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How can I prove that there is a non finite number of prime number ? I try to prove it by contradiction but it's not conclusive. Any idea ?

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    [Infinitely many proofs that there are infinitely many primes](http://www.cut-the-knot.org/proofs/InfinitudeOfProofs.shtml), by cut-the-knot.org. Just for instance... – 2017-02-05
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    Where are you stuck ? – 2017-02-05
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    Assume that primes are finite, let $\{p_1, p_2, \cdots, p_n\}$ be the set of all prime numbers. $p_1 \cdot p_2 \cdots p_{n-1} \cdot p_n + 1$ can't be prime as it is greater than any $p_i$, but it can't be composite as it's not divisible by any prime (it leaves remainder 1 for every $p_i$). The contradiction arises from assuming that there are finitely many primes, so there must be an infinite number of them. – 2017-02-05
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    You don't even have to assume that you have a finite number of primes, you can assume given any finite list of prime numbers then just do what chelivery did above with this list. What would follow is either the product all these primes plus one is a new prime that is not in your list, or it is composite but not divisible by any of the prime numbers. In either case you can add a new prime number to this list and repeat the process and you will find that you can constantly add new prime numbers. Thus, there must be an infinite number of prime numbers. – 2017-02-05
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    we have Fermat number $F_n=2^{2^n}+1$ then either $F_n$ prime or $\exists p; p|F_n$ – 2017-02-05

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Maybe you need to have the fundamental theorem of arithmetic proven first.

For every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes, uniquely up to the order in which they appear.

If there is a finite number of prime numbers, that means every composite number is a product of one or more of those prime numbers (with or without repetition).

This is where the classic proof comes in. The Tooth Fairy came up with it millennia before Euclid, but humans prefer to give Euclid the credit. I think there was also a Chinese mathematician who came up with the same thing a century or two before Euclid.

Multiply all the (finite) primes together $p_1 p_2 \ldots p_k$ ($k$ is the total number of primes that you think exist) and call this product $P$. Then what is the factorization of $P + 1$? In each instance of dividing $P + 1$ by one of the primes $p_i$ you will find that it leaves a remainder of $1$, that is $P + 1 \equiv 1 \pmod{p_i}$.

Then either $P + 1$ is some other prime that's not on your finite list of primes, and it's larger than all of them, or it's the product of primes not on your list, which may or may not be larger than the primes on your list. Update your list of primes and also update $k$ to reflect the enlarged list.

You can keep doing this for as long as you want to, or are able to, and you will always find at least one new prime at each iteration of the process.

However, given the limitations of your human brain, and of your computers as well, you could reach a point at which you're unable to determine whether $P + 1$ is prime or the product of primes not in your list. Nevertheless, rest assured that if you were to overcome that hurdle, you would have to enlarge your list of primes and update $k$ accordingly.

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This is something that Euclid proved centuries ago. But, to defend a dead mathematician's honor, some people here today will argue that his proof is constructive, not by contradiction. Whatever the case, I hope at least you find it convincing.

Assume that there the set of prime numbers is finite, and you know them all. For example, 2, 5, 11, 17. Since the set is finite, you can multiply its elements and add 1 to obtain another number, 1871, in the example. This other number is not divisible by any of the known primes, but it must either be itself prime or be divisible by a prime other than any in our list.

As it happens, 1871 is prime. So we amend our list of primes: 2, 5, 11, 17, 1871. We can multiply them again, and add 1 again, this time obtaining 3498771, which is composite but divisible by primes not already in our list: 3, 1033, 1129. So we amend our list of primes yet again, and so on and so forth.

This means that any time you assume a particular set of prime numbers is the set of all prime numbers, you can use the set to obtain at least one other prime number not already in your list.

More commonly people use the set of the first $n$ primes, and by multiplying them they obtain what are called "the primorials": 2, 6, 30, 210, 2310, etc. Add 1 to those and you get: 3, 7, 31, 211, 2311, etc., some of which are prime, some of which are composite but not divisible by primes encountered earlier on.

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    While I like your description of the process, you've missed the key part: how we *know* that we'll always get a new prime this way. – 2017-02-06
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    @Noah Feel free to add it to this community wiki answer or put it in your own answer. – 2017-02-06
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If the number of primes is finite, there is a larger one, let $p$. But then, the number $p!+1>p$ isn't divisible by any prime and is... prime.

(If $p!+1$ can be factored, the factors must be larger than $p$ and are themselves composite. By infinite descent, they must have a prime factor, which divides... $p!+1$.)

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    I'm not sure $p! + 1$ is prime. – 2017-02-05
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    It is not divisible by any of the primes below ,so if that number is composite , then it should have a prime factor which is not one of the primes we have . Thus a contradiction. – 2017-02-05
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    @pjs36: why not ? – 2017-02-05
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    Because for instance $5! + 1 = 121$ isn't. – 2017-02-05
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    @pjs36: indeed, $121=11^2$, but as $11>5$, $11$ must be composite and have a prime factor; then this prime factor divides $121$ ?! I will update. – 2017-02-05
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    @pjs36 **Under the assumption** that $p$ is the greatest prime, we can indeed conclude that $p!+1$ is prime. **In reality**, though, since there is no largest prime, the most we can conclude in general is that for all $p$, $p!+1$ is divisible by some prime $>p$. Now, I'd argue that for this reason the phrasing above isn't the clearest, but it is correct. – 2017-02-06