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標題: Prove n|(n − 1)! for all composite n > 4 [打印本頁]

作者: kingwinner 時間: 10-1-18 04:05 PM 標題: Prove n|(n − 1)! for all composite n > 4

Prove n|(n − 1)! for all composite n > 4.

只諗到要用 fundamental theorem of arithmetic, 但諗唔到點 prove...

請高手賜教, thx!

作者: calvinchong 時間: 10-1-19 01:47 PM

原帖由 kingwinner 於 10-1-18 04:05 PM 發表
Prove n|(n − 1)! for all composite n > 4.

只諗到要用 fundamental theorem of arithmetic, 但諗唔到點 prove...

請高手賜教, thx!

let n = p x q where n > p >= q > 1
since n-1 divides n iff n = 2 for all natural number n, p < n - 1
case 1 : p > q
since n-1 > p > q, and n-1, p and q are all natural number,
p and q will appear in the factorization of (n-1)!
thus pq | (n-1)!, n | (n-1)!

case 2 : p = q
since n > 4, n = p x q = p^2
p >= 3 (if p < 3, p = 1 or 2, then n = 1 or 4, contradicts n > 4)
n = p^2
p^2-2p-1 >= 0  for p > sqrt(2)+1 ~ 2.414
p^2-2p-1 >= 0  for p >= 3
p^2-1 >= 2p for p >= 3
n-1 = p^2-1
   >= 2p
thus n-1 > 2p > p > 3
p and 2p will appear in the factorization of (n-1)!
thus n | (n-1)!

作者: Naozumi 時間: 10-1-19 11:06 PM

樓上的網友的方法很好,只要稍稍改一改p^2-2p-1 >= 0  for p > sqrt(2)+1 ~ 2.414的argument便可

p^2-2p-1 > p^2 - 2p - p    (p>=3)
               = p^2 - 3p
               = p(p - 3)
               >=0                   (p>=3)

作者: kingwinner 時間: 10-1-21 01:29 PM

唔該哂!

請問點解要改 p^2-2p-1 >= 0 for p > sqrt(2)+1 ~ 2.414 呢個 argument?

作者: Naozumi 時間: 10-1-21 09:05 PM

只不過你要再證明點解p^2-2p-1 >= 0 for p > sqrt(2)+1,我廢事再解,用inequality算數

作者: 36661124 時間: 10-1-21 10:21 PM

咩topic 既數黎-,-點解睇唔明

作者: kingwinner 時間: 10-1-22 01:43 PM

number theory

作者: 36661124 時間: 10-1-22 04:15 PM

大學既數-,-?

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