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#1
Mar805, 01:48 PM

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There is (as far as I know) no prooffor 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
Mar805, 02:50 PM

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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
Mar805, 03:46 PM

P: 65

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
Mar805, 04:53 PM

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



#5
Mar805, 06:14 PM

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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. 


#6
Mar805, 06:19 PM

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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 [itex]Cx(\log{x})^{2}[/itex], 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
Mar805, 06:21 PM

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#8
Mar805, 06:25 PM

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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
Mar805, 06:29 PM

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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
Mar805, 06:43 PM

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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 [itex]\sum_{n\leq x}\mu(n)\leq x^{1/2}[/itex] where [itex]\mu[/itex] 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
Mar805, 06:53 PM

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I love this theorem: the Frivolous theorem of Arithmetic.



#12
Mar805, 07:27 PM

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#13
Mar805, 07:32 PM

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Mar805, 07:36 PM

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#15
May105, 11:48 AM

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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/...rticle7en.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 6n1, 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
May105, 12:27 PM

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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 6n1, 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
May105, 12:34 PM

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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 6m1. (e.g. 25) Also, A number of the form 6n1 can have factors of the form 6m+1. 


#18
May105, 12:48 PM

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"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 'presieving' 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. 


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