Mathematical Induction

👨🏻‍🏫 學術台數學

知識就是力量

2010-04-17 1:45 am

岩岩諗到條數..
Prove, by MI, that n^n - 1 is divisible by n-1 , where n =/= 0 and n=/= 1

唔知有冇錯...

更新1:

for all integers n.

更新2:

Sorry, typing mistake... It should be n^a - 1 is divisible by n-1 , where n =/= 0 and n=/= 1 and a is any integers.

更新3:

a is any positive integers.

回答 (3)

[匿名]

2010-04-17 5:12 am

✔ 最佳答案

圖片參考：http://i826.photobucket.com/albums/zz190/fai109/66.jpg?t=1271423462

,which is divisible by k

參考: http://i826.photobucket.com/albums/zz190/fai109/66.jpg

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So Much

2010-04-17 7:03 am

when a=2, n^2-1=(n+1)(n-2) , it is true for n=2 ,

Assume it is true for n=k , where k=2,3,....

i.e. n^k-1=(n-1)M, where M is a integer

consider n=k+1 ,

n^(k+1)-1

=n^(k+1)-n^k+n^k-1

=n^k(n-1)+(n-1)M ( by assumption)

=(n-1)(n^k-M) , it is true for n=k+1 ,

By mathematical induction , it is true for all integers greater than 1

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hung

2010-04-17 2:02 am

好似有少少問題 , n=k+1 果陣用直證都做到唔需要n=k既結果

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