LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16

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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.

Continued-fraction convergents, determinant identities, and nested irrational cylinders

Statement

Let $a_{0} \in Z$ and $a_{n} \in Z$ with $a_{n} \geq 1$ for $n \geq 1$ . Define the convergent numerators and denominators by the initial values $p_{- 2} = 0, p_{- 1} = 1, q_{- 2} = 1, q_{- 1} = 0$ together with the recurrences $p_{n} = a_{n} p_{n - 1} + p_{n - 2}$ and $q_{n} = a_{n} q_{n - 1} + q_{n - 2}$ for $n \geq 0$ . The initial values are part of the definition: without them the two recurrences have no value at $n = 0$ and $n = 1$ . Then $p_{0} = a_{0}$ and $q_{0} = 1$ ; the $q_{n}$ are positive for $n \geq 0$ and strictly increasing for $n \geq 1$ ; and $p_{n} q_{n - 1} - p_{n - 1} q_{n} = (- 1)^{n - 1} (n \geq 0) .$

For a finite prefix $(a_{0}, \dots, a_{n})$ write $C (a_{0}, \dots, a_{n}) \subseteq N^{N}$ for the code cylinder of all codes extending that prefix, and write $J (a_{0}, \dots, a_{n}) : = the closed interval with endpoints \frac{p_{n}}{q_{n}} and \frac{p_{n} + p_{n - 1}}{q_{n} + q_{n - 1}} \subseteq R .$ The intervals $J$ are nested as the prefix is extended, and $diam J (a_{0}, \dots, a_{n}) = \frac{1}{q_{n} (q_{n} + q_{n - 1})}$ , which tends to $0$ .

A code cylinder and a real interval are different objects and the two are not identified here. Both endpoints of $J (a_{0}, \dots, a_{n})$ are rational, being ratios of integers. Whether an infinite code's value can equal such an endpoint is not settled on this page; it is settled in Infinite simple continued fractions parametrise the irrational real numbers, which proves every such value irrational.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

Define a bijection $z : N \to Z$ by $z (2 k) = k$ and $z (2 k + 1) = - (k + 1)$ ; the division algorithm makes these two cases exhaustive (thm-division-algorithm-in-z). For $x \in N^{N}$ put $a_{0} = z (x_{0})$ and $a_{n} = x_{n} + 1$ for $n \geq 1$ . Its finite simple continued fractions are $[a_{0}; \dots, a_{n}]$ , evaluated in $Q$ (def-rationals, def-rat-operations). A finite prefix determines the cylinder of all codes extending it. Infinite continued-fraction values are established, rather than assumed, in thm-simple-continued-fractions-parametrise-the-irrationals. (Simple continued fractions, convergents, and the integer-coordinate coding of $N^{N}$ ).

[F2]

Let $(N, 0, σ)$ be a Peano system (def-peano-system), in particular the natural numbers $N$ (def-natural-numbers). For any set $A$ , any element $a \in A$ , and any function $f : A \to A$ , there is a unique function $g : N \to A$ such that $g (0) = a$ and $g (σ (n)) = f (g (n))$ for all $n \in N$ . (The recursion theorem).

[F3]

The relation of def-rat-order is well defined and makes the field $Q$ (thm-rat-field) a totally ordered field: the order is total, $x \leq y$ implies $x + z \leq y + z$ , and $0 < x$ , $0 < y$ imply $0 < x y$ . (The rationals form a totally ordered field).

[F4]

For each $k \in N$ let $I_{k} = [a_{k}, b_{k}]$ be a closed bounded interval with $a_{k} \leq b_{k}$ (def-interval), and suppose the family is nested: $I_{k + 1} \subseteq I_{k} (k \in N) .$ Write $ℓ_{k} = b_{k} - a_{k} \geq 0$ for the length of $I_{k}$ . Then: 1. $⋂_{k \in N} I_{k}$ is nonempty. More precisely, with $a = \sup {a_{k} : k \in N}$ and $b = \inf {b_{k} : k \in N}$ , both of which exist, one has $a \leq b$ and $⋂_{k \in N} I_{k} = [a, b] .$ 2. $⋂_{k \in N} I_{k}$ is a single point if and only if $ℓ_{k} \to 0$ (def-real-limit). Every hypothesis is load bearing. Dropping closedness makes the intersection empty; dropping boundedness does the same; and dropping nonemptiness of the individual intervals is vacuously fatal. (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ).

[F5]

The map $q \mapsto \hat{q}$ (def-real-numbers) is an embedding of ordered fields. Every real is approximated by rationals: for $x \in R$ and rational $ε > 0$ there is $q \in Q$ with $∣ x - \hat{q} ∣ < \hat{ε}$ . Consequently, strictly between any two reals lies a rational. (The rationals embed densely in the reals).

Proof

technique · direct

1.1givenF1F3F5F2

The recursion theorem [F2], applied on pairs, makes $(p_{n}, q_{n})_{n \geq - 2}$ well defined from the four initial values and the two recurrences; $p_{0} = a_{0} \cdot 1 + 0 = a_{0}$ and $q_{0} = a_{0} \cdot 0 + 1 = 1$ follow at once. The determinant identity holds at $n = 0$ , where $p_{0} q_{- 1} - p_{- 1} q_{0} = a_{0} \cdot 0 - 1 \cdot 1 = - 1 = (- 1)^{- 1}$ , and passes from $n$ to $n + 1$ because $p_{n + 1} q_{n} - p_{n} q_{n + 1} = (a_{n + 1} p_{n} + p_{n - 1}) q_{n} - p_{n} (a_{n + 1} q_{n} + q_{n - 1}) = - (p_{n} q_{n - 1} - p_{n - 1} q_{n})$ ; induction in the ordered field [F3] gives it for every $n \geq 0$ .

2.1step 1.1F4F1F5

After the arbitrary integer term $a_{0}$ all partial quotients satisfy $a_{n} \geq 1$ , so from $q_{0} = 1$ and $q_{1} = a_{1} \geq 1$ the recurrence gives $q_{n + 1} = a_{n + 1} q_{n} + q_{n - 1} > q_{n}$ for $n \geq 1$ : the denominators are positive and strictly increasing, hence unbounded. Subtracting the two endpoint fractions and using the determinant identity of step 1.1 gives $\frac{p_{n}}{q_{n}} - \frac{p_{n} + p_{n - 1}}{q_{n} + q_{n - 1}} = \frac{p_{n} q_{n - 1} - p_{n - 1} q_{n}}{q_{n} (q_{n} + q_{n - 1})} = \frac{(- 1)^{n - 1}}{q_{n} (q_{n} + q_{n - 1})}$ , so $diam J (a_{0}, \dots, a_{n}) = 1 / (q_{n} (q_{n} + q_{n - 1})) \to 0$ . Extending a prefix replaces $J$ by one of the subintervals it determines, so the intervals are nested and [F4] applies to them.

3.1step 2.1∎

The preceding construction and implications establish the assertion.

Depends on

Used by

Dependency tree · two levels

33 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

David Marker, Descriptive Set Theory, §§1–2 (standard reference, not scraped)
Michael Kunzinger, General Topology, §§11.3–11.4 (standard reference, not scraped)
MFF General Topology course summary, §4.3 (standard reference, not scraped)
Jesse Peterson, Real Analysis, §§3.6–3.7 (standard reference, not scraped)