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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)3 C; X" _* A1 v
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Proof: & P4 \4 d' t# |0 y5 l5 a. m/ b% ~
Let n >1 be an integer
0 a1 E) @) ]( w& M5 L! R6 s& lBasis: (n=2)# N0 \0 j) p: \
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3! r8 ~3 y% t% j& o7 i, x' g! h$ v
) U9 Y9 g: t: f, ?7 }2 T2 B' @Induction Hypothesis: Let K >=2 be integers, support that
! L4 [$ y9 k# A. D }; k1 L K^3 – K can by divided by 3.# [7 q& u5 D: r- r1 C) v
- b- _# J7 q b4 y: {Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
9 X- p# w4 {; J. I; ?9 G3 d" gsince we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
% I! J! d/ ?. M; ZThen we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)0 P3 N( ~3 F( \; K# W
= K^3 + 3K^2 + 2K8 f3 u/ j) Y$ j# C3 @6 o8 U
= ( K^3 – K) + ( 3K^2 + 3K)9 E9 h/ r6 a: t$ ~' C
= ( K^3 – K) + 3 ( K^2 + K)7 e0 `5 S1 _% Q6 x$ p# X8 S
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! D" v! f! M/ f; g
So we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K) n9 L7 w3 }0 C1 }6 B. O) A5 y1 k
= 3X + 3 ( K^2 + K)
/ e y& a5 J) e+ B4 {3 O = 3(X+ K^2 + K) which can be divided by 3
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# D; G6 f7 j/ M- T1 [% @+ dConclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers n >1.
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# `* p' p) }' o4 V9 S[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ] |
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