0

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!!

Community
  • 1
Soham
  • 9,990

3 Answers3

3

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.

S.C.B.
  • 22,768
  • What do you mean by all fermat numbers are co prime?Co prime to what? – Soham Feb 28 '16 at 05:55
  • @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. – S.C.B. Feb 28 '16 at 05:56
  • Does it necessarily mean all Fermat numbers are prime numbers? – Soham Feb 28 '16 at 05:57
  • @tatan No, it implies all Fermat numbers have distinct prime factors. Which is sufficient. – S.C.B. Feb 28 '16 at 05:58
  • 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. – TheRandomGuy Feb 28 '16 at 06:29
  • @DhruvSomani Fermat Numbers $\neq$ Fermat Primes. Fermat Numbers are $2^{2^{n}}+1$, or $a_{n}=a_{n-1}^2+1$ & $ a_{1}=2$. – S.C.B. Feb 28 '16 at 06:31
0

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.

GoodDeeds
  • 11,185
  • 3
  • 22
  • 42
TheRandomGuy
  • 4,014
  • 2
  • 20
  • 56
  • 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. – GoodDeeds Feb 28 '16 at 06:23
  • 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. – GoodDeeds Feb 28 '16 at 06:27
  • @GoodDeeds Thanks. – TheRandomGuy Feb 28 '16 at 06:27
0

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

Leox
  • 8,120