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Old 2018-03-02, 20:09   #34
Steve One
 
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Originally Posted by science_man_88 View Post
It's square roots not square routes.
Therefore you should have written the spelling, not word. Route is the same word as root.
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Old 2018-03-02, 20:19   #35
science_man_88
 
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Quote:
Originally Posted by Steve One View Post
Therefore you should have written the spelling, not word. Route is the same word as root.
No it's not, it's a homophone of it. Not the same thing as being the same.
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Old 2018-03-02, 20:38   #36
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Originally Posted by Steve One View Post
I keep having to repeat that the "if primes end, then....," is not something l believe. It is used to show that if it were true then the equation l wrote would have to equal zero. The equation CAN'T equal zero, therefore there can't be a prime that does this. How many times do l have to say this?
I fully understand this, and I never said you believed it. What you're doing is a proof by contradiction, where you assume that something is true (in this case, that there is a prime that "ends all possibilities for twin primes"), and derive a false result from it to show that it must in fact have been false.

I don't have any problems with the argument you used in your proof. The problem is that what it proves isn't the twin prime conjecture.

You seem to have ignored my "proof" that there are infinitely many primes of the form n2-1. I posted it because I thought it would be useful for you to look at it, work out why it doesn't actually prove what it sets out to prove, and then look at your own work in the same light.
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Old 2018-03-02, 23:51   #37
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Quote:
Originally Posted by 10metreh View Post
Here's a proof that there are infinitely many primes of the form n2-1:

Suppose there are finitely many primes of the form n2-1.
Then there is some prime p which "ends the possibilities" of primes of this form.
But then take N = p#, the product of all the primes up to p.
Now N2-1 is a possibility for a prime of the form n2-1 that is not excluded by primes up to p. This contradicts the statement that p ends all possibilities for primes of this form.
So there are infinitely many primes of the form n2-1.

Of course this can't be a valid proof, as the only prime of this form is 3. The mistake in it is exactly the same as the mistake in your proof.
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I keep having to repeat that the "if primes end, then....," is not something l believe. It is used to show that if it were true then the equation l wrote would have to equal zero. The equation CAN'T equal zero, therefore there can't be a prime that does this. How many times do l have to say this?
We all understand that you don't believe your RAA proposition. But you don't seem to have understood 10metreh. He's pointing out that your 'proof', which contains a thing you don't believe, has the same formal presentation as the invalid 'proof' he gives. The mistake you see in his supposed proof is the same mistake that exists in yours.
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Old 2018-03-03, 04:26   #38
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I think you all go too far with this guy, his real problem is that he has no idea why you used a sharp sign after the p, because this is not a music class, and he also has no idea why you used small n here and big N there (possibly a typo?). And you are trying to teach him math...
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Old 2018-03-03, 12:52   #39
Steve One
 
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Originally Posted by science_man_88 View Post
No it's not, it's a homophone of it. Not the same thing as being the same.
You are correct!
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Old 2018-03-03, 13:48   #40
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Originally Posted by CRGreathouse View Post
We all understand that you don't believe your RAA proposition. But you don't seem to have understood 10metreh. He's pointing out that your 'proof', which contains a thing you don't believe, has the same formal presentation as the invalid 'proof' he gives. The mistake you see in his supposed proof is the same mistake that exists in yours.
When 10metreh gets to ,"So that there are infinitely many primes of the form n squared -1" is where we differ. I am not saying at THAT moment, "therefore there are infinitely many twin primes". I am saying at THAT point, there can be NO prime number that would create a result of zero from the equation. Therefore the equation can only calculate spaces. This is not the same. The equation although true in giving the number of spaces/twin primes when primes 7 up to (n) are placed as lowest prime factors onto 29/31 59/61 89/91 etc does, by it's own failure to create an answer of zero, show that there CANNOT be a prime as lowest prime factor that does the job of ending twin primes.
Imagine,
2 3 2 5 2 7 2 3 2 11 2 13 2 3 2 17 2 19 2 3 2 23 2 5 2 3 2 29 2 31 2

3 2 5 2 37 2 3 2 41 2 43 2 5 2 47 2 7 2 3 2 53 2 5 2 3 2 59 2 61 2 3
These are the lowest prime factors of 2 to 63.
If you take out using 2s, 3s and 5s we get: (7s 11s etc 'become' lowest prime factors)
7 11 13 7 17 19 7 11 23 7 29 31
11 7 37 41 7 43 11 47 7 53 7 59 61
That is one 7, 11, 13, 17 etc repeated. Given the chance to use another 7, 11, 13, 17 etc, the same ""individual spacing""as is on (29/31 59/61 89/91 119/121 etc) to close twin prime possibilities, NEVER will the result produce zero gaps, not to 63 or any number, because (correct use of equation) 5/7 × 9/11 × 11/13 × 15/17.......× (n-2)/n CANNOT equal zero. If twin primes ended at some point, then after the last twin prime, the equation, which DOES calculate spaces left open would have to, using primes 7 up to (n), equal zero. It can't. Ever! The equation can use a prime with 15 gazillion digits. It can't equal zero.
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Old 2018-03-03, 14:03   #41
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Quote:
Originally Posted by Steve One View Post
[snip!]
Please don't bring in modulo anything. Everything is in base 10.
"Modulo" has nothing to do with what base you're using.
Quote:
All that is necessary for you or anyone to ruin my proof [snip!]
As of yet, you've given no proof to be ruined.
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Old 2018-03-03, 14:07   #42
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Quote:
Originally Posted by Steve One View Post
When 10metreh gets to ,"So that there are infinitely many primes of the form n squared -1" is where we differ. I am not saying at THAT moment, "therefore there are infinitely many twin primes". I am saying at THAT point, there can be NO prime number that would create a result of zero from the equation. Therefore the equation can only calculate spaces. This is not the same. The equation although true in giving the number of spaces/twin primes when primes 7 up to (n) are placed as lowest prime factors onto 29/31 59/61 89/91 etc does, by it's own failure to create an answer of zero, show that there CANNOT be a prime as lowest prime factor that does the job of ending twin primes.
Yes, I agree you've proved your expression (which you're calling an equation although it doesn't have an equals sign in it) cannot equal 0. But this doesn't imply there are infinitely many twin primes - just that for any prime p, there are infinitely many pairs of numbers that might be twin primes in that neither of them is divisible by a prime less than or equal to p.
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Old 2018-03-03, 15:20   #43
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For each prime P, there is a finite set of arithmetic progressions with common difference P# (the product of the primes up to P) giving all positive integers p for which neither p nor p + 2 is divisible by any prime less than or equal to P. You have given this for P = 5; there are three such pairs of AP's with common difference 2*3*5.

p = 30*k + 11, p + 2 = 30*k + 13
p = 30*k + 17, p + 2 = 30*k + 19
p = 30*k + 29, p + 2 = 30*k + 31


Each successive prime P multiplies the number of such pairs by P - 2. So appending the prime 7 multiplies the number of pairs by 5, giving a total of 15 such pairs of AP's with common difference 2*3*5*7, or 210. A pair of positive integers p, p + 2 are both indivisible by 2, 3, 5, or 7 if and only if it is of one of the following 15 forms:

p = 210*k + 11, p + 2 = 210*k + 13
p = 210*k + 17, p + 2 = 210*k + 19
p = 210*k + 29, p + 2 = 210*k + 31
p = 210*k + 41, p + 2 = 210*k + 43
p = 210*k + 59, p + 2 = 210*k + 61
p = 210*k + 71, p + 2 = 210*k + 73
p = 210*k + 101, p + 2 = 210*k + 103
p = 210*k + 107, p + 2 = 210*k + 109
p = 210*k + 137, p + 2 = 210*k + 139
p = 210*k + 149, p + 2 = 210*k + 151
p = 210*k + 167, p + 2 = 210*k + 169
p = 210*k + 179, p + 2 = 210*k + 181
p = 210*k + 191, p + 2 = 210*k + 193
p = 210*k + 197, p + 2 = 210*k + 199
p = 210*k + 209, p + 2 = 210*k + 211

This process can be continued. It may thus be shown that, for any prime P, there are infinitely many pairs p, p + 2, neither of which is divisible by any prime less than or equal to P.

Unfortunately, "not divisible by any prime less than or equal to P" doesn't mean "prime." The smallest composite number which is indivisible by any prime less than or equal to P, is the square of the least prime that is greater than P. Thus, e.g. the smallest composite number indivisible by 2, 3, or 5, is 49.

One can say more. It is well known (first proved by Dirichlet) that any individual arithmetic progression a*k + b where a and b are positive integers with gcd(a, b) = 1 contains infinitely many primes. Thus, for example, there are infinitely many primes of the form 30*k + 11. Thus, one can say with certainly that, in any of the pairs p, p + 2 as described above, there are infinitely many values of k for which at least one of the numbers p and p + 2 is prime.

Unfortunately (for the sake of proving the twin primes conjecture) there is no result known about when both p and p + 2 are simultaneously prime for the same value of k.

I would venture a guess that there may be known results that there are infinitely p for which neither p nor p + 2 have more than some fixed number of prime factors, perhaps as few as two prime factors.
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Old 2018-03-03, 18:57   #44
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Quote:
Originally Posted by Dr Sardonicus View Post
I would venture a guess that there may be known results that there are infinitely p for which neither p nor p + 2 have more than some fixed number of prime factors, perhaps as few as two prime factors.
Right -- Chen proved this just before (and published just after) the Cultural Revolution. There are various strengthenings that have been published since, for example Heath-Brown & Li show that i.o. p+2 has at most two prime factors and p+6 has a bounded number of prime divisors.
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