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This is a question from Mathematics for Computer Science by Lehman: Prove that ${\sqrt 2}^{\sqrt 2}$ can be rational.Prove by making cases. How can we write it by showing different cases?

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closed as off-topic by naslundx, Mike Haskel, Johannes Kloos, Namaste, Watson Aug 16 '16 at 14:05

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    $\begingroup$ Do you mean ${\sqrt 2}^{\sqrt 2}$ ? Doesn't the problem ask to prove the existence of irrational numbers $a,b$ such that $a^b$ is rational? $\endgroup$ – Watson Aug 16 '16 at 12:31
  • $\begingroup$ yes i mean that only..sry for the mistake..i edited it now $\endgroup$ – Shubham Jindal Aug 16 '16 at 12:34
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    $\begingroup$ Duplicate ? math.stackexchange.com/questions/446647/… $\endgroup$ – Robert Z Aug 16 '16 at 12:34
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    $\begingroup$ As far as I know, you can prove that it is rational or that it is not. Or perhaps you can even prove that its rationality can not be decided. But, proving that a number can be rational makes no sense. This would be modal logic. $\endgroup$ – ajotatxe Aug 16 '16 at 12:35
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    $\begingroup$ The number you ask about is irrational (not rational). That's a proven fact. (Yet out of scope for such a course.) Likely the question you intend to as is slightly different. Please clarify what you want to know. $\endgroup$ – quid Aug 16 '16 at 12:47
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T.P. Two irrational numbers $\\a,b\\$ such that $\\a^b\\$ is rational
Let $\\a = \sqrt{2}^\sqrt{2}\\$ and $\\b = \sqrt{2}\\$,
$\\a^b = \sqrt{2}^{\sqrt{2}^2} = \sqrt{2}^2 = 2\\$ which is clearly rational.

If $\sqrt{2}^\sqrt{2}$ is irrational, then we have found our example. If $\sqrt{2}^\sqrt{2}$ is however rational, then of course $a=b=\sqrt{2}$ is an example of irrationals $a,b$ such that $a^b$ is rational.

Henceforth, there exists two irrational numbers $\\a,b\\$ such that $\\a^b\\$ is rational.
I hope this was the correct implementation of the answer.

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  • $\begingroup$ But for the case a = ${\sqrt 2}$ and b=${\sqrt 2}$ it won't be so simple to prove it I think $\endgroup$ – Shubham Jindal Aug 16 '16 at 12:46
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    $\begingroup$ More to the point, $b = \sqrt2$ is clearly irrational. (There's a rather famous proof of that.) If $a = b^b$ was rational, then it would in itself be a valid example of an irrational power of an irrational number being rational; if it's not, then $a^b = 2$ clearly is. $\endgroup$ – Ilmari Karonen Aug 16 '16 at 12:47
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    $\begingroup$ I expanded the answer slightly, since I think otherwise it will cause confusion. If you do not like this just roll it back. $\endgroup$ – quid Aug 16 '16 at 12:53
  • $\begingroup$ Something is wrong in this answer: if $a = {\sqrt 2}^{\sqrt 2}$ and $b = \sqrt 2$, then $a^b$ is $\left( {\sqrt 2}^{\sqrt 2} \right) ^{\sqrt 2}$, not $\left( {\sqrt 2}^{\sqrt 2} \right) ^2$. $\endgroup$ – Alex M. Aug 17 '18 at 12:17
  • $\begingroup$ @AlexM. It says $\left(\sqrt2^\sqrt2\right)^\sqrt2 = \sqrt2^{\sqrt2\cdot\sqrt2} = \sqrt2^{\sqrt2^2} \text{(note: same as $\sqrt2^{\left(\sqrt2^2\right)}$}) = \sqrt2^2 = 2$, not $\left(\sqrt2^\sqrt2\right)^2$ $\endgroup$ – CherryDT Aug 18 '18 at 13:21

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