精華區beta NSYSU_math 關於我們 聯絡資訊
※ [本文轉錄自 Dirichlet 信箱] 作者: pretend.bbs@lalala.twbbs.org ("Weak and only weak") 標題: Re: e 為無理數 時間: Sat Jul 19 21:44:02 2014 作者: pretend (Big Lion) 站內: pretend 標題: Re: e 為無理數 時間: Fri Nov 5 07:48:14 2004 ※ 引述《pretend (Big Lion)》之銘言: > oo > Assume e is a rational number ==> e = Σ 1/k! = q/p , gcd(p,q)=1 , p,q belong > k=0 > p > to N . Hence e(p!) = (Σ p!/k!) + [p!/(p+1)! +p!/(p+2)! +...] = q[(p-1)!] > k=0 > But [p!/(p+1)! +p!/(p+2)! +...] < p!/(p+1)! [1 +1/(p+1) + 1/(p+1)^2 + ...] > 1 > = [p!/(p+1)!] --------- = 1/p < 1 > p/(p+1) > p > It is obvious that (Σ p!/k!) + [p!/(p+1)! +p!/(p+2)! +...] ≠ q[(p-1)!] > k=0 > So e is a irrational number ...... note : e > 2 is trivial .... e= 1 + 1 + 1/2! + ... < 1 + 1 + 1/2!(1 + 1/3 + 1/3^2 + ...) = 1 + 1 + (1/2!)(3/2) = 2 + 3/4 < 3 Since 2 < e < 3 , it follows that p ≠ 1 -- ===== === === =========================== === ╱ ▌==╱ ▌== ╱ ▌ === 中山lalala小站 ====== = ◢▄▄ ╱ —▌ ◢▄▄ ╱ —▌ ◢▄▄ ╱ —▌twbbsorg ============== From : 220-142-8-99.dynamic.hinet.net ※ 發信站: 批踢踢實業坊(ptt.cc) ※ 轉錄者: Dirichlet (111.255.113.174), 07/19/2014 21:45:56