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Topic: RE: Matheology 400: Quantifier Confusion
Replies: 188   Last Post: Dec 13, 2013 5:48 AM

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 Tucsondrew@me.com Posts: 1,157 Registered: 5/24/13
Re: Matheology 400: Quantifier Confusion
Posted: Dec 5, 2013 2:33 PM

On Thursday, December 5, 2013 2:45:33 AM UTC-7, WM wrote:
> Am Mittwoch, 4. Dezember 2013 20:57:14 UTC+1 schrieb Zeit Geist:
> > On Wednesday, December 4, 2013 11:12:36 AM UTC-7, WM wrote:
>
> > > That is true in the list of all FIS too
>
> > > d_1
> > > d_1, d_2
> > > d_1, d_2, d_3
> > > ...

>
> > > All these FIS however, are not capable of distinguishing themselves from all rationals. That requires more: A good portion of faith. And a completed infinity for that np larger index is possible. Obviously the largest index has been applied. An index which is not present in any rational number.
>
> > We need only show that for each Rational Number, r, there is One particular FIS(d_n) such that we have r ~= FIS(d_n). This is a Sufficient Condition to conclude that r =~ FIS(d_n).
>
> No, that is not enough, because each rational number for that you can show that, belongs to the first few lines of the list while infinitely many will follow. That is fact for each rational you can "take". In order to show that d differs from all rational numbers, you must not exchange quantifiers but prove that for all rational numbers.
>

You are highly confused and keep bring up the irrelevant fact that "infinitely many follow".

Let S be a non-empty Set.
We wish to prove that t ~e S.
All we need to do is show that For Any Arbitrarily Chosen s e S, we have t ~= s.
If s is arbitrary then it holds that for Any, All and Every Element of S.

Conjecture: Omega is Not an a Natural Number.

Proof: Choose an arbitrary integer, n. We know | n | is finite. Since omega is not finite, we know that omega ~= n. Hence, since n was chose in an arbitrary manner, we know o Eva is Not equal to any Natural Number. Thus, omega is Not a Natural Number.

> Further your task is futile since we know already that for each FIS(d_n) there are infinitely many rational numbers not differing. So why taking one or the other that differs? Do you intentionally try to cheat?
>

But d is Not equal to any.

> > > He did not avoid it. He knew fairly early, in the 1890s, that the set of all sets is impossible. Unfortunately he did not recognize that the set of all natural numbers that are followed by infinitely many natural numbers is impossible too. (I.e., the set of all numbers that is not all numbers.)
>
> > He didn't realize that, because it is not true.
>
> It is true.
>

N does Not exist? You say this is True?

> > How does N = { All Natural Numbers } --> N = { All Numbers that is not All Numbers }?
> > You exposition is bizarre.

>
> Your "All" says all but means every number that belongs to the 0 % of the first numbers. "Every" natural number is followed by infinitely many. If you think that all exist, then every natural number is followed by nearly all natural numbers.
>

Yes, they are and it does not Invalidate their Existence.

> > What we can prove, in Modern Mathematics,
>
> is ridiculous with respect to the above fact.
>

And yet Logically Valid.

> > > The usual reply of matheologians, on the question how they know that they have all numbers without having the last one, is: There is no last natural number. But I im interested to see their escape with respect to the fact that no digit d_i ad no FIS is sufficient to accomplish a distinction of d from all rationals of the rationals-complete list. Obviously d contains somewhat more than every digit d_i.
>
> > We "have them all" because there is a Set that contains "them all".
>
> > The "escape" is accomplished by knowing that in the Set of All FIS of d, there is (at least) one k, such that FIS(d_k) ~= r, where is an Arbitrarily Chosen Rational Number.
>
> You can only choose among the first 0 % of the listed rational numbers. If you disagree then let me know at least one chosen number that is not followed by infinitely many. But please do not infinitely repeat your counterfactual claims.
>

That doesn't matter, you fool.

Yes, there infinitely many following any. That's because it's an infinite well-ordered Set.
That doesn't prevent from defining a property that belongs to every and all of them.

Stop saying, "you can't count that high."!
That's what that statement boils down to.

> Regards, WM

ZG

Date Subject Author
12/4/13 Tucsondrew@me.com
12/4/13 wolfgang.mueckenheim@hs-augsburg.de
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