How many ways are there to prove that there is no largest prime?

Question

Is there any other proof by which I can show that there is no largest prime?

I saw an example where it is proved with contradiction.(Idea is basically that of Euclid's proof)

Imagine that the largest prime prime is $13$.So, total number of primes we know are-$2,3,5,7,11,13$.

Now,if I do $(2\times3\times5\times7\times11\times13)+1=30031$.So, we can see that $30031$ is not divisible by $2,3,5,7,11,13$ as they leave remainder $1$. Also,as it is formed by multiplying only primes it does not have any other composite factors.We also see that $30031=59\times 509$.Which are again two primes.Thus,$13$ is not the largest prime.

What are the other ways to prove that there is no largest prime?

Thanks for any proof!!

Answer 1

Note that all Fermat Numbers are coprime to each other.

Thus, if there are a finite number of prime numbers, this is a contradiction as there are infinite number of Fermat Numbers.

Thus, there are a infinite number of prime numbers.

And so there is no largest prime.

Answer 2

Well, the proof you have shown is not exactly the right way.

Proof: Assume there are a finite number of primes.

Let $s$ be the set of all primes possible. And let the primes be $p_1, p_2, p_3, \dots , p_n$.

Now by the Fundamental Theorem of Arithmetic, we know that every number is a prime or a unique product of primes.

Consider the number $p_1 p_2 p_3 \cdots p_n +1$. We know that it is not divisible by any of the numbers in the set $s$. Thus we get a number which is a prime or composed of primes not in the set $s$. That's a contradiction. Thus there are an infinite number of primes.

Answer 3

Let $p$ is the last prime. Then according to Bertrand's postulate the interval $(p,2p)$ consists a prime number. We get a contradiction.