the problem at hand is:
Prove that there is an irrational number $a$ such that $a{^\sqrt{3}}$ is rational.
Hint: Consider $\sqrt[3]{2}^\sqrt{3}$ and argue by cases.
Any help on this problem would be greatly appreciated!
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4If $\sqrt[3]2^\sqrt3$ is irrational, consider $\left(\sqrt[3]2^\sqrt3\right)^\sqrt3=\sqrt[3]2^3$ – Rushabh Mehta Jan 26 '20 at 17:28
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5Does this answer your question? Can an irrational number raised to an irrational power be rational? – Brian61354270 Jan 26 '20 at 17:29
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No I do not think so, I need to prove that there is an irrational number a such that a^sqr(3) is rational. – sparky123 Jan 26 '20 at 17:37
3 Answers
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Hint:
If $\sqrt[3]{2}^{\sqrt{3}}$ is rational, you are done. If it is irrational, then what is $\left(\sqrt[3]{2}^{\sqrt{3}}\right)^{\sqrt{3}}$?
Thorgott
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1I'm not sure I am following, the necessary answer has to be in a proof format. Disregard the "Hint", what would be the idea to prove the first line? – sparky123 Jan 26 '20 at 17:36
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Note that $\sqrt[3]{2}$ is an irrational number. If you don't understand the hint (be it the original or mine), then please say which part you do not understand and I will elaborate. Your comment is too vague for me to get what you're asking for. – Thorgott Jan 26 '20 at 17:41
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$\alpha=\sqrt[3]{2}^{\sqrt{3}}$ is trascendental (hence irrational) by the Gelfond-Schneider theorem, hence $\alpha^{\sqrt{3}}=2$ proves the statement, but a simple dichotomy is sufficient to answer the question. If $\alpha=\sqrt[3]{2}^{\sqrt{3}}$ is rational we're done, and if $\alpha=\sqrt[3]{2}^{\sqrt{3}}$ is irrational we're equally done by considering $\alpha^{\sqrt{3}}=2$.
Jack D'Aurizio
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Another approach is to compare the cardinalities of the rational numbers and irrational numbers. Assume $f(x)=x^{\sqrt{3}}$ is rational for every irrational number $x$ and define
$$A=\mathbb{R}^{+}-\mathbb{Q}\text{ (positive irrational numbers)}$$
$$B=\{r:f(x)=r\text{ for some }x\in A\}$$
Note that by our assumption $B\subseteq \mathbb{Q}$. Since $f(x)$ is increasing on the positive reals, there is a map from $B$ to $A$. Thus
$$A\subseteq B\subseteq \mathbb{Q}$$
This is a contradiction as $A$ is an uncountable set while the rationals are a countable set. We conclude there exists an irrational number $x$ such that $x^{\sqrt{3}}$ is rational.
QC_QAOA
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