|
hi tl i saw a few math topics recently and noticed there are some good mathematics here. i have a prime number problem and can't solve it.
lets look at this progression (is progression the right word?)
with ![[image loading]](http://s3.imgimg.de/uploads/latex2pngad1d8bebphpad1d8bebpng.png)
it produces:
2 3 7 127 ...
how can i prove that every number coming of this progression is prime? if that's not possible, how can i disprove it?
ty 
|
I haven't done mathematical proofs in donkey's years, but I recall the easiest way to disprove something is to find one specific scenario where the rule doesn't work. That's enough to can the whole formula.
|
the problem is that the numbers get really high really fast. after 127 comes 2^127-1 ... (which is prime)
|
|
Will do.
Anyone with a supercomp here? Need a5 - a100 :D
|
:\ can't you google your homework?
The proof is on wiki.
|
If his homework is interesting in and of itself and generates discussion I think he has the right to share it. Of course math isn't my strong suit so I can't say if its interesting or not :p
|
On April 22 2011 08:03 Starfox wrote:You are looking at http://en.wikipedia.org/wiki/Mersenne_prime basically. If you'd be abelt to proof that every step produces a prime you would be the candidate for the next fields medal for sure. :D
Isn't there a cash bounty for calculating new Mersenne primes?
|
Well...a tiny subset of Mersenne primes, anyway.
|
I was thinking about how to prove it using fermat's little theorem... but according to wiki it is still an open question
http://en.wikipedia.org/wiki/Catalan's_Mersenne_conjecture#Catalan-Mersenne_number
"Although the first five terms (up to M(127)) are prime, no known methods can decide if any more of these numbers are prime (in any reasonable time) simply because the numbers in question are too huge, unless a factor of M(M(127)) is discovered."
|
[edit] sorry, I don't know what I am talking about.
|
On April 22 2011 08:05 graNite wrote: Will do.
Anyone with a supercomp here? Need a5 - a100 :D I don't think any computer will help you to do that. a5 = M127 and is a prime, but above that you are too far out.
If you would be able to prove your statement, then you would have solved one of important unsolved mathematical problems and I do not think anyone here can really help you with that Basically if I see it correctly you would have proven that there is infinite number of Mersenne primes.
EDIT:typo
|
RECURSIVE INDUCTION FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU Just finished a course on this stuff, but I've forgotten most of it sorry man. Best of luck though
|
i tried ..
(01:24) gp > a = 2 %1 = 2 (01:24) gp > a = 2 ^ a - 1 %2 = 3 (01:24) gp > a = 2 ^ a - 1 %3 = 7 (01:24) gp > a = 2 ^ a - 1 %4 = 127 (01:24) gp > a = 2 ^ a - 1 %5 = 170141183460469231731687303715884105727 (01:24) gp > a = 2 ^ a - 1 *** length (lg) overflow
so .. a5 works, a6 does not work anymore^^ it's a little big.
so for a5:
(01:24) gp > a %6 = 170141183460469231731687303715884105727 (01:25) gp > factor(a) %7 = [170141183460469231731687303715884105727 1]
meaning a5 is still prime.
|
2^67 - 1 = 147,573,952,589,676,412,927 = 193,707,721 x 761,838,257,287
So 2^67 isn't prime : ]
Cheers!
|
Oh shit I misread your post, I thought you were asking about Marsenne primes in general. Agh I'm such a doof
|
Can we start nominating Day9 for the fields medal yet?
|
If there were a way to prove that this progression results in primes then people wouldn't be paid $100,000 to find the next largest prime, since 2^a_n - 1 where an is some arbitrarily large prime in that sequence is prime.... we could theoretically have infinitely many large primer numbers, bigger than the largest known prime number.... unless my logic doesnt follow
But what you might say is that youre looking for an upper bound on n for which this statement holds true?
In that case, run a fast fourier transform primality test on a supercomputer with this progession, and hope there exists an n which breaks this progression
|
omg i just screamed when i saw you on my thread, becuase i saw your math vid on youtube 
pls solve this one !! :D
|
2^8 = 256. (2^8)-1 = 255. 255 is not prime. Am i doing this right?
|
8 is not part of this progression.
|
United States3824 Posts
Not that such a sequence exists yet but 
You would want to use an inductive proof to show that the next element in the sequence satisfied your constraint.
which it doesn't.
|
Aren't those Mersenne primes?
|
On April 22 2011 08:35 Rtran10 wrote: 2^8 = 256. (2^8)-1 = 255. 255 is not prime. Am i doing this right? No, the start point is given.
|
You guys have me running Prime95 again
|
I miss when I still found enjoyment in math...
Grade 1-8? Easiest thing in the world.
Algebra 1, 2, and pre calc? Ok, this is a little challening, but very easy once you finally get it.
Calculus and AP calculus? urrgghh
College advanced math? Screw that, I am majoring in biology.
|
TL is not here to do your homework for you.
edit: reopened. apparently this is an unsolved problem, so even if it is his homework, no one can do it for him. :p
|
On April 22 2011 08:45 SonuvBob wrote: TL is not here to do your homework for you.
edit: reopened. apparently this is an unsolved problem, so even if it is his homework, no one can do it for him. :p
The OP needs to rephrase his question so its not so easy to misread. I had to read it like 5 times.
|
United States24667 Posts
Haha my initial insticts were right! I was gonna leave this open then when I saw SoB close it I was like "hmmmm it DOES sound like a hw thread" but... GOOD LUCK. Whoever solves it gets my <3
|
On April 22 2011 12:50 micronesia wrote: Haha my initial insticts were right! I was gonna leave this open then when I saw SoB close it I was like "hmmmm it DOES sound like a hw thread" but... GOOD LUCK. Whoever solves it gets my <3 I doubt anyone here has a Ph D in number theory, and even then, you'd break the world if this problem was solved.
On April 22 2011 12:43 Antisocialmunky wrote:Show nested quote +On April 22 2011 08:45 SonuvBob wrote: TL is not here to do your homework for you.
edit: reopened. apparently this is an unsolved problem, so even if it is his homework, no one can do it for him. :p The OP needs to rephrase his question so its not so easy to misread. I had to read it like 5 times.
Lol there's absolutely no ambiguity in this question, you're clearly trying to backpeddle this post.
|
On the subject of unsolved problems. (e^i*pi) + 1 = 0 <--- blew my mind when I first saw it. Mathematics is amazing.
|
^ Actually, (e^(i*pi)) + 1 = 0 isn't even a problem, it's Euler's formula. Euler's Formula on wiki. you learn the proof in first year calc.
|
It's fairly easy to make plausible conjectures in number theory that are hopelessly beyond the reach of modern techniques, but there's nothing especially interesting about doing so. 2^{2^{127}-1}-1 is much, much bigger than the largest known prime; its primality may never be known within our lifetimes.
|
If you solve a problem like this, i can guarantee you gonna have a fields medal. All this kind of problems "does this sequence have an infinite amount of prime number" are extremely difficult to solve and there are some massive research on many of them.
I think the op was trolling though.
Edit : If the question is, does this sequence give only prime number then it's pretty much sure that the answer is no, and the easiest way to prove is to find the first non prime number.
If the question is, does this sequence can give an infinite amount of prime number but not necessary all of them, then read what i wrote previously.
|
On April 22 2011 07:48 graNite wrote:![[image loading]](http://s3.imgimg.de/uploads/latex2png1deb35a6php1deb35a6png.png) with ![[image loading]](http://s3.imgimg.de/uploads/latex2pngad1d8bebphpad1d8bebpng.png) it produces: 2 3 7 127 ... how can i prove that every number coming of this progression is prime? if that's not possible, how can i disprove it? That one is easy. All that problem is saying is that a number X, if multiplied by itself a prime number of times minus 1 (itself) will always be a prime number.
It's obviously this will only be true if X = 48÷2(9+3)
Since 48÷2(9+3)=2 then all the output of that sequence are prime numbers.
Next?
|
On May 15 2011 08:11 VIB wrote:Show nested quote +On April 22 2011 07:48 graNite wrote:![[image loading]](http://s3.imgimg.de/uploads/latex2png1deb35a6php1deb35a6png.png) with ![[image loading]](http://s3.imgimg.de/uploads/latex2pngad1d8bebphpad1d8bebpng.png) it produces: 2 3 7 127 ... how can i prove that every number coming of this progression is prime? if that's not possible, how can i disprove it? That one is easy. All that problem is saying is that a number X, if multiplied by itself a prime number of times minus 1 (itself) will always be a prime number. It's obviously this will only be true if X = 48÷2(9+3) Since 48÷2(9+3)=2 then all the output of that sequence are prime numbers. Next?
Man what are you smoking? If you can find a sequence with only prime numbers it's fucking huge! I don't think you realise what you are saying...
|
On April 22 2011 08:40 cgrinker wrote:Not that such a sequence exists yet but  You would want to use an inductive proof to show that the next element in the sequence satisfied your constraint. which it doesn't.
worth just requoting this since the thread seems to be full of people with no mathmatical teaching and are just trying to count every prime out
|
On May 15 2011 08:15 Samhax wrote:Show nested quote +On May 15 2011 08:11 VIB wrote:On April 22 2011 07:48 graNite wrote:![[image loading]](http://s3.imgimg.de/uploads/latex2png1deb35a6php1deb35a6png.png) with ![[image loading]](http://s3.imgimg.de/uploads/latex2pngad1d8bebphpad1d8bebpng.png) it produces: 2 3 7 127 ... how can i prove that every number coming of this progression is prime? if that's not possible, how can i disprove it? That one is easy. All that problem is saying is that a number X, if multiplied by itself a prime number of times minus 1 (itself) will always be a prime number. It's obviously this will only be true if X = 48÷2(9+3) Since 48÷2(9+3)=2 then all the output of that sequence are prime numbers. Next? Man what are you smoking? If you can find a sequence with only prime numbers it's fucking huge! I don't think you realise what you are saying... I suppose he is trolling using the formula from the infamous "math" thread about notations.
|
On May 15 2011 08:20 mcc wrote:Show nested quote +On May 15 2011 08:15 Samhax wrote:On May 15 2011 08:11 VIB wrote:On April 22 2011 07:48 graNite wrote:![[image loading]](http://s3.imgimg.de/uploads/latex2png1deb35a6php1deb35a6png.png) with ![[image loading]](http://s3.imgimg.de/uploads/latex2pngad1d8bebphpad1d8bebpng.png) it produces: 2 3 7 127 ... how can i prove that every number coming of this progression is prime? if that's not possible, how can i disprove it? That one is easy. All that problem is saying is that a number X, if multiplied by itself a prime number of times minus 1 (itself) will always be a prime number. It's obviously this will only be true if X = 48÷2(9+3) Since 48÷2(9+3)=2 then all the output of that sequence are prime numbers. Next? Man what are you smoking? If you can find a sequence with only prime numbers it's fucking huge! I don't think you realise what you are saying... I suppose he is trolling using the formula from the infamous "math" thread about notations.
lol ok, i got trolled :p
|
it probably is homework, but it's an interesting question nonetheless past a5, computers can't store the number using standard datatypes if you have any programming experience, perhaps you could create a program to work it out and use some sort of overflow, but that would be a beast to test, not to mention having to factor it realistically thats as far as your going to get in the progression, and should base your theories off that proving it's another matter
|
Dammit. I learned something. But since the equation is posted care to explain how it works exactly?
I'm a lowly high school grade so I don't know things.
|
The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You could write a program which can store the ridiculously large integers produced. Each will need to be tested for divisibility using modulus. Probably fairly time consuming. And when do you stop testing it? You're basically hoping it fails.
|
On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You could write a program which can store the ridiculously large integers produced. Each will need to be tested for divisibility using modulus. Probably fairly time consuming. And when do you stop testing it?
I dont think the op has the computer to do it, but i'm pretty the answer is no but you need a super computer to prove it or if he is lucky the first non prime number is not that big.
|
On May 15 2011 08:46 Samhax wrote:Show nested quote +On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You could write a program which can store the ridiculously large integers produced. Each will need to be tested for divisibility using modulus. Probably fairly time consuming. And when do you stop testing it? I dont think the op has the computer to do it, but i'm pretty the answer is no but you need a super computer to prove it or if he is lucky the first non prime number is not that big.
That's basically my point.
|
On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You do realize "each case" is infinite right? If anyone ever solves this problem, it's not gonna be by making a computer bigger than infinite
|
On May 15 2011 08:52 VIB wrote:Show nested quote +On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You do realize "each case" is infinite right? If anyone ever solves this problem, it's not gonna be by making a computer bigger than infinite 
Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it.
|
I just proved that If a_n is the lowest non prime number in sequence A, then it's smallest dividor is larger that a_(n-1).
That should save you the bothering with checking if A_6 is divisible by a small normal number
|
On May 15 2011 08:48 dUTtrOACh wrote:Show nested quote +On May 15 2011 08:46 Samhax wrote:On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You could write a program which can store the ridiculously large integers produced. Each will need to be tested for divisibility using modulus. Probably fairly time consuming. And when do you stop testing it? I dont think the op has the computer to do it, but i'm pretty the answer is no but you need a super computer to prove it or if he is lucky the first non prime number is not that big. That's basically my point. There are other ways to prove things in math than to test every case, which in case of infinite sets is problematic
|
I feel so hopelessly stupid when i read these math threads
|
From the wikipedia entry on Mersenne primes...
[...]
In mathematics, a Mersenne number, named after Marin Mersenne (a French monk who began the study of these numbers in the early 17th century), is a positive integer that is one less than a power of two:
M_p=2^p-1
Some definitions of Mersenne numbers require that the exponent p be prime, since the associated number must be composite if p is.
A Mersenne prime is a Mersenne number that is prime. It is known[2] that if 2p − 1 is prime then p is prime, so it makes no difference which Mersenne number definition is used. As of October 2009, 47 Mersenne primes are known. The largest known prime number (243,112,609 − 1) is a Mersenne prime.[3] Since 1997, all newly-found Mersenne primes have been discovered by the "Great Internet Mersenne Prime Search" (GIMPS), a distributed computing project on the Internet.
[...]
Many fundamental questions about Mersenne primes remain unresolved. It is not even known whether the set of Mersenne primes is finite. The Lenstra–Pomerance–Wagstaff conjecture asserts that, on the contrary, there are infinitely many Mersenne primes and predicts their order of growth. It is also not known whether infinitely many Mersenne numbers with prime exponents are composite, although this would follow from widely believed conjectures about prime numbers, for example, the infinitude of Sophie Germain primes congruent to 3 (mod 4).
A basic theorem about Mersenne numbers states that in order for Mp to be a Mersenne prime, the exponent p itself must be a prime number. This rules out primality for numbers such as M4 = 24 − 1 = 15: since the exponent 4 = 2×2 is composite, the theorem predicts that 15 is also composite; indeed, 15 = 3×5. The three smallest Mersenne primes are
M2 = 3, M3 = 7, M5 = 31.
While it is true that only Mersenne numbers Mp, where p = 2, 3, 5, … could be prime, often Mp is not prime even for a prime exponent p. The smallest counterexample is the Mersenne number
M11 = 211 − 1 = 2047 = 23 × 89,
which is not prime, even though 11 is a prime number. The lack of an obvious rule to determine whether a given Mersenne number is prime makes the search for Mersenne primes an interesting task, which becomes difficult very quickly, since Mersenne numbers grow very rapidly. The Lucas–Lehmer primality test (LLT) is an efficient primality test that greatly aids this task. The search for the largest known prime has somewhat of a cult following. Consequently, a lot of computer power has been expended searching for new Mersenne primes, much of which is now done using distributed computing.
|
Well, in logic that would be a tautology check.
Example: (that's conjunction and implication if you don't get the notation used)
p ^ q -> p
Now, we asume that -> is false (0), and for it to be false the left hand side of it must be true (p ^ q) while the right hand side (p) must be false (you can't get false results from true things while you can get anything you want from false things).
Then we get:
0 ^ q -> 0
Now we get to the problem, since no matter what we put in stead of q, our implication is going to be true (1), since in no way conjunction can be true if one of its elements is false, thus proving our original statement ( p ^ q -> p) to be a tautology, since it can never produce false results.
Now all you need to do is create such test for your stuff.
|
Wikipedia give the larger Mersenne prime number 2^43 112 609 -1 discovered and 2^(2^127-1) -1 is bigger. So there is no way to prove it even with a super computer.
Close the post :p
|
|
On May 15 2011 09:03 Samhax wrote: Wikipedia give the larger Mersenne prime number 2^43 112 609 -1 discovered and 2^(2^127-1) -1 is bigger. So there is no way to prove it even with a super computer.
Close the post :p
Actually it's perfectly possible that this particular number can be proved as composed. It's just not known to be prime.
Was OP just a troll? So rude.
|
I seriously doubt you have given a formula that always produces prime number.
Thus, I suggest you should compute a few iterations and check if they are primes.
|
On May 15 2011 09:08 ZerGuy wrote:Show nested quote +On May 15 2011 09:03 Samhax wrote: Wikipedia give the larger Mersenne prime number 2^43 112 609 -1 discovered and 2^(2^127-1) -1 is bigger. So there is no way to prove it even with a super computer.
Close the post :p Actually it's perfectly possible that this particular number can be proved as composed. It's just not known to be prime.
I don't think so, because it's very hard top prove that a Mersenne number (2^p-1) is not prime when p is prime so if 2^127 -1 is prime, it's wil be tough.
I'm not saying it's impossible but he will be really hard.
|
On May 15 2011 09:12 Sufficiency wrote: I seriously doubt you have given a formula that always produces prime number.
Thus, I suggest you should compute a few iterations and check if they are primes.
The only iterations easily checkable have been checked, and they are prime.
Like I said, nobody knows if this sequence (Catalan-Mersenne primes) always produces primes. Hell, this sequence is a subset of Mersenne Primes and nobody even knows if there are an infinite number of Mersenne Primes.
|
On May 15 2011 09:08 ZerGuy wrote:
Was OP just a troll? So rude.
Probably a troll. By the time you discuss questions this hard, you know whether they're even solved. I have a hard time believing some high schooler/undergrad found this problem scribbled in their math textbook and wanted to know the solution. Solving this problem would be a huge achievement for a professional mathematician.
|
I have discovered a truly marvellous solution for this problem. Unfortunately, the post width here is too narrow to contain it.
|
On May 15 2011 09:20 aoeua wrote: I have discovered a truly marvellous solution for this problem. Unfortunately, the post width here is too narrow to contain it.
Haha Fermat quote, but Andrew Wiles did it!
|
On May 15 2011 09:21 Samhax wrote:Show nested quote +On May 15 2011 09:20 aoeua wrote: I have discovered a truly marvellous solution for this problem. Unfortunately, the post width here is too narrow to contain it. Haha Fermat quote, but Andrew Wiles did it! 
Hahaha at the reference.
And yes Andrew Wiles did it, using more than the margin . Also using math that wasn't invented until many, many years later.
To a Mod: Now that this problem is known to be unsolved, can this thread be closed?
|
On May 15 2011 08:57 Samhax wrote:Show nested quote +On May 15 2011 08:52 VIB wrote:On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You do realize "each case" is infinite right? If anyone ever solves this problem, it's not gonna be by making a computer bigger than infinite  Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it.
2^{127}-1 is prime.
|
On May 15 2011 10:32 blah_blah wrote:Show nested quote +On May 15 2011 08:57 Samhax wrote:On May 15 2011 08:52 VIB wrote:On May 15 2011 08:44 dUTtrOACh wrote: The only way to test if it produces only prime numbers is to run divisibility tests on the answers for each case. You do realize "each case" is infinite right? If anyone ever solves this problem, it's not gonna be by making a computer bigger than infinite  Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. 2^{127}-1 is prime.
that's unfortunate for the OP :p
|
If you could prove every number was a prime you be in line for hundreds of thousands of dollars
|
On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it.
This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility.
|
On May 15 2011 11:26 Wonders wrote:Show nested quote +On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility.
I didn't say for 2^127 -1 you need a super computer, this number is "relatively" small (~10^38). But for 2^(2^127-1) -1, show me how you can do it whitout a super computer, knowing that 2^127 -1 is prime. I really want to see that.
edit : when i talked about super computer, i was talking about the next iterations.
|
On May 15 2011 09:20 aoeua wrote: I have discovered a truly marvellous solution for this problem. Unfortunately, the post width here is too narrow to contain it.
I've always wondered if fermat was just a huge troll or if he actually thought he had the proof - although of course he could not have.
|
one of the rule of TL is do not ask homework here!
about the math problem, use induction (basic 1st year math)
|
On May 15 2011 11:51 NB wrote: one of the rule of TL is do not ask homework here!
about the math problem, use induction (basic 1st year math)
You are wrong. This problem is unsolved. He was trolling.
|
On May 15 2011 11:41 Samhax wrote:Show nested quote +On May 15 2011 11:26 Wonders wrote:On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility. I didn't say for 2^127 -1 you need a super computer, this number is "relatively" small (~10^38). But for 2^(2^127-1) -1, show me how you can do it whitout a super computer, knowing that 2^127 -1 is prime. I really want to see that. edit : when i talked about super computer, i was talking about the next iterations.
2^(2^127-1)-1 cannot be done even with today's super computers, although people have checked all possible factors up to 10^50 without success.
|
On May 15 2011 11:55 Mithrandir wrote:Show nested quote +On May 15 2011 11:41 Samhax wrote:On May 15 2011 11:26 Wonders wrote:On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility. I didn't say for 2^127 -1 you need a super computer, this number is "relatively" small (~10^38). But for 2^(2^127-1) -1, show me how you can do it whitout a super computer, knowing that 2^127 -1 is prime. I really want to see that. edit : when i talked about super computer, i was talking about the next iterations. 2^(2^127-1)-1 cannot be done even with today's super computers, although people have checked all possible factors up to 10^50 without success.
Yeah i know, but knowing the size of the number it's impossible to test it without computers. It's my point.
|
On May 15 2011 12:00 Samhax wrote:Show nested quote +On May 15 2011 11:55 Mithrandir wrote:On May 15 2011 11:41 Samhax wrote:On May 15 2011 11:26 Wonders wrote:On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility. I didn't say for 2^127 -1 you need a super computer, this number is "relatively" small (~10^38). But for 2^(2^127-1) -1, show me how you can do it whitout a super computer, knowing that 2^127 -1 is prime. I really want to see that. edit : when i talked about super computer, i was talking about the next iterations. 2^(2^127-1)-1 cannot be done even with today's super computers, although people have checked all possible factors up to 10^50 without success. Yeah i know, but knowing the size of the number it's impossible to test it without computers. It's my point. If you'd put all the computers in the world to work on this problem for the next 10 billion years, you wouldn't yet know if that number is really prime. Assuming it is prime.
That's how big that number is. Math has many simple ways to tell the people that some questions will never get responses.
|
On May 15 2011 12:07 arbiter_md wrote:Show nested quote +On May 15 2011 12:00 Samhax wrote:On May 15 2011 11:55 Mithrandir wrote:On May 15 2011 11:41 Samhax wrote:On May 15 2011 11:26 Wonders wrote:On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility. I didn't say for 2^127 -1 you need a super computer, this number is "relatively" small (~10^38). But for 2^(2^127-1) -1, show me how you can do it whitout a super computer, knowing that 2^127 -1 is prime. I really want to see that. edit : when i talked about super computer, i was talking about the next iterations. 2^(2^127-1)-1 cannot be done even with today's super computers, although people have checked all possible factors up to 10^50 without success. Yeah i know, but knowing the size of the number it's impossible to test it without computers. It's my point. If you'd put all the computers in the world to work on this problem for the next 10 billion years, you wouldn't yet know if that number is really prime. Assuming it is prime. That's how big that number is. Math has many simple ways to tell the people that some questions will never get responses.
hum not sure about that, do you know Quantum computer (search it on wikipedia if not). Maybe one day, it will be possible, who knows.
|
On a sidenote: I've noticed an increasing amount of posts starting with 'hi TL' or something like that. It seems similar to 'sub /b/' or whatever. Anyway, my point is, why the hell would you use 'hi TL' on a TL? Just 'Hi!' would be enough in my opinion. There's no need for redundant stuff, especially one that shares some similarities with 4chan...
|
On May 15 2011 18:08 Manit0u wrote: On a sidenote: I've noticed an increasing amount of posts starting with 'hi TL' or something like that. It seems similar to 'sub /b/' or whatever. Anyway, my point is, why the hell would you use 'hi TL' on a TL? Just 'Hi!' would be enough in my opinion. There's no need for redundant stuff, especially one that shares some similarities with 4chan... Similarly, 4chan.org and teamliquid.net both have advertisements. I don't think teamliquid.net should have advertisements because 4chan.org has them. Also, 4chan.org has moderators, so I think the banlings should surrender their powers. 4chan is a website, so I think teamliquid should become a restaurant.
Seriously. All he did was greet us, it's not the least bit rude.
|
On May 15 2011 12:07 arbiter_md wrote:Show nested quote +On May 15 2011 12:00 Samhax wrote:On May 15 2011 11:55 Mithrandir wrote:On May 15 2011 11:41 Samhax wrote:On May 15 2011 11:26 Wonders wrote:On May 15 2011 08:57 Samhax wrote: Maybe 2^127 -1 is not prime, if the op is lucky. But if 2^127 -1 is prime then you need for sure a super computer to prove it. This number was known to be prime in 1876, when they didn't have supercomputers. Even then there were much better techniques for testing primality than exhaustively checking divisibility. I didn't say for 2^127 -1 you need a super computer, this number is "relatively" small (~10^38). But for 2^(2^127-1) -1, show me how you can do it whitout a super computer, knowing that 2^127 -1 is prime. I really want to see that. edit : when i talked about super computer, i was talking about the next iterations. 2^(2^127-1)-1 cannot be done even with today's super computers, although people have checked all possible factors up to 10^50 without success. Yeah i know, but knowing the size of the number it's impossible to test it without computers. It's my point. If you'd put all the computers in the world to work on this problem for the next 10 billion years, you wouldn't yet know if that number is really prime. Assuming it is prime. That's how big that number is. Math has many simple ways to tell the people that some questions will never get responses. That assumes that checking divisibility is the only way to check this or even that our current methods for checking primality are the only ones. I agree that we won't find a solution to OP, but that does not mean that propositions here are the only way to get there.
EDIT: by "we" I mean this thread/TL
|
|
|
|