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.