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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!!

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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.

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  • $\begingroup$ What do you mean by all fermat numbers are co prime?Co prime to what? $\endgroup$ – tatan Feb 28 '16 at 5:55
  • $\begingroup$ @tatan co-prime to each other. The first Fermat number is $2$, the second is $5$, and the third is $17$. These numbers are all coprime to each other. $\endgroup$ – S.C.B. Feb 28 '16 at 5:56
  • $\begingroup$ Does it necessarily mean all Fermat numbers are prime numbers? $\endgroup$ – tatan Feb 28 '16 at 5:57
  • $\begingroup$ @tatan No, it implies all Fermat numbers have distinct prime factors. Which is sufficient. $\endgroup$ – S.C.B. Feb 28 '16 at 5:58
  • $\begingroup$ I don't find a reason to believe that there are an infinite number of Fermat Numbers unless there are an infinite number of primes. $\endgroup$ – TheRandomGuy Feb 28 '16 at 6:29
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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.

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  • $\begingroup$ There are two problems in this: 1. The number to be considered should be $p_1p_2p_3\cdots p_n+1$. 2. It does not necessarily follow that this number is a prime. It follows that either this number is a prime or there is its factors contain some other primes not included in $s$. Either way the conclusion holds. $\endgroup$ – GoodDeeds Feb 28 '16 at 6:23
  • $\begingroup$ Yes. So the contradiction is you got a prime outside the set $s$. But $p_1p_2\cdots p_n+1$ is not necessarily that prime. It may also be the case that there was some other prime missed out in $s$, which leads to the same contradicton. $\endgroup$ – GoodDeeds Feb 28 '16 at 6:27
  • $\begingroup$ @GoodDeeds Thanks. $\endgroup$ – TheRandomGuy Feb 28 '16 at 6:27
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Let $p$ is the last prime. Then according to Bertrand's postulate the interval $(p,2p)$ consists a prime number. We get a contradiction.

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