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Euclid's proof that they are infinity prime numbers look more as a statement then a proof for me. For thing to be even worse, it's example used every time to show the basic proof you can find. If I have understood it good, in nutshell proof is:

If this is a prime number it’s good, and we have a new one:

        N = (p0 · p1 · p2 · … · pn)


If not, adding 1 we will create a new one:

        N = (p0 · p1 · p2 · … · pn)+1



Example:

Let’s imagen we know only for first two primes 3 and 5. With this formula 

        N = 3·5 = 15 (15 is not prime, you can divide it with 3 and 5)

        N = (3·5)+1 = 16 (it’s not prime)

What I doing wrong?

Here is explanation already given, but can someone do relation between actual explanation and example i posted.

mathNoob
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1 Answers1

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The argument is the following. Suppose you have already $n$ primes $p_{1}, \dots, p_{n}$.

Then $$ A = p_{1} \cdot p_{2} \cdots p_{n} + 1 > 1 $$ is not divisible by $p_{1}, \dots, p_{n}$, as Euclidean division by any of them yields remainder $1$.

Since every natural number greater than $1$ is a product of primes, there must be a prime $p$, distinct from $p_{1}, \dots, p_{n}$, which divides $A$.

In your case, you have taken $n = 2$, $p_{1} = 3$, $p_{2} = 5$. You get $A = 16$, and the only choice for $p$ is $2$.

Had you taken $n = 2$, $p_{1} = 5$ and $p_{2} = 7$, then $A = 36$ so you have either $p = 2$ or $p = 3$.

Andreas Caranti
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