1. Chebychev’s Theorem on the Density of Prime Numbers Yaron Heger and Tal Kaminker

2. So far we only proved that there are ∞ primes. But can we say more than this? We will prove in this lecture a good estimation on the density of primes

3. Density of primes Let us define = number of primes till x Compare this with:

4. For large numbers… xpi(x)x/ln xx/(ln x -1) 1000168145169 10000122910861218 100000959286869512 1000000784987238278030 10000000664579620420661459 100000000576145554286815740304

5. Basically it is possible to prove that: The numbers of primes till n is asymptotically the same as n/ln(n) Meaning: This was proven in 1896 by Hadamard and de la Vallée Poussin

6. But we can’t prove it right now… So we will prove less powerful but still good estimation We will prove that: π(x) = Θ(x/ log x) This is a result of a work by Chebychev made in 1852.

7. Chebychev-type estimates We will now prove that:

8. Lower bound The heart of the proof is understanding some facts of the binomial coefficient Let us define:

9. Overview First we will prove that: Thus, we get that: Then we will prove that: For each

10. Assuming (1) and (2) we conclude… For N = 2n (N is even): For N = 2n+1 (N is odd): Proof:

11. Proof of Using the Newton's binomial: Thus

12. … So, if we manage to prove that is the largest coefficient we will get that: Using simple arguments (for example that the first and the last coefficients are 1) we can get that:

13. Proof that is the largest coefficient Both the numerator and the denominator have n-k elements. But any of the numerator’s elements is larger than any of the denominator elements, thus: This is enough since

14. A graph:

15. Now we know that Now we will prove that: For each Where:

16. Proof that: For each First define: For and prime p, define = the power to which p appears in the factorization of n. Thus, is the largest k≥0 such that And we also get: For example,

17. For each Proof that: Next, define the floor function: = a div b For example: Also, this equals to the number of multiplies b,2b,3b,4b,…mb that do not exceed a

18. Proof that: For each Legendre’s lemma Proof: First define: R p,n = {(i, k) | 1 ≤ i ≤ n and divides i }.

19. Legendre’s lemma R p,n = {(i, k) | 1 ≤ i ≤ n and divides i }. Example for p=2 and n=20: For every ● the pair (i,k) is in Rp,n

20. According to what we have seen on the blackboard, since both sums represents |R p,n | we have: Legendre’s lemma And this proves the Legendre’s lemma.

21. So far we have seen… We proved And now we will prove that: ● ● And we defined:

22. For each Proof that: According to the proof on the blackboard, we get And since, We conclude that:

23. Summary : We first proved that: Then we proved: Finally we proved: And from here we concluded the lower bound theorem. ● ● For each