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✓ 29 results · all verified · 24 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Construction of the Real Numbers via Cauchy Sequences

1 · Prerequisites

2 · Summary

Objective. This page constructs the real number field $\mathbb{R}$ via Cauchy sequences, starting from the fundamental definitions of the integers and the rational numbers, and proves that $\mathbb{R}$ is a totally ordered, complete field. The natural numbers $\mathbb{N}$ , with their arithmetic and order, are the assumed starting point.

Layer 1: from $\mathbb{N}$ to $\mathbb{Z}$ . Subtraction fails in $\mathbb{N}$ . The integers are defined as equivalence classes of pairs $(a,b)$ of naturals, where $(a,b)$ encodes the formal difference $a - b$ (The integers as equivalence classes of pairs of naturals), with arithmetic defined on representatives (Arithmetic on the integers). The result is a commutative ring (The integers form a commutative ring): a set with an addition and a multiplication, both associative and commutative, where multiplication distributes over addition, addition has an identity $0$ and an inverse for every element, and multiplication has an identity $1$ . The integers also carry a total order, defined on representatives by $(a,b) \le (c,d)$ exactly when $a + d \le b + c$ : any two integers are comparable, and the order is compatible with addition and with multiplication by nonnegative elements, so $\mathbb{Z}$ is a totally ordered commutative ring.

Layer 2: from $\mathbb{Z}$ to $\mathbb{Q}$ . Division fails in $\mathbb{Z}$ . The rationals are equivalence classes of pairs of integers with nonzero second component, encoding formal quotients $a/b$ . The result is a field: a commutative ring in which every nonzero element has a multiplicative inverse. The total order extends from $\mathbb{Z}$ : any two rationals remain comparable, and the order stays compatible with the arithmetic; a field equipped with such an order is a totally ordered field.

Layer 3: from $\mathbb{Q}$ to $\mathbb{R}$ . Limits fail in $\mathbb{Q}$ : sequences such as $1, 1.4, 1.41, 1.414, \dots$ cluster ever more tightly yet converge to no rational. A Cauchy sequence is one whose terms eventually stay within any prescribed distance of one another (Cauchy sequence of rationals). The reals are equivalence classes of Cauchy sequences of rationals, two sequences being identified when their difference is a null sequence (Null sequence). Arithmetic passes to the quotient, the null sequences form a maximal ideal so that $\mathbb{R}$ is a field, and the order extends from $\mathbb{Q}$ .

Destination. $\mathbb{R}$ is a totally ordered field that is complete: every Cauchy sequence of reals converges to a real. Completeness is the payoff of the construction, and the construction cannot be iterated to gain anything further: The reals are complete already gives every Cauchy sequence of reals a real limit.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-24Open item page →

The integers as equivalence classes of pairs of naturals

Definition

On the set $\mathbb{N} \times \mathbb{N}$ of pairs of natural numbers, define

$(a,b) \sim (c,d) \iff a + d = b + c.$

This is an equivalence relation (The integer-defining relation is an equivalence relation ↗). The integers are the quotient

$\mathbb{Z} := (\mathbb{N} \times \mathbb{N}) / \sim,$

and we write $[(a,b)]$ for the equivalence class of $(a,b)$ .

Remarks

The pair $(a,b)$ encodes the formal difference $a - b$ ; the defining relation $a + d = b + c$ is the equation " $a - b = c - d$ " restated using only addition, which is all $\mathbb{N}$ has.
$\mathbb{N}$ and its arithmetic (commutativity, associativity, distributivity, cancellation of addition, the order) are taken as given background throughout this construction.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The integer-defining relation is an equivalence relation

Statement

The relation $(a,b) \sim (c,d) \iff a + d = b + c$ on $\mathbb{N} \times \mathbb{N}$ (The integers as equivalence classes of pairs of naturals) is an equivalence relation.

Facts & Assumptions

Given: The relation $(a,b) \sim (c,d) \iff a + d = b + c$ on $\mathbb{N} \times \mathbb{N}$ .

[A1]

Addition on $\mathbb{N}$ is commutative and associative.

[A2]

Cancellation in $\mathbb{N}$ : if $x + z = y + z$ then $x = y$ .

Proof

technique · direct

1.1

Reflexivity: for any $(a,b)$ we have $a + b = b + a$ , so $(a,b) \sim (a,b)$ .

1.2

Symmetry: suppose $(a,b) \sim (c,d)$ , i.e. $a + d = b + c$ . Then $c + b = d + a$ , which is the defining equation for $(c,d) \sim (a,b)$ .

1.3

Suppose $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ , i.e. $a + d = b + c$ and $c + f = d + e$ .

given

2.1

Adding the two equations: $(a + d) + (c + f) = (b + c) + (d + e)$ .

step 1.3algebra

3.1

Regrouping both sides: $(a + f) + (c + d) = (b + e) + (c + d)$ .

step 2.1A1

4.1

Cancelling $(c + d)$ : $a + f = b + e$ , so $(a,b) \sim (e,f)$ ; the relation is transitive.

step 3.1A2

5.1

The relation is reflexive, symmetric, and transitive, hence an equivalence relation.

step 1.1step 1.2step 4.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-24Open item page →

Arithmetic on the integers

Definition

On $\mathbb{Z}$ (The integers as equivalence classes of pairs of naturals) define, on representatives,

$[(a,b)] + [(c,d)] := [(a+c,\; b+d)],$ $[(a,b)] \cdot [(c,d)] := [(ac+bd,\; ad+bc)],$ $-[(a,b)] := [(b,a)], \qquad 0 := [(0,0)], \qquad 1 := [(1,0)].$

These do not depend on the chosen representatives (Integer addition and negation are well defined ↗, Integer multiplication is well defined ↗).

Remarks

The product formula is forced by the intended meaning: if $(a,b)$ stands for $a-b$ and $(c,d)$ for $c-d$ , then $(a-b)(c-d) = (ac+bd) - (ad+bc)$ .
Negation swaps the components: $-(a-b) = b-a$ .

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Integer addition and negation are well defined

Statement

The operations $[(a,b)] + [(c,d)] = [(a+c, b+d)]$ and $-[(a,b)] = [(b,a)]$ on $\mathbb{Z}$ (Arithmetic on the integers) are independent of the chosen representatives.

Facts & Assumptions

Given: Pairs with $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ in the sense of The integers as equivalence classes of pairs of naturals.

[A1]

Addition on $\mathbb{N}$ is commutative and associative.

Proof

technique · direct

1.1

By hypothesis $a + b' = b + a'$ .

given

1.2

By hypothesis $c + d' = d + c'$ .

given

2.1

Adding the two equations: $(a + b') + (c + d') = (b + a') + (d + c')$ .

step 1.1step 1.2algebra

2.2

Commuting the equation of step 1.1: $b + a' = a + b'$ , which is the defining equation for $(b,a) \sim (b',a')$ ; negation is well defined.

step 1.1A1

3.1

Regrouping both sides: $(a+c) + (b'+d') = (b+d) + (a'+c')$ , which is the defining equation for $(a+c,\, b+d) \sim (a'+c',\, b'+d')$ ; addition is well defined.

step 2.1A1

4.1

Both operations are independent of representatives.

step 3.1step 2.2∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Integer multiplication is well defined

Statement

The operation $[(a,b)] \cdot [(c,d)] = [(ac+bd,\, ad+bc)]$ on $\mathbb{Z}$ (Arithmetic on the integers) is independent of the chosen representatives.

Facts & Assumptions

Given: Pairs with $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ in the sense of The integers as equivalence classes of pairs of naturals.

[A1]

Addition on $\mathbb{N}$ is commutative and associative.

[A2]

Multiplication on $\mathbb{N}$ is commutative.

[A3]

Distributivity in $\mathbb{N}$ : $x(y+z) = xy + xz$ .

Proof

technique · direct

1.1

By hypothesis $a + b' = b + a'$ ; write $s$ for this common value.

given

1.2

Regrouping and factoring: $(ac+bd) + (a'd+b'c) = c(a+b') + d(b+a')$ .

A1A2A3

1.3

Regrouping and factoring: $(ad+bc) + (a'c+b'd) = d(a+b') + c(b+a')$ .

A1A2A3

2.1

By step 1.1 both right-hand sides equal $cs + ds$ , so $(ac+bd) + (a'd+b'c) = (ad+bc) + (a'c+b'd)$ .

step 1.1step 1.2step 1.3

3.1

That equation is precisely the defining relation $(ac+bd,\, ad+bc) \sim (a'c+b'd,\, a'd+b'c)$ : the product class is unchanged when the first factor's representative changes.

step 2.1

4.1

The product formula is symmetric in its two arguments: swapping $(a,b) \leftrightarrow (c,d)$ sends $(ac+bd,\, ad+bc)$ to $(ca+db,\, cb+da)$ , the same pair. Hence, by the argument of steps 1.1–3.1 applied to the second factor, the product class is also unchanged when $(c,d)$ is replaced by $(c',d')$ .

step 3.1A1A2

5.1

Replacing first $(a,b)$ by $(a',b')$ and then $(c,d)$ by $(c',d')$ : $[(a,b)] \cdot [(c,d)] = [(a',b')] \cdot [(c,d)] = [(a',b')] \cdot [(c',d')]$ ; multiplication is well defined.

step 3.1step 4.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

The integers form a commutative ring

Statement

$(\mathbb{Z}, +, \cdot, 0, 1)$ with the operations of Arithmetic on the integers is a commutative ring with multiplicative identity, in which every element has the additive inverse $-[(a,b)] = [(b,a)]$ .

Facts & Assumptions

Given: $\mathbb{Z}$ with the operations of Arithmetic on the integers.

[L1]

Addition on $\mathbb{N}$ is commutative and associative, with zero as a two-sided identity (Addition is commutative, Addition is associative, Left identity for addition).

[L2]

Multiplication on $\mathbb{N}$ is commutative and associative, distributes over addition, and has the usual zero and identity laws (Multiplication is commutative, Multiplication is associative, Distributivity and the successor law for multiplication, Zero and one under multiplication).

[L3]

The integer operations are independent of representatives (Integer addition and negation are well defined, Integer multiplication is well defined).

Proof

technique · direct

1.1

By [L3], each axiom may be verified on arbitrary fixed representatives $(a,b), (c,d), (e,f)$ .

L3suffices

1.2

Associativity of $+$ : $((a,b)+(c,d))+(e,f)$ and $(a,b)+((c,d)+(e,f))$ both equal $(a+c+e,\; b+d+f)$ .

1.3

Commutativity of $+$ : $(a+c,\, b+d) = (c+a,\, d+b)$ .

1.4

Additive identity: $(a,b) + (0,0) = (a+0,\, b+0) = (a,b)$ .

1.5

Additive inverses: $(a,b) + (b,a) = (a+b,\, b+a) \sim (0,0)$ , since $(a+b) + 0 = (b+a) + 0$ .

1.6

Commutativity of $\cdot$ : swapping $(a,b) \leftrightarrow (c,d)$ sends $(ac+bd,\, ad+bc)$ to $(ca+db,\, cb+da)$ , the same pair.

L1L2

1.7

Multiplicative identity: $(a,b) \cdot (1,0) = (a \cdot 1 + b \cdot 0,\; a \cdot 0 + b \cdot 1) = (a,b)$ .

L1L2

1.8

Associativity of $\cdot$ : expanding, both $((a,b)(c,d))(e,f)$ and $(a,b)((c,d)(e,f))$ equal $(ace+adf+bcf+bde,\; acf+ade+bce+bdf)$ .

L1L2

1.9

Distributivity: $(a,b) \cdot ((c,d)+(e,f)) = (a(c+e)+b(d+f),\; a(d+f)+b(c+e)) = (ac+bd,\, ad+bc) + (ae+bf,\, af+be) = (a,b)(c,d) + (a,b)(e,f)$ .

L1L2

2.1

Steps 1.2–1.9 verify all axioms: $(\mathbb{Z}, +, \cdot, 0, 1)$ is a commutative ring with identity and additive inverses.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 1.8step 1.9∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-24Open item page →

Order on the integers

Definition

For integers $x = [(a,b)]$ and $y = [(c,d)]$ (The integers as equivalence classes of pairs of naturals) define

$x \le y \iff a + d \le b + c \quad \text{in } \mathbb{N}.$

Write $x < y$ when $x \le y$ and $x \ne y$ . An integer $x$ is positive when $0 < x$ ; on representatives, $0 < [(a,b)]$ exactly when $b < a$ .

Remarks

The defining inequality is " $a - b \le c - d$ " restated without subtraction.
Well-definedness, totality, and compatibility with the arithmetic are proved in The integers form a totally ordered ring ↗.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)Open item page →

The integers form a totally ordered ring

Statement

The relation of Order on the integers is well defined and is a total order on $\mathbb{Z}$ ; it is compatible with addition ( $x \le y$ implies $x + z \le y + z$ ) and positives are closed under multiplication ( $0 < x$ and $0 < y$ imply $0 < xy$ ). Thus $\mathbb{Z}$ is a totally ordered commutative ring.

Facts & Assumptions

Given: Integers $x = [(a,b)]$ , $y = [(c,d)]$ , $z = [(e,f)]$ with the operations of Arithmetic on the integers.

[L1]

Addition on $\mathbb{N}$ is commutative and associative (Addition is commutative, Addition is associative).

[L2]

The order on $\mathbb{N}$ is total, antisymmetric, and transitive ( $\le$ is a linear order on $\mathbb{N}$ ).

[L3]

Natural order is translation-invariant: $x\le y\iff x+z\le y+z$ (Order is compatible with addition).

[L4]

By the definition of natural order, $x\le y$ exactly when $y=x+k$ for some natural $k$ ; if $x<y$ , such a gap is nonzero (Order on the natural numbers).

[L5]

Natural multiplication is commutative and associative, distributes over addition, and has the usual zero and identity laws; multiplication by a nonzero natural preserves strict order (Multiplication is commutative, Multiplication is associative, Distributivity and the successor law for multiplication, Zero and one under multiplication, Order is compatible with multiplication).

[L6]

The integer operations make $\mathbb{Z}$ a commutative ring with multiplicative identity (The integers form a commutative ring).

Proof

technique · direct

1.1

Suppose $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ , i.e. $a + b' = b + a'$ and $c + d' = d + c'$ .

given

1.2

Suppose $a + d \le b + c$ .

given

1.3

Totality: in $\mathbb{N}$ either $a + d \le b + c$ or $b + c \le a + d$ , so $x \le y$ or $y \le x$ .

1.4

Antisymmetry: $x \le y$ and $y \le x$ give $a + d \le b + c$ and $b + c \le a + d$ , hence $a + d = b + c$ , which is $x = y$ as classes.

1.5

Transitivity: from $a + d \le b + c$ and $c + f \le d + e$ , adding gives $(a+d)+(c+f) \le (b+c)+(d+e)$ ; regrouping, $(a+f)+(c+d) \le (b+e)+(c+d)$ ; cancelling, $a + f \le b + e$ , i.e. $x \le z$ .

L1L2L3

1.6

Compatibility with addition: $x + z \le y + z$ reads $(a+e)+(d+f) \le (b+f)+(c+e)$ , which regroups to $(a+d) + (e+f) \le (b+c) + (e+f)$ , equivalent to $a + d \le b + c$ , i.e. $x \le y$ .

L1L3

1.7

Suppose $0 < x$ and $0 < y$ , i.e. $b < a$ and $d < c$ .

given

2.1

Adding $(a'+d')$ : $(a+d) + (a'+d') \le (b+c) + (a'+d')$ .

step 1.2L3

2.2

Regrouping with the equations of step 1.1: $(b+c) + (a'+d') = (b+a') + (c+d') = (a+b') + (d+c') = (a+d) + (b'+c')$ .

step 1.1L1

2.3

Write $a = b + k$ and $c = d + m$ with nonzero naturals $k,m$ .

step 1.7L4

3.1

So $(a+d) + (a'+d') \le (a+d) + (b'+c')$ ; cancelling $(a+d)$ gives $a'+d'\le b'+c'$ . Applying the same argument with primed and unprimed representatives exchanged proves the converse implication, so the relation is independent of representatives.

step 1.1step 2.1step 2.2L3

3.2

Expanding both components of $xy = [(ac+bd,\, ad+bc)]$ : $ac + bd = (b+k)(d+m) + bd = (bd + bm + kd + bd) + km$ and $ad + bc = (b+k)d + b(d+m) = bd + kd + bd + bm$ , so $ac + bd = (ad + bc) + km$ .

step 2.3L1L5

4.1

Since $k,m$ are nonzero, $0<k$ and $0<m$ by [L4]. Hence [L5] gives $0=0\cdot m<km$ , and step 3.2 yields $ad+bc<ac+bd$ , which is $0<xy$ .

step 2.3step 3.2L2L4L5

5.1

The order is well defined, total, compatible with addition, and positives are closed under multiplication; together with [L6], this makes $\mathbb{Z}$ a totally ordered commutative ring.

step 3.1step 1.3step 1.4step 1.5step 1.6step 4.1L6∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The integers have no zero divisors; multiplicative cancellation

Statement

If $x, y \in \mathbb{Z}$ are nonzero then $xy \ne 0$ . Consequently, if $xz = yz$ and $z \ne 0$ , then $x = y$ .

Facts & Assumptions

Given: Nonzero integers $x, y$ .

[L1]

$\mathbb{Z}$ is a totally ordered ring; in particular trichotomy holds (each integer is negative, zero, or positive) and positives are closed under multiplication (The integers form a totally ordered ring).

[L2]

Ring arithmetic in $\mathbb{Z}$ : $(-u)v = -(uv)$ , $-(-u) = u$ , and distributivity (The integers form a commutative ring).

Proof

technique · cases

1.1

By trichotomy, each of $x$ and $y$ is positive or negative; split on the sign pattern.

L1cases-exhaustive

1.2

Case both positive: $0 < xy$ by closure of positives, so $xy \ne 0$ .

assume-case bothL1

1.3

Case exactly one negative, say $y < 0$ (the other case is identical by commutativity): $0 < -y$ , so $0 < x(-y) = -(xy)$ , hence $xy < 0$ and $xy \ne 0$ .

assume-case mixedL1L2

1.4

Case both negative: $0 < -x$ and $0 < -y$ , so $0 < (-x)(-y) = xy$ , and $xy \ne 0$ .

assume-case negnegL1L2

2.1

In every case $xy \ne 0$ : the product of nonzero integers is nonzero.

step 1.2step 1.3step 1.4cases

3.1

Cancellation: if $xz = yz$ with $z \ne 0$ , then $(x-y)z = xz - yz = 0$ ; were $x - y \ne 0$ , the product would be nonzero, so $x - y = 0$ , i.e. $x = y$ .

step 2.1L2∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The naturals embed in the integers

Statement

The map $\iota(n) = [(n,0)]$ is injective and preserves addition, multiplication, and order. Its image is exactly the set of nonnegative integers, so every $x \ge 0$ in $\mathbb{Z}$ is $\iota(k)$ for a unique natural $k$ .

Facts & Assumptions

Given: The map $\iota : \mathbb{N} \to \mathbb{Z}$ , $\iota(n) = [(n,0)]$ .

[A1]

Arithmetic identities in $\mathbb{N}$ : $x + 0 = x$ , $0 + x = x$ , $x \cdot 0 = 0$ , $0 \cdot x = 0$ , $x \cdot 1 = x$ .

[A2]

The order on $\mathbb{N}$ , and: $x \le y$ iff $y = x + k$ for some $k$ .

Proof

technique · direct

1.1

Injectivity: $[(m,0)] = [(n,0)]$ means $m + 0 = 0 + n$ , i.e. $m = n$ .

1.2

Addition: $\iota(m) + \iota(n) = [(m+n,\, 0+0)] = \iota(m+n)$ .

1.3

Multiplication: $\iota(m)\,\iota(n) = [(mn + 0 \cdot 0,\; m \cdot 0 + 0 \cdot n)] = \iota(mn)$ .

1.4

Order: $\iota(m) \le \iota(n)$ reads $m + 0 \le 0 + n$ , i.e. $m \le n$ .

A1A2

1.5

Image: if $0 \le [(a,b)]$ then $b \le a$ , so $a = b + k$ and $(a,b) \sim (k,0)$ , giving $[(a,b)] = \iota(k)$ ; conversely $0 \le \iota(k)$ for every $k$ .

A2algebra

2.1

$\iota$ embeds $\mathbb{N}$ into $\mathbb{Z}$ , preserving arithmetic and order, with image the nonnegative integers.

step 1.1step 1.2step 1.3step 1.4step 1.5∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-24Open item page →

The rationals as equivalence classes of pairs of integers

Definition

On the set of pairs $(a,b)$ with $a, b \in \mathbb{Z}$ and $b \ne 0$ , define

$(a,b) \sim (c,d) \iff ad = cb \quad \text{in } \mathbb{Z}.$

This is an equivalence relation (The rational-defining relation is an equivalence relation ↗). The rationals are the quotient $\mathbb{Q}$ , and $[(a,b)]$ is written $a/b$ .

Remarks

The pair $(a,b)$ encodes the formal quotient $a/b$ ; the relation $ad = cb$ is " $a/b = c/d$ " cleared of denominators, using only the ring arithmetic of $\mathbb{Z}$ (The integers form a commutative ring).

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The rational-defining relation is an equivalence relation

Statement

The relation $(a,b) \sim (c,d) \iff ad = cb$ on pairs of integers with nonzero second component (The rationals as equivalence classes of pairs of integers) is an equivalence relation.

Facts & Assumptions

Given: Pairs $(a,b), (c,d), (e,f)$ of integers with $b, d, f \ne 0$ .

[L1]

$\mathbb{Z}$ is a commutative ring (The integers form a commutative ring).

[L2]

Multiplicative cancellation in $\mathbb{Z}$ : $ud = vd$ with $d \ne 0$ implies $u = v$ (The integers have no zero divisors; multiplicative cancellation).

Proof

technique · direct

1.1

Reflexivity: $ab = ba$ , so $(a,b) \sim (a,b)$ .

1.2

Symmetry: if $ad = cb$ then $cb = ad$ , which is the defining equation for $(c,d) \sim (a,b)$ .

1.3

Suppose $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ , i.e. $ad = cb$ and $cf = ed$ .

given

2.1

Multiplying the first equation by $f$ and the second by $b$ : $adf = cbf$ and $cfb = edb$ .

step 1.3L1

3.1

Chaining: $(af)d = adf = cbf = cfb = edb = (eb)d$ .

step 2.1L1

4.1

Cancelling the nonzero $d$ : $af = eb$ , so $(a,b) \sim (e,f)$ ; the relation is transitive.

step 3.1L2

5.1

The relation is reflexive, symmetric, and transitive, hence an equivalence relation.

step 1.1step 1.2step 4.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-24Open item page →

Arithmetic on the rationals

Definition

On $\mathbb{Q}$ (The rationals as equivalence classes of pairs of integers) define, on representatives,

$[(a,b)] + [(c,d)] := [(ad + cb,\; bd)], \qquad [(a,b)] \cdot [(c,d)] := [(ac,\; bd)],$ $-[(a,b)] := [(-a, b)], \qquad 0 := [(0,1)], \qquad 1 := [(1,1)],$

and, for $[(a,b)] \ne 0$ (equivalently $a \ne 0$ ), the inverse $[(a,b)]^{-1} := [(b,a)]$ .

Remarks

The denominators stay legal: $bd \ne 0$ because $\mathbb{Z}$ has no zero divisors (The integers have no zero divisors; multiplicative cancellation).
Independence of representatives: Rational arithmetic is well defined ↗ for sum, product, and negation; The reciprocal on the rationals is well-defined ↗ for the reciprocal.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Rational arithmetic is well defined

Statement

Sum, product, and negation of Arithmetic on the rationals are independent of the chosen representatives.

Facts & Assumptions

Given: Pairs with $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ (The rationals as equivalence classes of pairs of integers), all second components nonzero.

[L1]

$\mathbb{Z}$ is a commutative ring (The integers form a commutative ring).

Proof

technique · direct

1.1

By hypothesis $ab' = a'b$ and $cd' = c'd$ .

given

2.1

Product: $(ac)(b'd') = (ab')(cd') = (a'b)(c'd) = (a'c')(bd)$ , so $(ac,\, bd) \sim (a'c',\, b'd')$ .

step 1.1L1

2.2

Sum: $(ad + cb)(b'd') = (ab')(dd') + (cd')(bb') = (a'b)(dd') + (c'd)(bb') = (a'd' + c'b')(bd)$ , so $(ad+cb,\, bd) \sim (a'd'+c'b',\, b'd')$ .

step 1.1L1

2.3

Negation: from $ab' = a'b$ , $(-a)b' = (-a')b$ , so $(-a, b) \sim (-a', b')$ .

step 1.1L1

3.1

All three operations are independent of representatives.

step 2.1step 2.2step 2.3∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25Open item page →

The reciprocal on the rationals is well-defined

Statement

On $\mathbb{Q} \setminus \{0\}$ the reciprocal $[(a,b)]^{-1} := [(b,a)]$ (Arithmetic on the rationals) is independent of the chosen representative: if $(a,b) \sim (a',b')$ with $a, a', b, b' \ne 0$ , then $(b,a) \sim (b',a')$ . Hence the reciprocal is a well-defined function on $\mathbb{Q} \setminus \{0\}$ .

Facts & Assumptions

Given: Nonzero rationals represented by integer pairs $(a,b)$ and $(a',b')$ with $a, a', b, b' \ne 0$ , where $(x,y) \sim (z,w) \iff x w = z y$ in $\mathbb{Z}$ (The rationals as equivalence classes of pairs of integers).

[L1]

Multiplication in $\mathbb{Z}$ is commutative (The integers form a commutative ring).

Proof

technique · direct

1.1

By hypothesis $(a,b) \sim (a',b')$ , that is $a b' = a' b$ in $\mathbb{Z}$ .

given

2.1

Commuting each product by [L1], $b' a = b a'$ , hence $b a' = b' a$ .

step 1.1L1

3.1

The equation $b a' = b' a$ is exactly the defining relation $(b,a) \sim (b',a')$ , and since $a, a' \ne 0$ the pairs $(b,a), (b',a')$ are legal rational representatives; therefore $[(a,b)]^{-1} = [(b,a)] = [(b',a')] = [(a',b')]^{-1}$ , so the reciprocal is well-defined on $\mathbb{Q} \setminus \{0\}$ .

step 2.1given∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The rationals form a field

Statement

$(\mathbb{Q}, +, \cdot, 0, 1)$ with the operations of Arithmetic on the rationals is a field: a commutative ring with $1 \ne 0$ in which every nonzero element has a multiplicative inverse.

Facts & Assumptions

Given: Rationals with representatives $(a,b), (c,d), (e,f)$ , second components nonzero.

[L1]

The operations are independent of representatives (Rational arithmetic is well defined).

[L2]

$\mathbb{Z}$ is a commutative ring in which $1 \ne 0$ (The integers form a commutative ring; $1 \ne 0$ because $1 \ne 0$ in $\mathbb{N}$ ).

[L3]

$\mathbb{Z}$ has no zero divisors and admits cancellation (The integers have no zero divisors; multiplicative cancellation).

Proof

technique · direct

1.1

By [L1], each axiom may be verified on fixed representatives.

L1suffices

1.2

Associativity of $+$ : both $((a,b)+(c,d))+(e,f)$ and $(a,b)+((c,d)+(e,f))$ have representative $(adf + cbf + ebd,\; bdf)$ .

1.3

Commutativity of $+$ : $(ad+cb,\, bd)$ and $(cb+ad,\, db)$ are equal pairs.

1.4

Additive identity: $(a,b) + (0,1) = (a \cdot 1 + 0 \cdot b,\; b \cdot 1) = (a,b)$ .

1.5

Additive inverses: $(a,b) + (-a,b) = (ab - ab,\; b^2) = (0, b^2) \sim (0,1)$ .

1.6

Associativity and commutativity of $\cdot$ : $((ac)e,\, (bd)f) = (a(ce),\, b(df))$ and $(ac,\, bd) = (ca,\, db)$ .

1.7

Multiplicative identity: $(a,b)(1,1) = (a,b)$ ; and $1 \ne 0$ since $1 \cdot 1 \ne 0 \cdot 1$ .

1.8

Distributivity: $(a,b)\bigl((c,d)+(e,f)\bigr)$ has representative $(a(cf+ed),\; b(df))$ , while $(a,b)(c,d) + (a,b)(e,f)$ has representative $(acbf + aebd,\; b^2 df) = (b(acf + aed),\; b(bdf))$ ; cancelling the common nonzero factor $b$ these are equivalent.

L2L3

1.9

Inverses: $[(a,b)] = 0$ iff $a \cdot 1 = 0 \cdot b$ iff $a = 0$ ; for $a \ne 0$ , $(a,b)(b,a) = (ab,\, ba) \sim (1,1)$ .

2.1

All field axioms hold: $\mathbb{Q}$ is a field.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 1.8step 1.9∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-24Open item page →

Order on the rationals

Definition

Every rational has a representative with positive denominator (if $b < 0$ then $(a,b) \sim (-a,-b)$ with $-b > 0$ , Every rational has a positive-denominator representative ↗). For representatives with $b > 0$ and $d > 0$ define

$[(a,b)] \le [(c,d)] \iff ad \le cb \quad \text{in } \mathbb{Z}.$

A rational $x$ is positive when $0 < x$ ; on such representatives, $0 < [(a,b)]$ exactly when $0 < a$ .

Remarks

Well-definedness, totality, and compatibility with the arithmetic: The rationals form a totally ordered field.
The integer order used on the right is Order on the integers.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25Open item page →

Every rational has a positive-denominator representative

Statement

Every rational has a representative $(a,b)$ with $b > 0$ : for a class $[(a,b)]$ (where $b \ne 0$ ), if $b > 0$ take $(a,b)$ itself, and if $b < 0$ then $(a,b) \sim (-a,-b)$ with $-b > 0$ . Consequently the order on $\mathbb{Q}$ (Order on the rationals), which is stated on positive-denominator representatives, is defined on all of $\mathbb{Q}$ .

Facts & Assumptions

Given: A rational represented by $(a,b)$ with $a, b \in \mathbb{Z}$ , $b \ne 0$ , and the relation $(x,y) \sim (z,w) \iff x w = z y$ (The rationals as equivalence classes of pairs of integers).

[L1]

Trichotomy in $\mathbb{Z}$ : each nonzero integer is either $> 0$ or $< 0$ , and $b < 0$ iff $-b > 0$ (The integers form a totally ordered ring).

[L2]

In the commutative ring $\mathbb{Z}$ , $a(-b) = -(ab) = (-a)b$ (both products are the additive inverse of $ab$ , by distributivity) (The integers form a commutative ring).

Proof

technique · direct

1.1

Since $b \ne 0$ , by trichotomy [L1] either $b > 0$ or $b < 0$ .

givenL1

2.1

If $b > 0$ , the representative $(a,b)$ already has positive denominator.

step 1.1

2.2

If $b < 0$ , then $-b > 0$ by [L1], and $a(-b) = (-a)b$ by [L2] is exactly the defining relation $(a,b) \sim (-a,-b)$ ; so $(-a,-b)$ represents the same class and has positive denominator $-b$ .

step 1.1L1L2

3.1

In either case the class has a representative with positive denominator; hence the order Order on the rationals, stated on such representatives, is defined for every rational.

step 2.1step 2.2∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

The rationals form a totally ordered field

Statement

The relation of Order on the rationals is well defined and makes the field $\mathbb{Q}$ (The rationals form a field) a totally ordered field: the order is total, $x \le y$ implies $x + z \le y + z$ , and $0 < x$ , $0 < y$ imply $0 < xy$ .

Facts & Assumptions

Given: Rationals $x = [(a,b)]$ , $y = [(c,d)]$ , $z = [(e,f)]$ with $b, d, f > 0$ .

[L1]

$\mathbb{Z}$ is a totally ordered commutative ring; positives are closed under products (The integers form a totally ordered ring).

Proof

technique · direct

1.1

Order-scaling in $\mathbb{Z}$ : for $p > 0$ , if $u < v$ then $0 < (v-u)p$ , so $up < vp$ ; conversely if $up \le vp$ and $v < u$ then $vp < up$ , impossible; hence $u \le v$ iff $up \le vp$ .

L1algebra

1.2

Suppose $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ with $b', d' > 0$ , i.e. $ab' = a'b$ and $cd' = c'd$ ; suppose also $ad \le cb$ .

given

1.3

Totality: $ad \le cb$ or $cb \le ad$ in $\mathbb{Z}$ , so $x \le y$ or $y \le x$ .

1.4

Antisymmetry: $ad \le cb$ and $cb \le ad$ give $ad = cb$ , i.e. $x = y$ as classes.

1.5

Positive products: if $0 < x$ and $0 < y$ then $0 < a$ and $0 < c$ , so $xy = [(ac,\, bd)]$ has $ac > 0$ and $bd > 0$ , hence $0 < xy$ .

1.6

For transitivity, let $z = [(e,f)]$ with $f > 0$ and suppose additionally $y \le z$ , i.e. $cf \le ed$ .

given

2.1

Scaling the hypothesis $ad \le cb$ by $b'd' > 0$ : $(ad)(b'd') \le (cb)(b'd')$ .

step 1.1step 1.2L1

2.2

Rearranging both sides with $ab' = a'b$ and $cd' = c'd$ : $(ad)(b'd') = (ab')(dd') = (a'b)(dd') = (a'd')(bd)$ and $(cb)(b'd') = (cd')(bb') = (c'd)(bb') = (c'b')(bd)$ .

step 1.2L1

2.3

Transitivity: from $ad \le cb$ and $cf \le ed$ , scaling by $f > 0$ and $b > 0$ gives $(af)d = (ad)f \le (cb)f = (cf)b \le (ed)b = (eb)d$ ; cancelling $d > 0$ via order-scaling, $af \le eb$ , i.e. $x \le z$ .

step 1.1step 1.2step 1.6L1

2.4

Compatibility with addition: $x + z \le y + z$ reads $(af+eb)(df) \le (cf+ed)(bf)$ , which expands to $(ad)f^2 + (eb)(df) \le (cb)f^2 + (ed)(bf)$ ; the second terms are equal, so this is $(ad)f^2 \le (cb)f^2$ , equivalent by order-scaling with $f^2 > 0$ to $ad \le cb$ , i.e. $x \le y$ .

step 1.1L1

3.1

Combining: $(a'd')(bd) \le (c'b')(bd)$ with $bd > 0$ , so $a'd' \le c'b'$ by order-scaling: the order is independent of representatives.

step 2.1step 2.2step 1.1L1

4.1

The order is well defined, total, compatible with addition, and positives are closed under multiplication: $\mathbb{Q}$ is a totally ordered field.

step 3.1step 1.3step 1.4step 2.3step 2.4step 1.5∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

The integers embed in the rationals

Statement

The map $j(k) = [(k,1)]$ is injective and preserves addition, multiplication, and order. Composing with The naturals embed in the integers embeds $\mathbb{N}$ in $\mathbb{Q}$ ; we write $k$ for $j(k)$ throughout.

Facts & Assumptions

Given: The map $j : \mathbb{Z} \to \mathbb{Q}$ , $j(k) = [(k,1)]$ .

[L1]

Ring arithmetic in $\mathbb{Z}$ (The integers form a commutative ring).

[L2]

The order on $\mathbb{Q}$ (Order on the rationals), whose defining inequality is read in the order on $\mathbb{Z}$ (Order on the integers).

[L3]

Equality in $\mathbb{Q}$ : $[(a,b)] = [(c,d)]$ iff $ad = cb$ (The rationals as equivalence classes of pairs of integers).

Proof

technique · direct

1.1

Injectivity: $(k,1) \sim (m,1)$ means $k \cdot 1 = m \cdot 1$ , i.e. $k = m$ .

L3L1

1.2

Addition: $j(k) + j(m) = [(k \cdot 1 + m \cdot 1,\; 1)] = j(k+m)$ .

1.3

Multiplication: $j(k)\,j(m) = [(km,\, 1)] = j(km)$ .

1.4

Order: denominators are $1 > 0$ , so $j(k) \le j(m)$ reads $k \cdot 1 \le m \cdot 1$ , i.e. $k \le m$ .

L1L2

2.1

$j$ embeds $\mathbb{Z}$ into $\mathbb{Q}$ , preserving arithmetic and order.

step 1.1step 1.2step 1.3step 1.4∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The rationals are Archimedean

Statement

For every rational $x$ there is a natural number $n$ with $x < n$ . Consequently, for every rational $\varepsilon > 0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ .

Facts & Assumptions

Given: A rational $x = [(a,b)]$ with $b > 0$ .

[L1]

The order and arithmetic of $\mathbb{Q}$ (The rationals form a totally ordered field).

[L2]

Integer facts: positive integers are exactly $\iota(m)$ with $m \ge 1$ natural; nonnegative integers are the image of $\mathbb{N}$ ; the embeddings preserve arithmetic and order (The naturals embed in the integers, The integers embed in the rationals, The integers form a totally ordered ring).

Proof

technique · direct

1.1

Since $b > 0$ , $b$ lies in the image of $\mathbb{N}$ and $b \ge 1$ .

1.2

If $a \le 0$ set $k = 0$ ; otherwise $a$ is a positive integer, so $a = \iota(k)$ for some natural $k \ge 1$ . In both cases $a \le k$ (as integers).

2.1

Then $a < k + 1$ and, since $b \ge 1$ and $k+1 > 0$ , also $k + 1 \le (k+1)b$ .

step 1.1step 1.2L2

3.1

Hence $a < (k+1)b$ , and dividing by $b > 0$ (order-scaling in the definition of the rational order), $x = [(a,b)] < [( (k+1)b,\, b)] = k+1 =: n$ .

step 2.1L1

4.1

For rational $\varepsilon > 0$ : apply the above to $x = 1/\varepsilon$ to get $n$ with $1/\varepsilon < n$ , hence $1/n < \varepsilon$ .

step 3.1L1∎

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Absolute value and the triangle inequality

Statement

For $x \in \mathbb{Q}$ set $|x| = x$ if $x \ge 0$ and $|x| = -x$ otherwise. Then $|xy| = |x|\,|y|$ , $|x + y| \le |x| + |y|$ , and $\bigl||x| - |y|\bigr| \le |x - y|$ .

The proof uses only the axioms of a totally ordered field, so it applies verbatim to any such field (in particular to $\mathbb{R}$ once constructed).

Facts & Assumptions

Given: Elements $x, y$ of the totally ordered field $\mathbb{Q}$ .

[L1]

Ordered-field arithmetic: trichotomy, compatibility of the order with addition, products of positives are positive, $(-u)v = -(uv)$ (The rationals form a totally ordered field).

Proof

technique · cases

1.1

For every $u$ : $u \le |u|$ , $-u \le |u|$ , $|u| \ge 0$ , and $|-u| = |u|$ , directly from the definition and trichotomy.

1.2

Case $x$ and $y$ have the same sign (both $\ge 0$ or both $< 0$ ): $xy \ge 0$ and $xy = |x||y|$ (for both negative, $xy = (-x)(-y)$ ).

assume-case sameL1

1.3

Case $x, y$ have mixed signs, say $x \ge 0 > y$ (the other order is identical): $xy \le 0$ , so $|xy| = -(xy) = x(-y) = |x||y|$ .

assume-case mixedL1

2.1

In all cases $|xy| = |x|\,|y|$ .

step 1.2step 1.3cases-exhaustive

2.2

Adding $x \le |x|$ and $y \le |y|$ gives $x + y \le |x| + |y|$ ; adding $-x \le |x|$ and $-y \le |y|$ gives $-(x+y) \le |x| + |y|$ ; since $|x+y|$ is one of $\pm(x+y)$ , $|x+y| \le |x| + |y|$ .

step 1.1L1

3.1

From $x = (x - y) + y$ : $|x| \le |x-y| + |y|$ , so $|x| - |y| \le |x-y|$ ; symmetrically $|y| - |x| \le |y-x| = |x-y|$ ; hence $\bigl||x|-|y|\bigr| \le |x-y|$ .

step 2.2L1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-24Open item page →

Cauchy sequence of rationals

Definition

A sequence $(a_n)_{n \ge 1}$ of rational numbers is a Cauchy sequence if for every rational $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

$|a_m - a_n| < \varepsilon \qquad \text{for all } m, n \ge N.$

Remarks

The quantifier ranges over rational $\varepsilon$ only. This is deliberate: the definition is stated before the real numbers exist, which is exactly what allows the reals to be constructed from it.
Informally: the terms of the sequence eventually cluster arbitrarily tightly, without any reference to a limit value.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-24Open item page →

Null sequence

Definition

A sequence $(a_n)_{n \ge 1}$ of rational numbers is a null sequence if for every rational $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that

$|a_n| < \varepsilon \qquad \text{for all } n \ge N.$

Remarks

Equivalently: $(a_n)$ converges to $0$ (with rational tolerances).
In the construction of $\mathbb{R}$ , two Cauchy sequences will be declared equivalent exactly when their difference is a null sequence; the null sequences are the "zero" of the construction.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Every Cauchy sequence of rationals is bounded

Statement

Every Cauchy sequence $(a_n)_{n \ge 1}$ of rational numbers (Cauchy sequence of rationals) is bounded: there exists a rational $M \ge 0$ such that $|a_n| \le M$ for all $n \ge 1$ .

Facts & Assumptions

Given: A Cauchy sequence $(a_n)_{n \ge 1}$ of rational numbers.

[A1]

For every rational $\varepsilon > 0$ there exists $N \in \mathbb{N}$ with $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$ .

[L1]

Triangle inequality on $\mathbb{Q}$ : $|x| \le |x - y| + |y|$ (Absolute value and the triangle inequality).

Proof

technique · direct

1.1

Apply [A1] with $\varepsilon = 1$ : fix $N \in \mathbb{N}$ such that $|a_m - a_n| < 1$ for all $m, n \ge N$ .

1.2

For every $n \ge N$ : $|a_n| \le |a_n - a_N| + |a_N|$ .

2.1

For every $n \ge N$ : $|a_n| < |a_N| + 1$ .

step 1.1step 1.2algebra

2.2

Define $M = \max\{|a_1|, |a_2|, \dots, |a_{N-1}|, |a_N| + 1\}$ , a maximum of finitely many rationals, hence rational and $\ge 0$ .

step 1.1choose

3.1

For every $n < N$ : $|a_n| \le M$ , since $|a_n|$ appears in the maximum.

step 2.2algebra

3.2

For every $n \ge N$ : $|a_n| < |a_N| + 1 \le M$ .

step 2.1step 2.2algebra

4.1

For every $n \ge 1$ : $|a_n| \le M$ , so $(a_n)$ is bounded.

step 3.1step 3.2∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Null sequences are Cauchy

Statement

Every null sequence (Null sequence) is a Cauchy sequence (Cauchy sequence of rationals).

Facts & Assumptions

Given: A null sequence $(a_n)$ and a rational $\varepsilon > 0$ .

[A1]

Null: for every rational $\delta > 0$ there is $N$ with $|a_n| < \delta$ for all $n \ge N$ .

[A2]

Cauchy: for every rational $\delta > 0$ there is $N$ with $|a_m - a_n| < \delta$ for all $m,n \ge N$ .

[L1]

Field arithmetic in $\mathbb{Q}$ : $\varepsilon/2 > 0$ and $\varepsilon/2 + \varepsilon/2 = \varepsilon$ (The rationals form a field, The rationals form a totally ordered field).

[L2]

Triangle inequality (Absolute value and the triangle inequality).

Proof

technique · direct

1.1

$\varepsilon/2$ is a positive rational.

2.1

Fix $N$ with $|a_n| < \varepsilon/2$ for all $n \ge N$ .

step 1.1A1

3.1

For $m, n \ge N$ : $|a_m - a_n| \le |a_m| + |a_n| < \varepsilon/2 + \varepsilon/2 = \varepsilon$ .

step 2.1L1L2

4.1

$(a_n)$ is Cauchy.

step 3.1A2∎

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Cauchy sequences form a commutative ring

Statement

The set $\mathcal{C}$ of Cauchy sequences of rationals, with termwise addition and multiplication and the constant sequences $0$ and $1$ , is a commutative ring with identity.

Facts & Assumptions

Given: Cauchy sequences $(a_n), (b_n)$ and a rational $\varepsilon > 0$ .

[A1]

Cauchy: for every rational $\delta > 0$ there is an index beyond which any two terms differ by less than $\delta$ .

[L1]

$\mathbb{Q}$ is a field; ring axioms hold termwise (The rationals form a field).

[L2]

Triangle inequality and $|uv| = |u||v|$ (Absolute value and the triangle inequality).

[L3]

Cauchy sequences are bounded (Every Cauchy sequence of rationals is bounded).

[L4]

Ordered-field arithmetic: $\varepsilon/2$ , $\varepsilon/(A+B)$ are positive rationals for $A, B \ge 1$ (The rationals form a totally ordered field).

Proof

technique · direct

1.1

Termwise, all ring axioms (associativity, commutativity, identities, inverses for $+$ , distributivity) hold in $\mathcal{C}$ because they hold in $\mathbb{Q}$ at every index; the constant sequences $0, 1$ are Cauchy. What needs proof is closure of $\mathcal{C}$ under $+$ , negation, and $\cdot$ .

1.2

Fix $N_a$ with $|a_m - a_n| < \varepsilon/2$ for $m,n \ge N_a$ and $N_b$ likewise for $(b_n)$ .

A1L4

1.3

Fix rational bounds $A, B \ge 1$ with $|a_n| \le A$ and $|b_n| \le B$ for all $n$ .

2.1

step 1.2L2

2.2

Set $\delta = \varepsilon/(A+B) > 0$ and fix $M_a, M_b$ with $|a_m - a_n| < \delta$ for $m,n \ge M_a$ and $|b_m - b_n| < \delta$ for $m,n \ge M_b$ .

step 1.3A1L4

3.1

step 1.3step 2.2L2

4.1

$\mathcal{C}$ is closed under addition, negation, and multiplication, contains $0$ and $1$ , and satisfies the ring axioms termwise: a commutative ring with identity.

step 1.1step 2.1step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

Null sequences form an ideal

Statement

The set $\mathcal{N}$ of null sequences is an ideal of the ring $\mathcal{C}$ of Cauchy sequences (Cauchy sequences form a commutative ring): it is a subgroup under addition, and $c \cdot z \in \mathcal{N}$ whenever $c \in \mathcal{C}$ and $z \in \mathcal{N}$ .

Facts & Assumptions

Given: Null sequences $(z_n), (w_n)$ , a Cauchy sequence $(c_n)$ , and a rational $\varepsilon > 0$ .

[A1]

Null: beyond some index, $|z_n|$ is smaller than any prescribed positive rational.

[L1]

Ordered-field arithmetic in $\mathbb{Q}$ (The rationals form a totally ordered field).

[L2]

Triangle inequality and multiplicativity of $|\cdot|$ (Absolute value and the triangle inequality).

[L3]

Cauchy sequences are bounded (Every Cauchy sequence of rationals is bounded).

[L4]

Null sequences are Cauchy, so $\mathcal{N} \subseteq \mathcal{C}$ (Null sequences are Cauchy).

Proof

technique · direct

1.1

Fix $N_z, N_w$ with $|z_n| < \varepsilon/2$ for $n \ge N_z$ and $|w_n| < \varepsilon/2$ for $n \ge N_w$ .

A1L1

1.2

Fix $M \ge 0$ with $|c_n| \le M$ ([L3]) and set $B = \max(M, 1) \ge 1$ , so $|c_n| \le B$ for all $n$ .

2.1

Sum: for $n \ge \max(N_z, N_w)$ , $|z_n + w_n| \le |z_n| + |w_n| < \varepsilon$ ; negation: $|-z_n| = |z_n|$ . So $\mathcal{N}$ is a subgroup under addition.

step 1.1L2

2.2

Fix $N$ with $|z_n| < \varepsilon/B$ for $n \ge N$ .

step 1.2A1L1

3.1

Product: for $n \ge N$ , $|c_n z_n| = |c_n|\,|z_n| \le B\,|z_n| < \varepsilon$ ; so $(c_n z_n)$ is null.

step 1.2step 2.2L2

4.1

$\mathcal{N}$ is a nonempty additive subgroup of $\mathcal{C}$ absorbing multiplication by $\mathcal{C}$ : an ideal.

step 2.1step 3.1L4∎

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

A non-null Cauchy sequence is eventually bounded away from zero, with constant sign

Statement

If $(a_n)$ is Cauchy but not null, there are a rational $\delta > 0$ and an index $N_0$ such that $|a_n| > \delta$ for all $n \ge N_0$ ; moreover either $a_n > \delta$ for all $n \ge N_0$ , or $a_n < -\delta$ for all $n \ge N_0$ .

Facts & Assumptions

Given: A Cauchy sequence $(a_n)$ that is not null.

[A1]

Negation of null (Null sequence): there is a rational $\varepsilon_0 > 0$ such that for every $N$ some $n \ge N$ has $|a_n| \ge \varepsilon_0$ .

[A2]

Cauchy (Cauchy sequence of rationals).

[L1]

Ordered-field arithmetic: $\varepsilon_0/3 > 0$ , $\varepsilon_0 - \varepsilon_0/3 > \varepsilon_0/3$ (The rationals form a totally ordered field).

[L2]

Triangle inequality, in the form $|u| \ge |v| - |v - u|$ (Absolute value and the triangle inequality).

Proof

technique · direct

1.1

Fix $\varepsilon_0 > 0$ witnessing that $(a_n)$ is not null.

1.2

Fix $N_0$ with $|a_m - a_n| < \varepsilon_0/3$ for all $m, n \ge N_0$ .

A2L1

2.1

Pick $n_0 \ge N_0$ with $|a_{n_0}| \ge \varepsilon_0$ .

step 1.1step 1.2

3.1

For every $n \ge N_0$ : $|a_n| \ge |a_{n_0}| - |a_{n_0} - a_n| > \varepsilon_0 - \varepsilon_0/3 > \varepsilon_0/3 =: \delta$ ; so $|a_n| > \delta$ for all $n \ge N_0$ .

step 2.1step 1.2L2L1

4.1

Sign stability: if some $a_m > \delta$ and some $a_n < -\delta$ with $m, n \ge N_0$ , then $|a_m - a_n| > 2\delta = 2\varepsilon_0/3 > \varepsilon_0/3$ , impossible by step 1.2; so beyond $N_0$ all terms have one sign, and by step 3.1 either $a_n > \delta$ for all $n \ge N_0$ or $a_n < -\delta$ for all $n \ge N_0$ .

step 3.1step 1.2L1∎

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The null ideal is maximal

Statement

If $(a_n) \in \mathcal{C} \setminus \mathcal{N}$ , then the ideal of $\mathcal{C}$ generated by $\mathcal{N}$ and $(a_n)$ is all of $\mathcal{C}$ . Hence $\mathcal{N}$ is a maximal ideal.

Facts & Assumptions

Given: A Cauchy sequence $(a_n)$ that is not null.

[L1]

Away-from-zero: there are $\delta > 0$ and $N_0$ with $|a_n| > \delta$ for all $n \ge N_0$ (A non-null Cauchy sequence is eventually bounded away from zero, with constant sign).

[L2]

Triangle inequality and $|uv| = |u||v|$ (Absolute value and the triangle inequality).

[L3]

Cauchy definition and field arithmetic in $\mathbb{Q}$ : $\varepsilon\delta^2 > 0$ for $\varepsilon > 0$ (The rationals form a totally ordered field).

[L4]

Null definition; a sequence that is $0$ from some index on is null (Null sequence).

[L5]

Ideal arithmetic in $\mathcal{C}$ : an ideal containing $1$ is the whole ring (Cauchy sequences form a commutative ring, Null sequences form an ideal).

Proof

technique · direct

1.1

Fix $\delta > 0$ and $N_0$ with $|a_n| > \delta$ for all $n \ge N_0$ ; in particular $a_n \ne 0$ there.

2.1

Define $b_n = 1$ for $n < N_0$ and $b_n = 1/a_n$ for $n \ge N_0$ .

step 1.1choose

3.1

$(b_n)$ is Cauchy: for $m, n \ge N_0$ , $|b_m - b_n| = \dfrac{|a_n - a_m|}{|a_m|\,|a_n|} < \dfrac{|a_m - a_n|}{\delta^2}$ ; given $\varepsilon > 0$ , choosing the Cauchy index of $(a_n)$ at $\varepsilon\delta^2$ makes this $< \varepsilon$ .

step 2.1step 1.1L2L3

3.2

For $n \ge N_0$ , $a_n b_n = 1$ , so the sequence $(a_n b_n - 1)$ is $0$ from $N_0$ on, hence null.

step 2.1L4

4.1

Therefore $1_{\mathcal{C}} = (a_n)(b_n) - \bigl((a_n b_n) - 1\bigr)$ lies in the ideal generated by $(a_n)$ and $\mathcal{N}$ , so that ideal is all of $\mathcal{C}$ ; any ideal strictly containing $\mathcal{N}$ contains a non-null element and thus equals $\mathcal{C}$ : $\mathcal{N}$ is maximal.

step 3.1step 3.2L5∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-07-24Open item page →

The real numbers

Definition

The real numbers are the quotient ring

$\mathbb{R} := \mathcal{C} / \mathcal{N}$

of the ring of Cauchy sequences (Cauchy sequences form a commutative ring) by the ideal of null sequences (Null sequences form an ideal). The class of $(a_n)$ is written $[(a_n)]$ . Each rational $q$ maps to the class $\hat q$ of the constant sequence $(q, q, q, \dots)$ .

Remarks

Two Cauchy sequences define the same real exactly when their difference is a null sequence.
That $\mathbb{R}$ is a field is The reals form a field ↗; its order is Order on the reals.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The reals form a field

Statement

$\mathbb{R} = \mathcal{C}/\mathcal{N}$ (The real numbers) is a field.

Facts & Assumptions

Given: Classes $[(a_n)], [(b_n)] \in \mathbb{R}$ .

[L1]

$\mathcal{N}$ is an ideal of $\mathcal{C}$ (Null sequences form an ideal).

[L2]

$\mathcal{C}$ is a commutative ring with $1$ (Cauchy sequences form a commutative ring).

[L3]

Maximality construction: for non-null $(a_n)$ there is Cauchy $(b_n)$ with $(a_n b_n) - 1$ null (The null ideal is maximal).

[L4]

The constant sequence $1$ is not a null sequence (its terms stay at $1$ ), so $\mathcal{N}$ is a proper ideal and $\hat 1 \ne \hat 0$ (Null sequence).

Proof

technique · direct

1.1

Operations on classes via representatives are well defined: for $z, w \in \mathcal{N}$ , $\bigl((a+z)+(b+w)\bigr) - (a+b) = z + w \in \mathcal{N}$ and $(a+z)(b+w) - ab = aw + zb + zw \in \mathcal{N}$ , since ideals absorb products and sums.

L1L2

2.1

The ring axioms descend to the quotient because the operations are well defined and $\mathcal{C}$ is a ring, verified on representatives; $\hat 1 \ne \hat 0$ since the constant sequence $1$ is not null.

step 1.1L1L2L4

2.2

Inverses: a nonzero class has a non-null representative $(a_n)$ ; taking $(b_n)$ from the maximality construction, $(a_n)(b_n) - 1 \in \mathcal{N}$ , so $[(b_n)]$ is a multiplicative inverse of $[(a_n)]$ .

step 1.1L3

3.1

$\mathbb{R}$ is a commutative ring with $1 \ne 0$ in which every nonzero element is invertible: a field.

step 2.1step 2.2∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-24Open item page →

Order on the reals

Definition

A real $x$ is positive when it has a representative $(a_n)$ together with a rational $\delta > 0$ and an index $N$ such that $a_n > \delta$ for all $n \ge N$ . Define

$x \le y \iff x = y \;\text{ or }\; y - x \text{ is positive},$

and $|x| = x$ if $x \ge 0$ , $|x| = -x$ otherwise.

Remarks

Independence of the representative, trichotomy, and compatibility with the field operations: The reals form a totally ordered field ↗.
The triangle inequality for this $|\cdot|$ holds because the proof of Absolute value and the triangle inequality uses only the axioms of a totally ordered field.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The reals form a totally ordered field

Statement

The relation of Order on the reals is well defined and makes $\mathbb{R}$ (The reals form a field) a totally ordered field.

Facts & Assumptions

Given: Reals $x, y$ with representatives $(a_n), (b_n)$ .

[L1]

A sequence $(u_n)_{n \ge 1}$ of rational numbers is null if, for every rational $\varepsilon > 0$ , there is $N \in \mathbb{N}$ such that $|u_n| < \varepsilon$ for every $n \ge N$ (Null sequence).

[L2]

Ordered-field arithmetic in $\mathbb{Q}$ : $\delta/2 > 0$ ; sums and products of eventual lower bounds (The rationals form a totally ordered field).

[L3]

Dichotomy for non-null Cauchy sequences: eventually $> \delta$ or eventually $< -\delta$ (A non-null Cauchy sequence is eventually bounded away from zero, with constant sign).

[L4]

$\mathbb{R}$ is a field (The reals form a field).

[L5]

In $\mathbb{R} = \mathcal{C}/\mathcal{N}$ , $x = 0$ iff a representative is null; so $x \ne 0$ iff every representative is non-null (The real numbers).

Proof

technique · direct

1.1

Positivity is independent of the representative: if $a_n > \delta$ for $n \ge N$ and $(a'_n - a_n)$ is null, then beyond some $N' \ge N$ also $|a'_n - a_n| < \delta/2$ , so $a'_n > \delta/2$ : the defining property holds for $(a'_n)$ with $\delta/2$ .

L1L2

1.2

Trichotomy: if $x \ne 0$ , any representative is non-null, so by the dichotomy either $a_n > \delta$ eventually ( $x$ positive) or $a_n < -\delta$ eventually ( $-x$ positive); the two exclude each other, and exactly one of $x$ positive, $x = 0$ , $-x$ positive holds.

L1L3L5

1.3

Positives are closed under $+$ and $\cdot$ : from $a_n > \delta$ and $b_n > \delta'$ eventually, $a_n + b_n > \delta + \delta'$ and $a_n b_n > \delta\delta'$ eventually, with $\delta + \delta', \delta\delta' > 0$ .

2.1

Consequently $\le$ is a total order (trichotomy plus transitivity from closure under sums), compatible with addition (translation preserves the difference) and with multiplication by positives: $\mathbb{R}$ is a totally ordered field.

step 1.1step 1.2step 1.3L4∎

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

The rationals embed densely in the reals

Statement

The map $q \mapsto \hat q$ (The real numbers) is an embedding of ordered fields. Every real is approximated by rationals: for $x \in \mathbb{R}$ and rational $\varepsilon > 0$ there is $q \in \mathbb{Q}$ with $|x - \hat q| < \hat\varepsilon$ . Consequently, strictly between any two reals lies a rational.

Facts & Assumptions

Given: A real $x = [(a_n)]$ and a rational $\varepsilon > 0$ .

[L1]

The orders of $\mathbb{Q}$ and $\mathbb{R}$ ; ordered-field arithmetic (The rationals form a totally ordered field, The reals form a totally ordered field).

[L2]

Field arithmetic in $\mathbb{Q}$ : $\varepsilon/2, \delta/4$ are positive rationals, and every nonzero rational $q$ has a reciprocal $1/q$ with $q \cdot (1/q) = 1$ (The rationals form a field).

[L3]

Cauchy definition (Cauchy sequence of rationals).

[L4]

Real positivity via eventual rational lower bounds (Order on the reals).

[L5]

$\mathbb{R} = \mathcal{C}/\mathcal{N}$ is a field (The reals form a field), and $0_{\mathbb{R}} = \hat 0$ , $1_{\mathbb{R}} = \hat 1$ are the classes of the constant sequences (The real numbers). A multiplicative inverse there is unique: if $ab = 1_{\mathbb{R}} = ac$ then $b = b(ac) = (ba)c = c$ .

Proof

technique · direct

1.1

Embedding: constant sequences are Cauchy; $\hat q = \hat r$ iff the constant $q - r$ is null iff $q = r$ ; operations match termwise; and $q < r$ gives the constant lower bound $r - q > 0$ , so $\hat q < \hat r$ and order is preserved and reflected.

L1L4

1.2

Fix $N$ with $|a_m - a_n| < \varepsilon/2$ for all $m, n \ge N$ , and set $q = a_N$ .

L3L2

2.1

The difference $\hat q - x$ has representative $(a_N - a_n)$ with $|a_N - a_n| < \varepsilon/2$ for $n \ge N$ ; hence both $\hat\varepsilon - (x - \hat q)$ and $\hat\varepsilon - (\hat q - x)$ have representatives eventually $> \varepsilon/2$ , so both are positive: $|x - \hat q| < \hat\varepsilon$ .

step 1.2L4L1

2.2

Inverses: let $q$ be a nonzero rational. Then $\hat q \ne \hat 0 = 0_{\mathbb{R}}$ by the injectivity of step 1.1, and $1/q$ exists in $\mathbb{Q}$ by [L2]; since the operations match termwise (step 1.1), $\hat q \cdot \widehat{1/q} = \widehat{q \cdot (1/q)} = \hat 1 = 1_{\mathbb{R}}$ . Inverses in $\mathbb{R}$ are unique by [L5], so $(\hat q)^{-1} = \widehat{1/q}$ : the embedding preserves reciprocals.

step 1.1L2L5

3.1

Density: let $x < y$ ; pick $\delta > 0$ rational and $N$ with the representative of $y - x$ eventually $> \delta$ ; set $\varepsilon = \delta/4$ and pick $q$ with $|x - \hat q| < \hat\varepsilon$ ; then $q' = q + 2\varepsilon$ satisfies $\hat q' - x \ge -\hat\varepsilon + 2\hat\varepsilon = \hat\varepsilon > 0$ and $y - \hat q' \ge \hat\delta - \hat\varepsilon - 2\hat\varepsilon = \hat\delta/4 > 0$ , so $x < \hat q' < y$ .

step 2.1L4L1L2

4.1

The rationals embed as an ordered subfield — injectively, preserving the order in both directions, the ring operations, and reciprocals — and they approximate every real arbitrarily well and separate any two reals.

step 1.1step 2.2step 3.1∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-24Open item page →

Limits and Cauchy sequences of reals

Definition

A sequence $(x_k)$ of reals converges to $x \in \mathbb{R}$ when for every rational $\varepsilon > 0$ there is $K$ with $|x_k - x| < \hat\varepsilon$ for all $k \ge K$ . It is Cauchy when for every rational $\varepsilon > 0$ there is $K$ with $|x_k - x_l| < \hat\varepsilon$ for all $k, l \ge K$ .

Remarks

Quantifying over rational $\varepsilon$ loses nothing: below any real $\varepsilon > 0$ lies a positive rational (The rationals embed densely in the reals).
$|\cdot|$ is the absolute value of Order on the reals.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The reals are complete

Statement

Every Cauchy sequence of real numbers (Limits and Cauchy sequences of reals) converges to a real number. Together with The reals form a totally ordered field, this completes the construction: $\mathbb{R}$ is a complete totally ordered field.

Facts & Assumptions

Given: A Cauchy sequence $(x_k)_{k \ge 1}$ of reals.

[L1]

Rational approximation: for any real $z$ and rational $\eta > 0$ there is $q$ with $|z - \hat q| < \hat\eta$ (The rationals embed densely in the reals).

[L2]

Archimedean property: for rational $\varepsilon > 0$ there is $k$ with $1/k < \varepsilon$ (The rationals are Archimedean).

[L3]

Cauchy definitions in $\mathbb{Q}$ and $\mathbb{R}$ (Cauchy sequence of rationals, Limits and Cauchy sequences of reals).

[L4]

The embedding preserves and reflects order and arithmetic; triangle inequality in $\mathbb{R}$ (The rationals embed densely in the reals, The reals form a totally ordered field, Order on the reals).

[L5]

Reals are classes of rational Cauchy sequences (The real numbers).

Proof

technique · direct

1.1

For each $k \ge 1$ pick a rational $q_k$ with $|x_k - \hat q_k| < \widehat{1/k}$ .

L1choose

2.1

$(q_k)$ is Cauchy in $\mathbb{Q}$ : given rational $\varepsilon > 0$ , pick $k_0$ with $1/k_0 < \varepsilon/3$ and $K$ with $|x_k - x_l| < \widehat{\varepsilon/3}$ for $k, l \ge K$ ; then for $k, l \ge \max(k_0, K)$ , $\widehat{|q_k - q_l|} \le |\hat q_k - x_k| + |x_k - x_l| + |x_l - \hat q_l| < \widehat{1/k} + \widehat{\varepsilon/3} + \widehat{1/l} \le 3\,\widehat{\varepsilon/3} = \hat\varepsilon$ , and the embedding reflects order, so $|q_k - q_l| < \varepsilon$ .

step 1.1L2L3L4

3.1

Set $x := [(q_k)] \in \mathbb{R}$ , the class of this rational Cauchy sequence.

step 2.1L5

4.1

$x_k \to x$ : given rational $\varepsilon > 0$ , pick $k_1$ with $1/k_1 < \varepsilon/3$ and $K_2$ with $|q_k - q_l| < \varepsilon/3$ for $k, l \ge K_2$ ; for $k \ge \max(k_1, K_2)$ , the difference $\hat q_k - x$ has representative $(q_k - q_l)_l$ , whose absolute values $|q_k - q_l|$ are eventually below $\varepsilon/3$ , so $|\hat q_k - x| \le \widehat{\varepsilon/3}$ , and $|x_k - x| \le |x_k - \hat q_k| + |\hat q_k - x| < \widehat{1/k} + \widehat{\varepsilon/3} \le 2\,\widehat{\varepsilon/3} < \hat\varepsilon$ .

step 1.1step 2.1step 3.1L4

5.1

Every Cauchy sequence of reals converges in $\mathbb{R}$ : the reals are complete.

step 4.1∎

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

The sequence $1/n$ is null

Example

The sequence $(1/n)_{n \ge 1}$ of rationals is a null sequence (Null sequence): its terms become smaller than any prescribed positive rational. It is the standard witness that "eventually arbitrarily small" is a nonvacuous notion, and the rate $1/k$ used in the completeness proof.

Facts & Assumptions

Given: A rational $\varepsilon > 0$ .

[L1]

Archimedean property (The rationals are Archimedean).

[L2]

Order arithmetic: $0 < 1/n \le 1/N$ for $n \ge N \ge 1$ (The rationals form a totally ordered field).

[L3]

Null definition (Null sequence).

Verification

technique · direct

1.1

Pick a natural $N \ge 1$ with $1/N < \varepsilon$ .

2.1

For every $n \ge N$ : $0 < 1/n \le 1/N < \varepsilon$ , so $|1/n| < \varepsilon$ .

step 1.1L2

3.1

$(1/n)$ is null.

step 2.1L3∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

FALSE: some rational number squares to 2

Statement

False claim: there exists $x \in \mathbb{Q}$ with $x^2 = 2$ .

Facts & Assumptions

Given: The rationals as quotients of integer pairs (The rationals as equivalence classes of pairs of integers).

[A1]

Every rational can be written $a/b$ in lowest terms, with $a, b$ not both even (cancel common factors of $2$ ; the process terminates by well-ordering of $\mathbb{N}$ ).

[A2]

Parity in $\mathbb{Z}$ : if $k^2$ is even then $k$ is even (an odd $k = 2j+1$ has odd square $4j^2 + 4j + 1$ ).

Refutation

technique · contradiction

1.1

Assume, for contradiction, that some $x \in \mathbb{Q}$ has $x^2 = 2$ ; write $x = a/b$ in lowest terms, $a, b$ not both even.

assume-contraA1

2.1

Clearing denominators: $a^2 = 2b^2$ .

step 1.1algebra

3.1

$a^2$ is even, hence $a$ is even: $a = 2c$ .

step 2.1A2

4.1

Substituting: $4c^2 = 2b^2$ , so $b^2 = 2c^2$ is even, hence $b$ is even.

step 3.1A2algebra

5.1

Both $a$ and $b$ are even, contradicting the lowest-terms choice of step 1.1; the assumption fails: no rational squares to $2$ .

step 4.1step 1.1discharge-contradiction∎

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24Open item page →

FALSE: the rationals are complete

Statement

False claim: every Cauchy sequence of rationals converges to a rational (where $(a_n) \to q$ means $(a_n - q)$ is null).

This is precisely the defect the construction of $\mathbb{R}$ repairs.

Facts & Assumptions

Given: The decimal truncations of $\sqrt 2$ , built below.

[A1]

In $\mathbb{N}$ : for each $n$ there is a largest natural $k$ with $k^2 \le 2 \cdot 10^{2n}$ (only finitely many candidates, since $k > 2 \cdot 10^n$ already fails); and $n < 10^n$ .

[L1]

Ordered-field arithmetic in $\mathbb{Q}$ (The rationals form a totally ordered field, Absolute value and the triangle inequality).

[L2]

No rational squares to $2$ (FALSE: some rational number squares to 2).

[L3]

A constant sequence is null only if the constant is $0$ ; sums of null sequences are null; a Cauchy multiple of a null sequence is null (Null sequence, Null sequences form an ideal).

[L4]

Archimedean property, so $10^{-n}$ falls below any positive rational (The rationals are Archimedean).

Refutation

technique · direct

1.1

For each $n$ let $k_n$ be the largest natural with $k_n^2 \le 2 \cdot 10^{2n}$ , and set $s_n = k_n / 10^n$ ; then $s_n^2 \le 2 < (s_n + 10^{-n})^2$ , and $s_n \le 2$ (since $k_n^2 \le 2\cdot 10^{2n} < (2 \cdot 10^n)^2$ ).

A1L1

2.1

$(s_n)$ is Cauchy: for $m \ge n$ , $(10^{m-n} k_n)^2 \le 2 \cdot 10^{2m}$ gives $10^{m-n} k_n \le k_m$ , so $s_n \le s_m$ ; and $s_m^2 \le 2 < (s_n + 10^{-n})^2$ forces $s_m < s_n + 10^{-n}$ ; hence $0 \le s_m - s_n < 10^{-n}$ , and $10^{-n}$ eventually falls below any $\varepsilon > 0$ .

step 1.1L1L4

2.2

$(s_n^2 - 2)$ is null: $0 \le 2 - s_n^2 < (s_n + 10^{-n})^2 - s_n^2 = 10^{-n}(2 s_n + 10^{-n}) \le 5 \cdot 10^{-n}$ .

step 1.1L1L4

3.1

If $(s_n)$ converged to a rational $q$ , then $(s_n - q)$ would be null; since $(s_n + q)$ is Cauchy, $(s_n^2 - q^2) = (s_n - q)(s_n + q)$ would be null; adding the null $(2 - s_n^2)$ , the constant $2 - q^2$ would be null, forcing $q^2 = 2$ .

step 2.1step 2.2L3L1

4.1

No rational squares to $2$ , so $(s_n)$ is a Cauchy sequence of rationals with no rational limit: the claim is refuted.

step 3.1step 2.1L2∎

Sources

Standard references

Recommended treatments; not extraction sources.