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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)5 A1 j& _ c* L/ Z. v+ N# X" v7 s$ `
# Z! \% ?: q1 m( \% Q; L/ @: HProof: , m1 X# }$ k- N$ c% Z
Let n >1 be an integer
& S& w6 H; l7 M6 a3 ~4 TBasis: (n=2)1 k- U2 q: |6 y7 J' w! W# _
2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3/ c9 }8 X7 u: |0 x
, c' [* d, I3 h( x. i! d2 p; f6 {Induction Hypothesis: Let K >=2 be integers, support that
1 o: {. A/ [" N, J! J% P K^3 – K can by divided by 3.
8 L) ?; C% F" d% q/ H8 D @( k \: K( Q% g* E8 r) X8 f+ {. z- P
Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
1 G; k+ w1 B' C1 ~since we have (K+1)^3 = K^3 +3K^2 + 3K +1 by Binomial Theorem
" c4 r2 ]% b' @Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1): O) X& v& a# E) N0 {/ N/ e% G
= K^3 + 3K^2 + 2K r3 |. k1 a( a
= ( K^3 – K) + ( 3K^2 + 3K)1 t# q9 r1 x8 _, w- I
= ( K^3 – K) + 3 ( K^2 + K)
/ n2 ?4 @3 p+ @* \6 V/ ]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
3 w7 y" n& V4 M* ySo we have (K+1)^3 – ( K+1) = ( K^3 – K) + 3 ( K^2 + K): _6 K& @6 K' S( {7 _
= 3X + 3 ( K^2 + K)& n& R# N- r; j4 k# \
= 3(X+ K^2 + K) which can be divided by 3
5 ]# l8 w% {' ~6 V: R5 i( j$ e3 S3 m" M- C5 q: L
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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