Twin Primes

[Hugh]

Originally Posted by axn

After sieving by Pn and looking at Pn#, we can then further sieve from Pn to SQRT(Pn#) to get the full list of twin primes in the Primorial range. I don't see how you can formulate the argument to exclude the primes from Pn to SQRT(Pn#) from consideration.

Not sure I follow this bit. They are not being excluded from consideration. It is just a question of whether they can all subsequently be cast out.

[CRGreathouse]

Suppose you had been trying to prove this conjecture: Call an integer n "special" if n-1 is not an integer. Then there exist infinitely many positive special integers.

Your attempted proof works for this case, mutatis mutandis:

Assume there is not an infinite number of special numbers. If so, there must be a number above which there are no more special numbers.

We use the Very Special Sieve to cast out potential special numbers. For each integer n, cast out n+1 as a potential special number.

There must be a “last number” that, when added to the pattern, casts out the final remaining special candidates. (Meaning that no number larger than this is needed to cast out further special candidates).

Any single number pn can only remove 1 of the remaining special candidates in the pattern.

Therefore it is impossible for such a “last number” to exist.

Therefore there is an infinite number of special numbers.

But my 'proof' above is incorrect, as is yours. Can you see my 'mistake'? Can you see yours?

[Hugh]

Originally Posted by CRGreathouse

But my 'proof' above is incorrect, as is yours. Can you see my 'mistake'? Can you see yours?

OK, let me have a look

Originally Posted by CRGreathouse

Suppose you had been trying to prove this conjecture: Call an integer n "special" if n-1 is not an integer. Then there exist infinitely many positive special integers.

Well, we're in trouble here since this is self-contradictory. So let's not go through the rest of it. There are no such "special numbers". Whereas there are such things as primes and twin primes. Meaning it's not logically identical to my attempt.

I appreciate that you are trying to bring out the logical error though, so can you find a way of doing it that doesn't rely on suggesting my proof starts from an impossibility.

Short of a successful proof or disproof neither you nor I know whether there is an infinite number of twin primes whereas we both know there are no "special numbers".

Is there another better version you could suggest? I am genuinely interested and willing to admit defeat if need be.

[CRGreathouse]

Originally Posted by Hugh

Well, we're in trouble here since this is self-contradictory.

No, it's not. I did not assume that any special numbers existed (which would be incorrect), I merely reasoned about them.

Similarly, I could define special numbers as even Mersenne numbers, then prove as a theorem that no special numbers exist.

Originally Posted by Hugh

Short of a successful proof or disproof neither you nor I know whether there is an infinite number of twin primes whereas we both know there are no "special numbers".

I chose it to be that way so that it would be clear to you why your method doesn't work -- that you could just look at its analogue in my proof. If I used something unknown and 'proved' that there were infinitely many of them, I'd be afraid you'd take my 'proof' seriously instead of taking it for what it is: an intentionally flawed argument.

Having said that, you could change the first sentence to

Call a positive integer integer n "special" if n-1 is not a positive integer.

and you would have a finite set of special numbers for which my 'proof' claims to show infinitely many.

[Hugh]

Originally Posted by CRGreathouse

No, it's not. I did not assume that any special numbers existed (which would be incorrect), I merely reasoned about them.

Similarly, I could define special numbers as even Mersenne numbers, then prove as a theorem that no special numbers exist.




I chose it to be that way so that it would be clear to you why your method doesn't work -- that you could just look at its analogue in my proof. If I used something unknown and 'proved' that there were infinitely many of them, I'd be afraid you'd take my 'proof' seriously instead of taking it for what it is: an intentionally flawed argument.

Having said that, you could change the first sentence to

Call a positive integer integer n "special" if n-1 is not a positive integer.

and you would have a finite set of special numbers for which my 'proof' claims to show infinitely many.

No, because then again we know from the start that no such special numbers exist so it would be a nonsensical argument, based on dividing zero by infinity. There's a difference between a flawed argument and a nonsensical one.

If you're correct it is not because the argument is nonsensical in that way, but because it is a version of a Zeno-style paradox.

[CRGreathouse]

Originally Posted by Hugh

No, because then again we know from the start that no such special numbers exist

Uh, no... under the second definition there are one or more special numbers. Can you find them?

Originally Posted by Hugh

If you're correct it is not because the argument is nonsensical in that way, but because it is a version of a Zeno-style paradox.

So far you've misunderstood all three of the flawed proofs on this thread. This seems to show something about your understanding of mathematics!

Dare I suggest a fourth incorrect 'proof'?

A special number is a positive integer which cannot be written in the form 10n + k where n and k are positive integers and k <= n^2.

Assume there is not an infinite number of special numbers. If so, there must be a number above which there are no more special numbers.

We use the Very Very Special Sieve to cast out potential special numbers. Start with the positive integers. Remove 11, ..., 10 + 1^2, then remove 21, ..., 20 + 2^2, then remove 31, ..., 30 + 3^2, then remove 41, ..., 40 + 4^2, and so on.

There must be a “last number” that, when added to the pattern, casts out the final remaining special candidates. (Meaning that no number larger than this is needed to cast out further special candidates).

Any single number m can only remove at most m^2 of the remaining special candidates in the pattern.

Therefore it is impossible for such a “last number” to exist.

Therefore there is an infinite number of special numbers.

This proof fails for just the same reason as your proof. (And remember, its sieving steps are much weaker than yours -- yours remove an infinite number of candidates at each step, in fact a number of candidates with positive density; the VVSS removes candidates with density 0, in fact only finitely many at each step.)

[Hugh]

Originally Posted by CRGreathouse

Uh, no... under the second definition there are one or more special numbers. Can you find them?



So far you've misunderstood all three of the flawed proofs on this thread. This seems to show something about your understanding of mathematics!

OK, apologies for ignoring the trivial case of 1-1 = 0. Though it' s a tiny bit unfair to use that as a stick to beat me with, and also doesn't mean I didn't understand the previous example, to which my original objection still applies. And this one isn't a great deal more use as a counterexample.

Originally Posted by CRGreathouse

Dare I suggest a fourth incorrect 'proof'?

A special number is a positive integer which cannot be written in the form 10n + k where n and k are positive integers and k <= n^2.

Assume there is not an infinite number of special numbers. If so, there must be a number above which there are no more special numbers.

We use the Very Very Special Sieve to cast out potential special numbers. Start with the positive integers. Remove 11, ..., 10 + 1^2, then remove 21, ..., 20 + 2^2, then remove 31, ..., 30 + 3^2, then remove 41, ..., 40 + 4^2, and so on.

There must be a “last number” that, when added to the pattern, casts out the final remaining special candidates. (Meaning that no number larger than this is needed to cast out further special candidates).

Any single number m can only remove at most m^2 of the remaining special candidates in the pattern.

Therefore it is impossible for such a “last number” to exist.

Therefore there is an infinite number of special numbers.

This proof fails for just the same reason as your proof. (And remember, its sieving steps are much weaker than yours -- yours remove an infinite number of candidates at each step, in fact a number of candidates with positive density; the VVSS removes candidates with density 0, in fact only finitely many at each step.)

Thanks, this looks a more serious attempt to answer the question. I'll look at this properly later on today.

[Hugh]

Originally Posted by CRGreathouse

Uh, no... under the second definition there are one or more special numbers. Can you find them?



So far you've misunderstood all three of the flawed proofs on this thread. This seems to show something about your understanding of mathematics!

Dare I suggest a fourth incorrect 'proof'?

A special number is a positive integer which cannot be written in the form 10n + k where n and k are positive integers and k <= n^2.

Assume there is not an infinite number of special numbers. If so, there must be a number above which there are no more special numbers.

We use the Very Very Special Sieve to cast out potential special numbers. Start with the positive integers. Remove 11, ..., 10 + 1^2, then remove 21, ..., 20 + 2^2, then remove 31, ..., 30 + 3^2, then remove 41, ..., 40 + 4^2, and so on.

There must be a “last number” that, when added to the pattern, casts out the final remaining special candidates. (Meaning that no number larger than this is needed to cast out further special candidates).

Any single number m can only remove at most m^2 of the remaining special candidates in the pattern.

Therefore it is impossible for such a “last number” to exist.

Therefore there is an infinite number of special numbers.

This proof fails for just the same reason as your proof. (And remember, its sieving steps are much weaker than yours -- yours remove an infinite number of candidates at each step, in fact a number of candidates with positive density; the VVSS removes candidates with density 0, in fact only finitely many at each step.)

OK, so if I'm not being dozy, we know from n=3 that every successive value of n will remove the candidates between n and n+1, which is enough to mean there can't be an infinite number of special numbers.

Slight differences to primes and twin primes:

Firstly, in this case the sieve removes numbers higher than n and we quickly reach a range where no number can be missed by the sieve, meaning it is sieving 100% of remaining candidates. With twin primes, it is necessary that the sieve for pn eliminate some candidates lower than pn# and for it to always be able to do so before the sieve reaches pn#, so it is a more valid question whether this must always be possible.

Secondly, in your example, once we are past n=3, any number that is "cast out" by the sieve cannot be a special number. In the case of twin primes, it is not sufficient for us to know that all numbers will eventually be covered by the sieve, because some numbers that are within the sieve are nonetheless examples of the numbers we are searching for.

[Hugh]

Originally Posted by CRGreathouse

Dare I suggest a fourth incorrect 'proof'?

Out of curiosity, what is wrong with the "proof" below? The logic is closer to my attempt so it might help me get my head around this if you could look at this one for me.

A “special number” is a natural number that has exactly two distinct natural number divisors: 1 and itself.

Assume there is not an infinite number of special numbers. If so, there must be a number above which there are no more special numbers.

We use the Very Very Special Sieve to cast out potential special numbers.

There must be a “last number” that, when added to the pattern, casts out the final remaining special candidates. (Meaning that no number larger than this is needed to cast out further special candidates).

Any single number m removes a symmetrical repeating pattern of remaining candidates, and can only remove at most 1/p of the remaining special candidates in the pattern.

Therefore it is impossible for such a “last number” to exist.

Therefore there is an infinite number of special numbers.

[Hugh]

On reflection I think axn was right about the last prime. The existence of gaps means we need ever higher primes to fill the gaps, but not necessarily ever higher twin primes. Doh.

Thanks to all for responding, I do appreciate the help.

[Hugh]

Originally Posted by axn

After sieving by Pn and looking at Pn#, we can then further sieve from Pn to SQRT(Pn#) to get the full list of twin primes in the Primorial range. I don't see how you can formulate the argument to exclude the primes from Pn to SQRT(Pn#) from consideration

And coming back to this, I think the only way would be an illegimate "averaging" step - it is unlikely that the fraction 2/pn x 2/pn+1 .... 2px [the last prime before SQRT(Pn#)] can sieve out all the remaining gaps in the primorial range, if we assume it removes a reasonably consistent average of the remaining gaps. But clearly that's not a proof, just an observation, and one can't assume a consistent average in this particular range.

Or the other way (again, probably not feasible) would be to prove there is always a twin prime gap between a prime and its square.

Thanks again, took me a day or two but I see it more clearly now.