TheoremStatement: AI-adaptedProof: AI-generatedprecheck 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.

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 $\log 1 = 0$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources