Difference between revisions of "Twin Prime Conjecture"
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if <math>f</math> is a factor of a number in <math>\{m P_s + a_1\}</math> then there is an unique value in <math>\{mP_s + a_1\}</math> which <math>f</math> is a factor when <math>n \leq m<n+f</math>, <math>0 \leq n</math> | if <math>f</math> is a factor of a number in <math>\{m P_s + a_1\}</math> then there is an unique value in <math>\{mP_s + a_1\}</math> which <math>f</math> is a factor when <math>n \leq m<n+f</math>, <math>0 \leq n</math> | ||
− | Mark possible non-prime in pair values generated from arithmetic progression <math>m1P_s+a_1</math> and <math>m2P_s+a_1</math> where values are paired if <math> | + | Mark possible non-prime in pair values generated from arithmetic progression <math>m1P_s+a_1</math> and <math>m2P_s+a_1</math> where values are paired if <math>m_1=m_2</math>. |
The largest factor to eliminate <math>P_s-1</math> is smaller than the number of pairs elements generate by two arithmetic progressions in <math>\{mP_s+a\}</math> where <math>0 \leq m<P_s</math> and <math>a</math> in <math>A_s</math> | The largest factor to eliminate <math>P_s-1</math> is smaller than the number of pairs elements generate by two arithmetic progressions in <math>\{mP_s+a\}</math> where <math>0 \leq m<P_s</math> and <math>a</math> in <math>A_s</math> | ||
− | Can guarantee there are <math>P_s-1-2</math> elements without the factor of <math>P_s-1</math> in a consecutive sequence of <math>P_s-1</math> elements from arithmetic progression <math>\{mP_s+a\}</math> where the two numbers with factor of <math>P_s-1</math> are generated in two different arithmetic progression in two different pairs. Assume the remaining <math>P_s-1-2</math> without factor of <math>P_s-1</math> are in a consecutive sequence eliminate the next smaller odd number which differs by <math>2</math>. | + | Can guarantee there are <math>P_s-1-2</math> elements without the factor of <math>P_s-1</math> in a consecutive sequence of <math>P_s-1</math> elements from arithmetic progression <math>\{mP_s+a\}</math> where the two numbers with factor of <math>P_s-1</math> are generated in two different arithmetic progression in two different pairs. Assume the remaining <math>P_s-1-2</math> pairs without factor of <math>P_s-1</math> are in a consecutive sequence eliminate the next smaller odd number which differs by <math>2</math>. |
− | Assume the remaining <math>P_s-1-2-2</math> without factor of <math>P_s-1-2</math> are in a consecutive sequence eliminate the next smaller odd number which differs by <math>2</math>. | + | Assume the remaining <math>P_s-1-2-2</math> pairs without factor of <math>P_s-1-2</math> are in a consecutive sequence eliminate the next smaller odd number which differs by <math>2</math>. |
Repeat until the number of elements in consecutive sequence is <math>P_s-1-2n =3</math>. Removing numbers with factor of <math>3</math>. There must be a pair of numbers where both of them are prime numbers. | Repeat until the number of elements in consecutive sequence is <math>P_s-1-2n =3</math>. Removing numbers with factor of <math>3</math>. There must be a pair of numbers where both of them are prime numbers. | ||
There must be infinite number of twin primes. | There must be infinite number of twin primes. |
Revision as of 18:12, 19 May 2019
The Twin Prime Conjecture is a conjecture (i.e., not a theorem) that states that there are infinitely many pairs of twin primes, i.e. pairs of primes that differ by .
Contents
[hide]Failed Proofs
Using an infinite series
One possible strategy to prove the infinitude of twin primes is an idea adopted from the proof of Dirichlet's Theorem. If one can show that the sum
of the reciprocals of twin primes diverges, this would imply that there are infinitely many twin primes. Unfortunately, it has been shown that this sum converges to a constant , known as Brun's constant. This could mean either that there are finitely many twin prime pairs or that they are spaced "too far apart" for that series to diverge.
Yitang Zhang approach
A weaker version of twin prime conjecture was proved by Yitang Zhang in 2013. This version stated that there are infinitely many pairs of primes that differ by a finite number. The number Yitang chose was 7,000,000. Terence Tao and other people has reduced that boundary to 246 more numbers.
Elementary proof
Let be the multiplication of the first s prime numbers. Let be the sth prime number Let be the set of numbers relatively prime to and less than .
and where in and and
Pair up numbers generated from two arithmetic progression where
If it is not possible to generate a non-prime in each pair then there exist a twin prime.
The base case for numbers which differ by in is and . Induction there will always be two numbers which differ by in .
Let
and will propagate pairs of elements in which differ by where and and because only the unique values and in their respective arithmetic progression has the factor of when
and will propagate pairs of elements in which differ by where and and because only the unique values and in their respective arithmetic progression has the factor of when
All non-primes numbers generated by where in and can also be found in Therefore removing all numbers from with odd factors between and including to will either leave an empty set or a set only containing prime numbers.
Using the fact that there is a fix set of sequential numbers between numbers with the same factor f in arithmetic progression.
where and is an unique pick from if is a factor of a number in then there is an unique value in which is a factor when ,
Mark possible non-prime in pair values generated from arithmetic progression and where values are paired if .
The largest factor to eliminate is smaller than the number of pairs elements generate by two arithmetic progressions in where and in Can guarantee there are elements without the factor of in a consecutive sequence of elements from arithmetic progression where the two numbers with factor of are generated in two different arithmetic progression in two different pairs. Assume the remaining pairs without factor of are in a consecutive sequence eliminate the next smaller odd number which differs by . Assume the remaining pairs without factor of are in a consecutive sequence eliminate the next smaller odd number which differs by . Repeat until the number of elements in consecutive sequence is . Removing numbers with factor of . There must be a pair of numbers where both of them are prime numbers. There must be infinite number of twin primes.
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