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Let $a$ and $b$ be rational numbers, such that $\sqrt{a}$ and $\sqrt{b}$ are irrational.

Can $\sqrt{a}^\sqrt{b}$ be rational?

I found examples, where the irrational power of an irrational number is rational, but in those examples at least one of those numbers (base and exponent) has not been a square root of a rational.

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Since ${\sqrt{a}}^{\sqrt{b}}$ is expressed as an algebraic number not equal to $0$ or $1$ raised to an irrational algebraic power, the result will be transcendental (and hence irrational) by the Gelfond–Schneider theorem.

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    $\begingroup$ Any particular reason for rewriting the exponent? :) $\endgroup$ – Erick Wong Oct 2 '14 at 20:44
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    $\begingroup$ @Erick: Good point. No, there isn't. I'd forgotten exactly what Gelfond-Schneider said, and thought maybe it was rational to irrational algebraic, and didn't think about my conversion after I had looked up Gelfond-Schneider. I was doing this quick, thinking several others would have an answer in under a minute (and I'm hardly ever one of the "quick draws", for reasons I'd rather not go into here), and the result is that I didn't look as carefully at what I'd written as I usually do. $\endgroup$ – Dave L. Renfro Oct 2 '14 at 20:52

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