Originally Posted by

**Turiski** For the first problem, Tonio had the right idea, and so do you

Consider $\displaystyle \sqrt{2}^{\sqrt{2}}$. If it was rational, what would that imply?

If it was irrational, what would that imply?

...

OK, but this is incomplete. If $\displaystyle \sqrt2^{\sqrt2}$ is irrational (and, although I assume it is, how do we prove it?), all it proves is that there exist $\displaystyle a\, (=\sqrt2^{\sqrt2})$ and $\displaystyle b\,(=\sqrt2)$, both irrational, for which $\displaystyle a^b$ is rational. Which is not what the question asks for. We need to establish the existence of a rational, $\displaystyle a$, and an irrational, $\displaystyle b$, for which $\displaystyle a^b$ is irrational.

However, even if we *can* establish that $\displaystyle \sqrt2^{\sqrt2}$ is irrational, we would still need to go on to say:We have found $\displaystyle a,\, b$ such that $\displaystyle a^b = c$, where $\displaystyle a,b \notin \mathbb{Q}$ and $\displaystyle c \in \mathbb{Q}$

Now $\displaystyle a^b=c\Rightarrow a = c^{\frac1b}$

But if $\displaystyle b \notin \mathbb{Q}$, then $\displaystyle \frac1b \notin \mathbb{Q}$. So *now *we have the required result.

So, how about it? Why *is *$\displaystyle \sqrt2^{\sqrt2}$ irrational? And would it be just as easy to prove that $\displaystyle 2^{\sqrt2}$ is irrational, thus establishing the result much more simply, with $\displaystyle a = 2,\, b =\sqrt2$?

Or am I still missing something obvious? If so, please spell it out for me.

Grandad