# Ostrowski's theorem

Now, let {displaystyle a,nin mathbb {N} } with a > 1. Express bn in base a: {displaystyle b^{n}=sum _{i0.} Hence {displaystyle b^{n}geq a^{m-1},quad } so {displaystyle quad mleq n,{frac {log b}{log a}}+1.} Then we see, by the properties of an absolute value: {displaystyle |b|_{*}^{n}=|b^{n}|_{*}leq sum _{i1,} the above argument shows that {displaystyle |a|_{*}>1} regardless of the choice of a > 1 (otherwise {displaystyle |a|_{*}^{log _{a}!b}leq 1} , implying {displaystyle |b|_{*}leq 1} ). As a result, the initial condition above must be satisfied by any b > 1.

Thus for any choice of natural numbers a, b > 1, we get {displaystyle |b|_{*}leq |a|_{*}^{frac {log b}{log a}},} i.e.

{displaystyle {frac {log |b|_{*}}{log b}}leq {frac {log |a|_{*}}{log a}}.} By symmetry, this inequality is an equality.

Since a, b were arbitrary, there is a constant {displaystyle lambda in mathbb {R} _{+}} for which {displaystyle log |n|_{*}=lambda log n} , i.e. {displaystyle |n|_{*}=n^{lambda }=|n|_{infty }^{lambda }} for all naturals n > 1. As per the above remarks, we easily see that {displaystyle |x|_{*}=|x|_{infty }^{lambda }} for all rationals, thus demonstrating equivalence to the real absolute value.

Case (2) As this valuation is non-trivial, there must be a natural number for which {displaystyle |n|_{*}<1.} Factoring into primes: {displaystyle n=prod _{i

Si quieres conocer otros artículos parecidos a Ostrowski's theorem puedes visitar la categoría Theorems in algebraic number theory.

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