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October 31st, 2018, 02:06 AM   #21
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Even if you rephrase.....it is not enough.

Let's build successive sets of prime candidates by successive sieving of $p_n$.

First we take all positive natural numbers: $A_0=$N$^+\backslash\{1\}$ where you removed $p_0=1$.
$A_1$ is a new set where we remove any multiple of $p_1=2$ from the previous $A_0$ set. $\frac{1}{2}$ of all candidates from that set were removed.
$A_2$ is a new set where we remove any multiple of $p_2=3$ from the previous $A_1$ set. $\frac{1}{3}$ of all candidates from that set were removed (yes, multiple of $3$ is $\frac{1}{3}$ of all naturals but also $\frac{1}{3}$ of the remaining $A_1$ set).
$A_n$ is a new set where we remove any multiple of $p_n$ from the previous $A_{n-1}$ set. $\frac{1}{p_n}$ of all candidates from that set were removed.
Since at each step we build a new infinite set where we only remove a (smaller and smaller) fraction of the previous one, it is not possible to end up with an empty set (they are all infinite), therefore there must be infinitely many primes.

Well, this "proof" is flawed (or should I say, incomplete).

Now let's do the same with twins:
Let's build successive sets of twin candidates by successive sieving of $p_n$.

First we sieve $2$ and $3$ and we are left with $A_2=\{(5,7),(11,13),(17,19),(23,25),(29,31),(35,3 7),(41,43),(47,49),(53,55),(59,61),...\}$ the set of twin candidates.
$A_3$ is a new set where we remove any multiple of $p_3=5$ (and its "binomial") from the previous $A_2$ set. $\frac{2}{5}$ of all candidates from that set were removed. $A_3=\{(11,13),(17,19),(29,31),(41,43),(47,49),(59 ,61),...\}$
$A_n$ is a new set where we remove any multiple of $p_n$ (and its "binomial") from the previous $A_{n-1}$ set. $\frac{2}{p_n}$ of all candidates from that set were removed (no flaw here, this can be proved).
Since at each step we build a new infinite set where we only remove a (smaller and smaller) fraction of the previous one, it is not possible to end up with an empty set (they are all infinite), therefore there must be infinitely many twins.

Well, this "proof" has the same big flaw as the previous one.
Although it can be easily fixed for the primes, for the twins, this is another story.

Without "rigorous" math, can you fix the flaw, let even spot it?

Last edited by skipjack; October 31st, 2018 at 03:35 AM.
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