LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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.

Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$

Statement

Identify $\mathbb{Z}$ with its canonical copy inside $\mathbb{R}$ , along the embeddings $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$ (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers as equivalence classes of pairs of naturals). Then for every real $x$ there is exactly one integer $m$ with

$m \;\le\; x \;<\; m + 1 .$

It is written $\lfloor x \rfloor$ and called the integer part, or floor, of $x$ .

Two independent ingredients are needed and neither may be dropped. Existence is the Archimedean property (Every complete ordered field is Archimedean) together with the well-ordering of $\mathbb{N}$ (The well-ordering principle): the first says that $x$ is caught between two integers at all, the second picks the least integer above $x$ . Uniqueness is the discreteness of $\mathbb{Z}$ : no integer lies strictly between $m$ and $m+1$ .

This lemma is stated once here and reused. It is what turns "the nearest integer to $x$ " from a picture into an object, and the companion page's oscillator $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is computed from it in one line.

Facts & Assumptions

Given: A real $x$ . Naturals, integers and rationals are identified with their canonical copies in $\mathbb{R}$ along $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$ .

[L1]

The embeddings $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$ are injective and preserve $0$ , $1$ , addition, multiplication and order (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers as equivalence classes of pairs of naturals); $\mathbb{Z}$ is a totally ordered commutative ring (The integers form a totally ordered ring, The integers form a commutative ring); every integer $\ge 0$ is the image of a unique natural, that map being injective and order preserving (The naturals embed in the integers); and a natural $j \ne 0$ satisfies $j \ge 1$ (Discreteness: $\sigma(n)$ is the immediate successor, The natural numbers $\mathbb{N}$ (von Neumann)).

[L2]

The image of a natural $n \ge 1$ under the composite $\mathbb{N} \to \mathbb{R}$ is the canonical natural $n \cdot 1_{\mathbb{R}}$ of Canonical naturals are positive and strictly increasing. Indeed that composite preserves $1$ and addition by [L1], while $n \cdot 1_{\mathbb{R}}$ is defined by $1 \cdot 1_{\mathbb{R}} = 1_{\mathbb{R}}$ and $(n+1) \cdot 1_{\mathbb{R}} = n \cdot 1_{\mathbb{R}} + 1_{\mathbb{R}}$ , so the two agree at $1$ and satisfy the same recursion; induction on $n$ (The principle of mathematical induction) gives the identification.

[L3]

Archimedean property: for every real $t$ there is a natural $n \ge 1$ with $t < n \cdot 1_{\mathbb{R}}$ (Every complete ordered field is Archimedean, Complete ordered field (least-upper-bound property)).

[L4]

Well-ordering principle: every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L5]

Order arithmetic in $\mathbb{R}$ : the order is total, so the negation of $t < u$ is $u \le t$ ; trichotomy, so $t < u$ and $u \le t$ cannot both hold; translation invariance (Order is preserved by adding a constant and by adding inequalities); $t \le |t|$ and $-t \le |t|$ (Basic properties of the absolute value); and transitivity (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · constructive

1.1

Apply [L3] to the real $|x|$ : fix a natural $n \ge 1$ with $|x| < n$ . Since $x \le |x|$ and $-x \le |x|$ , this gives $-n < x < n$ .

L2L3L5choose

2.1

Put $S := \{\, k \in \mathbb{N} \ : \ x < k - n \,\}$ , where $k - n$ is formed in $\mathbb{Z}$ and read in $\mathbb{R}$ through [L1]. It is a subset of $\mathbb{N}$ , and it is nonempty: the natural $2n$ satisfies $2n - n = n > x$ by step 1.1, so $2n \in S$ .

step 1.1L1L2construct

3.1

By the well-ordering principle [L4] let $k_0$ be the least element of $S$ .

step 2.1L4choose

4.1

The index $k_0$ is not $0$ : for $k = 0$ the defining condition reads $x < 0 - n = -n$ , which trichotomy excludes since $-n < x$ by step 1.1. Hence $k_0 \ne 0$ , so $k_0 \ge 1$ by [L1], and $k_0 - 1$ is again a natural number.

step 1.1step 3.1L1L5

5.1

Set $m := (k_0 - 1) - n$ , an integer. Since $k_0 - 1 < k_0$ and $k_0$ is the least element of $S$ , the natural $k_0 - 1$ does not lie in $S$ , that is, $x < (k_0 - 1) - n$ fails; the order being total, $m = (k_0 - 1) - n \le x$ .

step 3.1step 4.1L1L5construct

6.1

On the other hand $k_0 \in S$ gives $x < k_0 - n = \bigl((k_0 - 1) - n\bigr) + 1 = m + 1$ . So $m \le x < m + 1$ , and existence is proved.

step 3.1step 5.1L1L5

7.1

Uniqueness: suppose an integer $m'$ also satisfies $m' \le x < m' + 1$ and $m' \ne m$ . The order of $\mathbb{Z}$ being total, one of $m < m'$ and $m' < m$ holds, and the two cases are the same with the roles of $m$ and $m'$ exchanged; so assume $m < m'$ . Then $m' - m$ is an integer $> 0$ , hence by [L1] the image of a natural $j \ne 0$ , so $j \ge 1$ and $m' - m \ge 1$ , that is $m + 1 \le m'$ . But then $x < m + 1 \le m' \le x$ , which trichotomy forbids. Hence $m' = m$ .

step 6.1L1L5

8.1

Therefore exactly one integer $m$ satisfies $m \le x < m + 1$ , and we write $m = \lfloor x \rfloor$ .

step 6.1step 7.1discharge-construct∎

Remarks

What the two halves of the proof really use. Step 1.1 is the only use of the Archimedean property, and it is indispensable: in a non-Archimedean ordered field (Not every ordered field is Archimedean) an element larger than every canonical natural has no integer part at all, since the set $S$ of step 2.1 would be empty. Step 3.1 is the only use of the well-ordering principle, and it is what makes the construction canonical: no choice is made anywhere, and $\lfloor x \rfloor$ is a function of $x$ .
Immediate consequences, used later. From $m \le x < m + 1$ one reads off $0 \le x - m < 1$ and $0 < (m+1) - x \le 1$ ; and $\lfloor x \rfloor = x$ exactly when $x$ is an integer, since an integer $m$ satisfies $m \le m < m + 1$ and uniqueness does the rest. The translation identity $\lfloor x + p \rfloor = \lfloor x \rfloor + p$ for an integer $p$ follows the same way: adding $p$ to $m \le x < m+1$ gives $m + p \le x + p < (m + p) + 1$ , and uniqueness identifies $m + p$ as the integer part of $x + p$ .
The ceiling is not defined here and is not needed on this page; it would be the least integer $\ge x$ , obtained from the same set $S$ without the shift by one.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 79 results over 31 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Floor and ceiling functions (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §5.4 (standard reference, not scraped)

Integer part: for every real xxx there is exactly one integer mmm with m≤x<m+1m \le x < m + 1m≤x<m+1