Deeper than primes

[doronshadmi]

Perhaps this is one root of the problem of your sub-optimal relationship with mathematics - that all specialist areas of knowledge must, and do, clearly define the semantics of the words they use, in order to avoid misunderstanding. In the process, the ill-defined everyday semantics of those words necessarily gives way to the more rigorous definitions. If you could bring yourself to accept this, and to define the semantics of your own 'custom' vocabulary, perhaps this discussion would be less frustrating for all.

Again.

First of all there must be notions.

Definitions without notions is nothing but a notionless maneuvers with symbols.

In order to get Organic Mathematics notions, you have no choice but to read all of http://www.geocities.com/complementarytheory/OMPT.pdf .

Furthermore, you can air your view about some part of http://www.geocities.com/complementarytheory/OMPT.pdf only if you first read all of it.

Organic Mathematics is not a serial step-by-step reasoning, and any part of it can be understood iff you first read all of it.

A step by step reasoning is actually a weak emergence reasoning, where the Whole is the sum of the Parts.

Organic Mathematics is a non-standard Strong Emergence Reasoning, where the Whole is greater than the sum of the Parts.

(In Standard Strong Emergence the Whole is greater than the sum of its Parts)

The members of this forum try to force Weak Emergence reasoning on Organic Mathematics, and as an obvious result, they don't understand it.

 

[jsfisher]

I don't know exactly what I was thinking when I wrote this:

By the way, the word is qualifier, not quantifier.

It is incorrect. Universal quantifier is the correct term.

 

[doronshadmi]

This is just sad, Doron. You don't comprehend your own words. You said:

jsfisher, please cut the BS, we both know that an interval is made of R members, and the universal quantification "for all" is used on it.

http://en.wikipedia.org/wiki/Interval_(mathematics)

In mathematics, a (real) interval is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. For example, the set of all numbers x satisfying 0 ≤ x ≤ 1 is an interval which contains 0 and 1, as well as all numbers between them.

As a result y of [x,y)(even if y is not included) or [x,y] must have an immediate predecessor.

There is no other option.

 

[jsfisher]

jsfisher, please cut the BS, we both know that an interval is made of R members

I have never denied this. You, however, said an interval was a member of R. That statement was bogus.

...and the universal quantification "for all" is used on it.

Where? I've already provided you with set equivalent to the interval [X,Y). I don't recall seeing [SIZE=+1]∀[/SIZE] anywhere nearby.

As a result y of [x,y) or [x,y] must have an immediate predecessor.

There is no other option.

How so? Just how does the appearance of the universal quantifier mandate the existence of an immediate predecessor?

 

[doronshadmi]

http://en.wikipedia.org/wiki/Interval_(mathematics)

In mathematics, a (real) interval is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. For example, the set of all numbers x satisfying 0 ≤ x ≤ 1 is an interval which contains 0 and 1, as well as all numbers between them.

As a result y of [x,y)(even if y is not included in the interval) or [x,y] must have an immediate predecessor.

There is no other option.

 

[jsfisher]

http://en.wikipedia.org/wiki/Interval_(mathematics)

In mathematics, a (real) interval is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. For example, the set of all numbers x satisfying 0 ≤ x ≤ 1 is an interval which contains 0 and 1, as well as all numbers between them.

Ok, I guess we can now add universal quantification to the list of things Doron doesn't understand. An (informal) statement in English that uses the word, all, is not automatically an example of universal quantification.

Here is the very first sentence from the very wiki article you cited not long ago:

In predicate logic, universal quantification formalizes the notion that something (a logical predicate) is true for everything

Your reference to intervals doesn't offer any statement alleged to be true for everything. If you think otherwise, express the statement as a formal logical predicate.

As a result y of [x,y)(even if y is not included) or [x,y] must have an immediate predecessor.

There is no other option.

So you continue to allege, but you haven't provided any connection between universal quantification and predecessors, just the bare allegation.

 

[doronshadmi]

Ok, I guess we can now add universal quantification to the list of things Doron doesn't understand. An (informal) statement in English that uses the word, all, is not automatically an example of universal quantification.

Here is the very first sentence from the very wiki article you cited not long ago:



Your reference to intervals doesn't offer any statement alleged to be true for everything. If you think otherwise, express the statement as a formal logical predicate.

This is a good example of my claim that you, the current notionless mathematicians hijacked the straightforward meaning of the word "all" and used it incorrectly as a part of the term "for all".

"for all" has an exact meaning which is "without exceptions".

"for all" members (which means that we are talking about the all members, without exceptions) of the interval [x,y] y must have an immediate predecessor, because all members < y are included in [x,y] (otherwise the word "all" of "for all" is meaningless).

 

[jsfisher]

This is a good example of my claim that you, the current notionless mathematicians hijacked the straightforward meaning of the word "all" and used it incorrectly as a part of the term "for all".

I see. Since you never bothered to really learn what the terms meant, you just guessed. When it's pointed out you guessed wrong, you blame the rest of the world.

"for all" members (which means that we are talking about the all members, without exceptions) of the interval [x,y] y must have an immediate predecessor, because all members < y are included in [x,y] (otherwise the word "all" of "for all" is meaningless).

Having a set of all real numbers less than some real number does not in any way require the set have a largest element. Why do you assume it must have a largest element?

 

[doronshadmi]

I see. Since you never bothered to really learn what the terms meant, you just guessed. When it's pointed out you guessed wrong, you blame the rest of the world.

Now you play the megalomaniac.

jsfisher, your community of current notionless mathematicians is not the rest of the world.

It is no more than a religious closed sect that plays with words and symbols without any notion behind it.

Having a set of all real numbers less than some real number does not in any way require the set have a largest element. Why do you assume it must have a largest element?

y is the largest member of all the members of the interval [x,y] and because all members < y are included in [x,y], then y has an immediate predecessor.

There is no other meaning to the word "all".

I see. Since you never bothered to really learn what the terms meant, you just guessed.

You never bothered to really learn what this word means , not even gussing.

 

[jsfisher]

y is the largest member of all the members of the interval [x,y] and because all member < y are included in [x,y], then y has an immediate predecessor.

No, you are completely wrong on this. You are implicitly assuming there must be a largest member included in that "all". The assumption is unwarranted; there is not largest member < Y. Without a largest member < Y, the immediate predecessor for Y does not exist.

There is no other meaning to the word "all" (you never bothered to really learn what this word means).

Messrs. Merriam and Webster would disagree.

 

[doronshadmi]

No, you are completely wrong on this. You are implicitly assuming there must be a largest member included in that "all". The assumption is unwarranted; there is not largest member < Y.

Why, because you want to ignore the meaning to the word "all"?

As I said, you force your arbitrary determinations, in order to get your requested artificial results, no more no less, remember?

Messrs. Merriam and Webster would disagree.

Also in the case of a non-finite interval?

 

[jsfisher]

Why, because you want to ignore the meaning to the word "all"?

Nothing you've put forth so far gets to a point where there must be a largest element in the set of reals < Y.

 

[doronshadmi]

Nothing you've put forth so far gets to a point where there must be a largest element in the set of reals < Y.

Just the term "all" on the non-finite interval [x,y].

We do not need more than that, jsfisher.

On the contrary, you did not provide anything that shows that this is not the case.

 

[Little 10 Toes]

Using OM, tell us what the predecessor of y is in the expression [x,y] when using a number line.

 

[doronshadmi]

Using OM, tell us what the predecessor of y is in the expression [x,y] when using a number line.

Organic Mathematic explicitly defines the immediate predecessor of y, as the non-local element that is both on y AND on any arbitrary member of [x,y] interval.

 

[jsfisher]

Just the term "all" on the non-finite interval [x,y].

That would be a finite interval, also known as a bounded interval.

We do not need more than that, jsfisher.

Yours is just an empty allegation. If that is all you have, you don't have anything. Don't you have anything to support your claim other than hand waving?

On the contrary, you did not provide anything that shows that this is not the case.

Your claim; your responsibility to support your claim.

 

[jsfisher]

Using OM, tell us what the predecessor of y is in the expression [x,y] when using a number line.

If you look at the post he'll cross-reference in response to this query, you'll may notice what he claims for the predecessor (at the very bottom of the post) isn't a real number and isn't in the interval.

ETA: Sigh. Doron has already rewritten his response to avoid the cross-reference.

 

[doronshadmi]

That would be a finite interval, also known as a bounded interval.

Why, because you want to ignore the fact that [x,y] is a non-finte interval?

This is just your artificial deretmination, no more no less.

Yours is just an empty allegation. If that is all you have, you don't have anything. Don't you have anything to support your claim other than hand waving?

Your artificial deretmination is not even a hand waving.

Your claim; your responsibility to support your claim.

there is not largest member < Y

Your claim; your responsibility to support your claim.

 

[Little 10 Toes]

Organic Mathematic explicitly defines the immediate predecessor of y, as the non-local element that is both on y AND on any arbitrary member of [x,y] interval.

So then, the predecessor of y is y since y is an arbitrary member of [x,y]. and now tell me how y can be a predecessor of itself.

See, jsfisher, there was some logic there.

 

[jsfisher]

Why, because you want to ignore the fact that [x,y] is a non-finte interval?

This is just your artificial deretmination, no more no less.

Learn what mathematical terms mean, Doron, and stop blaming others for your lack of knowledge. The interval [X,Y] is a finite interval, and that's a fact.

Your artificial deretmination is not even a hand waving.

If you can't support your claim, fine, but please have the decency to cease with your empty allegations, then.

For reference, your claim was that somehow the use of the word, all, you found in a wiki article about intervals requires that any real number Y must have an immediate predecessor.

You are unable to support this allegation, so we may reject it.

I separately alleged the set { X : X < Y } for some Y has no largest element. You seem hung up on this one, too, so I will support the claim.

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.