# Prime pairs

1. Mar 8, 2005

### 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?

2. Mar 8, 2005

### HallsofIvy

Staff Emeritus
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.

3. Mar 8, 2005

### nnnnnnnn

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

Anyways, I cant 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...

4. Mar 8, 2005

### matt grime

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

5. Mar 8, 2005

### 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.

Last edited: Mar 8, 2005
6. Mar 8, 2005

### 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).

7. Mar 8, 2005

### nnnnnnnn

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

8. Mar 8, 2005

### 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.

9. Mar 8, 2005

### 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.

10. Mar 8, 2005

### 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.

11. Mar 8, 2005

### Hurkyl

Staff Emeritus
12. Mar 8, 2005

### Icebreaker

13. Mar 8, 2005

### hypermorphism

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. :rofl:

14. Mar 8, 2005

### Icebreaker

What the...

Last edited by a moderator: Mar 8, 2005
15. May 1, 2005

### 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

16. May 1, 2005

### 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)

17. May 1, 2005

### Hurkyl

Staff Emeritus
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.

18. May 1, 2005

### shmoe

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.

Last edited: May 1, 2005
19. May 1, 2005

### 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.

20. May 1, 2005

### 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.

Last edited: May 1, 2005