This seemed like a very obvious thing. I was playing around with exponents and logarithms, and I just started wondering, "if $a$$\in$$\mathbb Z$, and $r$$\in$$\mathbb R \setminus \mathbb{Q}$, is $a^r$ irrational?"
As I thought about this, it was almost obivous, but I could not prove it. Is there a proof to the above statement? Namely, is every integer to the power of a real number a real number that is also irrational? (This is not true for real numbers though. Namely, real numbers to the power of real numbers could be rational, i.e., $\sqrt 2^{2log_22}$=$1$ which is an integer let alone a rational). How would also one go about proving such a statement?
EDIT: I have found a mistake. I meant to say if $a^r$ is not an integer, then it is real. (If $a$$\in$$\mathbb Z$, and $r$$\in$$\mathbb R$, and if $a^r$ is not an integer, is then $a^r$ irrational?)
[I am assuming you mean an integer to the power of an irrational number.] What about $3 = 2^{\log_2 3}$? We show that $\log_2 3$ is indeed irrational; if $2^{\frac{m}{n}}=3$ for some integers $n,m \ge 1$ then raising both sides to the $n$-th power gives $2^m = 3^n$ for some integers $m,n$. Which is impossible as $2^m$ is even and $3^n$ is odd.