ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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.

Thomae's function computed: $t (1 / 2) = 1 / 2$ , $t (2 / 3) = 1 / 3$ , $t (m) = 1$ at every integer $m$ , $t (x) = 0$ at every irrational, and $ω_{t} (c) = t (c)$ at every real $c$

Example

Let $t$ be Thomae's function (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ), so that $t (x) = 1 / ι (q (x))$ at a rational $x$ with least denominator $q (x)$ and $t (x) = 0$ at an irrational $x$ . Then:

$t (0) = 1$ and $t (m) = 1$ for every integer $m$ ;
$t (1 / 2) = 1 / 2$ , and more generally $t (1 / ι (q)) = 1 / ι (q)$ for every natural $q \geq 1$ ;
$t (2 / 3) = 1 / 3$ ;
$t (x) = 0$ at every irrational $x$ ;
$ω_{t} (c) = t (c)$ at every real $c$ (The oscillation $ω_{f} (S) = \sup {∣ f (x) - f (y) ∣ : x, y \in S}$ of $f$ on a set and the oscillation $ω_{f} (c) = \inf_{δ > 0} ω_{f} (A \cap N_{δ} (c))$ at a point, both taken in the extended reals), so $ω_{t}$ is $1$ at every integer, $1 / 2$ at every half-integer that is not an integer, and $0$ at every irrational.

Claim 5 is claim 2 of The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ evaluated at the points computed here; nothing new is proved about the oscillation, and the point of the example is to see the numbers.

Facts & Assumptions

Given: Thomae's function $t$ , with $q (x) = \min {q \in N : q \geq 1 and ι (q) x \in Z}$ for rational $x$ ; $N \subseteq Z \subseteq Q \subseteq R$ are the canonical copies and $ι (q)$ is the canonical natural (The rationals embed densely in the reals, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

$t (x) = 1 / ι (q (x))$ for $x \in Q$ and $t (x) = 0$ for $x \notin Q$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ).

[L1]

$ω_{t} (c) = t (c)$ for every real $c$ (The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ , claim 2); and $t$ is continuous at $c$ exactly when $ω_{t} (c) = 0$ ( $f : A \to R$ is continuous at $c \in A$ if and only if $ω_{f} (c) = 0$ , The oscillation $ω_{f} (S) = \sup {∣ f (x) - f (y) ∣ : x, y \in S}$ of $f$ on a set and the oscillation $ω_{f} (c) = \inf_{δ > 0} ω_{f} (A \cap N_{δ} (c))$ at a point, both taken in the extended reals).

[L2]

No integer lies strictly between $m$ and $m + 1$ ; equivalently a real of the form $k / ι (q)$ with $0 < k < q$ naturals is not an integer, lying strictly between $0$ and $1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ , Canonical naturals are positive and strictly increasing).

[L3]

There exist irrational reals, the irrationals being dense in $R$ (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

Verification

technique · direct

1.1

Claim 1: for an integer $m$ one has $ι (1) m = m \in Z$ , so $1 \in Q (m)$ and $q (m) = 1$ , the least element of a set of naturals $\geq 1$ containing $1$ ; hence $t (m) = 1 / ι (1) = 1$ . The case $m = 0$ is included.

A1L2

1.2

Claim 2: let $q \geq 1$ be a natural and put $x : = 1 / ι (q)$ . Then $ι (q) x = 1 \in Z$ , so $q \in Q (x)$ and $q (x) \leq q$ . Conversely, if $1 \leq k \leq q$ is a natural with $ι (k) x = ι (k) / ι (q) \in Z$ , then $k < q$ would put $ι (k) / ι (q)$ strictly between $0$ and $1$ , which no integer is; so $k = q$ . Hence $q (x) = q$ and $t (1 / ι (q)) = 1 / ι (q)$ . Taking $q = 2$ gives $t (1 / 2) = 1 / 2$ .

A1L2

1.3

Claim 3: put $x : = 2 / 3$ . Then $ι (3) x = 2 \in Z$ , so $q (x) \leq 3$ . Also $ι (1) x = 2 / 3$ lies strictly between $0$ and $1$ and so is not an integer, and $ι (2) x = 4 / 3$ lies strictly between $1$ and $2$ and so is not an integer. Hence $q (x) = 3$ and $t (2 / 3) = 1 / 3$ .

A1L2

1.4

Claim 4 is the second clause of the definition of $t$ , and irrational reals exist.

A1L3

2.1

Claim 5: $ω_{t} (c) = t (c)$ at every real $c$ . At an integer $m$ this is $1$ by step 1.1; at a real of the form $m + 1 / 2$ with $m$ an integer, the least denominator is $2$ , by the same computation as in step 1.2 applied to $ι (2) (m + 1 / 2) = 2 m + 1 \in Z$ together with $ι (1) (m + 1 / 2) = m + 1 / 2$ lying strictly between $m$ and $m + 1$ , so the value is $1 / 2$ ; and at an irrational it is $0$ .

step 1.1step 1.2step 1.4A1L1L2

3.1

In particular $t$ is continuous at every irrational, where $ω_{t} = 0$ , and discontinuous at every rational, where $ω_{t} = t > 0$ ; the numbers above are the sizes of those failures.

step 2.1A1L1∎

Remarks

The least denominator is what the values record. $t$ is large exactly at the rationals with small denominators, and those are sparse: every point with least denominator $q$ is a multiple of $1 / ι (q)$ , and consecutive multiples of $1 / ι (q)$ are $1 / ι (q)$ apart. The graph is the familiar picture of tall spikes at the integers, half as tall at the half-integers, and so on down.
Every value $1 / ι (q)$ is attained, by step 1.2, so the range of $t$ is exactly ${0} \cup {1 / ι (q) : q \in N, q \geq 1}$ ; the value $0$ is attained at every irrational.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

58 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

Thomae's function (Wikipedia) (standard reference, not scraped)
Dirichlet Function (MathWorld) (standard reference, not scraped)

Thomae's function computed: t(1/2)=1/2t(1/2)=1/2, t(2/3)=1/3t(2/3)=1/3, t(m)=1t(m)=1 at every integer mm, t(x)=0t(x)=0 at every irrational, and ωt(c)=t(c)ωt​(c)=t(c) at every real cc

Example

Facts & Assumptions

Verification

Remarks

Depends on

Used by

Dependency tree · two levels

Sources

Thomae's function computed: $t (1 / 2) = 1 / 2$ , $t (2 / 3) = 1 / 3$ , $t (m) = 1$ at every integer $m$ , $t (x) = 0$ at every irrational, and $ω_{t} (c) = t (c)$ at every real $c$