# Is $x^y$ always irrational if $x$ is rational and $y$ is irrational?

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?

• What is not true? Oct 16 '14 at 23:49
• sorry, I will edit it
– JLL
Oct 16 '14 at 23:50
• you just proved that the proposition is false. Oct 16 '14 at 23:51
• @mookid can you explain? in the format $P \arrow Q$ if possible?
– JLL
Oct 16 '14 at 23:54
• 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. Oct 16 '14 at 23:55

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

• Well if you declare 0 to be irrational, you can prove anything by ex falso quodlibet. Zero is rational. 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 $x=2$, $y=\log_23$.

• 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. Oct 17 '14 at 0:04
• @Asaf True, but (at the time I am writing) all the other solutions suggested are trivial. It's well worth knowing a bit more. Oct 17 '14 at 0:05
• 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! Oct 17 '14 at 0:06

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.

• @user46944 A "disproof" is something that "disproves" a statement. Oct 16 '14 at 23:59

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 $$\sqrt{2}^{\sqrt{2}}, \ (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$$

• $2^{\pi}$ isn't known to be irrational, so that example doesn't work. Jul 5 at 3:38