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If $1$ were admitted as a prime, uniqueness would fail: $6 = 2 \cdot 3 = 1 \cdot 2 \cdot 3 = 1 \cdot 1 \cdot 2 \cdot 3$ , lists of different lengths that no permutation matches

Statement refuted

Refuted claim. The clause $p > 1$ in Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ is an arbitrary convention: replacing it by $p \geq 1$ would leave The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ true as stated.

Write prime $^{*}$ for the modified notion — $p \geq 1$ and the only positive divisors of $p$ are $1$ and $p$ — so that $1$ is prime $^{*}$ and every prime is prime $^{*}$ . The claim under refutation is that for lists of prime $^{*}$ s, $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ still forces $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

Witness. The lists $(2, 3)$ , $(1, 2, 3)$ and $(1, 1, 2, 3)$ , of lengths $2$ , $3$ and $4$ , all consist of prime $^{*}$ s and all have product $6$ . So $r = s$ fails already between the first two, and no permutation can exist because the index sets have different sizes.

Numerals. For $k \in N$ the symbol $k$ inside $Z$ means $ι (k)$ , the embedding of The naturals embed in the integers.

Facts & Assumptions

Given: The integers $1$ , $2$ , $3$ and $6$ , and the three lists above.

[L1]

$p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L2]

$\prod_{i < 0} g_{i} = e$ and $\prod_{i < σ (m)} g_{i} = (\prod_{i < m} g_{i}) g_{m}$ , in the commutative monoid $(Z, \cdot, 1)$ (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ , Semigroup and monoid).

[L3]

$u ∣ 1$ exactly when $u = 1$ or $u = - 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ); $d ∣ a$ means $a = d c$ for some $c$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L4]

If $d ∣ a$ and $a \neq 0$ then $∣ d ∣ \leq ∣ a ∣$ (If $d ∣ a$ and $a \neq 0$ then $d \neq 0$ and $∣ d ∣ \leq ∣ a ∣$ ; hence the set of divisors of a nonzero integer is bounded above by $∣ a ∣$ ).

[L5]

For $a \in Z$ and $b > 0$ there is exactly one pair $(q, r)$ with $a = q b + r$ and $0 \leq r < b$ , and $b ∣ a$ exactly when $r = 0$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L6]

If $n \approx m$ with $n, m \in N$ then $n = m$ ; equivalently there is no bijection between two distinct natural numbers (The pigeonhole principle on $N$ , Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Injection, surjection, bijection).

[L7]

A permutation of the von Neumann natural $r$ is a bijection $r \to r$ (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition); $n = {m : m < n}$ (On $N$ the order is membership: $m < n ⟺ m \in n$ , The natural numbers $N$ (von Neumann)).

[L8]

$Z$ is a commutative ring with $x \cdot 1 = x$ ; its order is total, antisymmetric and transitive (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals, The integers form a totally ordered ring, Order on the integers); $ι$ is injective and order preserving with image the nonnegative integers (The naturals embed in the integers, Discreteness: $σ (n)$ is the immediate successor, Order on the natural numbers).

[L9]

Every $n \geq 1$ is a finite product of primes (Every integer $n \geq 1$ is a finite product of primes: there are $r \in N$ and a list $p : r \to Z$ of primes with $n = \prod_{i < r} p_{i}$ , the case $n = 1$ being the empty product).

Counterexample

technique · direct

1.1

$0 < 1 < 2 < 3$ , and every integer $y > 0$ satisfies $y \geq 1$ : $y = ι (t)$ with $t \neq 0$ , so $1 \leq y$ since $ι$ preserves the order.

1.2

All three products are $6$ . By [L2], $\prod_{i < 2} (2, 3)_{i} = (1 \cdot 2) \cdot 3 = 6$ ; $\prod_{i < 3} (1, 2, 3)_{i} = ((1 \cdot 1) \cdot 2) \cdot 3 = 6$ ; and $\prod_{i < 4} (1, 1, 2, 3)_{i} = (((1 \cdot 1) \cdot 1) \cdot 2) \cdot 3 = 6$ .

L2L8algebra

2.1

$1$ is prime $^{*}$ : $1 \geq 1$ , and a positive divisor $d$ of $1$ satisfies $d ∣ 1$ , hence $d = 1$ or $d = - 1$ by [L3], and positivity leaves $d = 1$ , which is $1$ itself.

step 1.1L3L8

2.2

$2$ and $3$ are prime, hence prime $^{*}$ . Each exceeds $1$ ; a positive divisor $d$ of such an $a$ satisfies $1 \leq d \leq a$ by [L4] and step 1.1, and the intermediate candidate for $a = 3$ is settled by $3 = 1 \cdot 2 + 1$ , whose remainder is nonzero. For $a = 2$ there is no intermediate candidate.

step 1.1L1L4L5algebra

3.1

The three lists consist of prime $^{*}$ s, by steps 2.1 and 2.2, and their lengths are $2$ , $3$ and $4$ , which are pairwise distinct natural numbers.

step 2.1step 2.2L8

4.1

So the uniqueness clause fails at its very first assertion: taking $p = (2, 3)$ and $q = (1, 2, 3)$ gives $\prod_{i < 2} p_{i} = \prod_{j < 3} q_{j}$ with $2 \neq 3$ , so " $r = s$ " is false.

step 1.2step 3.1

5.1

Nor could the clause be rescued by dropping " $r = s$ " and asking only for a bijection: a permutation in the sense of [L7] is a bijection between index sets, and by [L6] no bijection exists between the distinct naturals $2$ and $3$ .

step 3.1step 4.1L6L7

6.1

Existence, by contrast, survives the change: $6 = 2 \cdot 3$ is still a factorisation into prime $^{*}$ s, and every $n \geq 1$ still has one by [L9], since every prime is prime $^{*}$ . So it is precisely the uniqueness half that forces the convention $p > 1$ .

step 2.2step 1.2step 5.1L9∎

Remarks

The failure is not a technicality about lists. Under the modified definition an integer would have infinitely many factorisations, one for each number of padding factors $1$ , so no formulation of uniqueness survives: neither "the same length", nor "the same multiset of factors", nor "the same exponent vector", since $1$ would carry an arbitrary exponent.
Existence is what makes the convention a genuine choice rather than a necessity. Both notions give factorisations of every $n \geq 1$ , and the difference shows up only when one asks whether the factorisation is unique. That is exactly the asymmetry Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ appeals to when it excludes $1$ .
$0$ is excluded for a different reason. It is not that $0$ would break uniqueness; it is that $0$ has every positive integer as a divisor, so it fails the divisor clause outright, and a product containing the factor $0$ is $0$ rather than the integer being factored.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

71 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

Fundamental theorem of arithmetic (Wikipedia) (standard reference, not scraped)
Prime number (Wikipedia) (standard reference, not scraped)
Janssen and Lindsey, Rings with Inquiry: Primes and Factorization (standard reference, not scraped)

If 11 were admitted as a prime, uniqueness would fail: 6=2⋅3=1⋅2⋅3=1⋅1⋅2⋅36=2⋅3=1⋅2⋅3=1⋅1⋅2⋅3, lists of different lengths that no permutation matches