Prove or disprove:

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

I am stuck with this, these are my steps.

let $x=2$ and $y=\sqrt{2}$

$\implies$ $x^y = 2^{\sqrt{2}} $

now if $x^y = 2^{\sqrt{2}} $ is irrational then we are done. But if this is rational then we can say:

let $x=2^\sqrt{2}$ (since we assume its rational) and let $y=\sqrt{2}$

$\implies$ $x^y = 2^{\sqrt{2}^{\sqrt{2}}} $ $=2^2$

this shows that if $x$ is rational and $y$ is irrational then $x^y$ is rational.

But I know that this is not true. Where did I go wrong in this?

  • $\begingroup$ What is not true? $\endgroup$ – Git Gud Oct 16 '14 at 23:49
  • $\begingroup$ sorry, I will edit it $\endgroup$ – JLL Oct 16 '14 at 23:50
  • 2
    $\begingroup$ you just proved that the proposition is false. $\endgroup$ – mookid Oct 16 '14 at 23:51
  • $\begingroup$ @mookid can you explain? in the format $P \arrow Q$ if possible? $\endgroup$ – JLL Oct 16 '14 at 23:54
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    $\begingroup$ You should really write it as $(2^{\sqrt{2}})^{\sqrt{2}}$ instead of $2^{{\sqrt{2}^{\sqrt{2}}}}$. They mean different things. Exponentiation is not associative. $\endgroup$ – layman Oct 16 '14 at 23:55

Your "proof" is not really a proof. You pick a particular number, and you claim that if it's irrational then the statement is proved, and in the second part you pick another particular case claiming that it's a counterexample to the statement anyway. But the proof is that every rational number $x$ and irrational number $y$ satisfy this. Not just this particular pair.

In fact, the second part is almost a disproof by itself. It says "If $x$ was rational, then $x^y$ was rational as well", which is exactly what you need to disprove the statement. Although the details of that second part are sketchy, for example if $y=\sqrt2^{\sqrt2}$, then $2^y=2^{\sqrt2^{\sqrt2}}$ is not the same thing as $(2^{\sqrt2})^{\sqrt2}$, which is really what you're looking for.

  • $\begingroup$ @user46944 A "disproof" is something that "disproves" a statement. $\endgroup$ – Omnomnomnom Oct 16 '14 at 23:59

Hint (for an easy proof/disproof): what if $x = 1$?

  • $\begingroup$ Well if you declare 0 to be irrational, you can prove anything by ex falso quodlibet. Zero is rational. $\endgroup$ – Pseudonym Oct 17 '14 at 0:18

Your statement has implied "for all"s in it. That is,

For all $x$ and for all $y$, if $x$ is rational and $y$ is irrational, then $x^y$ is irrational.

You cannot prove an "all" statement true by giving an example. You could prove it is false by giving an example where $x$ is rational, $y$ is irrational and $x^y$ is rational. An example would be $x=2$, $y=\log_23$.

  • $\begingroup$ For this you'd need first to know that $\log_23$ is irrational. It's not a terribly difficult proof, but there are simpler solutions to the problem the OP is facing. $\endgroup$ – Asaf Karagila Oct 17 '14 at 0:04
  • $\begingroup$ @Asaf True, but (at the time I am writing) all the other solutions suggested are trivial. It's well worth knowing a bit more. $\endgroup$ – David Oct 17 '14 at 0:05
  • $\begingroup$ I agree with that, which is why I upvoted your answer. It's a good one, even if it probably goes beyond the scope of the question! $\endgroup$ – Asaf Karagila Oct 17 '14 at 0:06

Let $x,y\in \mathbb{R}-\{1,0\}.$ Consider $x^y.$

1) $x,y$ both are rational $$2^{\frac{1}{2}},\ 2^2$$
2) $x$ rational, $y$ irrational $$2^{\pi},\ 2^{\log_23}$$
3) $x$ irrational, $y$ rational $$e^2,\ (\sqrt{2})^2$$
4) $x,y$ both are irrational $$\sqrt{2}^{\sqrt{2}}, \ (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$$


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