Dirichlet Prime Number Theorem on Arithmetic Progression

Dirichlet's theorem is also called as 'Dirichlet prime number theorem'. This theorem illustrates that for any arithmetic progression, the sum of the reciprocals of the prime numbers in the progression diverges, and that different arithmetic progressions with the same modulus have approximately the same proportions of primes.

Dirichlet Prime Number Theorem on Arithmetic Progression – Proof,Example

Statement
    An arithmetic progression in the form of a + nd will possess unlimited prime numbers in case a and c are 'co prime'.

Given
    Arithmetic Progression (AP) = a, a+d, a+2d, a+3d, ...

To Prove
    AP of form 'a + nd' has infinite prime numbers, when both the a and d are relatively prime.

Proof
Case 1 : When a and d are co-prime
For example, a = 2, d = 3
Arithmetic Progression (AP) = 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38,
41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, ...
Prime Numbers of AP = 2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, ...
This series contains infinite prime numbers.

Case 2 : When a and d aren't co-prime
For example, a = 6, d = 3
Arithmetic Progression (AP) = 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39,
42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, ...
Prime Number of AP = NIL
No prime numbers are available in the AP series.

From case1 and case2, the Dirichlet's Theorem was proved in case1 which contains infinite prime numbers.
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