My Math Forum Proof of twin prime conjecture

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 January 3rd, 2019, 09:23 AM #41 Newbie   Joined: Sep 2017 From: Belgium Posts: 19 Thanks: 7 Take $N^+$ and instead of sieving all multiples of some prime $p$, you sieve all positive exponent of 2 of these special numbers which I'll call teta-primes: $a\cdot 2^0$ where $a$ is odd (and obviously the teta-prime is too). After sieving, the question is: "Is there an infinite number of teta-twin" (teta-twin are just 2 successive numbers, 1 odd and 1 even, or vice-versa). First pass, you sieve all teta-multiples of $1\cdot 2^0$: $1\cdot 2^1$, $1\cdot 2^2$, $1\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of teta-twins. second pass, you sieve all teta-multiples of $3\cdot 2^0$: $3\cdot 2^1$, $3\cdot 2^2$, $3\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of teta-twins. Third pass, you sieve all teta-multiples of $5\cdot 2^0$: $5\cdot 2^1$, $5\cdot 2^2$, $5\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of teta-twins. After $n$ passes, you sieve all teta-multiples of $(2n-1)\cdot 2^0$: $(2n-1)\cdot 2^1$, $(2n-1)\cdot 2^2$, $(2n-1)\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of teta-twins. Your answer: There are infinitely many teta-twins. My answer: There are none.
January 3rd, 2019, 09:24 AM   #42
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 Originally Posted by MrAwojobi I can't see what more I can say to make you see things from what to me is a straight forward point of view. Maybe I will try again. 5, for instance, eliminates all its infinite number of multiples from the infinite number line. Why then can you not all see that if the twin prime conjecture is false then it takes just one prime to eliminate all the un-eliminated, infinite number of 6n-1 and 6n+1 pairs? I repeat again that it will be absurd to say more than 1 prime is required to do this i.e. one might just as well make an absurd comment by saying that more than one prime is required to eliminate all the infinite number of multiples of 5 when it is clear that one prime,5, can do this.
OK, so assume there are finitely many twin primes. Then start from a number $N$, we have that at least one of the pairs $6N\pm 1$, $6(N+1)\pm 1$, $6(N+2)\pm 1$, etc MUST be eliminated? agreed?

Why can't it be, for example, that
$5$ eliminates one of $6N \pm 1$
$7$ eliminates one of $6(N+k)\pm 1$ for $k>0$
$11$ eliminates one of $6(N+k')\pm 1$ for $k'>k$
$13$ eliminates one of $6(N+k'')\pm 1$ for $k''>k'$
etc.

Why is this scenario impossible?

January 3rd, 2019, 09:30 AM   #43
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 Originally Posted by AplanisTophet It's proven earlier in the thread that there is no finite set E of eliminating primes.
I don't think that has been proven?

January 3rd, 2019, 09:31 AM   #44
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 Originally Posted by MrAwojobi Why then can you not all see that
1) because we are complete idiots,
or
2) you are wrong

In either case, we established that we don't see it. So stop asking this and prove it.

January 3rd, 2019, 09:37 AM   #45
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 Originally Posted by AplanisTophet It's proven earlier in the thread that there is no finite set E of eliminating primes.
Please show me where there is such a proof. Mr. Owojobi's prose is hard to parse.

If such a proof exists, why is Mr. Owojobi so intent on proving that if the twin primes conjecture is false, such a set has exactly one member?

January 3rd, 2019, 10:30 AM   #46
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 Originally Posted by JeffM1 Please show me where there is such a proof. Mr. Owojobi's prose is hard to parse. If such a proof exists, why is Mr. Owojobi so intent on proving that if the twin primes conjecture is false, such a set has exactly one member?
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 Originally Posted by AplanisTophet Every prime greater than 3 will "hop over" (when sequencially eliminating its multiples on a number line after having eliminated all the multiples of previous primes under a standard ordering, and where "hop over" implies "fail to eliminate") an infinite number of 6n-1 and 6n+1 pairs, yes. Proof: $$\text{Let } x = p_1*p_2*p_3*...*p_i$$ $$\text{Then } x+1 \text{ and } x-1 \text{ will fail to be eliminated by the primes } \leq p_i$$ $$\text{The same goes for } 2x+1 \text{ and } 2x-1, 3x+1 \text{ and } 3x-1, ...$$
The set E cannot be finite because if it were and $p_i$ was its greatest element, then $x = p_1*p_2*p_3*...*p_i$ where $x+1$ and $x-1$ have yet to be eliminated would result in a contradiction.

MrAwojobi is failing to consider the possibility of an infinite E. That's why he's looking for the "one member" that eliminates all twin primes not eliminated by previous primes.

 January 3rd, 2019, 11:06 AM #47 Banned Camp   Joined: Aug 2010 Posts: 170 Thanks: 4 You are all making a very simple explanation of the truth of the twin prime conjecture over complicated probably because the maths gurus like Field medalist, Terence Tao have made the twin prime conjecture a big deal when really it is not. My proof is just as logical and as simple as proving the infinitude of primes. My proof uses the sieve of Eratosthenes, which by the way is the algorithmic generator of the primes, to show the infinitude of twin primes. I have shown that the sieve of Eratosthenes cannot 'hit' all the 6n-1 and 6n+1 pairs using a simple proof. This sieve operates by giving every prime greater than 3 opportunities to eliminate 6n-1 and 6n+1 pairs. It is agreed that each prime in its infinite hopping will hop over an infinite number of un-eliminated 6n-1and 6n+1 pairs and therefore no one prime can eliminate all infinite number of 6n-1 and 6n+1 pairs. This means that the twin primes will never run out.
January 3rd, 2019, 11:07 AM   #48
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 Originally Posted by AplanisTophet The set E cannot be finite because if it were and $p_i$ was its greatest element, then $x = p_1*p_2*p_3*...*p_i$ where $x+1$ and $x-1$ have yet to be eliminated would result in a contradiction. MrAwojobi is failing to consider the possibility of an infinite E. That's why he's looking for the "one member" that eliminates all twin primes not eliminated by previous primes.
I think that works. Sorry. I missed it, but it was not in Mr. Awojabi's arguments, which is what I was focusing on.

And his failure to consider that E was infinite was exposed when he said "last" and "final" way back when.

So his uniqueness argument is not merely unproven, but unprovable and unnecessary. ROFL

January 3rd, 2019, 11:13 AM   #49
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 Originally Posted by MrAwojobi You are all making a very simple explanation of the truth of the twin prime conjecture over complicated probably because the maths gurus like Field medalist, Terence Tao have made the twin prime conjecture a big deal when really it is not. My proof is just as logical and as simple as proving the infinitude of primes. My proof uses the sieve of Eratosthenes, which by the way is the algorithmic generator of the primes, to show the infinitude of twin primes. I have shown that the sieve of Eratosthenes cannot 'hit' all the 6n-1 and 6n+1 pairs using a simple proof. This sieve operates by giving every prime greater than 3 opportunities to eliminate 6n-1 and 6n+1 pairs. It is agreed that each prime in its infinite hopping will hop over an infinite number of un-eliminated 6n-1and 6n+1 pairs and therefore no one prime can eliminate all infinite number of 6n-1 and 6n+1 pairs. This means that the twin primes will never run out.
Blablabla... Lots of talk, no proofs. This is forum with top mathematicians you know, not St Columbia boy school.

January 3rd, 2019, 11:14 AM   #50
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 Originally Posted by MrAwojobi no one prime can eliminate all infinite number of 6n-1 and 6n+1 pairs. This means that the twin primes will never run out.
It does not follow that what one prime cannot do an infinite set of primes cannot do. It is not as though we are going to run out of primes to work with.

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