The Infinite Factorization Glitch
Meditations on Oneness
Local folio mark · 11
Page 1 · Find the missing stop
Rook's unfinished rule card
Rook is testing an old Archive instruction. It begins with this draft definition:
A positive whole number is a prime candidate if it cannot be written as a product of two smaller positive whole numbers.
Under this draft definition, is a prime candidate: there are no smaller positive whole numbers available to multiply.
Whenever a factor label for works, Rook may stamp a longer label if he can add another prime candidate without changing the product. The instruction contains no stopping rule.
Stretch the label
Write factors in the diagram. Use scratch paper for your explanation.
-
Under the draft definition, write as a product of exactly 3 prime candidates and as a product of exactly 11 prime candidates. Then explain how to write as a product of exactly prime candidates for every whole number , and explain why this proves that Rook never reaches a last valid label.
Page 2 · Repair the rule
Separate units from primes
Now widen the number system: every definition below is about the integers, so negative factors are allowed.
Unit
An integer is a unit if some integer satisfies .
Associates
Integers and are associates if for some unit .
Irreducible
A nonzero nonunit integer is irreducible if forces or to be a unit.
Granted for this problem: are irreducible.
Repair the comparison
Use separate paper for your proofs and explanation.
-
Work now in the integers. Prove that the only units are and . Then let be any positive prime. Prove that and are associates.
-
Compare the two irreducible factorizations and . Explain why changing order alone is not enough and how associates repair the comparison. Then read the Integer Factorization Theorem below. Explain why your examples motivate its qualifiers but do not prove either existence or uniqueness.
Integer Factorization TheoremEvery nonzero integer that is not a unit is a finite product of irreducible integers. If two such products have the same value, then they have the same number of factors and, after reordering, corresponding factors are associates.