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This can by done by Induction
Show that for all integers n >1, n^3 - n can be divided by 3. (Note: n^3 stands for n*n*n)" o& p! r) v. g! G. j
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Proof: 4 M. g" K8 V9 ?1 m
Let n >1 be an integer
6 A6 r) V; b+ R$ R2 t mBasis: (n=2)
1 P+ ]# e6 x% q# p4 ?' ^) @ 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3+ t/ ?, Y7 D* m g7 a/ N* J+ A
; D: u7 K+ S5 k1 q: @1 j2 C+ U3 x
Induction Hypothesis: Let K >=2 be integers, support that8 ~/ a, R9 J8 ]* Z$ v# }' `
K^3 – K can by divided by 3.
7 ]8 {6 m: T# K$ q% R5 j; ]$ x+ n7 O3 ]0 A
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
, M. }$ O8 {1 j! b9 O, Z( S# Jsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem; K0 M: Z3 x$ E! g% @# Z
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)$ W+ [" ?- C( L2 W6 t- a3 p/ w5 }
= K^3 + 3K^2 + 2K( ~( A* i$ l2 T8 A' i; T
= ( K^3 – K) + ( 3K^2 + 3K)
3 B8 d9 F# ]4 f = ( K^3 – K) + 3 ( K^2 + K); f% G* ~+ K0 Z! B1 \, |3 {# B
by Induction Hypothesis, we know that k^3 – k can by divided by 3 which means that k^3 – k = 3 X for some integer X>0
0 Z$ @/ F! {% w# I5 s eSo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)
: f' }+ O5 ]2 G = 3X + 3 ( K^2 + K)! S7 O7 A) O! O7 }; Z
= 3(X+ K^2 + K) which can be divided by 33 G- L* R0 W7 k
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Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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