TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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The rational-supremum construction of real powers agrees with the exponential construction

Statement

For every $a>0$ and $x\in\mathbb R$ , the rational-supremum value $a^{[x]}$ of Real powers from suprema of rational powers, with the reciprocal convention below base one equals the exponential real power $a^x$ of Real powers for positive bases, with the zero-base positive-exponent convention.

Facts & Assumptions

Given: A real $x$ and a positive base $a$ .

[L1]

For $a>1$ , $S_a(x):=\{a^q:q\in\mathbb Q,\ q<x\}$ and $a^{[x]}:=\sup S_a(x)$ ; for $0<a<1$ , $a^{[x]}:=1/\bigl((a^{-1})^{[x]}\bigr)$ ; and $1^{[x]}:=1$ (Real powers from suprema of rational powers, with the reciprocal convention below base one).

[L2]

Rational powers agree with exponential real powers, and $t\mapsto a^t$ is continuous; if $a>1$ , then $\log a>0$ , so $t\mapsto a^t=\exp(t\log a)$ is strictly increasing (The exponential definition of real powers agrees with the existing rational powers, Continuity and derivatives of positive-base real powers, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The exponential function is strictly increasing, Real powers for positive bases, with the zero-base positive-exponent convention).

[L3]

Rational numbers are dense in $\mathbb R$ , and the epsilon characterisation identifies a supremum of a nonempty bounded-above set (The rationals embed densely in the reals, Epsilon characterisation of the supremum).

[L4]

For $a,b>0$ and $r,s\in\mathbb R$ : $a^{r+s}=a^ra^s$ , $(ab)^r=a^rb^r$ , $(a/b)^r=a^r/b^r$ , and $(a^r)^s=a^{rs}$ (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

Proof

technique · direct

1.1

Assume $a>1$ . For every rational $q<x$ , strict increase of $t\mapsto a^t$ gives $a^q<a^x$ , so $a^x$ is an upper bound of $S_a(x)$ .

L1L2

1.2

Given $\varepsilon>0$ , continuity of $t\mapsto a^t$ at $x$ supplies $\delta>0$ such that $|t-x|<\delta$ implies $|a^t-a^x|<\varepsilon$ ; density supplies rational $q$ with $x-\delta<q<x$ , hence $a^x-\varepsilon<a^q\in S_a(x)$ .

L2L3choose

2.1

By the supremum characterisation, steps 1.1 and 1.2 give $a^{[x]}=a^x$ when $a>1$ .

step 1.1step 1.2L3

3.1

For $a=1$ both values are $1$ . For $0<a<1$ the base $a^{-1}$ exceeds $1$ , so step 2.1 applied to that base and the same exponent $x$ gives $(a^{-1})^{[x]}=(a^{-1})^{x}$ ; the subunit clause of [L1] then gives $a^{[x]}=1/\bigl((a^{-1})^{x}\bigr)$ . By [L4] with $r=-1$ and $s=x$ , $(a^{-1})^{x}=a^{-x}$ ; and by [L4] again, $a^{-x}a^{x}=a^{-x+x}=a^{0}=1$ , so $1/a^{-x}=a^{x}$ . Hence $a^{[x]}=a^{x}$ .

step 2.1L1L2L4∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 105 results over 28 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)