Irrational Number
An irrational number is an element $$x \in \mathbb{ R } \setminus \mathbb{ Q }$$
irrational to the power of irrational can be rational
Prove that there exists two irrational numbers $$a , b$$ such that $$a^{b}$$ is rational
Consider $$a = \sqrt{2}$$ and $$b = \sqrt{2}$$, then there are two cases, either $$\sqrt{2}^{\sqrt{2}}$$ is rational, and the proof is over, or $$\sqrt{2}^{\sqrt{2}}$$ is irrational and in that case consider $$a = \sqrt{2}^{\sqrt{2}}$$ and $$b = \sqrt{2}$$, then $$a^{b} = \left ( \sqrt{2}^{\sqrt{2}} \right )^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^{2} = 2$$, so in any case we've found an $$a , b$$ that are irrational such that $$a^{b}$$ is rational.
Square Root of a Non Perfect Square is Irrational
Suppose that $$n \in \mathbb{ N } _ 1$$ is not a perfect square, then $$\sqrt{ n }$$ is irrational.
Suppose for the sake of contradiction that $$\sqrt{ n }$$ was rational so that $$\sqrt{ n } = \frac{p}{q}$$ therefore $$q ^ 2 n = p ^ 2$$, since $$n$$ is not a perfect square then we know that $$n = \prod _ { i \in \mathbb{ N } _ 1 } p _ i ^ { \alpha _ i }$$ where there is some $$j \in \mathbb{ N } _ 1$$ such that $$\alpha _ j$$ is odd, now we also know that $$q ^ 2$$ is equal to $$\prod _ { i \in \mathbb{ N } _ 1 } p _ i ^ { 2 \beta _ i }$$ since odd plus even is still odd then we know that in the prime factorization of $$q ^ 2 n$$ the power on the $$j$$-th prime is still odd, thus $$q ^ 2n$$ is not a perfect square, but on the other hand $$p ^ 2$$ is, which is a contradiction, therefore $$\sqrt{ n }$$ must be irrational.
There is a Rational Between any Two Reals
For any $$x, y \in \mathbb{ R }$$ such that $$x \lt y$$ there exists a $$p \in \mathbb{ Q }$$ such that $x \lt p \lt y$

Since $$x \lt y$$ we have $$0 \lt y - x$$ therefore by the archimediean property we have an $$n \in \mathbb{ N } _ 1$$ such that $$n \left( y - x \right) \gt 1$$

Consider the set $$X = \left\{ z \in \mathbb{ Z } : z \gt nx \right\}$$, this set is not empty because $$nx + 1 \in X$$, additionally it is bounded below, and so as a subset of $$\mathbb{ Z }$$ it has a least element say $$m$$ such that $$m \gt nx$$ . Since it is the least element we know that $$m - 1 \le nx$$, which implies that $$m \le nx + 1$$.

Recall that our $$n$$ satisfies $$n \left( y - x \right) \gt 1$$ therefore $$m \le nx + 1 \le nx + \left( n \left( y - x \right) \right) = ny$$, thus we have $$nx \lt m \lt ny$$ and since $$n \neq 0$$ we have $$x \lt \frac{m}{n} \lt y$$ as needed.

There is an Irrational Number between any two Real Numbers
Suppose that $$x, y \in \mathbb{ R }$$ such that $$x \lt y$$ then there is a $$p \in \mathbb{ R } \setminus \mathbb{ Q }$$ such that $x \lt p \lt y$
If $$x \lt y$$ then $$\sqrt{ 2 } + x \lt \sqrt{ 2 } + y$$, and thus there is a rational number $$r \in \mathbb{ Q }$$ such that $\sqrt{ 2 } + x \lt r \lt \sqrt{ 2 } + y$ and therefore $$x \lt r - \sqrt{ 2 } \lt y$$, but $$r - \sqrt{ 2 }$$ is irrational, because the sum of a rational and irrational is irrational, then we are done.