Power to Real Number by Decimal Expansion is Uniquely Defined

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Let $r \in \R_{> 1}$ be a real number greater than $1$, expressed by its decimal expansion:

$r = n \cdotp d_1 d_2 d_3 \ldots$

The power $x^r$ of a (strictly) positive real number $x$ defined as:

$(1): \quad \displaystyle \lim_{k \mathop \to \infty} x^{\psi_1} \le \xi \le x^{\psi_2}$


\(\displaystyle \psi_1\) \(=\) \(\displaystyle n + \sum_{j \mathop = 1}^k \frac {d_1} {10^k} = n + \frac {d_1} {10} + \cdots + \frac {d_k} {10^k}\)
\(\displaystyle \psi_2\) \(=\) \(\displaystyle \psi_1 + \dfrac 1 {10^k}\)

is unique.


If $r$ is rational this has already been established.

Let $d$ denote the difference between $x^{\psi^1}$ and $x^{\psi^2}$:

\(\displaystyle d\) \(=\) \(\displaystyle x^{\psi^2} - x^{\psi^1}\)
\(\displaystyle \) \(=\) \(\displaystyle x^{\psi^1} \left({x^{\frac 1 {10^k} } - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle x^{\psi^1} \left({x^{\frac 1 {10^k} } - 1}\right)\)

It follows from Nth Root of 1 plus x not greater than 1 plus x over n that:

$d < \dfrac {x^{n + 1} \left({x - 1}\right)} {10^k}$

Thus as $k \to \infty$, $d \to 0$.

The result follows from the Squeeze Theorem.