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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)9 |+ Z" m) r, ]6 n
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Proof: + _/ Y0 L7 p" |1 v9 I, c1 j
Let n >1 be an integer
% _% T% H$ s7 P j+ nBasis: (n=2)- {0 ?% X3 `# S( o$ p# M9 W
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3% p- }2 c$ ], L; Y `6 z
3 ~7 ~2 P" w$ u6 [& oInduction Hypothesis: Let K >=2 be integers, support that0 F% F% N9 A6 [4 G! E
K^3 – K can by divided by 3.
- [( A. N7 Z6 i, T! e1 P- f9 P5 `0 m' p/ {" {7 a3 I
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
: M. P" s }) U& x: B$ A6 Qsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
$ q- g4 ~1 y4 {( e) }7 h* eThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
' K- i1 x l4 s0 o, N6 [ = K^3 + 3K^2 + 2K, h, `! U- f4 ~2 p
= ( K^3 – K) + ( 3K^2 + 3K)
/ @9 ^0 X! G- `) p E9 l0 t4 K' E = ( K^3 – K) + 3 ( K^2 + K)
5 {3 p% P2 ?7 [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; u* ?% E- p5 q! D5 W0 @& }) J1 f1 h
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)" P7 Y9 f7 K! \. [: u1 n$ O! [( x
= 3X + 3 ( K^2 + K)
+ W! I/ h+ K G = 3(X+ K^2 + K) which can be divided by 3
3 F o1 k) C; ^! C* k0 q* R- X) Y( o4 g! J; L* R8 K0 s
Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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, v- q$ h, j" f* X |# O[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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