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
