CMPSCI 250 Discussion #4: Infinitely Many Primes

David Mix Barrington

27 February/1 March 2006

Notation: S is a finite set of naturals, all other variables denote naturals, P(x) means "x is prime", P3(x) means "x is prime and x is congruent to 3 (mod 4)", D(u,v) means "u divides v".
  1. I: (proved on the board) ∀a:∃b:(a<b)∧P(b)
  2. II: ∀S[∀x:(x∈S)→P(x)]→ [∃y:P(y)∧(y∉S)]
  3. III: ∀S[∀x:(x∈S)→P3(x)]→ [∃y:P3(y)∧(y∉S)]
    • Let S be an arbitrary finite set.
    • Assume that S consists entirely of P3 primes.
    • Let d be the product of all the elements of S.
    • First assume that there are an even number of elements of S.
      • By arithmetic, d ≡ 1 (mod 4).
      • Let z be d + 2.
      • So z ≡ 3 (mod 4)
      • We know (as informally argued in lecture) that z is the product of primes.
      • If any of these primes were 2, we could not have z ≡ 3 (mod 4).
      • If all the primes dividing z were P1 primes, we would have z ≡ 1 (mod 4).
      • So there exists a P3 prime dividing z -- call it p.
      • As above in II, p cannot be an element of S since none of those elements divide z. (If x is in S, z ≡ 2 (mod x) and since x is a P3 prime, ¬D(x,z).)
      • So we have ∃y:P3(y) ∧ (y∉S) in this case.
    • Now, for the other case, assume that there are an odd number of elements in S.
      • Now d ≡ 3 (mod 4).
      • Let z = d + 4, so z ≡ 3 (mod 4).
      • Just as in the other case, we prove that there exists p such that P3(p) and (p∉S). (This time, we have z ≡ 4 (mod x) for any x in S.)
      • So in this case as well we have ∃y:P3(y) ∧ (y∉S).
    • By Generalization, ∀S:∃y:P3(y) ∧ (y∉S).

    Last modified 1 March 2006