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Similar work by others has been obtained. In particular, Ramanujan gave an argument for Bertrand's Hypothesis and noted that there are at least 5 primes in (x; 2x] for x 20:5. Our next theorem is a variation of Chebyshev's Theorem. The proof below is due to Erd}os. Theorem 33. If n is a su ciently large positive integer, then 1 n < (n) < 3 n : 6 log n log n Proof. Let m be a positive integer. We begin with the inequalities 2m < 4m : m The rst of these inequalities follows from noting that one can choose m objects from a collection of 2m objects by rst randomly deciding whether each of the rst m objects is to be included in the choice or not.

M) log m: Hence, (2m) ? (m) (log 4) logmm : We consider positive integers r and s satisfying 2r n < 2r+1 and 2s n19=20 < 2s+1 . Observe that s tends to in nity with n. Taking m = 2j above, we deduce 2j 2j (2j +1 ) ? (2j ) (log 4) log(2 (log 4) for j 2 fs; s + 1; : : : ; rg: j) log(2s ) Summing over j , we obtain ? r+1 ? log 4 2s + 2s+1 + + 2r (n)? n19=20 2 ? 2s log(2 s) (log 4)2r+1 2(log 4)n = 2(log 4)n s + 1 1 s log(2 ) s log 2 s (s + 1) log 2 1 log 4 s + 1 n < 2:92 s + 1 n : ? 2(log 4)n s +s 1 = 40 19 19 = 20 s log n s log n log n For n and, hence, s su ciently large, we deduce ?

We give two more examples. Theorem 40. The number of squarefree numbers x is asymptotic to (6= 2 )x. Proof. We make use of the identity Y () 1 ? p12 = 62 : p = x] ? 1 Y 1 ? p12 = 1 + p12 + p14 + p p Y = 1 X 1 = 2: 2 6 n=1 n Denote by A1 (z; x) the number of n x that are not divisible by p2 for every p z . Let A2 (z; x) denote the number of such n that are not squarefree. In other words, A1 (z; x) = jfn x : p2 jn =) p > z gj and A2 (z; x) = jfn x : p2 jn =) p > z; 9p such that p2 jngj: By the sieve of Eratosthenes, A1 (z; x) = X n x 1?

### Elementary Number Theory (Math 780 instructors notes)

by Steven

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