出两道中学数学题

扁舟 · 发表于 2005-2-22 07:58

1。一个整数的立方减去这个数一定能被3整除吗？能否证明？

2。下边证明有没有毛病？

设 a=b

则有： a*a-a*b=a*a-b*b
两边因式分解（左边提取公因式，右边平方差公式）：

a(a-b)=(a+b)(a-b)
a=a+b
a=2a
1=2

证毕，结论，１＝２

三思 · 发表于 2005-2-22 09:14

我幼儿园毕业啊（我劳工倒是初中毕业），不过闲着也闲着，我瞎猫试试

1）不能。比如1
2)a,b不能是0

lilian · 发表于 2005-2-22 20:58

1。 n.n.n - n = n(n.n - 1) = n(n+1)(n-1), 连着三个数相乘，当然能被3 整除。
2。因为a=b, 所以a-b=0, 0乘以任何数都等于0。也就是等式两边不可以约掉a-b, 0 不可以做商。所以证明有误。

三思 · 发表于 2005-2-22 21:30

Originally posted by lilian at 2005-2-22 09:58 PM:1 U% x* b r9 t# |
1。 n.n.n - n = n(n.n - 1) = n(n+1)(n-1), 连着三个数相乘，当然能被3 整除。, B: C. i( W' o" U( z4 n
2。因为a=b, 所以a-b=0, 0乘以任何数都等于0。也就是等式两边不可以约掉a-b, 0 不可以做商。所以证明有误。

看！有高中毕业的！

悟空 · 发表于 2005-2-22 22:16

Originally posted by lilian at 2005-2-22 09:58 PM:
/ K k7 O6 z( v% S; k8 E7 X! f6 S1。 n.n.n - n = n(n.n - 1) = n(n+1)(n-1), 连着三个数相乘，当然能被3 整除。
( P( y i X( {# L2 r

为证明扁同志的题目, 你需要证明 n(n+1)(n-1)能被3 整除

悟空 · 发表于 2005-2-22 22:21

Show that for all integers  n >1, n^3 - n  can be divided by 3. (Note: n^3 stands for n*n*n)

Proof:
Let n >1 be an integer
Basis: (n=2)
      2^3 - 2 = 2*2*2 –2 = 6 which can be divided by 3

Induction Hypothesis: Let K >=2 be integers, support that
                                 K^3 – K can by divided by 3.

Now, we need to show that ( K+1)^3 - ( K+1) can be divided by 3
since we have  (K+1)^3  =  K^3 +3K^2 + 3K +1 by Binomial Theorem
Then we have (K+1)^3 – ( K+1) = K^3 +3K^2 + 3K +1 –(K+1)
      = K^3 + 3K^2 + 2K
      = ( K^3 – K)  + ( 3K^2 + 3K)
      = ( K^3 – K)  + 3 ( K^2 + K)
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
So we have (K+1)^3 – ( K+1) = ( K^3 – K)  + 3 ( K^2 + K)
= 3X + 3 ( K^2 + K)
= 3(X+ K^2 + K)  which can be divided by 3

Conclusion: By the Principle of Mathematics Induction, n^3 - n can be divided by 3 For all integers  n >1.

[ Last edited by 悟空 on 2005-2-23 at 10:06 AM ]

醉酒当歌 · 发表于 2005-2-22 22:24

第一题估计用数学归纳法很容易解决。

第二题应该很简单

醉酒当歌 · 发表于 2005-2-22 22:26

这个题估计现在在国内属于小学数学了。

悟空 · 发表于 2005-2-23 09:15

Originally posted by 悟空 at 2005-2-22 11:21 PM:$ x X' @6 u/ x3 [- ]
Show that for all integers n >1, n^3 can be divided by 3.

SORRY, 严重笔误, 改过来了: n^3 应为 n^3 - n.

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出两道中学数学题

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