CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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Every rational approximation to a real exponent gives the same limiting real power

Statement

Let $a>0$ , let $x\in\mathbb R$ , and let $(q_n)$ be a rational sequence converging to $x$ . Then $a^{q_n}\to a^x$ . Hence the limit is independent of the rational approximating sequence and equals the rational-supremum value $a^{[x]}$ .

Facts & Assumptions

Given: $a>0$ , $x\in\mathbb R$ , and rational $q_n\to x$ .

[L1]

The rational-supremum and exponential constructions agree (The rational-supremum construction of real powers agrees with the exponential construction).

[L2]

Rational powers agree with real powers, and $t\mapsto a^t$ is continuous (The exponential definition of real powers agrees with the existing rational powers, Continuity and derivatives of positive-base real powers).

Proof

technique · direct

1.1

For every $n$ , rational-exponent agreement gives $a^{q_n}=\exp(q_n\log a)$ .

1.2

Continuity of $t\mapsto a^t$ gives $a^{q_n}\to a^x$ .

2.1

Since $a^{[x]}=a^x$ , this limit is independent of the chosen rational sequence and has the asserted supremum value.

step 1.2L1∎

Depends on

Used by

Two to the square root of two from rational suprema and from exp(sqrt(2) log 2) Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 68 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)