CMPSCI 250 Discussion #4: Infinitely Many Primes

David Mix Barrington

27 February/1 March 2006

1. I: (proved on the board) ∀a:∃b:(a<b)∧P(b)
   • Let a be an arbitrary natural.
   • Let z be a! + 1 (where "!" is the factorial function)
   • Claim 1: ∀q:(2≤q≤a)→¬D(q,z)
     • Proof of Claim 1: Let q be arbitrary and assume 2≤q≤a.
     • Since q is one of the numbers multiplied to make a!, a! is congruent to 1 (mod q) and thus a!+1 is congruent to 1 (mod q)
     • Since q≥2, we know D(q,z) is false.
   • Claim 2: ∀y:(y>1)→[∃p:P(p)∧D(p,z)]
     • We have argued this informally in lecture. Note p=z is possible.
   • Specialize Claim 2 to z and let p be the resulting natural. So P(p) and D(p,z) are true. Note z>1 because a!≥1.
   • Specialize Claim 1 to p. So (2≤p≤a)→¬D(p,z).
   • Since D(p,z) is true, by Modus Tollens, ¬(2≤p≤a).
   • Since p<2 is ruled out by p being prime, we have that p>a.
   • By Existence, ∃b:(a<b)∧P(b).
   • By Generalization, we are done.
2. II: ∀S[∀x:(x∈S)→P(x)]→ [∃y:P(y)∧(y∉S)]
   • Let S be an arbitrary finite set.
   • Assume that all elements of S are primes.
   • Let d be the product of all the elements of S.
   • Note that for every x in S, d ≡ 0 (mod x) because x is one of the numbers multiplied together to get d.
   • Let z = d+1.
   • By Claim 2 above, let p be a natural such that P(p) and D(p,z).
   • For any x in S, z ≡ 1 (mod x).
   • Thus p cannot be any of the numbers in S.
   • By Existence, ∃y:P(y) ∧ (y∉S).
   • By Generalization, ∀S:∃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