What kind of number is $2^\sqrt{2}$ ? It is real and obviously is not integer nor rational. Is it real algebraic number?
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6This is the Gelfond Schneider constant. It is transcendental. – lulu Dec 30 '16 at 15:37
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@pasaba por aqui Try to use logarithm. – Harsh Kumar Dec 30 '16 at 15:42
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4@HarshKumar How is that going to help? – Simply Beautiful Art Dec 30 '16 at 15:43
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@SimpleArt I think it will bring the power in multiplication $log 2^\sqrt2=\sqrt2 \times log2 $ – Harsh Kumar Dec 30 '16 at 15:49
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@HarshKumar Sure, but what does that tell us? Nothing is so clear. – Simply Beautiful Art Dec 30 '16 at 15:51
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@SimpleArt It will tell us that $2^{\sqrt2}$ is irrational. – Harsh Kumar Dec 30 '16 at 15:52
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@HarshKumar No, it tells us that the log of that is irrational, depending on which base you used for the log. To show that $2^{\text{irrational}}\implies\text{irrational}$ is a whole 'nother step. – Simply Beautiful Art Dec 30 '16 at 15:54
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@SimpleArt Can't I take antilog – Harsh Kumar Dec 30 '16 at 15:55
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1@HarshKumar But what does that tell you about the number? Without the Gelfon-Schneider theorem, it tells us nothing. – Simply Beautiful Art Dec 30 '16 at 15:56
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According to the Gelfond–Schneider theorem, since $2$ and $\sqrt2$ are both algebraic with $a\ne0,1$ and $\sqrt2$ irrational, it follows that $2^{\sqrt2}$ is irrational and transcendental.
Your exact number is also known as the Gelfond-Schneider constant, and it has its own Wikipedia full of such information.
Simply Beautiful Art
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It doesn't have its own Wikipedia. It has, however, its own page on Wikipedia :) – Wojowu Dec 30 '16 at 20:58
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