The natural numbers are 0, 1, 2 and so on; for further details, see Peano.
The positive integers are natural numbers ≥ 1, i.e. 1, 2, 3 and so on.
The multiples are natural numbers ≥ 2, i.e. 2, 3, 4 and so on. The product of two multiples is larger than each of those two, in formula
(A x, y : x ≥ 2, y ≥ 2 : x.y > x ∧ x.y > y ) | (1) |
Multiples that are the product of two multiples are called composite, multiples that are not composite are called prime, in formula
prime m = m ≥ 2 ∧ (A x, y : x ≥ 2, y ≥ 2 : x.y ≠ m) | (2) | |
comp m = m ≥ 2 ∧ (E x, y : x ≥ 2, y ≥ 2 : x.y = m) | (3) |
comp m = (E x, y : 2 ≤ x < m, 2 ≤ y < m : x.y = m) | (3′) |
From (3′) it is clear that it is decidable whether a specific m is composite.
Note that we have ¬(prime 1) ∧ ¬ (comp 1) .
With PF —for Prime Factorization— defined by
PF n = | n is the product of (0 or more) multiples, all of which are prime (the empty product being defined as 1) |
Theorem 0. (A n : n ≥ 1 : PF n)
Proof. We shall prove Theorem 0 by mathematical induction, i.e. we shall prove
(A n : n ≥ 1 : (PF n) ∨ (E i : 1 ≤ i ≤ n : ¬(PF i))) | (4) |
(A n : n ≥ 1 : (comp n) ∨ (PF n)) | (5) |
(A x, y : x ≥ 1, y ≥ 1 : ¬(PF x) ∨ ¬(PF y ) ∨ PF(x.y ))
from which we deduce with (3')
(A n : n ≥ 1 : ¬(comp n) ∨ (PF n) ∨ (E i : 1 ≤ i ≤ n: ¬ (PF i ))) | (6) |
Theorem 1. (A p : prime p : (E q : q > p : prime q ))
(This is Euclid's famous theorem that no prime is the largest one.)
Proof. Consider for arbitrary prime p the value Q defined by
Q = 1 + the product of all primes ≤ p.
Theorem 0 allows us to conclude that Q is the product of a bag of primes, and, because Q > 1, that bag is not empty. By virtue of Q’s construction, such a bag contains no prime ≤ p. Hence it contains at least one prime > p, hence at least one prime > p exists. (End of proof of Theorem 1.)
With UPF —for Unique Prime Factorization— defined by
UPF n = The bag of prime multiples whose product equals n is unique
we can formulate
Theorem 2. (A p, x, y : p ≥ 1, x ≥ 1, y ≥ 1 : ¬(prime p) ∨ ¬(p|(x.y )) ∨ p|x ∨ p|y ∨ ¬(UPF(x.y )))
(Here “a|b ” should be read as “a divides b”.)
Proof. When the first two terms are false, x.y is the product of a bag of primes containing p ; when the next two terms are false, x.y is also the product of a bag of primes not containing p. Those two bags being different, we deduce ¬(UPF(x.y )). (End of proof of Theorem 2.)
We are now ready to state and prove
Theorem 3. (A n : n ≥ 1 : UPF n)
Proof. Theorem 3 is proved by mathematical induction, i.e. by proving
(A n : n ≥ 1 : (UPF n) ∨ (E i : 1 ≤ i < n : ¬(UPF i ))) | (7) |
(A n : n ≥ 1 : (comp n) ∨ (UPF n)) | (8) | |
(A n : n ≥ 1 : ¬(comp n) ∨ (UPF n) ∨ (E i : 1 ≤ i < n : ¬(UPF i ))) | (9) |
Consider a value of n, m say, such that the first and last terms of (9) are false, i.e. such that
(comp m) ∧ (A i : 1 ≤ i < m : UPF i ) | (10) |
m = p.P = q.Q with | (11) | |
1) | prime p | |
2) | p ≤ q | |
3) | P standing for the product of a non empty bag of primes all of which are ≥ p | |
4) | Q standing for the product of a bag of primes all of which are ≥ q. |
We obviously have 2 ≤ P < m, 1 ≤ Q < m, and P ≥ Q.
The conclusion UPF m, required to prove (9), can be drawn from
p = q ∨ (comp p) | (12) |
Establishing (12) is now our only remaining obligation. It is trivial in the case p = q. In the case p < q we consider value the M defined by
M = (q−p) . Q |
p | (q−p) ∨ p|Q | (13) |
Since 1 ≤ Q < m, we conclude from (10) UPF Q, i.e. all the primes in the unique bag of primes whose product equals Q are ≥ q, hence > p. The bag being unique, we conclude ¬(P|Q ). In combination with (13) we conclude p|(q−p); since we were considering the case p < q, we can conclude that q is composite. Having thus established relation (12), we have fulfilled our last proof obligation. (End of Proof of Theorem 3.)
Corollary of theorems 2 and 3.
(A p, x, y : p ≥ 1, x ≥ 1, y ≥ 1 : ¬(prime p) ∨ ¬(p|(x.y )) ∨ p|x ∨ p|y ) .
The above was written in reaction to the presentation in [1] of essentially the same proof of Theorem 3. I found that presentation contorted and notationally clumsy. They write (for (11))
m = p1 p2 ... pr = q1 q2 ... qs |
[1] Courant, Richard & Robbins, Herbert, What is mathematics? Oxford University Press paperback, 1978.
Plantaanstraat 5 | 20 October 1980 |
5671 AL NUENEN | prof. dr. Edsger W. Dijkstra |
The Netherlands | Burroughs Research Fellow |
Transcribed by Martin P.M. van der Burgt
Last revision 10-Nov-2015
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