# Unusual metrics on the rationals

Suppose one day you get bored of the triangle inequality. You have done so much real analysis that you cannot see anything new and interesting in $|a-b| \le |a-z| + |z+b|$ anymore. But you want to keep doing analysis.

What can we do about this? One solution may be making this inequality more strict, and see where it takes us. For example, you may consider the following: $|a-b| \le \max{\{|a-z|,|z-b|\}}$. Note that this inequality does not hold under our usual absolute value, for example: $5 = |8 - 3| \not \le \max{ \{ |8-4|, |4-3| \} } = \max{ \{4,1\} } = 4$. But does this tweaking of the triangle inequality add anything new? It may not look like it at first, but it is a dream come true:  a sequence is Cauchy if and only if distance of consecutive terms goes to zero! More mathematically: the sequence $(a_n)_{n \in \mathbb{N}}$ is Cauchy $\iff$ $|a_n-a_{n+1}| \to 0$ as $n \to \infty$

This is not true in the case of our usual absolute value. For example, the infamous sequence $H_n = \sum_{k = 1}^{n} \frac{1}{k}$ is not Cauchy in the reals $\mathbb{R}$, for if it were Cauchy, it would converge since the reals numbers are complete. However, the difference between consecutive terms is $|1/n|$ which clearly tends to zero.

Before we dive into the proof of the previous claim, we need to note something: we are talking about convergence in $\mathbb{Q}$ however this would work in any metric space $(X,d)$ where the triangle inequality satisfies $d(a,b) \le \max{\{d(a,z), d(z,b) \}}$ for any $a,b,z \in X$. This is because an absolute value on $\mathbb{Q}$ induces a metric by setting $d(a,b) = |a-b|$.

Theorem: Let $| \cdot |$ be an absolute value on $\mathbb{Q}$ satisfying $|a+b| \le \max{ \{ |a|,|b|\} }$. Then any sequence $(a_n)_{n \in \mathbb{N}}$ of rational numbers is Cauchy if and only if $|a_n-a_{n+1}| \to 0$ as $n \to \infty$.

Proof: $( \implies)$ Let $\epsilon > 0$. Since $(a_n)_{n \in \mathbb{N}}$ is Cauchy there exists $N \in \mathbb{N}$ such that for all $n,m > N$ we have $|a_n - a_m| < \epsilon$. In particular, whenever $m = n+1$, we obtain  $|a_n - a_{n+1}| < \epsilon$, which means that $|a_n - a_{n+1}| \to 0$ as $n \to \infty$.

$( \impliedby )$ Let $\epsilon > 0$. There exists $N \in \mathbb{N}$ such that $|a_n - a_{n+1}| < \epsilon$ for any $n > N$.
Then, for all $m \ge n > N$, applying our strict triangle inequality:

\begin{aligned} |a_n - a_m| &= |a_n - a_{n+1} + a_{n+1} - \dots + a_{m-1} -a_{m}| \\ &\le \max{\{|a_n - a_{n+1}|, \dots, |a_{m-1} -a_{m}| \} } < \epsilon \end{aligned}.

Therefore the sequence is Cauchy.   $\square$

This is all well and good, however, it makes us wonder whether there is any absolute value on $\mathbb{Q}$ with this strict triangle inequality property. The answer is yes, in fact, there are infinitely many of them, but this is the topic for another post. Until then, to satisfy your curiosity I will write down one such absolute value and I will leave it up to you to check that it is indeed an absolute value:

$|x|_3 = \frac{1}{3^{v(x)}}$
where $v(x) = n$ is the exponent of $3$ whenever we write the rational number as $x = 3^n \frac{p}{q}$, where $p,q$ are coprime to $3$.

Can you suggest a different absolute value?