Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents

Statement

For a,b>0a,b>0 and r,s∈Rr,s\in\mathbb R, ar+s=aras,(ab)r=arbr,(a/b)r=ar/br,(ar)s=ars.a^{r+s}=a^ra^s,\qquad (ab)^r=a^rb^r,\qquad (a/b)^r=a^r/b^r,\qquad (a^r)^s=a^{rs}.

Facts & Assumptions

Given: Positive reals a,ba,b and real exponents r,sr,s.

[L1]

au=exp⁡(ulog⁡a)a^u=\exp(u\log a) for a>0a>0 (Real powers for positive bases, with the zero-base positive-exponent convention).

[L2]

log⁡(ab)=log⁡a+log⁡b\log(ab)=\log a+\log b, log⁡(a/b)=log⁡a−log⁡b\log(a/b)=\log a-\log b, and log⁡(exp⁡u)=u\log(\exp u)=u (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L3]

exp⁡(u+v)=exp⁡(u)exp⁡(v)\exp(u+v)=\exp(u)\exp(v) and exp⁡(−u)=1/exp⁡(u)\exp(-u)=1/\exp(u) (The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y)\exp(x+y)=\exp(x)\exp(y), The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x)\exp(-x)=1/\exp(x)).

Proof

technique · direct
1.1

Expanding ar+sa^{r+s} by [L1] and applying [L3] gives ar+s=exp⁡(rlog⁡a)exp⁡(slog⁡a)=arasa^{r+s}=\exp(r\log a)\exp(s\log a)=a^ra^s.

L1L3
1.2

Expanding (ab)r(ab)^r and using log⁡(ab)=log⁡a+log⁡b\log(ab)=\log a+\log b gives (ab)r=arbr(ab)^r=a^rb^r.

L1L2L3
1.3

The same calculation with log⁡(a/b)=log⁡a−log⁡b\log(a/b)=\log a-\log b and [L3] gives (a/b)r=ar/br(a/b)^r=a^r/b^r.

L1L2L3
2.1

Since log⁡(ar)=log⁡(exp⁡(rlog⁡a))=rlog⁡a\log(a^r)=\log(\exp(r\log a))=r\log a, expanding (ar)s(a^r)^s gives arsa^{rs}.

L1L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 results over 20 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