In mathematics, the Sieve of Eratosthenes is a simple, ancient algorithm for finding all prime numbers up to a specified integer.

It was created by Eratosthenes, an ancient Greek mathematician.

Consider a contiguous list of numbers from two to some maximum.

Strike off all multiples of 2 greater than 2 from the list.

The next lowest, uncrossed off number in the list is a prime number.

Strike off all multiples of this number from the list. The crossing-off of multiples can be started at the square of the number, as lower multiples have already been crossed out in previous steps.

Repeat steps 3 and 4 until you reach a number greater than the square root of the highest number in the list; all the numbers remaining in the list are prime.

Suppose we want to determine all primes less than 100.

First we write the integers from 2 to 100. we know that 2 is prime; so we encircle it,

and cross out the remaining even numbers on the list.

Now, the lowest number is 3; so we cross out again every 3rd number thereafter.

We continue the same process until we reach the last number that hasn’t been crossed out.

In order to find all the primes up to n, we need only to seive out multiples of primes ≤ √n.

To find the primes up to 100, we need only to cross out multiples of 2,3,5 and 7.

The operation of the Sieve of Eratosthenes, suggests that primes becomes rarer as the integer goes up.

For example, in finding for the prime numbers bet. 100-150.

It consists of crossing out the multiple of 5 primes ( 2,3,5,7,11) not exceeding √150 ≈ 12.2

The sieve suggests that at some point, the primes will become inexixtent. This is in contrary to what euclid said, “that primes are infinite”.

Euclid’s proof that primes is infinite

Q= p1p2..........pk + 1

Where: Q – either prime or has a prime factorp1p2..........pk – prime numbers.

Find all primes between 5- and 100. Find all primes between 100 and 150. Find all primes between 1000 and

1025. Find ( P ,n ) for n = 11 , 12 , . . . , 20 Find ( P , n ) for n = 21 , 22 , . . . , 30.

Τ (n)So, going back to the previous topics on

the tau τ (n).Suppose,

n = p1k1 p2

k2 . . . . . Pt

kt ,

Where ,0 ≤ j1 ≤ k1 for i = 1,2 . . . , t

Example: n= 63 = 32 7, then the divisors are

30 70 31 70 32 70

30 71 31 71 32 71

When wrtting for the divisors of a certain imteger, say 63, we find it convenient to organize them into a rectangular array.

This array reminds us of a multiplication table, because it is indeed.

1 3 9 1 1 3 9

7 7 21 63

As shown in the table, we have written the positive divisors of 63 ( 9 , 7 ).

Each entry in the table is the product of a divisor of 9 and one of 7. Thus in this case

Τ (n) = Τ (a) Τ (b) = a ∙ b = abΤ (63) = Τ (9) Τ (7) = 3 ∙ 2 = 6

This suggests that we can prove that Τ (n) = Τ (a) Τ (b) = a ∙ b = ab by showing the divisors of ab are just all products of a divisor of a with one of b.

Theorem 2.11. if a and b are relatively prime positive integers, then τ (ab) = τ (a) τ (b).

This is the theorem that served as the basis for our computations used in finding primes.

Theorem 2.12. if a and b are relatively prime positive integers, then the set of positive dicvisors of ab consists exactly of all products de, where d is a positive divisor of a and e is a positive divisor of b. Furthermore, these products are all distinct.

Find τ ( n ) by means of the formula presented in this discussion:

n = 75 n = 45 n = 30Make a multiplication table of products of

positive divisors of a times positive divisor of b.

a = 8 , b = 15a = 28 , b = 21

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