Please verify my proof: If $a$ and $b$ are irrational, then $a^b$ is irrational. I would like to prove or disprove the following statement:

If $a$ and $b$ are both irrational, then $a^b$ must be irrational.

I disproved the statement by giving a counter-example. It follows:

Let $a = \sqrt{10}$ and $b = \log(4)$. In this case, both $a$,$b$ $∉$ $\mathbb {Q}$.
  So $a^b = (\sqrt{10})$$^{(\log(4))}=2\in\mathbb {Q}$. Therefore, the statement is false.

Am I answering this in a right way? Please help!
 A: Your proof is just fine! What warrants statement here though is the classic non-constructive proof that your statement is false. We start with the number
$$\sqrt{2}^\sqrt{2}$$
Let us call this number $c$. There are two things $c$ could be: irrational or rational. If $c$ is rational we are done; if it is irrational we now raise $c$ to the power of $\sqrt{2}$ to get
$$c^\sqrt{2} = \left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2} = \sqrt{2}^{\sqrt{2}\sqrt{2}} = \sqrt{2}^2 = 2$$
We note that this number is rational, and since we are assuming $c$ is irrational and we know $\sqrt{2}$ is irrational we are done.  
Edit:
I should note that $c^2$ is known as the Gelfond–Schneider constant and it was later proved that $c$ is irrational
A: If you are using $\log $ as the base 10 logarithm, then your computations are fine. A more immediate example would come to mind though:
$$a = e, b = \ln{2},\\
a^b = e^{\ln{2}} = 2, 2\in \Bbb{Q} $$
Where $e $ is Neper's number, the base of the natural logarithm, or whatever you want to call it.
A: This works perfectly. The only thing I would do other than that is to generalize it to $a=\sqrt[n]{m}$ and $b=n\log_mk$ and from there prove that $a^b=k$. You could also just do $a\notin\mathbb{Q}$ and $b=\log_ak$, where $k\in\mathbb{Q}$. It isn't really necessary to prove that all irrational numbers produce irrational numbers, but it does show where you're coming from and make it look less like you just stumbled upon random numbers that worked.
