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32 votes
3 answers
2k views

A funny metric over $\mathbb{N}$

$\DeclareMathOperator{\lcm}{lcm}$ Fiddling with numbers I realized that for positive integers $x$ and $y$, the quantity $$\Vert x,y \Vert=\frac{\lcm(x,y)}{\gcd(x,y)}$$ has these properties: $\Vert x,...
35 votes
6 answers
2k views

Trigonometry / Euclidean Geometry for natural numbers?

Let $d(a,b) = 1 - \frac{2\gcd(a,b)^3}{ab(a+b)}$ be a metric on natural numbers without $0$. The metric space $X = \{x_0,x_1,\cdots,x_n\},n>2$ is isometric embeddable in $\mathbb{R}^n$ if and only ...
10 votes
0 answers
793 views

Two questions around the $abc$-conjecture

Let $d(a,b) = 1-\frac{2 \gcd(a,b)}{a+b}$, $d_{ABC}(a,b) = 1-\frac{2\gcd(a,b)^3}{ab(a+b)}$ be two metrics on natural numbers. The abc-conjecture can be formulated using these two metrics as: For ...
5 votes
1 answer
337 views

Greedy simplices in an ultrametric space (generalized Bhargava $p$-orderings)

Let $\left(U, d\right)$ be a finite ultrametric space -- that is, $U$ is a finite set, and $d : U \times U \to \mathbb{R}_{\geq 0}$ is a metric on $U$ such that every $x, y, z \in U$ satisfy $d\left(x,...
24 votes
4 answers
2k views

A reinterpretation of the $abc$ - conjecture in terms of metric spaces?

I hope it is appropriate to ask this question here: One formulation of the abc-conjecture is $$ c < \text{rad}(abc)^2$$ where $\gcd(a,b)=1$ and $c=a+b$. This is equivalent to ($a,b$ being ...