math.
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Unique Factorization

A simlpe proof for the unique factorization theorem.

Today’s post we are going to see an important theorem in arithmetic, “The unique factorization”. This theorem is used in various application and it states that : every integer strictly positive is either prime itself or is the product of a set of prime factors in an unique way, the order of its prime factors.

Proof by induction :

Let n be an integer strictly positive.

  • Existence:

For n = 1, 1 is the product of a set of prime numbers (the empty set)

We shall assume that the result hold for every integer <= n, and we will prove it for n+1.

Either n+1 is prime, and we are done.

Or n+1 = l * k, where l, k < n. From the induction assumption we can imply that both k and l could be written as the product of prime factors. And therefor, their product too. Making n+1 the product of a set of prime factors.

  • Uniqueness:

Suppose that n is the products of prime factors in two ways:

  • n = P1P2 … *Pu
  • n = Q1Q2 … *Qv

We have to prove that u = v, and that ∀ i, ∃ j such Pi = Qj.

We shall use the Euclid’s lemma that states, if p prime number divides a product x*y, p must divide x or/and y.

Let i, 1 <= i <= u. We know that Pi divides n, hence Pi divides one of Qjs.

Let j such index, since Qj is prime we conclude that Pi is either 1 or Qj. Pi is prime means that Pi = Qj. We will rearrange Pis and Qjs in a way that makes P1 = Q1.

we have now :

  • n/P1 = P2P3 … *Pu
  • n/P1 = Q2Q3 … *Qv

Following the same reasoning, we will find that :

  • n/(P1P2 … *Pu) = 1
  • n/(P1P2 … *Pu) = Qu+1 * … * Qv

Which means Qu+1 * … * Qv = 1, which is impossible. Therefor, u = v and every Pi = Qi.