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标题: 整数的幂还是整数 [打印本页]

作者: constant 时间: 2005-12-7 19:52
标题: 整数的幂还是整数

题目有点像流浪者

难度：++++

设r是正实数，并且满足对每个正整数N，N^r都是整数。证明r是整数。

作者: QL 时间: 2005-12-10 05:10
标题: 回复：整数的幂还是整数

I will prove that r can not be between (2,3), similar idea can be used to prove that r can not be between (k,k+1) for any non negative integer k.

Suppose 2

(N+1)^r = N^r(1+r*1/N+1/2 r*(r-1)*1/N^2+O(1/N^3))

(N+2)^r = N^r(1+r*2/N+1/2 r*(r-1)*4/N^2+O(1/N^3))

so,

N^r-2*(N+1)^r+(N+2)^r = N^r (r*(r-1)*1/N^2+O(1/N^3)) = r*(r-1)*N^(r-2)+O(1/N^(3-r))

Notice that, O(1/N^(3-r)) can be arbitrarily small, but the fraction part of N^(r-2) becomes dense as N goes to infinity, hence the set of the fraction part of r*(r-1)*N^(r-2) for all positive integer N is dense in [0,1], specifically, there is a subsequence of N, such that the fraction part of r*(r-1)*N^(r-2) approach 0.5. Hence N^r-2*(N+1)^r+(N+2)^r is not integer for some N.

Contradiction!

作者: QL 时间: 2005-12-10 05:19
标题: 回复：整数的幂还是整数

Can the conclusion be make stronger? e.g.:

Is there a non integer r, such that both 2^r and 3^r are integers?

Is there a non integer r, such that for any rational number q, q^r is alo rational?

作者: constant 时间: 2005-12-10 17:04
标题: 回复：回复：整数的幂还是整数

Right idea but why "the fraction part of N^(r-2) becomes dense "? It is not obvious, and actually not needed in the proof.

作者: QL 时间: 2005-12-10 18:18
标题: 回复：回复：回复：整数的幂还是整数

Another way to present the idea: r*(r-1)*N^(r-2) tends to infinity, yet the increment when N increase by 1, i.e. r*(r-1)*(N+1)^(r-2) -r*(r-1)*N^(r-2) tends to 0. Hence there exists a subsequece, whose fraction part converges to 0.5.

作者: constant 时间: 2005-12-10 22:21
标题: 回复：回复：回复：回复：整数的幂还是整数

Now you have proved "the fraction part of N^(r-2) becomes dense", and hence r cannot be between 2 and 3. But it is not very easy to present the idea to any r. Will you try it?

作者: QL 时间: 2005-12-10 23:46
标题: 回复：回复：回复：回复：回复：整数的幂还是整数

I think the following should work for general cases:

For any non negative integer k, we prove that r can not be between (k, k+1). Otherwise, suppose r is in (k, k+1), consider the Taylor's expansion:

(N+1)^r = N^r(1+r*1/N+1/2 r*(r-1)*1/N^2+...+O(1/N^(k+1)))

(N+2)^r = N^r(1+r*2/N+1/2 r*(r-1)*2^2/N^2+...+O(1/N^(k+1)))

...

(N+k)^r = N^r(1+r*k/N+1/2 r*(r-1)*k^2/N^2+...+O(1/N^(k+1)))

and by solving a system of linear equations, we can find integers A0,A1,A2,...Ak, s.t.

A0 N^r +A1 (N+1)^r +... Ak (N+k)^r = C* r* (r-1)...*(r-k)*N^(r-k) + O(1/N^(k+1-r))

and similar argument will leads to contradiction. (Notice that 0

本贴由[QL]最后编辑于：2005-12-10 15:53:25

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