TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

How statement and proof provenance work

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.

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

Statement

The function $\log : (0, \infty) \to R$ is continuous and strictly increasing, is onto $R$ , and satisfies, for $x, y > 0$ , $\log (x y) = \log x + \log y, \log (x / y) = \log x - \log y, \log (1 / x) = - \log x .$ Also $\log 1 = 0$ .

Facts & Assumptions

Given: Positive reals $x, y$ .

[L1]

The exponential is a continuous strictly increasing bijection from $R$ onto $(0, \infty)$ , and $\log$ is its inverse (The natural logarithm as the inverse of the exponential function, The exponential function is strictly increasing, Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ ).

[L2]

For all reals $u, v$ , $\exp (u + v) = \exp (u) \exp (v)$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

[L3]

$\exp (- u) = 1 / \exp (u)$ and $\exp (u) > 0$ for every real $u$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Proof

technique · direct

1.1L1

Since it is the inverse of the continuous strictly increasing exponential, $\log$ is continuous, strictly increasing, and maps $(0, \infty)$ onto $R$ .

1.2L1L2

The equality $\exp (\log x + \log y) = \exp (\log x) \exp (\log y) = x y = \exp (\log (x y))$ and injectivity of $\exp$ give $\log (x y) = \log x + \log y$ .

2.1step 1.2L3

Since $x / y = x (1 / y)$ and $\exp (- \log y) = 1 / y$ by [L3], step 1.2 gives $\log (x / y) = \log x - \log y$ and $\log (1 / x) = - \log x$ .

3.1L1L2L3algebra∎

Taking $u = v = 0$ in [L2] gives $\exp (0) = \exp (0)^{2}$ . Since $\exp (0) > 0$ by [L3], division by $\exp (0)$ gives $\exp (0) = 1$ . The inverse identity then gives $\log 1 = 0$ .

Depends on

Used by

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…and 250 more results.

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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)