Question 9176: Least Divisor at Least Two and Primality

Original question and givens

Use natural-number primality: 0 and 1 are not prime. Trial division must cover all candidates through the square-root bound; an incomplete sample is not a primality certificate. n=0. Give (n, least divisor d≥2 or not-applicable when n<2, prime?). For a prime input the least such divisor is n itself.

• G001 — Use natural-number primality: 0 and 1 are not prime. Trial division must cover all candidates through the square-root bound; an incomplete sample is not a primality certificate. n=0. Give (n, least divisor d≥2 or not-applicable when n<2, prime?). For a prime input the least such divisor is n itself.

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

• mko-prime-number

Step-by-step derivation

natural-number prime protocol;0 and1 are not prime

a failed candidate witness alone is not a primality proof

phase=given n;n=0

n>=2=No

Reverse checks

• S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

(0, not-applicable, No)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.