LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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.

$x^{2} + 1$ is irreducible over $R$

The polynomial $x^{2} + 1$ is irreducible in $R [x]$ .

Given: The polynomial $x^{2} + 1 \in R [x]$ .

[F1]

The real numbers form an ordered field (The reals form a totally ordered field).

[F2]

In an ordered field, the square of every nonzero element is positive (Squares of nonzero elements are positive).

[F3]

A polynomial of degree two or three over a field is irreducible if and only if it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

technique · direct

1.1F1F2

For $r \in R$ , either $r = 0$ , so $r^{2} = 0$ , or [F2] gives $r^{2} > 0$ . Thus $r^{2} + 1 > 0$ by the ordered-field laws in [F1].

2.1step 1.1

Hence $x^{2} + 1$ has no real root.

3.1F3step 2.1∎

Since it has degree two, [F3] now gives its irreducibility over $R$ .

Depends on