Arithmetic neighbourhoods of numbers

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dc.creatorTyszka, Apoloniusz
dc.date2006-02-14
dc.date2007-06-11
dc.date.accessioned2026-07-07T09:23:49Z
dc.date.available2026-07-07T09:23:49Z
dc.descriptionLet K be a ring and let A be a subset of K. We say that a map f:A \to K is arithmetic if it satisfies the following conditions: if 1 \in A then f(1)=1, if a,b \in A and a+b \in A then f(a+b)=f(a)+f(b), if a,b \in A and a \cdot b \in A then f(a \cdot b)=f(a) \cdot f(b). We call an element r \in K arithmetically fixed if there is a finite set A \subseteq K (an arithmetic neighbourhood of r inside K) with r \in A such that each arithmetic map f:A \to K fixes r, i.e. f(r)=r. We prove: for infinitely many integers r for some arithmetic neighbourhood of r inside Z this neighbourhood is a neighbourhood of r inside R and is not a neighbourhood of r inside Z[\sqrt{-1}]; for infinitely many integers r for some arithmetic neighbourhood of r inside Z this neighbourhood is not a neighbourhood of r inside Q; if K=Q(\sqrt{5}) or K=Q(\sqrt{33}), then for infinitely many rational numbers r for some arithmetic neighbourhood of r inside Q this neighbourhood is not a neighbourhood of r inside K; for each n \in (Z \cap [3,\infty)) \setminus {2^2,2^3,2^4,...} there exists a finite set J(n) \subseteq Q such that J(n) is an arithmetic neighbourhood of n inside R and J(n) is not an arithmetic neighbourhood of n inside C.
dc.description15 pages, LaTeX2e
dc.identifierhttps://arxiv.org/abs/math/0602310
dc.identifierhttp://arxiv.org/abs/math/0602310
dc.identifierInternational Mathematical Forum 3 (2008), no. 8, pp. 349-362
dc.identifier.urihttp://salesiana.dossiersoluciones.com/handle/123456789/155876
dc.subjectNumber Theory
dc.subject12E99, 12L12, 11D99
dc.titleArithmetic neighbourhoods of numbers
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