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Prove/Disprove that if $x$ is irrational, then $2^x$ is also irrational.

My attempt for the proof: Suppose $2^x>0$ is a rational number, then $2^x=\frac{a}{b}$ for some natural numbers $a$ and $b$. Taking logarithm with base $2$ on both sides to get, $x=\log_2 \frac{a}{b}$. Here I stuck! how to reach at $x$ is rational?

Riaz
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1 Answers1

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This is false in general, take $x=\frac{\ln 3}{\ln 2}$, then $2^x=e^{x\ln 2}=e^{\ln 3}=3\in\mathbb{Q}$ but $x\notin\mathbb{Q}$. Otherwise, there would exist $p,q\geqslant 1$ coprimes such that $x=\frac{p}{q}$, that is to say $q\ln 3=p\ln 2$ and thus $3^q=2^p$ which is not because $2$ and $3$ are coprimes.

Tuvasbien
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    But it's true if $x$ is an algebraic irrational: https://en.wikipedia.org/wiki/Hilbert%27s_seventh_problem – bof Jul 26 '20 at 02:33
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    when you take $x = \frac{\ln 3}{\ln 2}$, you would also have to argue that $\frac{\ln 3}{\ln 2}$ is irrational – p_square Aug 21 '21 at 05:31