Pure Maths 一問?

咁似某年past paper入面given左既Theorem (*)既……

2009-01-30 13:13:34 補充：
其實要理解呢條定理唔難，我舉一個例子：

π= 3.141592653589793238462643383279502884...

Construct a sequence {x_n} as follow:
x_1 = 3
x_2 = 3.1
x_3 = 3.14
x_4 = 3.141
...

In the above sequence, all terms are clearly rational numbers,
and you can easily see that x_n→π as n→∞.

For all real number x, just define the rational sequence as above,
then the theorem are clear.


2009-01-31 00:02:23 補充：
Claim: Assume x>=0 be any real number. Then for every integer n>=1 there is a finite decimal r_n=(a_0).(a_1)(a_2)...(a_n) such that r_n <= x < r_n + 1/10^n.

2009-01-31 00:02:27 補充：
Proof. Let S be the set of all non-negative integers <=x. Then S is non-empty, since 0 is in S, and S is bounded above by x. Therefore S has a supremum, say a_0 = sup S.

2009-01-31 00:04:25 補充：
It is easily verified that a_0 is in S, so a_0 is a non-negative integer. We called a_0 the greatest integer in x, and we write a_0 = [x]. Clearly, we have a_0 <= x < a_0 + 1.

2009-01-31 00:07:23 補充：
Now let a_1 = [10x - 10*a_0], the greatest integer in 10x - 10*a_0. Since 0 <= 10x - 10*a_0 = 10(x - a_0) < 10, we have 0 <= a_1 < 9 and a_1 <= 10x - 10*a_0 < a_1 + 1. In other words, a_1 is the largest integer satisfying the inequalities a_0 + a_1/10 <= x < a_0 + (a_1 + 1)/10.

2009-01-31 00:10:09 補充：
More generally, having chosen a_1, a_2, ..., a_{n-1} with 0 <= a_i < 9, let a_n be the largest integer satisgying the inequalities a_0 + a_1/10 + ... + a_n/10^n <= x < a_0 + a_1/10 + ... + (a_n + 1)/10^n. Then 0 <= a_n < 9 and we have r_n <= x < r_n + 1/10^n, where r_n = a_0.(a_1)(a_2)...(a_n).

2009-01-31 00:11:31 補充：
In this case, x is actually the supremum of the set of rational numbers r_1, r_2, ...
and {r_n} is the required rational sequence that converges to x.

EMK:

Your answer is suitable for F.6-7 student!

97 年 paper 2 Q 10 (c) part