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 Number theory and hyper operators MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 05/29/2013, 09:24 PM (This post was last modified: 05/31/2013, 08:49 AM by MphLee.) Yea, you're a genius! Now I understand what you tried to do. But I don't know if it will be really fruitfull, but at least sound interesting. Tell me if I got it right: You was right about the bijection (only if you first statements hold). Anyways we have that the sets $\mathbb{I}_ {m,n}$ and $\mathbb{I}_ {n}$ are really differents but they have interesting links. Lets re-start with a new version of your definitons: let $m, n \in \mathbb{N}$ Ia) $\mathbb{I} _{m,n} = \{ s: m \bigtriangleup_{s} n \in\mathbb{N} \}$ for fixed $m, n$ IIa) $\mathbb{I} _{n} = \{ s: m \bigtriangleup_{s} n \in\mathbb{N} \}$ for fixed $n$ second step: Ib) $\Pi _{m,n} (s) = m \bigtriangleup_{s} n$ IIb) $I _{n} (m) = s_m \text{ only if } (m \bigtriangleup_{s _m} n) \in\mathbb{N}$ First: $\Pi _{m,n}$ maps $\mathbb {N}$ on a subset of the naturals that now we call $P _{m,n}$ and like with the primes there are naturals numbers cutted off. We want $\Pi _{m,n} : \mathbb{N} \rightarrow \mathbb{N}$ to be injective (that is your condition $\Pi _{m,n} (s_1)=\Pi _{m,n} (s_2) \Longleftrightarrow s_1=s_2$ ) but is not invertible because is restricted to $P _{m,n}$: in fact $\Pi _{m,n} : \mathbb{N} \rightarrow P _{m,n}$ is bijective. Your idea is to extend $\Pi _{m,n}(s)$ to make it invertible adding more elements in the domain, in other words extending it to $\mathbb{I} _{m,n}$ (but you want do it adding real numbers) Finally we have a bijection, in fact $\mathbb{I} _{m,n} \approx^{\Pi_{m,n}} \mathbb{N}$ holds! for example using $0$ for the addition we have: $P _{2,3}=\{ \Pi _{2,3}(0); \Pi _{2,3}(1); \Pi _{2,3}(2); \Pi _{2,3}(3); ... \} = \{ 5; 6; 8; 16; ...\}$ then we build $\mathbb{I} _{2,3}$ inverting the naturals using $\Pi ^{\circ -1}_{2,3}$ $ \mathbb{I} _{2,3}= \{\Pi ^{\circ -1}_{2,3}(0);\Pi ^{\circ -1}_{2,3}(1);\Pi ^{\circ -1}_{2,3}(2);\Pi ^{\circ -1}_{2,3}(3);\Pi ^{\circ -1}_{2,3}(4); \Pi ^{\circ -1}_{2,3}(5); \Pi ^{\circ -1}_{2,3}(6); \Pi ^{\circ -1}_{2,3}(7); \Pi ^{\circ -1}_{2,3}(\text{8}); \Pi ^{\circ -1}_{2,3}(9); \Pi ^{\circ -1}_{2,3}(10); \Pi ^{\circ -1}_{2,3}(11); \Pi ^{\circ -1}_{2,3}(12); \Pi ^{\circ -1}_{2,3}(13); \Pi ^{\circ -1}_{2,3}(14); \Pi ^{\circ -1}_{2,3}(15);\Pi ^{\circ -1}_{2,3}(16); ... \} $ $\Pi ^{\circ -1}_{2,3}$ maps the naturals only if the domain is $P_{2,3}$ then in our set we have numbers that are not naturals $\mathbb{I} _{2,3}= \{r_0; r_1; r_2; r_3; r_4; 0; 1; r_7; 2; r_9; r_{10} ;r_{11} ;r_{12} ;r_{13} ;r_{14} ;r_{16} ; 3; ... \}$ To be honest I don't know if the extension of the naturals to $\mathbb{I} _{m,n}$ is a subset of the reals (as you defined) but if we have that $\Pi _{m,n} (s)$ preserves the order we can find this result $m \bigtriangleup_{s_0} n \lt m \bigtriangleup_{s_1} n \Leftrightarrow s_0 \lt s_1$ in our example this lead us to this $1 \lt r_7 \lt 2$ so the rank is not a natural number. My question is: is it rational, irrational? Maybe it is trascendental, but this is beyond my limits. Second: About the relations with $\mathbb{I} _{n}$ now the differencese are clear. We have that $\forall m,n \in\mathbb {N}$ $P_{m,n} \subset \mathbb{N} \subset \mathbb{I} _{m,n} \subset \mathbb{I} _{n}$ because $I_n [ \{ k \} ]= \mathbb{I} _{k,n}$ in other words the function IIb) is multivalued, and the set $\mathbb{I} _{n}$ is the set of all possibles solution, in symbols: $I_n [ \mathbb{N} ]= \mathbb{I} _{n}$ that is equivalent to $\mathbb{I} _{n}= \bigcup _{i\in \mathbb{N} } \mathbb{I} _{i, n}$ MathStackExchange account:MphLee « Next Oldest | Next Newest »

 Messages In This Thread Number theory and hyper operators - by JmsNxn - 08/30/2012, 02:49 AM RE: Number theory and hyper operators - by tommy1729 - 08/30/2012, 01:45 PM RE: Number theory and hyper operators - by JmsNxn - 08/30/2012, 05:24 PM RE: Number theory and hyper operators - by MphLee - 05/27/2013, 01:18 PM RE: Number theory and hyper operators - by MphLee - 05/25/2013, 10:15 PM RE: Number theory and hyper operators - by JmsNxn - 05/27/2013, 11:33 PM RE: Number theory and hyper operators - by MphLee - 05/28/2013, 10:40 AM RE: Number theory and hyper operators - by MphLee - 05/29/2013, 09:24 PM

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