Power to Real Number by Decimal Expansion is Uniquely Defined
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Theorem
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:
- $(1): \quad \ds \lim_{k \mathop \to \infty} x^{\psi_1} \le \xi \le x^{\psi_2}$
where:
\(\ds \psi_1\) | \(=\) | \(\ds n + \sum_{j \mathop = 1}^k \frac {d_1} {10^k} = n + \frac {d_1} {10} + \cdots + \frac {d_k} {10^k}\) | ||||||||||||
\(\ds \psi_2\) | \(=\) | \(\ds \psi_1 + \dfrac 1 {10^k}\) |
is unique.
Proof
If $r$ is rational this has already been established.
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Let $d$ denote the difference between $x^{\psi^1}$ and $x^{\psi^2}$:
\(\ds d\) | \(=\) | \(\ds x^{\psi^2} - x^{\psi^1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^{\psi^1} \paren {x^{\frac 1 {10^k} } - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^{\psi^1} \paren {x^{\frac 1 {10^k} } - 1}\) |
It follows from Nth Root of 1 plus x not greater than 1 plus x over n that:
- $d < \dfrac {x^{n + 1} \paren {x - 1} } {10^k}$
Thus as $k \to \infty$, $d \to 0$.
The result follows from the Squeeze Theorem.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: $(7)$