ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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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.
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$\sqrt{2}$ and $(1 + \sqrt{5}) / 2$ are algebraic integers, while $1 / 2$ is not

Example

The numbers $\sqrt{2}$ and $(1 + \sqrt{5}) / 2$ are algebraic integers, whereas $1 / 2$ is not. See Integral elements over a commutative ring and algebraic integers.

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

Let $A \to B$ be a homomorphism of commutative rings. An element $b \in B$ is integral over $A$ when it is a root of a monic polynomial in $A [X]$ . The extension is integral when every element is integral. An algebraic integer is a complex number integral over $Z$ . (Integral elements over a commutative ring and algebraic integers).

[L2]

A rational number is an algebraic integer if and only if it is an integer. (The rational algebraic integers are exactly the integers).

[L3]

Let $F$ be a complete ordered field (def-complete-ordered-field). Then every $a \in F$ with $a \geq 0$ has a unique $s \in F$ with $s \geq 0$ and $s^{2} = a$ ; we write $s = \sqrt{a}$ . Consequently the positive elements of $F$ are exactly the nonzero squares: $x > 0$ if and only if $x = y^{2}$ for some $y \neq 0$ . (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · direct

1.1

The number $\sqrt{2}$ is a root of the monic polynomial $X^{2} - 2$ , and $(1 + \sqrt{5}) / 2$ is a root of the monic polynomial $X^{2} - X - 1$ ; both are therefore algebraic integers.

L1L2L3givenalgebra

2.1

The rational algebraic-integer criterion already excludes $1 / 2$ . Directly, a monic equation of degree $n$ at $1 / 2$ would, after multiplication by $2^{n}$ , read $1 + 2 a_{n - 1} + \dots + 2^{n} a_{0} = 0$ , whose left side is odd. This proves the stated claim.

step 1.1givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 7 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

Eloisa Grifo, Commutative Algebra I, Section 1.4 (standard reference, not scraped)

22​ and (1+5)/2(1+5​)/2 are algebraic integers, while 1/21/2 is not