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

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.

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$

Statement

Let $R$ be a complete ordered field (Complete ordered field (least-upper-bound property)). For every $a \in R$ with $a \geq 0$ and every $n \in N$ with $n \geq 1$ there is a unique $s \in R$ with $s \geq 0$ and $s^{n} = a$ (Integer powers $a^{m}$ ); we write

$s = a^{1 / n} = \sqrt[n]{a} .$

Moreover $a^{1 / n} > 0$ when $a > 0$ , and $a^{1 / 1} = a$ .

This generalises the published Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , and the case $n = 2$ is not new. That theorem already produces the unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ , and it is cited as such throughout the library; the notation $a^{1 / 2}$ introduced here is the same number. What is new is the passage to general $n$ : the completed square that drives the $n = 2$ argument has no direct analogue, and its place is taken by the factorisation of $b^{n} - a^{n}$ and the resulting Lipschitz estimate (Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate).

Facts & Assumptions

Given: A complete ordered field $R$ (Complete ordered field (least-upper-bound property)), that is, an ordered field (Ordered field) with the least-upper-bound property; an element $a \geq 0$ ; and a natural $n \geq 1$ , with field coefficients $n$ denoting $n \cdot 1_{R}$ while the root order remains natural (Canonical naturals are positive and strictly increasing, The unique embedding of ℚ into an ordered field).

[L1]

Least-upper-bound property: every nonempty subset of $R$ that is bounded above (Lower bound, bounded below, bounded set) has a least upper bound, and it is unique, so the notation $\sup S$ is legitimate (Complete ordered field (least-upper-bound property), Suprema and infima are unique).

[L2]

Epsilon characterisation of the supremum: if $S$ is nonempty and bounded above and $u = \sup S$ , then for every $ε > 0$ there is $t \in S$ with $u - ε < t$ (Epsilon characterisation of the supremum).

[L3]

Monotonicity of powers (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ): $x \mapsto x^{n}$ is strictly increasing on ${x \geq 0}$ for $n \geq 1$ , hence injective there; $x \geq 0$ implies $x^{n} \geq 0$ and $x > 0$ implies $x^{n} > 0$ ; and $x \geq 1$ implies $x^{n} \geq x$ .

[L4]

Lipschitz estimate (Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate): if $0 \leq x \leq y \leq B$ and $n \geq 1$ then $0 \leq y^{n} - x^{n} \leq n B^{n - 1} (y - x)$ .

[L5]

Order arithmetic: adding a constant preserves the order and for $c > 0$ , $x < y ⟺ c x < c y$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication), both stated there for the STRICT order alone, so where a $\leq$ is added or scaled below the move is that statement together with the case of equality, settled by trichotomy (Ordered field); $1 > 0$ , which is proved in The multiplicative identity is positive and stated by none of those three, hence $2 = 1 + 1 > 0$ ; and $0 < 2^{- 1} < 1$ , since $0 < 1 < 2$ gives $0 < 2^{- 1} < 1^{- 1} = 1$ (Inverses of positives are positive, and reciprocation reverses order, claim 2).

[L6]

Trichotomy: for $x, y \in R$ exactly one of $x < y$ , $x = y$ , $y < x$ holds; consequently any two elements have a minimum (Ordered field).

[L7]

A product with a zero factor vanishes: $0 \cdot x = x \cdot 0 = 0$ (Multiplication by zero: $0 \cdot a = 0$ ).

Proof

technique · contradiction

1.1givenL3L7algebra

If $a = 0$ then $s = 0$ satisfies $s \geq 0$ and $s^{n} = 0 = a$ , since $0^{n} = 0^{n - 1} \cdot 0 = 0$ for $n \geq 1$ ; so existence holds in that case and we assume $a > 0$ from here on.

1.2L3L6

Uniqueness holds as soon as a root exists: if $u, v \geq 0$ satisfy $u^{n} = v^{n}$ , then strict monotonicity of $x \mapsto x^{n}$ on the nonnegatives rules out both $u < v$ and $v < u$ , so $u = v$ by trichotomy.

1.3givenL3L7

Define $S = {t \in R : t \geq 0 and t^{n} \leq a}$ ; then $0 \in S$ , because $0 \geq 0$ and $0^{n} = 0 \leq a$ , so $S$ is nonempty.

1.4givenL3L5

The element $M : = 1 + a$ is an upper bound of $S$ : since $a > 0$ and $1 > 0$ we have $M > 1$ and $M > a$ , so any $t > M$ satisfies $t > 1$ , whence $t^{n} \geq t > M > a$ and $t \notin S$ .

2.1L1step 1.3step 1.4

By the least-upper-bound property $s : = \sup S$ exists in $R$ ; moreover $s \geq 0$ because $0 \in S$ , and $s \leq M$ because $M$ is an upper bound and $s$ is the least one.

3.1step 2.1L3L5

Put $B : = M + 1 = a + 2$ ; then $B > M \geq 1 > 0$ , so $B^{n - 1} > 0$ and $n B^{n - 1} > 0$ , and every $x$ with $0 \leq x \leq M + 1$ satisfies $0 \leq x \leq B$ .

3.2assume-contrastep 2.1L6

Assume, for contradiction, that $s^{n} \neq a$ ; by trichotomy either $s^{n} < a$ or $s^{n} > a$ .

4.1assume-case lowstep 3.2step 2.1step 3.1L4L5L6choose

(Case $s^{n} < a$ .) Put $δ : = \min {1, (a - s^{n}) (n B^{n - 1})^{- 1}}$ , which is $> 0$ since $a - s^{n} > 0$ and $n B^{n - 1} > 0$ , and put $h : = 2^{- 1} δ$ , so that $0 < h < δ \leq 1$ and $h < (a - s^{n}) (n B^{n - 1})^{- 1}$ ; then $0 \leq s \leq s + h \leq M + 1 = B$ , so the Lipschitz estimate gives $(s + h)^{n} - s^{n} \leq n B^{n - 1} h < a - s^{n}$ , hence $(s + h)^{n} < a$ and $s + h \in S$ , while $s + h > s$ contradicts that $s$ is an upper bound of $S$ .

4.2assume-case highstep 3.2step 2.1step 3.1L2L3L4L5L6L7choose

(Case $s^{n} > a$ .) Here $s > 0$ , since $s = 0$ would give $s^{n} = 0 \leq a$ ; put $δ : = \min {s, (s^{n} - a) (n B^{n - 1})^{- 1}} > 0$ and $h : = 2^{- 1} δ$ , so that $0 < h < s$ and $h < (s^{n} - a) (n B^{n - 1})^{- 1}$ ; then $0 \leq s - h \leq s \leq B$ , so the Lipschitz estimate gives $s^{n} - (s - h)^{n} \leq n B^{n - 1} h < s^{n} - a$ , hence $(s - h)^{n} > a$ ; applying the epsilon characterisation with $ε = h$ produces $t \in S$ with $t > s - h \geq 0$ , whence $t^{n} > (s - h)^{n} > a$ by strict monotonicity, contradicting $t \in S$ .

5.1step 4.1step 4.2step 3.2step 1.2step 1.1L3L7casesdischarge-contradiction∎

Both cases of the disjunction in step 3.2 are impossible, so the assumption fails and $s^{n} = a$ ; this $s$ is the unique nonnegative $n$ -th root of $a$ by step 1.2, it satisfies $s > 0$ when $a > 0$ because $s = 0$ would force $a = s^{n} = 0$ , and at $n = 1$ the element $a$ itself is a nonnegative solution of $s^{1} = a$ , so $a^{1 / 1} = a$ ; writing $a^{1 / n} = \sqrt[n]{a}$ for it, the case $n = 2$ recovers the already published $\sqrt{a}$ of Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ .

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Existence and uniqueness of nn-th roots: a unique a1/n≥0a1/n≥0 with (a1/n)n=a(a1/n)n=a

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$