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I need to prove the following is a metric space over the integers:

$b \geq 2$. For distinct integers $x, y$. Let $N(x,y)$ be the greatest integer $n$ such that $b^n$ divides $(x - y)$. Let $d(x,y) = b^{-N(x,y)} $. Also assume for $x = y$, $d(x,y) = 0$.

I have proven the positive definite axiom, and the symmetry axiom fairly easily. However, as often happens, I am struggling severely with proving the triangle inequality. I am aware we need to thus show that:

$$b^{-N(x,y)} \leq b^{-N(x,z)} + b^{-N(z,y)}$$

I have tried to do it by trying to in some way factorise $b$ but the equations were getting far too confusing. My next approach was to try and rewrite the equation as follows (where $s,t,u \in \mathbb{Z}$):

If $b^{n_1} = s(x-y)$. Then $d(x,y)$ = $b^{-n_1} = \frac{1}{s(x-y)}$.

Similarly If $b^{n_2} = t(x-z)$. Then $d(x,z)$ = $b^{-n_2} = \frac{1}{t(x-z)}$.

If $b^{n_3} = u(z-y)$. Then $d(z,y)$ = $b^{-n_3} = \frac{1}{u(z-y)}$.

I am then hoping to use the triangle inequality on the usual metric to argue $$\frac{1}{s(x-y)} \leq \frac{1}{t(x-z)} + \frac{1}{u(z-y)}$$ however I am having problems because we don't know what $s,t,u$ are. Am I missing something about the definition of $n$ that can tell me how these three relate, or should I be going about this some other way entirely? Thanks for your help

yhu
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Assume that $n\le N(x,z)$ and $n\le N(z,y)$. Then $b^n\,|\,x-z$ and $b^n\,|\,z-y$, hence also $b^n\,|\,x-y$, i.e. $n\le N(x,y)$.

This yields $\ \min((N(x,z),\,N(z,y))\le N(x,y)$.
By that we get that $-N(x,y)$ is less or equal to one of $-N(x,z)$ and $-N(z,y)$.

So, as $b>1$, we will have $b^{-N(x,y)} \le b^{-N(x,z)}+b^{-N(z,y)}$.

Berci
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