Q: Does there exist an algebraic number $x,$ s.t. $f(x)$ is also an algebraic number?

$f(x)=\exp\bigg(\frac{1}{\ln(x)}\bigg)$ for $x\ne0,1.$

I would like to prove that the set of points $(x,f(x))$ is the empty set, for algebraic $x.$

Are there any promising approaches to solve this?

  • $\begingroup$ If $x\ne1$ then there is a positive answer $x=0 \implies f(0)=1 $ $\endgroup$ – Ultradark Nov 8 at 15:50
  • $\begingroup$ When did the logarithm become defined on $\{0\}$? In fact, this $f$ is maximally defined on $(0,\infty)$ if we are thinking of it as a real function. If we are thinking of it as a complex function (valid since the setting is algebraics), then there is freedom to choose a branch cut, but not the branch point: $0$. $\endgroup$ – Eric Towers Nov 8 at 15:52
  • $\begingroup$ wolfram alpha says that $f(0)=1$ $\endgroup$ – Ultradark Nov 8 at 15:53
  • 1
    $\begingroup$ Then apparently, Wolfram Alpha has misled you by silently taking $\lim_{x \rightarrow 0^+} f(x)$ instead of correctly observing $\ln 0$ is undefined, so $f$ is undefined at $x = 0$. $\endgroup$ – Eric Towers Nov 8 at 15:56
  • $\begingroup$ Always be wary of trusting a computer. Programmers make errors too. $\endgroup$ – Paul Sinclair Nov 9 at 3:58

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