Power Function on Base Greater than One is Strictly Increasing/Real Number

Theorem
Let $a \in \R$.

Let $a > 1$.

Let $f : \R \to \R$ be the real-valued function defined as:
 * $f \left({ x }\right) = a^x$

where $a^x$ denotes $a$ to the power of $x$.

Then $f$ is strictly increasing.

Proof
Let $x, y \in \R$ be such that $x < y$.

Let $\delta = y - x$.

From Rational Sequence Decreasing to Real Number, there is some rational sequence $\left\langle{ x_n }\right\rangle$ that  decreases to $x$.

From Rational Sequence Increasing to Real Number, there is some rational sequence $\left\langle{ y_n }\right\rangle$ that  increases to $y$.

From Convergent Real Sequence is Bounded:
 * $\exists N_1 \in \N : n \geq N_1 \implies x - \delta < x_n < x + \delta$

Since $\left\langle{ x_n }\right\rangle$ is decreasing:
 * $n \geq N_1 \implies x \leq x_n < x + \delta$

From Convergent Real Sequence is Bounded:
 * $\exists N_2 \in \N : n \geq N_1 \implies y - \delta < y_n < y + \delta$

Since $\left\langle{ y_n }\right\rangle$ is increasing:
 * $n \geq N_2 \implies y - \delta < y_n \leq y$

Let $N = \max \left\{ { N_1, N_2 } \right\}$

Thus:

From Exponential with Arbitrary Base is Continuous/Rational Number:
 * $x_n \to x \land y_n \to y \implies a^{x_n} \to a^{x} \land a^{y_n} \to a^{y}$

From Exponential with Base Greater than One is Strictly Increasing/Rational Number, $\left\langle{ a^{x_n} }\right\rangle$ decreases to $a^{x}$ and $\left\langle{ a^{y_n} }\right\rangle$ increases to $a^{y}$.

Suppose $n \geq N$.

Then:

Hence the result.