Only countable many reals -> 0=1 !

11/05/2014 - 08:37 von karl | Report spam
Assumption:

There are only countable many real numbers which can be enumerated in a list.
Easy to show that this is true using the binary from WM ;-).

There is also only a countable number of real numbers in the unit interval which can be listed.
Let the numbers in this list denoted by (r_n)_{n \in IN}.

So now we take for each of these numbers r_n an interval containing I_n it as its center point with length:

l_n= d* 2^(-n)

where d is an arbitrary positive number. So:

I_n=[r_n - d* 2^(-n-1) , r_n + d* 2^(-n-1)]

So now the length of all these intervals must be larger than the length of the unit interval since there
are no other numbers in it than these countable set of numbers:

[0 , 1] \subset \cup_{n \in IN} I_n

For the length of the union of these set L we get an upper bound by summing the length l_n of all intervals I_n:

length([0, 1]) <= L <= \sum_{n=1}^{oo} l_n = \sum_{n=1}^{oo} d * 2^(-n) = 2 * d

Since d can be chosen as an arbitrary positive number, L can be made smaller than any positive number eps
wwe obtain that the length of the unit interval is less than any
positive number:

length([0, 1]) < eps

for all eps > 0.

From this we conclude, since the length of the unit interval is known to be unity, that:

0=1

Now, this can be done for all intervals of an arbitrary length a > 0 , we conclude therefore that:

0=a

So we have shown now that all positive real numbers are equal to zero if the assumption that there are only countable many
real numbers holds. The same then holds for negative numbers.

But since WM has shown without any doubt using the binary tree that the assumption is true, maths collapses into playing
around with zero.

Only zero exists!!
 

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#1 WM
12/05/2014 - 22:24 | Warnen spam
On Sunday, 11 May 2014 08:37:17 UTC+2, karl wrote:


Since d can be chosen as an arbitrary positive number, L can be made smaller than any positive number eps

wwe obtain that the length of the unit interval is less than any

positive number:



length([0, 1]) < eps



for all eps > 0.



From this we conclude, since the length of the unit interval is known to be unity, that:



0=1



Now, this can be done for all intervals of an arbitrary length a > 0 , we conclude therefore that:



0=a



So we have shown now that all positive real numbers are equal to zero if the assumption that there are only countable many

real numbers holds. The same then holds for negative numbers.



But since WM has shown without any doubt using the binary tree that the assumption is true, maths collapses into playing

around with zero.



Only zero exists!!



Nein. Abzàhlbarkeit ist kein mathematischer Begriff, da er auf vollendeter Unendlichkeit beruht, also auf barem Unsinn.

Wollte man jedoch mit vollendeten Unendlichkeiten rechnen, so könnte man ebensogut sagen: Jedes noch so kleine endliche Intervall der Folge, das eine rational Zahl überdeckt, ist größer als der Abstand zwischen zwei rationalen Zahlen (welcher kleiner als jedes epsilon ist). Oder auch: *Jedes* noch so kleine Intervall der Folge enthàlt unendlich viele rationale Zahlen. Oder auch: Es gibt 1/epsilon mal mehr rationale Zahlen als natürliche Zahlen.

Weiterhin, so könnte man sagen, wissen wir, dass jede reelle Zahl eine endliche Definition besitzt und dass die Menge aller endlichen Definitionen abzàhlbar ist. Zwar ist "Definition" nicht formal definierbar, aber jede mathematisch verwendbare reelle Zahl ist durch einen endlichen Ausdruck definiert. Die Menge dieser Zahlen kann demnach nicht größer sein als die Menge der endlichen Ausdrücke - und diese ist abzàhlbar.

Gruß, WM

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