The Notebook
İstanbul, Turkey

Proof on Infinity of Prime Numbers

In this post, I show that we have infinitely many prime numbers. To do so, I use a proof from 300 BC which provided by Greek Mathematician. The one that people usually call as “Father of Geometry” or Euclid of Alexandria.

Let's write prime numbers in an ascending manner,

p_1 = 2, p_2 = 3, p_3 = 5, p_4 = 7

When we check the numbers which are in a pattern of:

p_1.p_2 + 1 = 2.3 + 1 = 7 p_1.p_2.p_3 + 1 = 2.3.5 + 1 = 31 p_1.p_2.p_3.p_4 + 1 = + 1 = 211

we oftenly come up with a prime number.

But how can we be sure about primality of number A = p_1.p_2 \dots p_n + 1 for all values of n?

Here we use contradiction technique to check: We intuitively think that A is a prime number, since it empirically is. Now we set up an equation which assumes that A is a non-prime number and try to show that this equation would not hold --hence it contradicts.

Following equation assumes that A is a divisor of some number.

We clearly know that, every number -except 1- is at least multiple of one prime number.

For this reason, let's say that p_r is a divisor of A.

A = p_1.p_2.p_3 \dots p_r \dots p_n + 1 \rightarrow \frac{A}{p_r} = p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n + \frac{1}{p_r}.

Consequently we have

\frac{A}{p_r} - p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n = \frac{1}{p_r}

Let's concentrate on this last equation.

\frac{A}{p_r} has to be an integer, since we assume A is a multiple of p_r.
Second term p_1.p_2.p_3 \dots p_{r-1}.p_{r+1} \dots p_n is also integer, nevertheless \frac{1}{p_r} is not.
So, our non-primality assumption of A was wrong. Consequently, this wrong assumption proves us that, A would always be prime for any value of n.