it is done at http://www.internationalskeptics.com/forums/showpost.php?p=4762726&postcount=3234 and you did not give anything that shows otherwise.

 

[jsfisher]

No jsfisher, it is done at http://www.internationalskeptics.com/forums/showpost.php?p=4762726&postcount=3234 and you did not give anything that shows otherwise.

You might want to pay attention to the post immediately following yours. It challenges both your premise and your conclusion, and until you address the premise defect, it really doesn't matter what you conclude from it.

You may also notice (well, the rest of use notice, but you probably won't) you aren't very good at making connected arguments. Remember, you are the one that insists the word all means non-finite regression. That doesn't fit so well into the sentence, "All Greeks are liars."

 

[doronshadmi]

You might want to pay attention to the post immediately following yours. It challenges both your premise and your conclusion, and until you address the premise defect, it really doesn't matter what you conclude from it.

You may also notice (well, the rest of use notice, but you probably won't) you aren't very good at making connected arguments. Remember, you are the one that insists the word all means non-finite regression. That doesn't fit so well into the sentence, "All Greeks are liars."

Jsfisher (you have challenged nothing and) you do not get what you read. As a result you change the context where the term "all" is used, so here it is again:

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

d=0 if "all" is used in the context of the non-finite universe between X and Y (there is no non-finite regression).

d>0 if "Y has no immediate predecessor" is used in the context of the non-finite universe between X and Y (there is a non-finite regression).

Since Standard Math insists to use "all" AND "Y has no immediate predecessor" in the context of the non-finite universe between X and Y, then Standard Math actually claims that d > AND = 0, which is a contradiction under Standard Math framework.

"All Greeks are liars."

It is revevant to this dialog only if you first define the order of your Greeks.

 

[jsfisher]

You might want to pay attention to the post immediately following yours. It challenges both your premise and your conclusion, and until you address the premise defect, it really doesn't matter what you conclude from it.

You may also notice (well, the rest of use notice, but you probably won't) you aren't very good at making connected arguments. Remember, you are the one that insists the word all means non-finite regression. That doesn't fit so well into the sentence, "All Greeks are liars."

Jsfisher (you have challenged nothing and) you do not get what you read. As a result you change the context where the term "all" is used

You provided no context. That is really my point. You may have all sorts of grand ideas floating around in your head, but you aren't so good about expressing them when it comes time to forming those sentences. Whole paragraphs of context get left out as you leap from ice floe to ice floe.

...so here it is again:

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

See? No context. Just a bare assertion that all and non-finite regression are synonyms. To make matters worse, you don't seem to be using the term, regression, correctly. What do you think it means?

d=0 if "all" is used in the context of the non-finite universe between X and Y (there is no non-finite regression)

d>0 if "Y has no immediate predecessor" is used in the context of the non-finite universe between X and Y (there is a non-finite regression)..

Again, no context, just a d=0 dropped in.

Fill in the spaces, Doron. Communicate.



...but first things first: What do you think regression means as you have used the term?

 

[The Man]

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"

No you have missed it, I wasn’t talking about ‘standard math’ just your claims and assertions, which usually have nothing to do with ‘standard math’ even when you claim to be talking about ‘standard math’.

Once again you can not determine if Y has an immediate predecessor until you first define what constitutes an immediate predecessor. If we define it like Y_P < Y such that the interval ( Y_P, Y) is empty or the interval [ Y_P, Y] contains only Y_P and Y. Then the value of an immediate predecessor for a given Y can be determined in the integers. This is specifically because there are gaps or the integers are discontinuous and discrete, such that you can have an empty interval or an interval that includes only the boundaries. With no gaps the reals are continuous, no explicitly defined value for an ‘immediate predecessor’ of a given Y can be established that meets those given requirements. Again it is precisely because there are no gaps that an empty interval or an interval that contains only the boundaries can not be established in the reals. As usual you’ve got it all backwards it is because the interval between X and Y is completely filled, continuous and with no gaps that no explicitly defined value, and thus no real number, for an immediate predecessor can be established for any given value of Y in the reals.

 

[doronshadmi]

No you have missed it, I wasn’t talking about ‘standard math’ just your claims and assertions, which usually have nothing to do with ‘standard math’ even when you claim to be talking about ‘standard math’.

Once again you can not determine if Y has an immediate predecessor until you first define what constitutes an immediate predecessor. If we define it like Y_P < Y such that the interval ( Y_P, Y) is empty or the interval [ Y_P, Y] contains only Y_P and Y. Then the value of an immediate predecessor for a given Y can be determined in the integers. This is specifically because there are gaps or the integers are discontinuous and discrete, such that you can have an empty interval or an interval that includes only the boundaries. With no gaps the reals are continuous, no explicitly defined value for an ‘immediate predecessor’ of a given Y can be established that meets those given requirements. Again it is precisely because there are no gaps that an empty interval or an interval that contains only the boundaries can not be established in the reals. As usual you’ve got it all backwards it is because the interval between X and Y is completely filled, continuous and with no gaps that no explicitly defined value, and thus no real number, for an immediate predecessor can be established for any given value of Y in the reals.

These "no gaps" of X < Z < Y hold iff (d > 0) AND (d ≠ 0).

Since Standard Math claims that the non-finite universe between X and Y is completely filled, then d must be the minimal value, which is 0.

So Standard Math in its own framework, has to decide between "completely filled" (and in this case d must be = 0) and "Y has no immediate predecessor" (and in this case d must be > 0).

As long as Standard Math insists to use both "completely filled" AND "Y has no immediate predecessor" on the same non-finite universe between X and Y, Standard Math is based on the reasoning that claims that d is both > AND = 0, which is a contradiction under Standard Math framework.

You have asked me to show a contradiction within the framework of Standard Math.

I provided such a contradiction (d is both > AND = 0).

Now, instead of face the facts, you behave like any fanatic religious community of people, which does not wish to face the facts about its own failure.

 

[jsfisher]

These "no gaps" of X < Z < Y hold iff (d > 0) AND (d not = 0).

Just to fill the gap that exists here, I'm going to assume, doron, you are using d to represent the difference between X and Y. That is, d = Y - X.

Since Standard Math claims that the non-finite universe between X and Y is completely filled, then d must be the minimal value, which is 0.

It is a fact, actually, not simply a claim. Be that as it may, though, no, it does not then require that d = 0. Your conclusion is faulty.

Your faulty conclusion, here, then becomes a premise for the rest of your post. Faulty premises lead to irrelevant conclusions, so there is not point continuing with your post.

 

[doronshadmi]

Just to fill the gap that exists here, I'm going to assume, doron, you are using d to represent the difference between X and Y. That is, d = Y - X.

If X < Z < Y is used, then d = abs(Z-X).

It is a fact, actually, not simply a claim. Be that as it may, though, no, it does not then require that d = 0. Your conclusion is faulty.

It does not require that d = 0, iff the interval between X AND Y is not completly filled ( in this case (d > 0) AND (d ≠ 0) ).

Your faulty conclusion, here, then becomes a premise for the rest of your post. Faulty premises lead to irrelevant conclusions, so there is not point continuing with your post.

Your faulty determination is the result of (d > 0) AND (d = 0).

 

[jsfisher]

If X < Z < Y is used, then d = abs(Z-X).

Ok, so be it. d = Z - X. (The absolute value function is redundant.)

It does not require that d = 0, iff the interval between X AND Y is not completly filled ( in this case (d > 0) AND (d ≠ 0) ).

You are still just asserting things. Where's your proof?

(By the way, "AND (d ≠ 0)" is also redundant.)

 

[doronshadmi]

Ok, so be it. d = Z - X. (The absolute value function is redundant.)



You are still just asserting things. Where's your proof?

(By the way, "AND (d ≠ 0)" is also redundant.)

Yes I know, but in order to avoid any misunderstanding, redundancy were used in these cases.

d=0 if the non-finite universe between X and Y is "completely filled".

It is obvious, and any other determination must be artificial, in this case, because there is no reason of why d=0 is excluded.

Again:

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) .


If you disagree with it, you have to provide the reason of why d > 0 (d does not have the minimal value 0) even if the non-finite universe between X and Y is "completely filled" (or in other words "why the case of d=0 is excluded, in this case?").

 

[jsfisher]

d=0 if the non-finite universe between X and Y is "completely filled".

It is obvious, and any other determination must be artificial, in this case, because there is no reason of why d=0 is excluded.

This is your proof? "It is obvious"? Do you not see why that comes up lacking?

Again:

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) .

I see the assertion...you provided no proof.

If you disagree with it, you have to provide the reason of why d > 0 even if the non-finite universe between X and Y is "completely filled" (or in other words "why the case of d=0 is excluded, in this case?").

No, I don't have to provide anything. You are the one asserting things contrary to established Mathematics; the burdern of proof falls to you.

But be that as it may, if you will drop the figurative completely filled phrase and stop trying to misunderstand the infinite right out of the gate, maybe you can follow this:

Given any two real numbers X and Y where X < Y, there is another number Z between them. (That is, X < Z and Z < Y).​

Will you accept that as fact, or do we need to (again) show you how we can set Z to be the mid-point between X and Y?

Ok, that's the only fact that's needed to conclude (1) no real number has an immediate predecessor or immediate successor, and (2) the real numbers are dense.

Both proofs are simple. The proof of the former is a simple proof-by-contradiction. For the latter, just pull out the applicable definition for dense and observe for any neighborhood of some real number, the above fact guarantees there is at least one real in that neighborhood.

 

[doronshadmi]

Given any two real numbers X and Y where X < Y, there is another number Z between them. (That is, X < Z and Z < Y).​

This is not a proof.

This is an agreed game with notations, that avoids any research of the notions that stand at the basis of the notations.

Again it is shown how your framework is limited to "how to define and\or use?" questions, and explicitly avoids "how it is?" , "what is this?" , "Why it is?" etc … questions.

In other words, you totally ignored the notions that are expressed in http://www.internationalskeptics.com/forums/showpost.php?p=4763491&postcount=3256 in order to fit the facts to your agreed game with notations.

EDIT:

by the way, it is easy to show the inconsistency of Standard Math, in this case:

1) From one hand it claims that Y of X<Y has no immediate predecessor, and for that d must be > 0.

2) On the other hand it claims that 0.999…[base 10] = 1 , and for that d must be 0.

I know that (2) is a sum of non-finite Q members, but it has no significance in this case, because both R and Q are dense, by Standard Math.

 

[jsfisher]

This is not a proof.

Perhaps you missed the part where I queried:

Will you accept that as fact, or do we need to (again) show you how we can set Z to be the mid-point between X and Y?

Seriously, Doron, if you are going to so willingly ignore simple questions whose purpose is to help foster understanding while you so willingly add the imaginary to other simple statements, what's the point?

 

[doronshadmi]

Perhaps you missed the part where I queried:


Seriously, Doron, if you are going to so willingly ignore simple questions whose purpose is to help foster understanding while you so willingly add the imaginary to other simple statements, what's the point?

You willingly ignored the rest of http://www.internationalskeptics.com/forums/showpost.php?p=4763566&postcount=3258 and the questions that can be found there and in likned post.

I wonder why?