Deeper than primes

[doronshadmi]

Your evidence for this would be?

Your inability to write this expression in simple English.

I give you 20 minutes for this, after that I'll write it down.

So I am waiting ...

 

[zooterkin]

Your inability to write this expression in simple English.

Says the person who wrote "No, you quated it without undertand it."

 

[doronshadmi]

Says the person who wrote "No, you quated it without undertand it."

I give you 20 minutes for this, after that I'll write it down.

So I am waiting ...

 

[zooterkin]

Write what down?

 

[doronshadmi]

Write what down?

This:

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​

Write it in plain English, 10 minutes are over, you have only 10 minutes to do that, before I do thet.

 

[zooterkin]

This:

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​

Write it in plain English, 10 minutes are over, you have only 10 minutes to do that, before I do thet.

A is less than B if for all x and y, where x is a member of A and y is a member of B, x is less than y.


Where's your version?

 

[doronshadmi]

You seem to want to totally ignore the interval in favor of its elements. I, on the other hand, ignore neither the interval nor its membership. The object under consideration is the interval, but its properties are shaped by its elements.

Here's what I (and everyone else here) mean by "interval A precedes interval B". What do you mean?

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​

This is quite pathetic jsfisher, you are using any possible trick in order to not admit that you made a fundamental mistake by claim that Y number is an immediate successor of [X,Y) interval.

Furthermore, let us write

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​

in simple English and show how you are using the elements of A and B in order to show that "interval A precedes interval B".

(A < B) iff (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y))

It is clearly seen that A<B because we compare between each element of A (called x) and each element of B (called y).

So A<B cannot be found in this expression, independently of the elements of A and B.

Also in your expression above there is no mixing between interval and an element of an interval, as you do here:

In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 you clearly say:

"Y is in fact an immediate successor to [X,Y)".

Since [X,Y) is an interval and Y is an element of an interval, then "Y is in fact an immediate successor to [X,Y)" is an utter gibberish, because you mix between different types (between an interval and an element of an interval, in this case).

EDIT:

Since you show that A<B expression depends of the elements of A (called x) and the elements of B (called y), and since no y is an immediate successor of any x, then (by following this reasoning) you cannot use

[latex]$$ (A \prec B) \, \Leftrightarrow \, (\forall x \, \forall y \, (x \in A \wedge y \in B) \Rightarrow (x \prec y)) $$[/latex]​

in order to claim that B is an immediate successor of A, because by your expression above A<B has a meaning only according the relations of A and B elements (notated as x and y, where no y element of interval B is an immediate successor of any x element of interval A).

 

[doronshadmi]

A is less than B if for all x and y, where x is a member of A and y is a member of B, x is less than y.


Where's your version?

This is wrong, it has to be:

A is less than B iff (if and only if) for all x and y, where x is a member of A and y is a member of B, x is less than y.

EDIT:

But more importent, you do not understand what follows from this expression, about "[3,5) < 5" gibberish expression, as clearly explained in http://www.internationalskeptics.com/forums/showpost.php?p=4791330&postcount=3547 .

 

[jsfisher]

Nothing is different here, immediate successor has a one and only one meaning in Standard Math which is:

If a and b are two whole numbers and b immediately follows a such that a<b, then b is called the immediate successor of a.

What ever gave that idea? There is nothing to limit the immediate successor construct to whole numbers.

(you can't take a and b as two intervals, because any use of "<" relation has a meaning only if it used between the elements of a and b intervals).

This is another misunderstanding on your part. Nothing restricts the ordering relation (symbolized by "<") to be just the conventional numeric less-than comparator. Moreover, I have been very clear what I am taking to be the ordering relation, so any confusion is of your own manufacturing.

Any other use of the term 'immediate successor' under Standard Math is a load of crap.

Ok, so you are failing at Order Theory, too. What branch will you be failing at next?

 

[doronshadmi]

What ever gave that idea? There is nothing to limit the immediate successor construct to whole numbers.



This is another misunderstanding on your part. Nothing restricts the ordering relation (symbolized by "<") to be just the conventional numeric less-than comparator. Moreover, I have been very clear what I am taking to be the ordering relation, so any confusion is of your own manufacturing.



Ok, so you are failing at Order Theory, too. What branch will you be failing at next?

The revevant posts for this case are

http://www.internationalskeptics.com/forums/showpost.php?p=4791128&postcount=3537

and

http://www.internationalskeptics.com/forums/showpost.php?p=4791330&postcount=3547

Please reply to them.

 

[jsfisher]

In http://www.internationalskeptics.com/forums/showpost.php?p=4721582&postcount=2864 jsfisher clearly says:

"Y is in fact an immediate successor to [X,Y)".

By doing that he mixes between an interval and an element of an interval, and the result is an utter gibberish.

And I have explained how the ordering would apply to a domain of real numbers and real intervals combined. It would seem every has been able to follow my explanation except you, Doron.

No matter. I can restrict it entirely to intervals, still consistent with the already-presented ordering relation:

[Y, Y] is an immediate successor to [X, Y).

Is that better for you, Doron?

 

[zooterkin]

This is wrong, it has to be:

A is less than B iff (if and only if) for all x and y, where x is a member of A and y is a member of B, x is less than y.

Now explain to me in what way that is significantly different from what I said.

But more importent, you do not understand what follows from this expression, about "[3,5) < 5" gibberish expression, as clearly explained in http://www.internationalskeptics.com/forums/showpost.php?p=4791330&postcount=3547 .

One of us doesn't understand it, that much is clear.

 

[doronshadmi]

EDIT: jsfisher, you are the one who supported [3,5) < 5 in http://www.internationalskeptics.com/forums/showpost.php?p=4788779&postcount=3458 .

And I have explained how the ordering would apply to a domain of real numbers and real intervals combined. It would seem every has been able to follow my explanation except you, Doron.

No matter. I can restrict it entirely to intervals, still consistent with the already-presented ordering relation:

[Y, Y] is an immediate successor to [X, Y).

Is that better for you, Doron?

As long as you ignore the elements of [X,Y) and [Y,Y].

Now please show us how [X,Y)< [Y,Y] by ignoring the elements of [X,Y) and the elements of [Y,Y].

 

[doronshadmi]

Now explain to me in what way that is significantly different from what I said.

One of us doesn't understand it, that much is clear.

EDIT: iff is = , (A = B)

If is a part of "if,then" , (B implies A)

 

[zooterkin]

iff is =

If is a part of "if,then"

Now explain to me in what way that is significantly different from what I said.

 

[doronshadmi]

Now explain to me in what way that is significantly different from what I said.

EDIT: (A < B) = (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y))

you wrote:

IF (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y)) THEN (A < B)

 

[zooterkin]

you wrote:

IF (for all x and for all y (such that x is a member of B AND y is a member of B) implies (x < y)) THEN (A < B)

No, I didn't. I wrote in plain English, as requested.


I also don't think that 'iff' means the same as 'equals'.

ETA:
Regardless, you are quibbling about a trivial point.

What is the successor to [3, 5)?

 

[doronshadmi]

No, I didn't. I wrote in plain English, as requested.


I also don't think that 'iff' means the same as 'equals'.

You are right, I mean =, and = in plain English is "if and only if".

Let us fix it in the previous posts.

ETA:
Regardless, you are quibbling about a trivial point.

What is the successor to [3, 5)?

Any number that is greater than eny number of [3,... interval.

 

[zooterkin]

Any number that is greater than eny number of [3,... interval.

The successor, not a successor.

 

[doronshadmi]

The successor, not a successor.

By "the successor" do you mean "an immediate successor"?