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If $\alpha \ne 0$ has degree $d$, then at every Archimedean embeddingIndeed, the factor of the defining product belonging to that Archimedean place shows $\max (1,|\alpha |{)}^{1/d}\le H(\alpha )$; using the exact local degree only improves this estimate. This proves the upper bound. Apply it to ${\alpha }^{-1}$ and use $H({\alpha }^{-1})=H(\alpha )$ to obtain the lower bound. This is the Liouville height inequality.

Solved by gpt-5.6-sol high.

Take distinct $x,y\in A_n$. Their difference has the formwhere $P\in \mathbb{Z}[X]$ has degree at most $n-1$ and polynomial length at most $nh$. By the height bound for a polynomial evaluation,The algebraic number $x-y$ is nonzero and has degree at most $d$, so the Liouville height inequality gives the separation

All elements of $A_n$ lie in an interval of length at mostSince $H(1-\alpha )\le 2H(\alpha )$, another application of the Liouville height inequality givesThe number of points in an interval is at most one plus its length divided by their minimum separation. ConsequentlyThus the requested statement holds, for example, with the absolute constant $C=2$.

Solved by gpt-5.6-sol high.

For $\beta \in K$, the nonzero algebraic number $\gamma =\alpha -\beta$ has degree at mostThe basic height inequality givesThe Liouville height inequality therefore yieldsThus one may take the explicit constant

Solved by gpt-5.6-sol high.