LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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Canonical naturals are positive and strictly increasing

Statement

Let $F$ be an ordered field, with field structure as in Field, and for $n \geq 1$ let $n \cdot 1_{F}$ denote the canonical natural number $\underset{n}{\underset{⏟}{1_{F} + \dots + 1_{F}}}$ , defined recursively by $1 \cdot 1_{F} = 1_{F}$ and $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ . Then for every $n \geq 1$ :

$n \cdot 1_{F} > 0$ and $(n + 1) \cdot 1_{F} > n \cdot 1_{F}$ ;
the map $n \mapsto n \cdot 1_{F}$ is strictly increasing, hence injective, on ${1, 2, 3, \dots}$ ;
$(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ and $(m n) \cdot 1_{F} = (m \cdot 1_{F}) (n \cdot 1_{F})$ for all $m, n \geq 1$ .

Statement references

Field

Facts & Assumptions

Given: An ordered field $F$ whose canonical naturals are defined recursively by $1 \cdot 1_{F} = 1_{F}$ and $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ , with $0 \cdot 1_{F} = 0$ .

[L1]

$0 < 1_{F}$ in every ordered field (The multiplicative identity is positive).

[L2]

The order is compatible with addition and positives are closed under addition: $a < b$ implies $a + c < b + c$ , and $a > 0$ , $b > 0$ imply $a + b > 0$ (Order is preserved by adding a constant and by adding inequalities).

[L3]

Natural addition satisfies $m + 0 = m$ and $m + (n + 1) = (m + n) + 1$ (Addition of natural numbers); it is associative (Addition is associative).

[L4]

Natural multiplication satisfies $m \cdot 0 = 0$ and $m (n + 1) = m n + m$ (Multiplication of natural numbers).

[L5]

Induction on the natural numbers holds (The principle of mathematical induction).

Proof

technique · induction

1.1L1base

Base case $n = 1$ : by definition $1 \cdot 1_{F} = 1_{F}$ , and $1_{F} > 0$ .

1.2ih

Inductive hypothesis: assume $n \cdot 1_{F} > 0$ for some fixed $n \geq 1$ .

1.3givenL1

By the recursion, $(n + 1) \cdot 1_{F} - n \cdot 1_{F} = 1_{F} > 0$ , so $(n + 1) \cdot 1_{F} > n \cdot 1_{F}$ .

1.4givenL3L5algebra

Additive identity: fix $m$ and induct on $n$ ; at $n = 1$ , $(m + 1) \cdot 1_{F} = m \cdot 1_{F} + 1_{F}$ by the recursion, and if $(m + n) \cdot 1_{F} = m \cdot 1_{F} + n \cdot 1_{F}$ then associativity of natural addition gives $m + (n + 1) = (m + n) + 1$ , whence $(m + (n + 1)) \cdot 1_{F} = (m + n) \cdot 1_{F} + 1_{F} = m \cdot 1_{F} + n \cdot 1_{F} + 1_{F} = m \cdot 1_{F} + (n + 1) \cdot 1_{F}$ .

2.1step 1.2L1L2

Inductive step: $(n + 1) \cdot 1_{F} = n \cdot 1_{F} + 1_{F}$ is a sum of two positives, by the hypothesis and $1_{F} > 0$ , hence $(n + 1) \cdot 1_{F} > 0$ .

2.2step 1.3L2

Iterating 1.3, for $m < n$ one has $m \cdot 1_{F} < n \cdot 1_{F}$ , so $n \mapsto n \cdot 1_{F}$ is strictly increasing and therefore injective on ${1, 2, 3, \dots}$ .

2.3L3L4L5step 1.4algebra

Multiplicative identity: fix $m$ and induct on $n$ ; at $n = 1$ , $m \cdot 1 = m \cdot 0 + m = m$ , so $(m \cdot 1) \cdot 1_{F} = m \cdot 1_{F} = (m \cdot 1_{F}) (1 \cdot 1_{F})$ . In the step, $m (n + 1) = m n + m$ , and therefore $(m (n + 1)) \cdot 1_{F} = (m n + m) \cdot 1_{F} = (m n) \cdot 1_{F} + m \cdot 1_{F} = (m \cdot 1_{F}) (n \cdot 1_{F}) + (m \cdot 1_{F}) (1_{F}) = (m \cdot 1_{F}) ((n + 1) \cdot 1_{F})$ by 1.4 and field distributivity.

3.1step 2.1step 2.2step 1.4step 2.3discharge-induction∎

By induction on 1.1, 1.2, 2.1, $n \cdot 1_{F} > 0$ for every $n \geq 1$ ; with the strict increase 2.2 and the identities 1.4, 2.3, every claim holds.

Canonical naturals are positive and strictly increasing

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources