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Jul 19, 2019 at 5:48 vote accept Dominic van der Zypen
Jul 18, 2019 at 21:51 answer added Pietro Majer timeline score: 5
Jul 4, 2019 at 13:14 comment added Dominic van der Zypen Sorry forgot to specify this. I mean the Euclidean norm, i.e. $|(x,y)| = (x^2+y^2)^{1/2}$ for all $x,y \in \mathbb{Z}$. Equivalently we could take the "taxi norm" $|(x,y)| = |x| + |y|$ as $|(x,y)|= 1$ (in the Euclidean norm) iff $|(x,y)|= 1$ in the taxi norm for $x,y\in \mathbb{Z}$.
Jul 4, 2019 at 12:57 comment added Pietro Majer And which norm is $|(x,y)|$ ?
Jul 4, 2019 at 10:55 history asked Dominic van der Zypen CC BY-SA 4.0