LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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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.

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Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$

Statement

Let $a \in R$ with $a > 0$ and let $r, s \in Q$ with $r < s$ (Order on the rationals), with rational powers as in Rational powers $a^{r}$ of a positive base.

In the exponent. If $a > 1$ then $a^{r} < a^{s}$ ; if $a = 1$ then $a^{r} = a^{s} = 1$ ; if $0 < a < 1$ then $a^{r} > a^{s}$ .
In the base. If $r \in Q$ with $r > 0$ and $0 < a < b$ , then $a^{r} < b^{r}$ ; so $a \mapsto a^{r}$ is strictly increasing on ${a \in R : a > 0}$ .
Comparison with $1$ . For rational $t > 0$ : $a > 1$ implies $a^{t} > 1$ , and $0 < a < 1$ implies $a^{t} < 1$ .

Facts & Assumptions

Given: A real $a > 0$ and rationals $r < s$ ; write $t : = s - r$ , a rational with $t > 0$ .

[L1]

Positive rationals have positive representatives: $t > 0$ can be written $t = m / N$ with $m, N$ naturals and $m \geq 1$ , $N \geq 1$ . Every rational has a representative with positive denominator (Every rational has a positive-denominator representative); on such a representative $m / N > 0$ holds exactly when $m > 0$ in $Z$ (Order on the rationals, Order on the integers); and a positive integer is the image of a unique natural $\geq 1$ (The naturals embed in the integers), which is what turns both the numerator $m > 0$ and the denominator $N > 0$ into naturals $\geq 1$ . The last passage is a genuine step and is what Rational powers do not depend on the representative uses at its own [L5].

[L2]

Rational power laws (Laws of rational exponents, Rational powers do not depend on the representative, Rational powers $a^{r}$ of a positive base): $a^{r} > 0$ ; $a^{r + t} = a^{r} a^{t}$ ; and $a^{t} = (a^{1 / N})^{m}$ for the representative $t = m / N$ .

[L3]

Roots (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ): $a^{1 / N}$ is the unique $x \geq 0$ with $x^{N} = a$ , and it is $> 0$ when $a > 0$ .

[L4]

Integer power monotonicity (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ): for $N \geq 1$ , $x \mapsto x^{N}$ is strictly increasing on ${x \geq 0}$ (claim 2); for $m \geq 1$ , $x > 1$ implies $x^{m} \geq x > 1$ (claim 3), while $0 \leq x < 1$ implies $x^{m} < 1^{m} = 1$ , which is claim 2 again with $1$ as the larger base and NOT claim 3, whose nonstrict $x^{m} \leq 1$ would not suffice; and $1^{m} = 1$ for every integer $m$ , by claim 4 for natural $m$ together with $1^{- k} = (1^{k})^{- 1} = 1$ (Laws of integer exponents).

[L5]

Order arithmetic: for $c > 0$ , $x < y ⟺ c x < c y$ ; and trichotomy, exactly one of $a < 1$ , $a = 1$ , $a > 1$ holds (Sign rules for products and monotonicity of multiplication, Ordered field).

Proof

technique · cases

1.1L1L2L5

Write $t = s - r$ , so $t$ is rational with $t > 0$ , and fix a representative $t = m / N$ with naturals $m, N \geq 1$ ; then $a^{s} = a^{r + t} = a^{r} a^{t}$ with $a^{r} > 0$ , so the comparison of $a^{s}$ with $a^{r}$ is exactly the comparison of $a^{t}$ with $1$ .

1.2L2L3L4

Claim 2, which needs no case split: let $r > 0$ be rational with representative $r = m / N$ , $m, N \geq 1$ , and let $0 < a < b$ ; then $a^{1 / N} < b^{1 / N}$ , since $b^{1 / N} \leq a^{1 / N}$ would give $b = (b^{1 / N})^{N} \leq (a^{1 / N})^{N} = a$ ; raising to the power $m$ preserves the strict inequality between nonnegatives, so $a^{r} = (a^{1 / N})^{m} < (b^{1 / N})^{m} = b^{r}$ .

2.1assume-case bigstep 1.1L3L4L5

Case $a > 1$ : then $a^{1 / N} > 1$ , because $a^{1 / N} \leq 1$ would give $a = (a^{1 / N})^{N} \leq 1$ ; hence $a^{t} = (a^{1 / N})^{m} \geq a^{1 / N} > 1$ , and multiplying $1 < a^{t}$ by $a^{r} > 0$ gives $a^{r} < a^{r} a^{t} = a^{s}$ .

2.2assume-case onestep 1.1L2L3L4

Case $a = 1$ : then $1^{N} = 1$ shows $a^{1 / N} = 1$ by uniqueness of the nonnegative $N$ -th root, so $a^{u} = 1^{m^{'}} = 1$ for every rational $u$ with representative $m^{'} / N^{'}$ ; in particular $a^{r} = a^{s} = 1$ .

2.3assume-case smallstep 1.1L3L4L5

Case $0 < a < 1$ : then $0 \leq a^{1 / N} < 1$ , because $a^{1 / N} \geq 1$ would give $a = (a^{1 / N})^{N} \geq 1$ ; hence $a^{t} = (a^{1 / N})^{m} < 1^{m} = 1$ by strict monotonicity of $x \mapsto x^{m}$ on the nonnegatives, $m \geq 1$ , and multiplying $a^{t} < 1$ by $a^{r} > 0$ gives $a^{s} = a^{r} a^{t} < a^{r}$ .

3.1cases-exhaustivestep 2.1step 2.2step 2.3step 1.2L5∎

The three cases $a > 1$ , $a = 1$ , $0 < a < 1$ exhaust the possibilities for $a > 0$ by trichotomy, so claim 1 holds; claim 3 is the comparison of $a^{t}$ with $1$ established inside the first and third cases; and claim 2 is step 1.2.

Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Monotonicity of r↦arr↦ar and of a↦ara↦ar

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$