Fix a prime $p$ and a positive rational number $r$ can be written uniquely in a form $r=p^e\frac{b}{c}$, with $e\in \Bbb Z$ and $p,b,c \in \Bbb N$ being pairwise coprime.
Define $|\cdot|_p$ as
- $|r|_p=0$, if $r=0$,
- $|r|_p=|-r|_p$, if $r<0$,
- $|r|=p^{-e}$, if $r=p^eb/c$.
The special metric is defined as $d(r_1,r_2)=|r_1-r_2|_p$.
I meet this problem in exercises, I want to construct a Cauchy sequence such that the difference of two items is of the form $p^e$, and show its limit is irrational. But I am confused with how to show a limit is irrational.
Are my thinkings right? I just want some hints (solution better).
This is the $p$-adic metric on $\Bbb Q$.
Textbooks describe how if $p$ is odd and $a$ is a quadratic residue modulo $p$ then $x^2=a$ is soluble in the $p$-adic numbers by using a method called "Hensel lifting". This involves constructing a sequence of integers converging to a square root of $a$ modulo $p$, so this sequences of integers is a non-convergent Cauchy sequence within $\Bbb Q$.
But we can proceed more naively using fixed point iteration. Let $p=7$ and $a=2$. Take $x_0=3$, so that $x_0^2\equiv2\pmod 7$. Define a sequence $(x_n)$ recursively by $$x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right).$$ It's easy to show that $x_n\to\sqrt2$ inside $\Bbb R$ but we want this to be true in our metric, more precisely $|x_n^2-2|_7\to0$. In fact $|x_n^2-2|_7=7^{-n-1}$ and $|x_{n+1}-x_n|_7=7^{-n}$ (exercises for reader). From this one deduces that $(x_n)$ is Cauchy, and if $x_n\to a$ then $x_n^2\to a^2$, but $x_n^2\to2$.