Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)
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.

R is a vector space over itself, over the embedded copy of Q by restriction of scalars, and over Q itself via the embedding

Example

Let R be the real numbers (The real numbers), a field (The reals form a field) and an ordered field (The reals form a totally ordered field, Ordered field), and let Q be the rationals, a field (The rationals form a field).

  1. R is a vector space over itself (Vector space over a field): the vectors are the reals, the vector addition is the field addition, the zero vector is 0, and the scalar multiplication is the field multiplication.
  2. Let ι:Q→R be the unique field homomorphism (The unique embedding of ℚ into an ordered field, Field homomorphism and embedding), which is injective and order-preserving. Its image QR:=ι[Q]={ ι(q):q∈Q } is a subfield of R (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations), and by restriction of scalars (A field is a vector space over itself, and over any subfield K⊆F every F-vector space is a K-vector space by restricting the scalars) R is a vector space over QR, the scalar multiplication being the field multiplication restricted to QR×R.
  3. Setting q⋅x:=ι(q) x for q∈Q and x∈R makes R a vector space over Q itself.

The subfield is the image, not Q. Q is not a subset of R in this library: a rational is a class of pairs of integers and a real is a class of Cauchy sequences of rationals (The real numbers). What sits inside R as a subfield is the image QR of the embedding, and claim 2 is a statement about that image. Claim 3 is the statement about Q itself, and it is proved directly rather than by restricting scalars, because restriction of scalars requires a subfield.

Facts & Assumptions

Given: The field R, the field Q, and the map ι:Q→R of The unique embedding of ℚ into an ordered field.

[L1]

R is a field (The reals form a field, The real numbers) and is an ordered field with positive cone as in Ordered field (The reals form a totally ordered field).

[L2]
[L3]

There is a unique field homomorphism ι:Q→R, and it is injective and order-preserving (The unique embedding of ℚ into an ordered field).

[L4]

A field homomorphism φ satisfies φ(x+y)=φ(x)+φ(y), φ(xy)=φ(x)φ(y) and φ(1)=1, and consequently φ(0)=0, φ(−x)=−φ(x) and φ(x−1)=φ(x)−1 for x≠0 (Field homomorphism and embedding).

[L5]

A subfield of a field F is a subring of F containing x−1 for each of its nonzero elements; equivalently, a subset containing 1F and closed under a−b and ab, and containing x−1 for each of its nonzero elements. Such a subset contains 0F and is closed under addition and additive inverses (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

[L6]

A field is a vector space over itself, and over any subfield K of F every F-vector space is a K-vector space by restricting the scalar multiplication to K (A field is a vector space over itself, and over any subfield K⊆F every F-vector space is a K-vector space by restricting the scalars).

[L7]

The vector space axioms (V1)–(V5) (Vector space over a field), and the field axioms of R: (R,+,0) is an abelian group, multiplication is associative and commutative with identity 1, and it distributes over addition (Field).

Verification

technique · direct
1.1

R is a field, so it is a vector space over itself with the field addition as vector addition and the field multiplication as scalar multiplication; this is claim 1.

L1L6
1.2

Since R is an ordered field, there is a unique field homomorphism ι:Q→R, and it is injective.

L1L2L3
1.3

QR=ι[Q] is a subfield of R: it contains ι(0)=0 and ι(1)=1; for p,q∈Q it contains ι(p)+ι(q)=ι(p+q), −ι(q)=ι(−q) and ι(p)ι(q)=ι(pq); and if ι(q)≠0 then q≠0, since ι(0)=0, so ι(q)−1=ι(q−1)∈QR.

L2L3L4L5
1.4

The assignment (q,x)↦ι(q) x is a map Q×R→R, since ι(q)∈R and the field multiplication of R takes values in R.

L3L7
2.1

Applying restriction of scalars to the R-vector space of step 1.1 and the subfield of step 1.3 shows that R is a vector space over QR, with the field multiplication restricted to QR×R as scalar multiplication; this is claim 2.

step 1.1step 1.3L6
2.2

The operation of step 1.4 satisfies the five axioms over Q. (V1) holds because (R,+,0) is an abelian group. For p,q∈Q and x,y∈R: q⋅(x+y)=ι(q)(x+y)=ι(q)x+ι(q)y=q⋅x+q⋅y by distributivity, which is (V2); (p+q)⋅x=ι(p+q)x=(ι(p)+ι(q))x=ι(p)x+ι(q)x=p⋅x+q⋅x by additivity of ι and distributivity, which is (V3); (pq)⋅x=ι(pq)x=(ι(p)ι(q))x=ι(p)(ι(q)x)=p⋅(q⋅x) by multiplicativity of ι and associativity, which is (V4); and 1⋅x=ι(1)x=1 x=x, which is (V5).

step 1.4L3L4L7
3.1

Claim 1 is step 1.1, claim 2 is step 2.1, and claim 3 is step 2.2, so R carries all three structures at once: over itself, over the embedded copy of Q inside it, and over Q.

step 1.1step 2.1step 2.2∎

Remarks

  • Three structures on one set. The vectors are the same reals throughout and the addition is the same in all three cases; what changes is which scalars are allowed to act. Claims 2 and 3 differ only in bookkeeping: the scalars are the elements of QR in one and the elements of Q in the other, and ι matches them up bijectively, being injective onto its image.

  • Nothing here is said about size. How big R is as a vector space over Q is a question about bases and dimension, which are developed on a later page; no claim about either is made above, and the verification uses neither.

  • Why the ordered field hypothesis appears at all. The embedding ι is supplied by The unique embedding of ℚ into an ordered field, which is stated for an ordered field, and R is one. The order plays no further role: once ι is in hand, every step above uses only that it is a field homomorphism.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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