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Lemma

Every number contains an atom

Every whole number bigger than 1 has at least one prime hiding inside it as a factor.
Formally: Every integer > 1 has a prime divisornumber theoryModeratedepth 3 in the graph

Take any number > 1 and keep splitting it into smaller factors; the process must stop (numbers can't shrink forever — that's well-ordering), and where it stops is a prime. So no number escapes the primes: each one is either prime itself or divisible by one.

Keep cutting a molecule into smaller pieces and you must eventually hit an atom.

Take 60 → 6 × 10 → the smallest divisor > 1 of 60 is 2, and 2 is prime. Take 49 → smallest divisor > 1 is 7 — prime. It never fails.

This is the entry point of both landmark theorems here: Euclid's proof needs it to find a NEW prime, and the Fundamental Theorem uses it to start every factorization.

Level 1 The precise statement

Every integer n > 1 has at least one prime divisor.

Level 2 The proof
Consider the set of divisors of n that are greater than 1; it is nonempty (it contains n). By well-ordering it has a least element p. If p were composite it would have a divisor 1 < q < p also dividing n, contradicting minimality. Hence p is prime.
Level 3 What it stands on (3 direct)
Level 4 The verified record

This page is generated from a machine-checked node. The kernel confirms its dependencies resolve, nothing is circular, and it grounds in axioms (foundation: peano). The content hash below makes tampering evident.

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