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Yaakov Baruch
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At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:

is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is q$q$ in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?

I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.

At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:

is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is q in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?

I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.

At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:

is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is $q$ in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?

I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.

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Source Link
Yaakov Baruch
  • 5.1k
  • 2
  • 31
  • 43

At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:

is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is q in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?

I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.