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Prove that $2^{\sqrt{3}}$ is irrational or even it is transcendental? I am wondering if there exists the method that describes its property and its irrationality of these kinds (or even with some numbers of the form like $5^{\sqrt{2}+\dfrac{1}{1+\sqrt[4]{2}}}$), which involves only with the algebraic number

W. Wongcharoenbhorn
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1 Answers1

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Gelfond-Schneider theorem: if $a,b$ are algebraic, $a \ne 0,1$, $b$ irrational, then all values of $a^b$ are transcendental.

So: $2^{\sqrt{3}}$ is transcendental.

To show $$ 5^{\displaystyle \left(\sqrt{2}+\frac{1}{1+\sqrt[4]{2}}\right)} \tag1$$ is transcendental, it suffices to show $$ \sqrt{2}+\frac{1}{1+\sqrt[4]{2}} \tag2$$ is irrational. I leave that to you, using the rational root theorem.

Note: Numerically evaluating $(2)$ we get $1.8709999$ so we may mistakenly think it is rational. In fact with more digits it is $1.8709999455101501528$.

GEdgar
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  • Thank you and if i'm not wrong let $x=\sqrt{2}+\dfrac{1}{1+\sqrt[4]{2}}$ is a root of $(x^2+2x-3)^2=18 \Longrightarrow x^4+4x^3-2x^2-12x-9=0$ and the rational root theorem says that, if $x=\dfrac{p}{q}$, $x$ is an integer since $q|1$ and $4^2<(x^2+2x-3)^2=18<5^2$ which is not possible. – W. Wongcharoenbhorn May 21 '20 at 13:12
  • I would say: from $x^4+4x^3-2x^2-12x-9=0$, look at the $9$ constant term (and the $1$ coefficient of $x^4$). If it has a rational root, it must be one of $\pm 1, \pm 3, \pm 9$. Trying all these, I see that none of them is a solution. So it has no rational roots. – GEdgar May 21 '20 at 13:16
  • Interesting aside. $(2)$ is equal to $\displaystyle -1+\sqrt{4+3\sqrt{2}}$ – GEdgar May 21 '20 at 13:19
  • Definitely, professor. – W. Wongcharoenbhorn May 21 '20 at 13:20