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[数论] $(1+\sqrt5)^n$到最近整数的距离可任意小?

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hbghlyj posted 2024-4-21 21:10 |Read mode
给定$ε>0$,存在$n\inN$,$(1+\sqrt3)^n$到最近整数的距离小于$ε$.
证明:注意到$(1+\sqrt3)^n+(1-\sqrt3)^n\inZ$,$|1-\sqrt3|<1$,取$|(1-\sqrt3)^n|<ε$即可。

给定$ε>0$,存在$n\inN$,$(1+\sqrt5)^n$到最近整数的距离小于$ε$.
证明:

给定$ε>0$,存在$n\inN$,$(1+\sqrt{-5})^n$到最近的$\Bbb Z[i]$的元素的距离小于$ε$.
证明:

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睡神 posted 2024-4-23 18:32 from mobile
更一般地,$\forall a\in N^*,0<b<1,f(n)=${$(a+b)^n$}无最小值?其中,{$x$}为高斯函数,表示取$x$的小数部分,例如:{$1.25$}$=0.25$
除了不懂,就是装懂

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