# Blog Archives

## Infinite number of primes :)

In today’s post , I will present a very well known proof. What makes this proof especially appealing is that it is not too complex. Even so, it is very powerful. This theorem was first presented in Euclid’s Elements (Book IX, Proposition 20).

Theorem: There are an infinite number of primes.

Proof:

(1) Assume that there is only a finite number of primes.

(2) Then, there exists a prime pn that is the largest prime.

(3) Let p1, p2, …, pn be the list of all primes that exist.

(4) Let x = p1*p2*…*pn + 1.

(5) By the fundamental theorem of arithmetic (see Theorem 3, here), we know there is at least one prime that divides x. Let us call this prime p*.

(6) But none of the primes p1 … pn divide x since x ≡ 1 (mod pi) for any of the primes p1 … pn

(7) Therefore, we have a contradiction. We have a prime p* that is not in the complete list of primes.

(8) So, we reject our assumption in step#1 and conclude that there are an infinite number of primes that exist.

QED

## Fundamental Theorem Of Arithmetic

Before, we can provide this theorem, we will need an initial theorem on division and one lemma which will we borrow from Euclid.

Theorem 1: Division Algorithm for Integers

This theorem shows that given any two integers, there exist a unique divisor and a unique remainder.

(1) Let S be a set such that { a – bc : c is an integer and a-bc ≥ 0}

(2) S is not empty.

Case I: b divides a

In this case, then there exists a c where a – bc = 0. Therefore, 0 is an element of this set.

Case II: b doesn’t divide a

In this case, there exists an c where a – bc has a remainder is greater than 0. This remainder is then an element of this set.

(3) Then, then there exists d such that d is the smallest element of the set. [Well Ordering Principle]

(4) We know that d is less than b.

(a) Assume that d is greater than or equal to b.

(b) Then d – b is greater than or equal to 0.

(c) Since

d – b = (a – bc) – b = a – bc – b = a – b(c+1)

(d) We conclude that a – b(c+1) must be an element of S.

(e) But a – b(c+1) is less than d

(f) Which is a contradiction since d is the smallest element from step #3.

(5) Finally, for any a,b, the lowest value d and the remainder c are unique.

(a) Assume C,D are integers that can be derived from a,b.

a = bc + d, d is greater than 0 and less than b.
a = bC + D, D is greater than 0 and less than b.

(b) So

bc + d = bC + D –> b(c – C) = d – D

(c) Which means that b divides (d – D).

(d) Since d – D is less than b, we know d – D = 0 and that d = D.

(e) Which gives us b(c – C) = 0 or that c – C = 0.

(f) We can therefore conclude that c = C.

QED

Lemma 2: Euclid’s Lemma

This lemma shows that given a prime that divides a product, either it divides one value or it divides the other.

(1) Let’s assume a prime p divides a product ab.

(2) Let’s assume that p does not divide a.

(3) Then gcd(p,a)=1. [See blog on Greatest Common Denominators for details]

(4) There exists s,t such that 1 = as + pt. [See blog on Greatest Common Denominators]

(5) Multiplying both sides with b gives us:

b(1) = b(as + pt) = b = abs + ptb

(6) Since p divides ab, there exists k such that ab = pk.

(7) Combining with (5), we get

b = pks + ptd = p(ks + td).

QED

We can also tag on a corollary.

Corollary 2.1: Generalized Euclid’s Lemma

The idea here is that if a prime divides a product of n elements, then it is necessarily divides at least one of those elements.

(1) Let’s assume that we have a product of n elements.

a = a1 * a2 * … * an

(2) We can take any one of these elements and get:

a = a1 * (a2 * … * an)

(3)So, by Euclid’s Lemma (see Lemma 2 above), the prime either divides a1 or one of the rest of the elements.

(4) So, either it divides a1 or a2 etc.

(5) And we get to the last two elements, we are done by Euclid’s Lemma.

QED

Theorem 3: Fundamental Theorem of Arithmetic

This theorem proves that each integer greater than 1 is the product of a unique set of primes.

(1) Let S be a set S of {primes or products of primes > 1}.

(2) 2 is an element of S since 2 is prime.

(3) From (1), (2) there exists a value n which is 3 or greater where all values less than n and greater or equal to 2 are in S.

(4) If n is a prime, then it is a member of S.

(5) If n is not a prime, then it is also a member of S.

(a) If n is not a prime, then there exist a,b such that n=ab where a,b are greater than 1 and less than n. [Definition of a prime]

(b) But a,b must be elements of S since they are less than n. [From step #3 above]

(c) Then, n must also be a member of S since n is a product of two numbers which are either primes or products of primes.

(6) Now, all the primes that make up n are necessarily unique.

(a) Let n = p1 * p2 * … pr = q1 * q2 * … qs where p,q are all primes.