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If $x$ is a rational number and $y$ is an irrational number then $x^y$ is irrational.

I have tried to prove it,

let $x= 2$ and $y=\sqrt{2}$ then $x^y = 2^{\sqrt{2}}$,

if it is an rational number then let $x=2^{\sqrt{2}}$ and $y=\sqrt{2}$, then, $x^y= 2^{\sqrt{2} *\sqrt{2}}=4$ which is rational,

therefore $x^y$ is rational. Is it correct?

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  • $\begingroup$ Tidy up your proof a little, but I'm pretty sure that's all there is. $\endgroup$ – ConMan Sep 27 '16 at 0:23
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    $\begingroup$ You have proved (almost, if you tidy up it will be proved) that there is a rational $x$ and an irrational $y$ such that $x^y$ is rational. Not what your title asks for. $\endgroup$ – Jean-Claude Arbaut Sep 27 '16 at 0:25
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    $\begingroup$ Not following...are you trying to prove that $2^{\sqrt 2 }$ is rational? $\endgroup$ – lulu Sep 27 '16 at 0:25
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    $\begingroup$ Simple counting shows that there must be an irrational $y$ such that $2^y$ is irrational (there are uncountably many irrationals, but only countably many rationals). $\endgroup$ – lulu Sep 27 '16 at 0:27
  • $\begingroup$ Didnvt you say you wanted to prove it was irational? $\endgroup$ – fleablood Sep 27 '16 at 0:28
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very close!

$\color{white}{\text{I like the try though +1}}$

Let $x=2$ and $y=\sqrt{1/2}$.

Either $2^{\sqrt{1/2}}$ is irrational, or $\left(2^{\sqrt{1/2}}\right)^{\sqrt{1/2}}$ is irrational

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  • $\begingroup$ Classic annoying example by Hilbert? Can't remember. $\endgroup$ – Jacky Chong Sep 27 '16 at 0:38
  • $\begingroup$ @JackyChong I don't know, but I decided to play along with the OP's attempted method. $\endgroup$ – Simply Beautiful Art Sep 27 '16 at 0:38
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As a concrete example of $(\text{rational})^{\text{irrational}}=\text{rational}$: $$10^{\log_{10}3}=3$$ To see that $\log_{10} 3$ is irrational, suppose it equaled $\frac ab$ with $a,b\in \mathbb N$. We'd then have $$3=10^{\frac ab}\implies 3^b=10^a$$ But this contradicts Unique Factorization.

As a concrete example of $(\text{rational})^{\text{irrational}}=\text{irrational}$: $$10^{\log_{10}\pi}=\pi$$ To see that $\log_{10}\pi$ is irrational, suppose it equaled $\frac ab$ $a,b\in \mathbb N$. We'd then have $$\pi=10^{\frac ab}\implies \pi^b=10^a$$ Whence $\pi $ would be a root of the equation $X^b-10^a=0$. But $\pi$ is transcendental, hence is not the root of any polynomial with rational coefficients.

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  • $\begingroup$ A simpler example of the former is $1^{\sqrt{2}} = 1 = \mbox{rational}$. $\endgroup$ – Michael Sep 28 '16 at 3:10
  • $\begingroup$ @Michael True, of course...but in this sort of question, it's usual to exclude $0,1$ from the base (as the examples aren't very interesting). For example, we have the powerful Gelfond Schneider Theorem which tells us that, if $a,b$ are algebraic with $b\notin \mathbb Q$ then $a^b$ is transcendental. But of course this must exclude $a=0,1$. $\endgroup$ – lulu Sep 28 '16 at 10:49
  • $\begingroup$ Given the countability argument and the Gelfond Schneider theorem you mention, there are abundant examples of $(\mbox{rational})^{\mbox{irrational}}= \mbox{irrational}$. So, it surprises me that it seems difficult to come up with one concrete example. Yours uses the transcendental nature of $\pi$ which is not a basic fact (i.e. proof is advanced). Are there simpler examples? Of course, I suppose you can replace $\pi$ with any transcendental, but then you have to prove existence of such, which perhaps goes back to countability arguments? $\endgroup$ – Michael Sep 28 '16 at 14:28
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    $\begingroup$ liouville's theorem let's you produce some "simple" transcendental numbers. Like $\sum 10^{-n!}$. $\endgroup$ – lulu Sep 28 '16 at 22:49
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Changing the letters for convenience, let's show that there's a rational number $x$ such that $2^x$ is irrational. For each rational number $r,$ the curve $y=2^x$ meets the vertical line $x=r$ in just one point, and it meets the horizontal line $y=r$ in just one point. Since the set of rational numbers is countable, there are only countably many points on the curve $y=2^x$ with at least one rational coordinate. Since there are uncountably many points on the curve, there are uncountably many points with both coordinates irrational; i.e., there are uncountably many (in fact continuum many) irrational values of $x$ such that $2^x$ is also irrational. The same goes with $2$ replaced by any positive rational number different from $1.$

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