Quod Erat Demonstrandum

2012/01/03

某質數題

Filed under: NSS,Pure Mathematics — johnmayhk @ 3:01 下午

答網友:

問題

p_n 為第 n 個質數。試以數學歸納法(Mathematical Induction)證明,對任意正整數 n,必有

p_n < 2^{2^n}

解答

n=1p_1=2 < 2^{2},故命題成立。

設 p_n < 2^{2^n} for n=1,2,\dots ,k

考慮

p_{k+1}

\le p_1p_2\dots p_k + 1 (為何?)

< 2^{2^1}2^{2^2}\dots 2^{2^k} + 1

= 2^{2^{k+1}-1} + 1

< 2^{2^{k+1}}

(注:利用 Bertrand’s postulate 可得比 2^{2^n} 好的上限 2^n for n > 1

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