Deeper than primes

[doronshadmi]

You've contradicted an earlier claim. Previously you said a point had no sides. Now, you are claiming it has one.

Y and the arbitrary element < Y are sides on the non-local element, and not points on the non-local element.

 

[jsfisher]

Please read all of http://www.internationalskeptics.com/forums/showpost.php?p=4736711&postcount=3010 .

I read the original. Your comments were addressed and dismissed point by point. If, as is your custom, you have re-edited it with new and different content, don't expect me to go back and consider it again to figure out what you may have changed or added and then dismiss your misunderstandings once again.

Forums are for dialog, Doron, not this rude behavior of yours to rewrite your posting history.

 

[jsfisher]

Y and the arbitrary element < Y are sides on the non-local element, and not points on the non-local element.

Y is a point. Y is a side. You said so. It must be true.

 

[doronshadmi]

Y is a point. Y is a side. You said so. It must be true.

No. Please look at http://www.internationalskeptics.com/forums/showpost.php?p=4736064&postcount=2973 .

A location can be a point XOR a side.

 

[doronshadmi]

I read the original. Your comments were addressed and dismissed point by point. If, as is your custom, you have re-edited it with new and different content, don't expect me to go back and consider it again to figure out what you may have changed or added and then dismiss your misunderstandings once again.

Forums are for dialog, Doron, not this rude behavior of yours to rewrite your posting history.

So don't read it. I don't care.

 

[doronshadmi]

In a deductive framework, you can't change your initial conditions.

You did not obey this rule, by changing the general initial Z < Y construction, by the general initial Z < h < Y construction.

 

[jsfisher]

So don't read it. I don't care.

Geez. A seven word post, and even that required re-editing.

 

[doronshadmi]

Geez. A seven word post, and even that required re-editing.

http://www.internationalskeptics.com/forums/showpost.php?p=4736824&postcount=3026

 

[jsfisher]

In a deductive framework, you can't change your initial conditions.

You did not obey this rule, by changing the initial general Z < Y by the initial Z < h < Y construction.

I didn't change any initial condition.

The premise was that Z was the largest element of the set. The premise was unchanged throughout the proof until the contradiction was reached.

That Z < Y was a conclusion deduced from the construction of the set.

That Z < h < Y was also a conclusion.

Moreover, Z < Y and Z < h < Y are not incompatible statements. In fact the former follows directly from the latter based solely on the transitive property of the comparison operator.

Do you not understand what Z < Y and Z < h < Y mean?

 

[doronshadmi]

In fact the former follows directly from the latter based solely on the transitive property of the comparison operat.

If you use google in order to search "the transitive property of the comparison operator" you get this:

http://www.google.com/search?hl=en&...on+operator"&btnG=Google+Search&aq=f&oq=&aqi=

So we look for "The transitive property of equality" and get http://209.85.229.132/search?q=cach...tive+property+of+equality"&cd=5&hl=en&ct=clnk :

The transitive property of equality states for any real numbers a, b, and c:

If a = b and b = c, then a = c.

For example, 5 = 3 + 2. 3 + 2 = 1 + 4. So, 5 = 1 + 4.

Another example: a = 3. 3 = b. So, a = b.

Or http://www.mathwords.com/t/transitive_property_inequalities.htm:

Transitive Property of Inequalities

Any of the following properties:

If a < b and b < c , then a < c.
If a ≤ b and b ≤ c , then a ≤ c.
If a > b and b > c , then a > c.
If a ≥ b and b ≥ c , then a ≥ c.

There is nothing do do here with Z < Y because it is based on construction a,b and not on construction a,b,c.

Moreover, Z < Y and Z < h < Y are not incompatible statements.

The fact is that Z < Y construction and Z < h < Y construction are simply different constructions that are not derived from each other, no more no less.

Do you not understand what Z < Y or Z < h < Y mean?

 

[integral]

Do yourself a favor, doronshadmi...check out MIT openCourseWare on mathematics.

 

[doronshadmi]

Do yourself a favor, doronshadmi...check out MIT openCourseWare on mathematics.

Its foundations is based on the same false determinations, shown by jsfisher.

If you disagree, then please explicitly show that this is not the case.

 

[Little 10 Toes]

If you use wikipedia to search for transitive property you'll find:

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

if a = b and b = c then a = c
if a < b and b < c then a < c
if a > b and b > c then a > c

 

[doronshadmi]

If you use wikipedia to search for transitive property you'll find:

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

if a = b and b = c then a = c
if a < b and b < c then a < c
if a > b and b > c then a > c

It does not matter, please look again at http://www.internationalskeptics.com/forums/showpost.php?p=4737591&postcount=3030 .

 

[jsfisher]

If you use google in order to search "the transitive property of the comparison operator" you get....

Thanks for confirming everyone's suspicion of your knowledge base: Post hoc google searches.

The fact is that Z < Y construction and Z < h < Y construction are simply different constructions that are not derived from each other, no more no less.

If you are trying to make a point, you really need to correct your vocabulary. The fact Z < Y is a consequence of the proof's initial assumption. It is not a construction. The fact Z < h < Y is also a consequence, not a construction.

The latter was not derived directly from the former, nor was any claim made the two were equivalent. (However, the former can be derived from the latter since Z < h < Y means Z < h and h < Y, and by the transitive property for the inequality, Z < Y.)

Are you really this confused by two separate facts being deduced from the initial assumption?

Do you not understand what Z < Y or Z < h < Y mean?

Clearly I do. Nobody here is sure about you, though.

 

[doronshadmi]

Thanks for confirming everyone's suspicion of your knowledge base: Post hoc google searches.



If you are trying to make a point, you really need to correct your vocabulary. The fact Z < Y is a consequence of the proof's initial assumption. It is not a construction. The fact Z < h < Y is also a consequence, not a construction.

The latter was not derived directly from the former, nor was any claim made the two were equivalent. (However, the former can be derived from the latter since Z < h < Y means Z < h and h < Y, and by the transitive property for the inequality, Z < Y.)

Are you really this confused by two separate facts being deduced from the initial assumption?



Clearly I do. Nobody here is sure about you, though.

Your post is a notionless game with words and symbols.

The simple fact is this:
http://www.mathwords.com/t/transitive_property_inequalities.htm:

Transitive Property of Inequalities

Any of the following properties:

If a < b and b < c , then a < c.
If a ≤ b and b ≤ c , then a ≤ c.
If a > b and b > c , then a > c.
If a ≥ b and b ≥ c , then a ≥ c.

There is nothing do do here with Z < Y because it is based on construction a,b and not on construction a,b,c.

I didn't change any initial condition.

The premise was that Z was the largest element of the set. The premise was unchanged throughout the proof until the contradiction was reached.

That Z < Y was a conclusion deduced from the construction of the set.

That Z < h < Y was also a conclusion.

Call it what ever you like. The fact is that you can't change a,b (Z < Y) to a,b,c (Z < h < Y) during the proof, and you did it.

 

[jsfisher]

Your post is a load of nonesense.

This is the fact:

http://www.mathwords.com/t/transitive_property_inequalities.htm:


There is nothing do do here with Z < Y because it is based on construction a,b and not on construction a,b,c.

I have no idea what relationship you think this has to the proof presented.

From the proof, Z < Y was a fact deduced from the fact Z was a member of the set {X : X < Y}.

Z < h < Y was a fact deduced from the fact h was selected from the interval (Z,Y).

Are you disputing those two deductions? If so, please do so directly. Irrelevant observations about the number of variables appearing in the two inequalities doesn't establish any cause to discredit the proof.

 

[doronshadmi]

I have no idea what relationship you think this has to the proof presented.

Because you force without notion a,b,c construction (Transitive property) on a,b construction (not the Transitive property).

Z < h < Y was a fact deduced from the fact h was selected from the interval (Z,Y)

Let us take the interval [X,Y] (X<Y).

If you are talking about [Z,Y] (Z<Y) it is clear that Z of [Z,Y] is not the immediate predecessor of Y exactly as X of [X,Y] is not the immediate predecessor of Y.

Since [X,Y] or [Z,Y] are intervals of the all members between X and Y or Z and Y, there is an immediate predecessor to Y.

Standard Math cannot explicitly define it and cannot explicitly disprove its existence because if [Z,Y] (Z<Y) is used, then Z is not the immediate predecessor of Y.

This is exactly my claim, Standard Math has no ability to explicitly define or disprove the existence of the immediate predecessor of Y, even if its existence must be true because [X,Y] or [Z,Y] are intervals of the all members between X and Y or Z and Y.

We are in the same state of Godel's incompleteness theorems, where things must be true but cannot be proved or disproved within the deductive framework that deals with the non-finite.

 

[jsfisher]

Because you force without notion a,b,c construction (Transitive property) on a,b construction (not the Transitive property).

I did no such thing. By the way, are you saying it is impossible for some h, Y, and Z for Z < Y and Z < h < Y? Is that really what you are saying?

Let us take the interval [X,Y] (X<Y).

If you are talking about [Z,Y] (Z<Y) it is clear that Z of [Z,Y] is not the immediate predecessor of Y exactly as X of [X,Y] is not the immediate predecessor of Y.

Why are you changing the subject to "immediate predecessors"? The proof was whether {X : X<Y} had a largest element.

Since [X,Y] or [Z,Y] are intervals of the all members between X and Y or Z and Y, there is an immediate predecessor to Y.

Nope. You have failed to show there is any requirement an immediate predecessor must exist. All you have done, all you continue to do is make baseless allegations.

Standard Math cannot explicitly define it and cannot explicitly disprove its existence because if [Z,Y] (Z<Y) is used, then Z is not the immediate predecessor of Y.

This is exactly my claim, Standard Math has no ability to explicitly define or disprove the existence of the immediate predecessor of Y, even if its existence must be true because [X,Y] or [Z,Y] are intervals of the all members between X and Y or Z and Y.

Your claim - another of your baseless allegations - is unsupported by you.

You have fabricated something that does not exist, then blamed Mathematics for being unable to find it.

We are in the same state of Godel's incompleteness theorems, where things must be true but cannot be proved or disproved within the deductive framework that deals with the non-finite.

Your claim - another of your baseless allegations - is unsupported by you.

 

[jsfisher]

While Doron is off googling to find things he can misinterpret into a post, permit me to say this has been a productive week. Doron, as he shared his wisdom, provided the following additional insights of doronetics:

• An interval, [X,Y) for example, is a real number.
• A collection with cardinality > 1 is equivalent to a collection of all distinct elements is equivalent to an interval.
• If you have A < C, then you cannot have A < B < C.
• For any real number Y, the immediate predecessor of Y is Y.

This brings my list up to an even 20. Did I miss any from this week?