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July 22nd, 2015, 05:25 AM   #1
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strange (??) proof

I'm 30 years computer programist from Poland.
My hobby is cybernetic, AI, math.
My English is not perfect, but I can read well.
Please tell me where I'm wrong.


We have $\displaystyle \mathbb{N}$={1,2,3,...}
We have $\displaystyle \mathbb{N}_{0}$={0,1,2,3,...}

from definition:
$\displaystyle \forall n \in \mathbb{N}_{0}$ n=0+1+1+...+1
$\displaystyle \forall n \in \mathbb{N}_{0}$ n-1-1-1-...-1=0

We can create length function

$\displaystyle F_{L} : \mathbb{N}_{0} \to \mathbb{N} \cup \lbrace \infty \rbrace $

$\displaystyle F_{L}(0)=1$
$\displaystyle F_{L}(1)=1$
$\displaystyle F_{L}(2)=1$
...
$\displaystyle F_{L}(10)=2$
...
$\displaystyle F_{L}(123)=3$
...

$\displaystyle \forall n \in \mathbb{N}_{0} F_{L}(n) < \infty$

proof

how we can create x that $\displaystyle F_{L}(x)=\infty$

we have y=$\displaystyle \Pi - 3$=0.1415...
we create x=...5141

we have Y$\displaystyle \subset(0,1>$ that Y is set of all irrational numbers or (I can't write what ??)

exist Lech Kaczynski function $\displaystyle F_{LK}: Y \to X$

that $\displaystyle F_{LK}(0.1415...)=...5141$

because $\displaystyle F_{LK}$ is bijection

|Y|=|X|=c

because |X|>|$\displaystyle \mathbb{N}_{0}$ $\displaystyle X \not\subset \mathbb{N}_{0}$

perhaps exist $\displaystyle A_{\infty} \subset X: A_{\infty} \subset \mathbb{N}_{0} $ and $\displaystyle |A_{\infty}| \le \aleph_{0}$

if $\displaystyle A_{\infty} \neq \emptyset$ then exist $\displaystyle x_{a} \in A_{\infty}$

$\displaystyle x_{a}=...x_{2}x_{1}x_{0}$

if $\displaystyle x_{0} > 0$ then $\displaystyle x_{a}-1=...x_{2}x_{1}(x_{0}-1)$

$\displaystyle F_{L}(x_{a})=F_{L}(x_{a}-1)=\infty$

if $\displaystyle x_{0}=0$ then $\displaystyle x_{a}=...x_{q+2}x_{q+1}x_{q}0...000$ and $\displaystyle x_{a}=...x_{q+1}(x_{q}-1)9...999$

$\displaystyle F_{L}(x_{a})=F_{L}(x_{a}-1)=\infty$

because $\displaystyle \forall x_{a} \in X : x_{a} - 1-1-1... \neq 0$ that $\displaystyle A_{\infty} = \emptyset$

end of proof

we can create

$\displaystyle A_{1} \subset \mathbb{N}_{0} : \forall n \in \mathbb{N}_{0} F_{LK}(n)=1$
$\displaystyle A_{1}=${0,1,2,3...9}
$\displaystyle |A_{1}|=10$


$\displaystyle A_{2} \subset \mathbb{N}_{0} : \forall n \in \mathbb{N}_{0} F_{LK}(n)=2$
$\displaystyle A_{2}=${10,11,12,...,99}
$\displaystyle |A_{1}|=90$

...

if x>1 we can create

$\displaystyle A_{x} \subset \mathbb{N}_{0} : \forall n \in \mathbb{N}_{0} F_{LK}(n)=x$
$\displaystyle A_{x}=${$\displaystyle 10^{x-1},10^{x-1}+1,...,10^{x}-1$}
$\displaystyle |A_{x}|=9*10^{x-1}$

using induction, we can prove that
$\displaystyle A_{1} \cup A_{2} \cup A_{3}... = \mathbb{N}_{0}$
$\displaystyle |A_{1} \cup A_{2} \cup A_{3}... |= |\mathbb{N}_{0}| = \aleph_{0}$


we can create A={$\displaystyle A_{1},A_{2},A_{3},...$}
|A|=$\displaystyle \aleph_{0}$
$\displaystyle A_{\infty} \not\in A$

we create

$\displaystyle A_{n}=$($\displaystyle A_{1},A_{2},A_{3},...$)

$\displaystyle S_{n}=$($\displaystyle S_{1}=A_{1},S_{2}=S_{1} \cup A_{2},S_{3}=S_{2}\cup A_{3},...$)

$\displaystyle P_{n}=$($\displaystyle |S_{1}|,|S_{2}|,|S_{3}|,...$)

$\displaystyle P_{n}$={10,100,1000,...}

using induction, we prove that
$\displaystyle |A_{1} \cup A_{2} \cup A_{3}... |< \aleph_{0}$

for n=1 $\displaystyle P_{1}=10 < \aleph_{0}$

if is true that for n is true $\displaystyle P_{n} < \aleph_{0}$

$\displaystyle P_{n+1}=10*P_{n} $

and $\displaystyle P_{n+1} < \aleph_{0} $

Last edited by skipjack; July 22nd, 2015 at 06:06 AM.
michalmozejko is offline  
 
July 22nd, 2015, 07:11 AM   #2
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If I understand correctly, your claim is that every nonnegative integer has a length (in decimal digits) which is finite. This is correct.
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July 22nd, 2015, 09:40 AM   #3
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But your statement that "we can create x such that $\displaystyle F_L(n)= \infty$" is wrong. Your function $\displaystyle F_L$ is only defined for integers and the "length" of any integer is finite. You have "we have y=Π−3=0.1415...
we create x=...5141" which, I think, means that you have reversed the decimal part of $\displaystyle \pi$. But there is no such integer- every integer has a "first" or "highest signicance" digit while "...5141" does not.
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July 22nd, 2015, 09:52 AM   #4
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Quote:
Originally Posted by Country Boy View Post
But there is no such integer- every integer has a "first" or "highest signicance" digit while "...5141" does not.
Right. There are other kinds of objects, most famously the 10-adic numbers, which allow numbers like ...5141 though.
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