Do Infinitely Many Prime Pairs Exist?

[nnnnnnnn]

There is (as far as I know) no proof-for or against- that there are infinately many prime pairs such as 3, 5 or 29, 31...

Anyway, is it intuitive to assume that there should be infinitely many pairs just b/c of the fact that there are infinitely many numbers? or does this have nothing to do with it?

 

[HallsofIvy]

Well, "intuitive" is not a very good "mathematics" term!

Is it "intuitive to assume" that there are infinitely many even primes b/c of the fact that there are infinitely many numbers.

 

[nnnnnnnn]

Its funny that you say that because talking about math is the only time I would say intuitive...

Anyways, I can't think of a good example but I can think of an example:
in a class for real numbers I had to prove that 1>0. I knew this to be true because it was intuitive but it was tricky to prove...

 

[matt grime]

You can't prove that 1>0, unless you assume certain things...

 

[mathwonk]

My intuition is that there are indeed infinitely many prime pairs, but it is based on nothing I can describe clearly.

I.e. to me it would just be odd for there to exist a largest prime pair. There is a tendency of mathematical facts to be "natural" and not so odd.

To me at least it would seem less surprizing or odd for there to be an infinite number of prime pairs.

I.e. for there to be a largest one, I would thionk there needs to be a "reason" for that. Whereas if there are infinitely many, then there is no special one, and that is more expected to me.

But all mathematicians have different intuition, so no one need agree with me.

 

[shmoe]

You're right that it's currently unkown whether or not there are infinitely many prime pairs.

There is the twin prime conjecture which claims that the number of prime pairs less than x is asymptotic to Cx(\log{x})^{-2}, where the C is explicit (about 1.32..). There are heuristic arguments to support this, but of course no one can prove it yet.

There's a partial victory by J.R. Chen which implies there are either infinitely many prime pairs, or there are infinitely many primes p where p+2 is the product of two primes (possibly both are true).

 

[nnnnnnnn]

You can't prove that 1>0, unless you assume certain things...

thats not the point but we were working with real numbers so didn't need to assume anything- just to follow the established rules...

 

[mathwonk]

I like shmoe's asymptotic formula. it gives substance to a prediction there are infinitely many.

I.e. if one has actual data up to a few billion billion billion... or so, that there is a pattern to the density of prime pairs, then it seems believable that the density will not suddenyl go to zero after some point.

 

[mathwonk]

try this: in the spirit of proving that 1>0, prove that any function f defined on the real numbers and satisfying f(x+y) = f(x)f(y), (think exponential function), is either identically zero, or always positive.

 

[shmoe]

Billions and billions of data points can look convincing, but can really come back to bite you in number theory. Like Merten's conjecture (that |\sum_{n\leq x}\mu(n)|\leq x^{1/2} where \mu is the mobius function), or the conjecture that the prime counting function is strictly bounded above by the logarithmic integral. Both were proven false, but the first counterexamples are huge (afaik, none are known explicitly in either case, just some scary upper bounds). These are a bit different then the twin primes though, I don't think there was really much to support these false conjectures besides computations. The twin prime conjecture has other convincing evidence.

 

[Hurkyl]

I love this theorem: the Frivolous theorem of Arithmetic.

 

[Icebreaker]

I love this theorem: the Frivolous theorem of Arithmetic.

What the...

 

[hypermorphism]

What the...

Let b be the natural number such that b + n for any natural n is considered to be a very, very, very large number. Let B be the set of all naturals less than or equal to b. Then the cardinality of B is finite, while the cardinality of the complement of B within the set of all naturals is the same as the cardinality of the set of all naturals. QED.

 

[Icebreaker]

http://mathworld.wolfram.com/StrongLawofSmallNumbers.html

"The first strong law of small numbers (Gardner 1980, Guy 1988ab, Guy 1990) states 'There aren't enough small numbers to meet the many demands made of them.' "

"The second strong law of small numbers (Guy 1990) states that 'When two numbers look equal, it ain't necessarily so.' "

What the...

 

[marteinson]

Actually I think we can prove there are infinitely many prime pairs. But I'm very rusty on formal proofs, so maybe one of you experts could formalize on what I'm saying will work.

There are three types of primes: (a) 2 and 3, (b) those which operated on by mod(6) = 5, and (c) those which under mod (6) = 1.

In other words, every multiple of six, 6n, has a pair of potential primes at 6n +/- 1, as noticed by eratosthenes.

However, no one seems to have used modular arithmetic as I suggest in my paper
http://www.chass.utoronto.ca/french/as-sa/ASSA-14/article7en.html
to separate the primes above 3 into two series, equalling 1 and 5 in mod6, or, you could think of them as equalling 7 and 5 in mod6. There is no interdependency between the primeness of the terms of the two series, 6n+1 and 6n-1, and both series display the only candidates for primeness, and contain all primes, and all their members -are- primes unless factorizable by an inferior member of the same series.

See the new "modulus 6 clock spiral" which I propose to replace Ulam's spiral, in the article, and you'll see what I mean.

Peter

 

[shmoe]

Considering primes mod 6, and indeed primes in more general arithmetic progressions, is an old concept.

That all prime pairs (except 3 and 5) are of the form 6n-1, 6n+1 is nothing new either, nor does it show there are infinitely many prime pairs. It just tells you (vaguely) where to look for them.

"...and all their members -are- primes unless factorizable by an inferior member of the same series."

This is false, 25=1 mod 6 but 25=5*5, and 5 is not 1 mod 6.

The other way is true, if n=5 mod 6 and n is composite then it has a prime divisor congruent to 5 mod 6 (though it may have prime divisors congruent to 1 mod 6 as well)

 

[Hurkyl]

Looking at primes of various modulo classes is done, and not just modulo 6.

You've made a mistake, BTW -- A number of the form 6n+1 can have all of its nontrivial factors of the form 6m-1. (e.g. 25) Also, A number of the form 6n-1 can have factors of the form 6m+1.

 

[shmoe]

http://www.chass.utoronto.ca/french/as-sa/ASSA-14/article7en.html

I've only skimmed some of it, an excerpt:

"Conversely, it can easily be demonstrated that each of the three even series on the spiral can be generated by some combination of two primes, either both in the five o'clock series, both in the seven o'clock series, or one in each, without exception, using simple modular arithemtic. I leave the formally correct proof to real mathematicians, however."

There are certain things that you can wave away with "can easily be demonstrated". Goldbach's conjecture is not one of them.


From your "Simple Algorithm":

" c) test each candidate by dividing it by each prime ≤√m, and by each previously rejected candidate ≤√m"

This is just the sieve of Eratosthenes, after 'pre-sieving' by 2 and 3, except you've added this unnecessary bit that I've highlighted in bold. If m is composite that it has a prime divisor less than or equal to it's square root, so it's sufficient (and faster) to only consider primes less that sqrt(m) here.

 

[marteinson]

I appreciate your insights, and it's helpful that real mathematicians can correct me when I'm wrong. But I don't see why the math community makes such a big deal out of Ulam's spiral's "strikingly non-random appearance!" when its non-randomness can be explained in terms of the 6n+/-1 observation by Eratosthenes, as I have done in the graphical illustration of the mod6 clock spiral.

Clearly, the literature is missing the forest for the tree, in failing to recognize that primes are indivisible precisely because they are adjacent to highly divisible numbers I have nicknamed 'prim' numbers, i.e. such things as multiples of 2 and 3, or 2 and 3 and 4, and so on.

And the two series, if you explore the modular arithmetic of all six series, still do demonstrate the Goldbach conjecture, just loook at them.

Once again, thanks for all your good points.

 

[marteinson]

As to hurkyl's pointing out my first 'error', I think he's incorrect. I never said a number of the form 6n+1 can't have factors on the series 6n-1. On the contrary, I said 6n+1, when factorizable and therefore not prime, either has both non-trivial factors in the form 6n-1 or both in the form 6n+1, or one from each series. On the second error, the "unnecessary bit", he is absolutely right and I stand corrected. It's easy to lose track of common sense when thinking in the abstract, and vice versa. I have taken that part out of the algorithm, which is, as he rightly says, just an Eratosthene sieve with 2 and 3 already taken out.

Many thanks.

 

[marteinson]

And thanks to smoe for his comment, but it's a misunderstanding due to the fact that "series" has the same spelling in the singular and plural. When I write "from the same series" I mean the "same two series".

But thanks for the feedback, it's humbling but useful.

 

[marteinson]

Here is a link to the modulus-6 spiral I drew, for people who want to have a quick look at it.
http://www.chass.utoronto.ca/french/as-sa/ASSA-14/modulus6-spiral90.jpg

Maybe you can see directly from it the properties in mod 6 that I'm trying to point out.

 

[shmoe]

I appreciate your insights, and it's helpful that real mathematicians can correct me when I'm wrong. But I don't see why the math community makes such a big deal out of Ulam's spiral's "strikingly non-random appearance!" when its non-randomness can be explained in terms of the 6n+/-1 observation by Eratosthenes, as I have done in the graphical illustration of the mod6 clock spiral.

They don't really make what I'd consider a big deal out of it. How exactly do you think it's explained by the 6n+/-1?

Clearly, the literature is missing the forest for the tree, in failing to recognize that primes are indivisible precisely because they are adjacent to highly divisible numbers I have nicknamed 'prim' numbers, i.e. such things as multiples of 2 and 3, or 2 and 3 and 4, and so on.

This is false. There will be numbers adjacent to multiples of 2, 3, 4, 5, etc. (any number of factors you like) that are not prime. In otherwords, being adjacent to a 'highly divisible' number (for whatever definition of 'highly divisible' you like) does not make your number prime. Pick any 'highly divisible' number k you like (actually any nonzero number at all that you like), the sequence kn+1 (and also kn-1) will contain infinitely many composite numbers.

And the two series, if you explore the modular arithmetic of all six series, still do demonstrate the Goldbach conjecture, just loook at them.

Again, "just look at them" doesn't even begin to resemble a proof. Give some details on why you think this is true.

And thanks to smoe for his comment, but it's a misunderstanding due to the fact that "series" has the same spelling in the singular and plural. When I write "from the same series" I mean the "same two series".

This would go for Hurkyl's objection too. When you said the "same series" it really looks like the 'same' is there to distinguish between the series 6n+1 and 6n-1, saying which one the factors of 6n+1 (or 6n-1) would have to come from.

Though if you've factored a number of the form 6n+1 into 2 factors, you can't have one factor congruent to 1 and the other congruent to 5, they are both 5 or both 1.

 

[neurocomp2003]

The question above is twin primes ...not paired primes...i thought paries primes was
primes of : (p,p+x)

 

[matt grime]

Here's one for the amateur Goldbach sleth.

Offer a definition of "highly divisible" and I almost certain I can guarantee a string of n "highly divisible" consecutive numbers (ie all the ones lying close to it, again in any sense you care to actually define, are not prime).

 

[marteinson]

Well, only he that knows the way can see that there's a path, and when you don't want to accept something you are inclined not to see it. I NOWHERE said that being adjacent to highly divisible integers automatically makes numbers prime, the point is that prime numbers are prime because they are adjacent to highly divisible integers, and those in that position which are not prime nevertheless have very few factors, also for the very reason that they are adjacent to highly divisible numbers. I suggest you read the entire text thoughtfully and speculatively before disagreeing. The posts here are brief paraphrases, intended to signal what is in the article, nothing more, and you have now, several times, misinterpreted them by skimming and shooting.

The point I'm making, which you don't want to see, is that highly divisible numbers may be thought of as depriving their immediate neighbors of factors, by what I described as a "displacement principle." The neighbors of multiples of six, for instance, are frequently prime, and in fact are the only places you can find any primes above 3, and even when they are not prime, they frequently have only one pair of non-trivial factors, even at orders of magnitude where most integers have several, even dozens, of non-trivial factors. In simple terms, the neighbors of the multiples of six are prime or just missed being prime -- when they have a pair of prime factors, which may be regarded as a coincidence.

So, in essence, you're reading the posts quickly and turning what I am saying in the article around, almost backwards, in order to "hastily" refute it.

Thanks for contributing, however. I'm not surprised to find resistance to the idea. I'll think about the answers to your other questions as well.

 

[Data]

In simple terms, the neighbors of the multiples of six are prime or just missed being prime -- when they have a pair of prime factors, which may be regarded as a coincidence.

This is just not true. I can find you a number with the form 6n \pm 1 with arbitrarily many prime factors (as a matter of fact, every composite number [and, of course, every integer in general] not divisible by 2 or 3 has this form).

The contrapositive also helps: The reason that every prime greater than 3 has the form above is precisely that every positive integer without that form is divisible by either 2 or 3. This says nothing about how many prime factors such integers have (besides having more than one as long as they're greater than 3 !).

 

[Hurkyl]

The neighbors of multiples of six, for instance, are frequently prime

False. As the numbers grow larger, the odds that a neighbor of a multiple of 6 is prime decreases to zero.

Remember the Frivolous Theorem of Arithmetic: almost all natural numbers are very, very, very large. You've only looked at small numbers, (and will ever only look at small numbers!) and have no reason to think large numbers behave like small numbers.

 

[matt grime]

Let's do it in more depth shall we?

As best we can tell:

1) n is a prim number if it is a multiple of 6 and has more factors than n-3,n-2,n-1,n+1,n+2,n+3.

2) Numbers adjacent to prim's are "likely" to be prime.

Though there is no proof of this, and you've only looked at small examples. This almost surely isn't going to hold in general. Obvisouly in the part you've graphed the result is merely a result of the smallness of the numbers you're looking at.

You offer no actual proof that studying "prim" numbers will lead to primes, only some very verbose argument that effectively states that all primes are +-/1 mod 6 except 2 and 3, and that modular arithmetic is under used. This isn't true, and can be used to provide a compellingly short proof of the statement all primes greater than 3 are congruent to +/-1 mod 6.

Simply put you are working with numbers that are far too small to have any interesting behaviour.

For instance, on that chart of pink and yellow highlighted primes and prims, it is no surprise that the "prims" have more factors than their immediate neighbours owing to the size of the numbers involved: it is difficult for numbers less than 32 to have lots of factors if they aren't mutliples of 6.


The "displacement principle" section

"Stated otherwise, all integers adjacent to multiples of 6 have zero factors and are therefore prime, unless by ‘coincidence’ they themselves have a prime pair by which they are divisible. All numbers in this particular n ±1 position either have zero (or very few) factors other than one and themselves."

"very few"? What does that mean? And "coincidence"?

Would you like to test your hypothesis?

the number n=25,194=2^3*3^3*13*17*19

(nb I've not checked the divisors of n+1, n-1, but I can, given enough time, find n a multiple of 6 such that n, n-1 and n+1 have at least as many prime divisors you care to give me via the chinese remainder theorem, or by trial and error.)


is such that n-1 is divisible by 7 and n+1 by 5. Do you think that is going to be a prim number? It has a lot of factors but what about its neighbours? Do your small examples give you any feel for the large ones?

You also state that "multiples of 6 are the most highly divisible numbers" What does that mean? why not multiples of 10 or of 15? What degree of multiplicity are you measuring that by?

 

[shmoe]

I NOWHERE said that being adjacent to highly divisible integers automatically makes numbers prime,...

Funny, because you say this exact thing in the next part of this sentence:

... the point is that prime numbers are prime because they are adjacent to highly divisible integers, and those in that position which are not prime nevertheless have very few factors, also for the very reason that they are adjacent to highly divisible numbers.

Earlier you said: "...primes are indivisible precisely because they are adjacent to highly divisible numbers...", I can't see any way to interpret this besides a belief that being next to a 'highly divisible' number will make you prime. (Of course 'highly divisible number' is a vague poorly defined term here, but that's a different issue.). This isn't the first time the words you've said don't match what you claim to mean mathematically.

The point I'm making, which you don't want to see, is that highly divisible numbers may be thought of as depriving their immediate neighbors of factors, by what I described as a "displacement principle."

Something solid can be made of this. If r>1 is a divisor of m, then r is not a divisor of m-1 or m+1. This is not new or in anyway complicated. However, for any \epsilon>0, the number of divisors* of a number m is O_{\epsilon}(m^\epsilon), while the number of primes less than m is about m/log(m), so as m grows this idea of numbers hogging all the prime factors becomes insignifigant because there are so many more other primes available.

*edit-I had meant to add that the number of distinct prime divisors of m is at most log(m)/log(2). So m 'uses up' an even smaller proportion of the primes.

The neighbors of multiples of six, for instance, are frequently prime, and in fact are the only places you can find any primes above 3, and even when they are not prime, they frequently have only one pair of non-trivial factors, even at orders of magnitude where most integers have several, even dozens, of non-trivial factors.

There's a reason that all the neighbours of multiples of 6 have at most two factors in your little spreadsheet. It's quite simple, 5*7*11>300. What you're observing will go away as you look out furthur and the impact of not being able to be divisible by 2 or 3 diminishes.

In simple terms, the neighbors of the multiples of six are prime or just missed being prime -- when they have a pair of prime factors, which may be regarded as a coincidence.

Data and matt have already mentioned being able to find numbers in 6n+/-1 that have arbitrarily many factors. Let me also mention that you can find an integer k where the numbers 6k+1, 6(k+1)+1, 6(k+2)+1,...6(k+1000000)+1 and 6k-1, 6(k+1)-1, 6(k+2)-1,...6(k+1000000)-1 are all composite. The choice of 1,000,000 here was arbitrary-you can find such a string as long as you like. I'd hardly call something which can occur billions upon billions of times in a row a "coicidence".

So, in essence, you're reading the posts quickly and turning what I am saying in the article around, almost backwards, in order to "hastily" refute it.

By the time of my last post in this thread I had read your article, so I find this insinuation unwarranted, false, and a little insulting. Don't take the fact that I'm not rushing around the streets shouting "Goldbach's has fallen!" as evidence that I don't understand or haven't read it, take it as evidence that I've read it and found it lacking anything interesting (that isn't a trivial observation) or new and correct. You'll often hear mathematicians call the sequence of primes an untameable beast. While from their perspective this is true, the primes are still understood in ways that you haven't begun to imagine. Mathematicians just aren't satisified yet, but this doesn't mean some pretty powerful results aren't known.


I've said that primes in arithmetic progressions have been studied before. You should look into Dirichlet's theorem on the matter to see exactly what it says. So far you've only been discussing 6n+/-1 but you can look at qn+r. For each choice of q these sequences divide up the integers into q 'bins' (depending on your choice of r). Dirichlet will give you an asymptotic relation for the number of primes in each bin and say how they are distributed amongst them. If you consider what happens in the case q=12 for example, you might find it suprising that primes are in fact no more likely to be found in 12n+/-1 than they are in 12n+/-5. I say this might be suprising under the belief that you might feel 12 is 'more highly divisible' than 6.

I don't want to discourage you from investigating prime numbers (or any mathematics at all), but you have to realize that there is a wealth of background information that you haven't seen yet. Without enough background you should consider the possibility that what you're doing has already been done before or is just rubbish. Actually this is something to keep in mind regardless of your background, but it gets easier to tell as you progress. Also, an inability to actually put in precise mathematical terms the things you wish to discuss is only going to cause confusion and frustration.