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I learned that in ancient, people believed that any number is a ratioal, but later found that the third length (i.e., $\sqrt2$) of a right triangle with two sides of length $1$ is not rational. I am wondering if there is a metric on $\mathbb Q^n$ so that this phenomenon does not appear; that is, the distance between any two points is a rational number.

If there does exist, how close can such a metric $d'$ be to the usual metric $d$ on $\mathbb R^n$, in the sense that how small can $\delta>0$ be so that $|d(x,y)-d'(x,y)|<\delta$ for all $x,y\in\mathbb Q^n$.

I think it for a long time while still have no idea.

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    What about the taxicab metric? $d'(x, y) = \sum_{k=1}^n |y_k - x_k|$ The difference, absolute value, sum of rational numbers is rational so this is always rational for $x,y \in \mathbb{Q}^n$. – Anakhand Apr 12 '22 at 16:09
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    The discrete metric $d(x,y) = 1$ when $x \not = y$ and $d(x,x)=0$ for all $x\in \mathbb{Q}^n$ seems to work. But it is not really useful on your second point. – Auke Schaap Apr 12 '22 at 16:14
  • Thanks you two! –  Apr 13 '22 at 00:19

1 Answers1

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Fix an $\delta>0$, and choose $m\in\mathbb N$ so large that $2^{-m}<\delta/2$. For each $x,y\in\mathbb Q^n$, let $$d'(x,y)=\inf\left\{ a_k\cdots a_0.a_{-1}\cdots a_{-m}\ge d(x,y)\,\middle|\,a_i\in\{0,1\} \right\}.$$ Here $a_k\cdots a_0.a_{-1}\cdots a_{-m}$ is a binary representation. It is readily to check that $d'$ is a metric which takes only rational numbers as values. Also, $|d'(x,y)-d(x,y)|<\delta$ for all $x,y$.

Remark: Note that this metric not only takes rational value on $\mathbb Q^n$ but in fact on the whole $\mathbb R^n$.

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