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 July 22nd, 2015, 05:25 AM #1 Newbie   Joined: Jul 2015 From: Poland Posts: 1 Thanks: 0 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.
 July 22nd, 2015, 07:11 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms If I understand correctly, your claim is that every nonnegative integer has a length (in decimal digits) which is finite. This is correct.
 July 22nd, 2015, 09:40 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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.
July 22nd, 2015, 09:52 AM   #4
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Quote:
 Originally Posted by Country Boy 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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