第3頁P402 易到爆
14588563 發表於 13-11-2012 23:08 
做完了 =,= (真係秒殺  )
但寫proof 就好麻煩
Analysis: If we can prove that at most half of the elements of S are composite numbers, we can then prove at least half of the elements of S are prime.
Proof: To make sure that more composite numbers are the elements of S, the composite numbers should be in the form p^n where p, n ∈ ℕ1, 1\< p^n \< 2012 and p is a prime
thus the composite numbers in S: 2^n, 3^n, 5^n, 7^n, ... , 37^n, 41^n and 43^n
example: 4, 9, 25, 49, ... , 1369, 1681, 1849 (for all the n = 2)
(Note: We can replace 2² by 2³ or 2^4, 2^5... since all the numbers stated here meet the statements: in the form p^n where p, n ∈ ℕ1, 1\< p^n \< 2012 and p is a prime)
We add number 1 to set S as it is not a prime.
there are 15 elements in S now: {1, 2^n1, 3^n2, 5^n3, 7^n4, 11^n5, 13², 17², 19², 23², 29², 31², 37², 41², 43²}
if there exists an element q which is relatively prime with all the elements in S now and it is a composite number, i.e. q = mn where m, n ∈ ℕ1, q must not be a multiple of all the primes within 2 - 43, and, m and n must be prime, thus the least possible value of q would be 47 times 53 = 2491 which is >/2012. Therefore, q does not exist.
Thus the 15 elements left in set S must be prime.
i.e. at least half of the elements of S are prime |
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