CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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The common divisors of $(0,0)$ are all of $\mathbb{Z}$ and have no greatest element in the order of $\mathbb{Z}$ , so $\gcd(0,0)$ cannot be defined as a maximum and is fixed by convention

Statement refuted

Refuted claim: for every pair of integers $(a,b)$ the set

$D(a,b) \;=\; \{\, d \in \mathbb{Z} \;:\; d \mid a \ \text{ and } \ d \mid b \,\}$

of common divisors has a greatest element, so that $\gcd(a,b)$ can be defined as that maximum at every pair (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ , Common divisor, and the greatest common divisor $\gcd(a,b)$ , with the convention $\gcd(0,0) := 0$ ).

Witness: $(a,b) = (0,0)$ . Every integer divides $0$ , so $D(0,0) = \mathbb{Z}$ ; and $\mathbb{Z}$ has no greatest element, since $m < m+1$ for every $m$ . So there is no maximum to take, and $\gcd(0,0)$ is fixed by the convention $\gcd(0,0) := 0$ of Common divisor, and the greatest common divisor $\gcd(a,b)$ , with the convention $\gcd(0,0) := 0$ rather than computed.

This does not contradict A nonempty set of integers bounded above has a greatest element, and a nonempty set of integers bounded below has a least element: that lemma requires the set to be bounded above, and $\mathbb{Z}$ is not.

Facts & Assumptions

Given: The set $D(0,0)$ of common divisors of $0$ and $0$ .

[L1]

$\mathbb{Z}$ is a commutative ring: $x \cdot 0 = 0$ , $x + 0 = x$ , $x \cdot 1 = x$ , $x \cdot (-1) = -x$ , and every $x$ has an additive inverse (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L2]

The order on $\mathbb{Z}$ is total, antisymmetric and transitive and is compatible with addition; positives are closed under multiplication; $x < y$ means $x \le y$ together with $x \ne y$ (The integers form a totally ordered ring, Order on the integers).

[L3]

$d \mid u$ means $u = dq$ for some $q$ ; in particular $d \mid 0$ for every $d$ , since $0 = d \cdot 0$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L4]

A nonempty set of integers bounded above has a greatest element (A nonempty set of integers bounded above has a greatest element, and a nonempty set of integers bounded below has a least element).

[L5]

$\iota : \mathbb{N} \to \mathbb{Z}$ is injective with image the nonnegative integers, and $\iota(1) = 1$ , $\iota(0) = 0$ (The naturals embed in the integers).

[L6]

$d = \gcd(u,v)$ exactly when $d \ge 0$ , $d \mid u$ , $d \mid v$ , and every common divisor of $u$ and $v$ divides $d$ (Every common divisor of $a$ and $b$ divides $\gcd(a,b)$ ; consequently $d = \gcd(a,b)$ exactly when $d \ge 0$ , $d \mid a$ , $d \mid b$ , and every common divisor of $a$ and $b$ divides $d$ — a characterisation that holds at $(a,b) = (0,0)$ as well).

Counterexample

technique · direct

1.1

$D(0,0) = \mathbb{Z}$ : every integer $d$ satisfies $d \mid 0$ , so every integer is a common divisor of $0$ and $0$ .

1.2

$0 < 1$ : $1 = \iota(1)$ lies in the image of $\iota$ , hence $0 \le 1$ ; and $1 \ne 0$ because $\iota$ is injective and $1 \ne 0$ in $\mathbb{N}$ .

2.1

For every $m \in \mathbb{Z}$ , $m < m+1$ : adding $m$ to $0 < 1$ gives $m \le m+1$ , and $m = m+1$ would give $0 = 1$ after adding $-m$ , contrary to step 1.2.

step 1.2L1L2

3.1

$\mathbb{Z}$ has no greatest element: if $m$ were one, then $m + 1 \in \mathbb{Z}$ would give $m+1 \le m$ , while $m < m+1$ by step 2.1, contradicting antisymmetry.

step 2.1L2

3.2

[L4] is not contradicted, since its hypothesis fails: $\mathbb{Z}$ is not bounded above, because for any candidate bound $b$ the integer $b+1$ exceeds it by step 2.1.

step 2.1L4

4.1

By steps 1.1 and 3.1 the set $D(0,0)$ has no greatest element, so the refuted claim fails at $(0,0)$ and no maximum defines $\gcd(0,0)$ .

step 1.1step 3.1

5.1

What survives at $(0,0)$ is the divisibility characterisation [L6]: $0 \ge 0$ , $0 \mid 0$ , and every common divisor of $(0,0)$ divides $0$ by [L3], so $0$ is the value that characterisation returns — which is exactly the convention adopted in Common divisor, and the greatest common divisor $\gcd(a,b)$ , with the convention $\gcd(0,0) := 0$ .

step 4.1L3L6∎

Remarks

The failure is only at $(0,0)$ . For every other pair one of the two arguments is nonzero, and then the common divisors are bounded above by its absolute value (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$ ; hence the set of divisors of a nonzero integer is bounded above by $|a|$ ), so A nonempty set of integers bounded above has a greatest element, and a nonempty set of integers bounded below has a least element applies and the maximum exists.
The convention is not a patch over an inconvenience but over an absence. There is no integer that could serve as "the greatest common divisor of $0$ and $0$ " in the order of $\mathbb{Z}$ , so a value has to be supplied; that it is $0$ is forced by the identities $\gcd$ is required to satisfy ( $\gcd$ at the boundary: $\gcd(a,0) = |a|$ , $\gcd(0,0) = 0$ , and the convention $\gcd(0,0) = 0$ is exactly what makes $\gcd(ca,cb) = |c|\gcd(a,b)$ true at $c = 0$ ).

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Greatest common divisor (Wikipedia) (standard reference, not scraped)
Divisor (Wikipedia) (standard reference, not scraped)

The common divisors of (0,0)(0,0)(0,0) are all of Z\mathbb{Z}Z and have no greatest element in the order of Z\mathbb{Z}Z, so gcd⁡(0,0)\gcd(0,0)gcd(0,0) cannot be defined as a maximum and is fixed by convention