The Conjecture
'Twin primes' are are prime numbers with a difference of 2. So, any 2 primes <P1> and <P2> such that
P1 + 2 = P2
would be twin primes. These are a special subset of primes, discovered by Euclid, around 300 BC. Since then, mathematicians have been pondering over a still unproven conjecture, called the 'Twin Prime Conjecture'. This conjecture states that there are infinitely many twin primes. This means there is no end to the number of twin primes, there is always such a pair larger then the one which we know.
This relatively easy to state problem has turned out be remarkably tough to prove.
Proof
Let <P> denote a prime number, and the subscript, <1, 2, 3, 4 ........> denote the order of the prime. Then <P1> is the first prime, 2, <P2> is the second prime, 3 and so on....
P1 = 2
P2 = 3
P3 = 5
.
.
.
It has already been proved that there are an infinite number of primes, so if we pick the 'nth' prime it would be <Pn>. multiplying all the primes lesser then <Pn> would yeild a composite number, <Q>.
P1 * P2 * P3 ........ * Pn = Q
Here, it is notable that the numbers <Q - 1> and <Q + 1> would be 'coprime' to <P1>, <P2 >, <P3> and so on, all the way till <Pn>. It is also notable that if these two numbers are prime, they would also be 'Twin Primes', and they would be prime if there exists no prime factor for <Q - 1>, and <Q + 1>.
Now, <Q - 1> and <Q + 1>, we already know are 'coprime' with all the primes from <P1> to <Pn>. So if they are also 'coprime' with all numbers larger than <Pn>, and smaller than <Q>, then they will be prime, consequently twin primes, and [since there are infinite possible <Pn>'s and <Q>'s] the twin prime conjecture would be proven true.
Lets leave this aside for some time.
PROPOSITION
Let us take 2 arbitrary numbers: <x>, and <y>, where we assume that <x> is smaller than <y>. It is obvious that for any number larger than these <z>, it is more probable for it [<z>] to be a multiple of <x>, than <y>.
if
x < y < z,
then
chances for z to be a multiple of x > chances for z to be a multiple of y
[ chances for z to be a multiple of x = 1/x ]
[ chances for z to be a multiple of y = 1/y ]
[ and it is given that 1/x > 1/y, as x < y ]
(Explained in the Comments)
Leaving this proposition aside, lets resume from where we left off.
Let <Pf> be the number larger than <Pn>, and smaller than <Q>, and also be a factor of <Q - 1>.
Pn < Pf < Q
a * Pf = Q-1
By definition, <P1> is smaller than <Pn>, which is smaller than <Pf>, which is smaller than <Q>.
Which is similar to the situation in the PROPOSITION, where <x> is <P1> [a constant: the smallest prime number - 2], <y> is <Pf>, and <z> is <Q - 1>.
P1 < Pn < Pf < Q
So the chances of the 'existence' of <Pf> are 1 / <Pf>, which we know is smaller than 1 / 2. So, by saying "the chances of the 'existence' of <Pf> are 1 / 2", we can be sure that we are increasing the chances of the 'existence' of <Pf>. In other words, more than half the time, <Pf> doesn't exist [ther is no <Pf>]!
Here, we said "more than half the time", but how many times is "time"? The answer: infinity, because there are infinite different values which <Q - 1> might have [ which is because there are infinite different values which <Pn> might have-as there are infinite prime numbers ].
Therefore, the probability of the 'existence' of Pf < 1/2.
if we increase it to 1/2 to be on the safe side,
even then the number of cases where Pf doesn't exist remains INFINITY, because,
1/2 of INFINITY is also INFINITY.
But how do we know that for these cases[cases of a <Q>, such that <Pf> doesn't exist], even <Q + 1> will be prime? There is a very an argument for this question, which is very similar to the above:
Let <Pg> be the number larger than <Pn>, and smaller than <Q>, and also be a factor of <Q + 1>.
Pn < Pg < Q
b * Pg = Q+1
By definition, <P1> is smaller than <Pn>, which is smaller than <Pg>, which is smaller than <Q>.
Which is similar to the situation in the PROPOSITION, where <x> is <P1>, <y> is <Pg>, and <z> is <Q + 1>.
P1 < Pn < Pg < Q
So the chances of the 'existence' of <Pg> are 1 / <Pg>, which we know is smaller than 1 / 2. So, by saying "the chances of the 'existence' of <Pg> are 1 / 2", we can be sure that we are increasing the chances of the 'existence' of <Pg>. In other words, more than half the time, <Pg> doesn't exist [there is no <Pg>]!
Here we again referred to "more than half the time" time refers to infinity here as well, this is the infinity of the cases where <Pf> doesn't exist.
Therefore, the probability of the 'existence' of Pg < 1/2,
if we increase it to 1/2 to be on the safe side,
even then the number of cases where Pg doesn't exist remains INFINITY, because
1/2 of 1/2 of INFINITY
=1/4 of INFINITY
= INFINITY.
Please Comment.
Kaivalya
Please check the following links, there are other claims to the proof, though they seem much more complicated.
http://arxiv.org/ftp/math/papers/0309/0309103.pdf
http://www.recoveredscience.com/primes1ebook02.htmKaivalya
Please check the following links, there are other claims to the proof, though they seem much more complicated.
http://arxiv.org/ftp/math/papers/0309/0309103.pdf
http://www.mersenneforum.org/showthread.php?t=12029
Explanation for the proposition:
ReplyDeleteconsider a number line such
0...1...2...3...4...5...6...7...8...9...10...11...
now, if x = 3,
and z is some number larger than 3
we can highlight all the multiples of 3, in the number line.
0...1...2...[3]...4...5...[6]...7...8...[9]...10...11...[12].......
if we continue this till infinity, 1/3 numbers in this case, are multiples of 3, hence the probability of a number[z] between 3 and infinity i.e. z>3 being a multiple of 3 is 1/3.
This is just for the case where x=3, but the same thing can be generalized to the statements in the proposition.
the probability of there being infinite twin primes is:
ReplyDeleteprobability of non - existence of pr that divides (Q+1) or (Q-1)
divided by
probability of existence of pr that divides(Q+1) or (Q-1)
which holds true because the probability of existence is inversely related/proportionate to the probability of existence of the prime pr that divides (Q+1) or (Q-1)
And you are right when u say half of the time(infinity) it doesn't exist therefore infinity of the time twin primes are infinity/x for some x and thus twin primes are infinite.
but even though less than half the time a prime exists that divides your (Q-1) or (Q+1) time is still infinite therefore the chances of such existences is still infinite and therefore by the formula above the x in it is also infinity.
in other words:
since probability of nonexistence is inf and the probability of existence is inf
probability of inf existence of twin primes = inf/inf = undefined.
so its not proven yet(easily)
Dr. Shailesh Shirali will be looking into this in a little bit of time
ReplyDeletethere seems to be an infinite chance of both existence and non existence of infinite twin primes by your proof
ReplyDeleteNo, it just says that at least 1/4th of the times Q-1 and Q+1 are twin primes. This means if you generate them this way, then a quarter of the time you will get twin primes. Yes, there will be infinite cases where generation this way will not work, but for an infinite number of cases it will work, hence proving the conjecture.
ReplyDeleteno if u take an infinite group of (Q-1) and (Q+1)s, and find out the pr that divides them i.e:
ReplyDeleteshort example:
{5,7 , 11,13 , .... ....}
now for the first 2 sets here as shown, pr is nonexistent and at some later point it will exist
right?
now suppose that pr doesn't exist for x values and exists for y values where:
x > y (so pr doesn't exist most of the time)
also probability of existence of twin primes are:
x/y
by my previous comment.
now as u traverse through the list increasing x and y, even if the rate of increase of x is trillion times gr8er than that of y x will not reach infinity before y bcoz at any point on the number line there is an infinite distance to travel.
and the rate of increase of x cannot be infinitely greater than y.
so as the list approaches it's end,
x approaches inf
at the same time as y approaches inf
Note that i use the word approaches and not is or becomes.
and therefore u end up with an undefined probability for twin prime's inf existence.
time never quite ends with 2 values increasing with relation to time unless they both reach inf
ReplyDeletebut if y approaches infinity, the conjecture is proved
ReplyDeletehowever, yes, it is true that this generation algorithm isn't certified to generate twin primes
ReplyDeletealso note:
ReplyDeletex is no. of times PF(Pr) exists, y is the no. of times it doesn't. then:
x + y = INFINITY
x = INFINITY - y
x = INFINITY
which prooves the conjecture!
but:
ReplyDeletex+y = inf
y = inf - x
y = inf
if it doesn't exist inf number of times then it never exists.
this disproves it.
u get 2 conflicting answers which is what results in the invalidity of your proof
yes but, here the 'inf' we are referring to is larger than x or y (by definition)
ReplyDeletei agree that x = y, and both can be shown to be infinity.
so this means that, upon generating twin primes this way, an infinite no. of times twin primes will be generated, and an infinite no. of times they won't.
so this generation algorithm has no guarantee of generating twin primes. However, it is still true that there are an infinite no. of twin primes. (x = infinity).
if you want to argue in that manner:
ReplyDeleteit is also true that there are not inf twin primes(y = inf).
and you can't say if y reaches inf first or x does bcuz they reach at the same time.
ReplyDeleteno, but this is a little complicated, so let me introduce some notation.
ReplyDeleteX is no. of times PF(Pr) exists
Y is the no. of times it doesn't
X + Y = INFINITY ....... [by definition]
X = INFINITY - Y ....... [transposing]
X = infinity ........... [inf - a = inf]
note that i use two different kinds of infinities here. This is obvious as we know that there are many more numbers other than X between 0 and INFINITY, so even if X = infinity, there are many numbers left out which can still be used to prove the twin prime conjecture. Some of these are in Y, which is also infinity.
So, even if X = infinity, and Y = infinity.
The twin prime conjecture is proved.
The next comment has an example illustrating these diffenr kinds of infinities.
INFINITY is the set of positive integers.
ReplyDeleteand X is the set of all the multiples of 10.
and Y is the set of all the multiples of 9.
we know that X and Y are infinite sets, however both are smaller than INFINITY.
so, if X = infinity, that doesn't disprove tha conjecture, because
INFINITY - X = infinity
and the elements in Y are form this infinity
(not exactly, because 9 and 10 have common multiples, but still).
so, recapping
X = infinity
Y = infinity
and yet the twin prime conjecture is proved (because all it requires is that Y have an infinite no. of elements, which it does.)
ok look here:
ReplyDeletethe number of primes under x is asymptotic to x/log(x).
this is to base e.
this means that the higher x is, the more accurate x/log(x) is altho it is very inaccurate for very small x.
(for proof search for: how many primes under x in google or sumthin).
now lets take your (Q-1) as the x here and therefore the number of primes below (Q-1) is approx = (Q-1)/log(Q-1).
let us also assume that the components of Q are:
(P1 x P2 x P3 ... x Pf).
then the number of primes between (Q-1) and Pf is approx =
[{(P1 X P2 ... X Pf) - 1}/log(Q-1)] - Pf.
this log is to (base e).
and as we see this is an increasing value with increase in Q as in the higher the Q we take, the more no. of primes exist between (Q-1) and Pf.
This implies that the chances of there being a prime Pg that divides (Q-1) or (Q+1) such that
Pf < Pg < (Q-1)
increases with Q and thus the chances of y(the count of times Pg divides) increasing increases and the chances of x(the vice versa of y) increasing decreases. (plz read thoroly).
And thus the higher the Q, lower the x AND higher the y.
so, the chances of x increasing reduces with higher Qs because of higher possibilities of Pg due to the increasing number of primes available as candidates for such a Pg. and the chances of y increasing increases(plz read thoroly).
now as we approach inf, the chances of x increasing approaches 0 and chances of y increasing approaches inf.
so at some point the chances of x increasing may reach 0 at which case there won't be anymore twin primes.
Now if you can prove that the chances of x increasing will never reach 0, you've proven the conjecture, if not it's still in doubt
If what u r saying is right, you have dis-proven the conjecture.
ReplyDeleteHowever, your argument starts of differently in itself, and has no relation whatsoever with my previous comment (not that it matters).
no i haven't dis-proven it.
ReplyDeletelook your mistake was not in your last post that was ok but you made a mistake in your original post, u said:
the probability of an existence of a Pg that divides (Q-1) or (Q+1) may be more than half but over an infinity of Qs the amount of times the primes don't divide(even if less than 1/2 probability) there will be inf such cases which would prove the conjecture.
but by the earlier post i say that the probability of primes not dividing (Q-1) or (Q+1), decreases with higher Qs.
now this gives possibilities to 3 results:
1. if one can prove that the chances of x increasing reach 0 at a finite number, the conjecture is dis-proved. i have not done this.
2. if one can prove that the chances of x increasing never reach 0, the conjecture is proved.
3. A very interesting result (and i have a feeling the most probable from my statements above): the chances of increase in x reach 0 at inf. this seems so absurd imagine increasing with a decreasing rate of change a number, whose rate of change approaches 0 at inf, its a bit like:
1/x is the rate of change since 1/inf is practically 0 it seems to suit the 3rd condition.
so now one of these results are left to prove. (i suppose)
Clarification: you have said that
ReplyDelete.... "the number of primes between (Q-1) and Pf is approx =
[{(P1 X P2 ... X Pf) - 1}/log(Q-1)] - Pf.
this log is to (base e)." ......
We know that the no. primes under Q-1 = R =
Q-1/log(Q-1)
and the primes under Pf = T =
Pf/log(Pf)
So, the number of primes between them = S =
R - T
this would be:
[Q-1/log(Q-1)] - Pf/log(Pf), not
[Q-1/log(Q-1)] - Pf, as you have stated.
This could also be written as
[Q-1/log(Q-1)] - f
However, this has little effect on what you have said following these statements
ReplyDeleteyea thats right sorry also that gives the function an even higher value than my incomplete one.
ReplyDelete