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world:boule [2025/10/03 01:58] rdouc [Proof] |
world:boule [2025/10/03 02:00] (current) rdouc [Remarks:] |
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* For $D = \mathbb{Z}^d$, one can construct $c$ explicitly, for example $c=(\pi,\ldots,\pi^d)$. Then, for $a,b \in \mathbb{Z}^d$, the equality $\|c-a\|^2=\|c-b\|^2$ becomes a polynomial equation with integer coefficients having $\pi$ as a root. This is impossible unless $a=b$, since $\pi$ is not algebraic. However, this explicit construction of $c$ does not generalize (or at least not simply) when $D$ is not $\mathbb{Z}^d$. | * For $D = \mathbb{Z}^d$, one can construct $c$ explicitly, for example $c=(\pi,\ldots,\pi^d)$. Then, for $a,b \in \mathbb{Z}^d$, the equality $\|c-a\|^2=\|c-b\|^2$ becomes a polynomial equation with integer coefficients having $\pi$ as a root. This is impossible unless $a=b$, since $\pi$ is not algebraic. However, this explicit construction of $c$ does not generalize (or at least not simply) when $D$ is not $\mathbb{Z}^d$. | ||
* Obviously, when $D = \mathbb{Z}^d$, one does not need to prove that $D$ is countable, nor that $A \cap [0,r]$ is finite, which significantly shortens the proof. | * Obviously, when $D = \mathbb{Z}^d$, one does not need to prove that $D$ is countable, nor that $A \cap [0,r]$ is finite, which significantly shortens the proof. | ||
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