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PROOF 11

Prove.

There are infinitely many primes

On the naturals ℕ.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

b | a means that a natural number k exists satisfying a = bk.

Proof.

This renders 'b is a divisor of a' as a single multiplication.convention

∎

Definition D2

p is prime means that p > 1 and the only divisors of p are 1 and p.

Proof.

There are two conditions: greater than 1, and no divisors besides those two.convention

∎

Lemma L1

If p | a and a | m, then p | m

Proof.

(1)There are k, l with a = pk and m = al.assumption, D1
(2)m = (pk)l(1), E
(3)  = p(kl)A4
(4)Therefore p | m.(3), D1

∎

Lemma L2

Every natural number greater than 1 has a prime divisor.

Proof.

(1)Let S be the set of all natural numbers greater than 1 with no prime factor.convention
(2)Suppose S ≠ ∅.assumption
(3)S has a least element m.A2
(4)Since m = m × 1, m | m.A6, D1
(5)If m were prime, then by (4) m would be a prime factor of m, so m ∉ S.D2, (4), (1)
(6)Therefore m is not prime, and has a divisor a that is neither 1 nor m.D2, (5)
(7)There is a k with m = ak, and since a ≠ m, k ≠ 1.(6), D1, A6
(8)Since 1 ≤ k, we have 1 < k, and a = a × 1 ≤ ak = m.(7), A1, A6, A7
(9)a is neither 1 nor m, so 1 < a < m.(6), (8), A1
(10)Since m is the least element of S, a ∉ S. That is, a has a prime factor p.(3), (9), (1)
(11)Since p | a and a | m, p | m.(10), L1
(12)m has a prime factor, so m ∉ S. This contradicts (3).(11), (1)
(13)Therefore S = ∅.(12)

∎

Theorem T

There are infinitely many primes.

Proof.

(1)Suppose the primes are finite in number, and let them all be p₁, p₂, …, pₙ.assumption
(2)Set P = p₁p₂⋯pₙ and N = P + 1.convention
(3)For each i, P can be grouped as pᵢ times the product of the remaining primes.A4, A5
(4)Therefore there is a t with P = pᵢ t, and pᵢ | P.(3), D1
(5)N = pᵢ t + 1, and since pᵢ > 1, 0 ≤ 1 < pᵢ.(2), (4), D2
(6)Quotient and remainder are unique, so the remainder of N divided by pᵢ is 1.A3, (5)
(7)If pᵢ | N, then N = pᵢ s + 0, so the remainder is 0.D1, A3
(8)Since 1 ≠ 0, no pᵢ divides N.(6), (7)
(9)Since 1 ≤ P, N = P + 1 > 1.A1, (2)
(10)Therefore N has a prime factor q.(9), L2
(11)Since q | N, by (8) q is not equal to any pᵢ.(10), (8)
(12)q is a prime not on the list in (1). This contradicts (1).(11), (1)
(13)Therefore the primes are not finite in number.(12)

∎

Remark

Can A2 be removed from this proof?

Well-ordering gives the existence of a least element without counting. Remove it and see which lines lose their justification.