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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.
  • 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.

Binary residue carrier and operations

Definition

Set

F2:=Z/2Z={[0]2,[1]2},

with addition and multiplication induced by the quotient operations

[r]2+[s]2=[r+s]2,[r]2[s]2=[rs]2

from Addition and multiplication on Z/n by [a]n+[b]n=[a+b]n and [a]n[b]n=[ab]n. This item fixes the two residue classes and their operations. It does not assert the field axioms; those must be established separately before an argument uses field properties.

Facts & Assumptions

Given: The quotient-set and arithmetic definitions for the modulus 2.

[F1]

The congruence class [a]n and the quotient set Z/n: two residue classes modulo 2 are equal exactly when their representatives are congruent modulo 2.

[F2]

Addition and multiplication on Z/n by [a]n+[b]n=[a+b]n and [a]n[b]n=[ab]n: addition and multiplication are the representative-independent operations [r]2+[s]2=[r+s]2 and [r]2[s]2=[rs]2.

Verification

1.1givenF1F3inductionalgebra

Every integer r has the form r=2q+ε for some q∈Z and ε∈{0,1}. For r≥0, induction starts with 0=2⋅0+0; from r=2q obtain r+1=2q+1, and from r=2q+1 obtain r+1=2(q+1)+0. For r<0, apply the nonnegative case to −r=2q+ε: if ε=0, then r=2(−q), and if ε=1, then r=2(−q−1)+1. By [F3], r≡ε(mod2), so [F1] gives [r]2=[ε]2. Hence every quotient class is one of [0]2,[1]2.

2.1givenF1F3step 1.1algebra

The classes [0]2 and [1]2 are distinct: equality would give 0≡1(mod2) by [F1], hence 2∣−1 by [F3], which is impossible because no integer q satisfies −1=2q. Together with step 1.1, this proves that the quotient has exactly the two listed classes.

3.1givenF1F2F3step 1.1step 2.1algebra∎

Applying [F2] to representatives 0 and 1 gives the addition table [0]+[0]=[0], [0]+[1]=[1], [1]+[0]=[1], [1]+[1]=[0], and the multiplication table [0][0]=[0], [0][1]=[1][0]=[0], [1][1]=[1]. Since steps 1.1 and 2.1 show that these are all quotient classes, the tables are exhaustive; both operations take values in this carrier and are exactly the quotient operations.

Depends on

Used by

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Sources