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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)7 b: \3 l. Q9 k/ y" {% N v% ]
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Proof: . Y. N6 O- d6 s$ n! G
Let n >1 be an integer
3 ^5 ^- w! H- ?1 ^8 ?0 iBasis: (n=2)
0 f8 @! j# \' W. y9 O 2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3
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Induction Hypothesis: Let K >=2 be integers, support that
9 q/ j R( u: O' k K^3 – K can by divided by 3.3 O, B6 r9 {0 K- n: b( K3 F
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Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
" T; X% ^! Z) t* @since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
6 G4 x" X% O. y4 b; pThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)+ t2 U# R ?, P4 E) r& O
= K^3 + 3K^2 + 2K$ }1 p" |9 B7 ?2 h
= ( K^3 – K) + ( 3K^2 + 3K)
3 p3 B, `" q' `! F+ D& M = ( K^3 – K) + 3 ( K^2 + K)( S5 B v$ ~/ A" S: l. J6 ~8 P
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>03 P1 ~8 d' j* c: {' f
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K)$ D& W; v9 q. T9 V
= 3X + 3 ( K^2 + K)
1 [/ j& x6 O" k) Y6 T+ z+ ^ = 3(X+ K^2 + K) which can be divided by 34 d! K% x: ~2 ~
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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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