無理數的觀點

Definitions:
(i) A number is rational if it has the form q / p, for some integers p and q.
(ii) A number is irrational if it is not rational.

Theorem:
A number is rational if and only if one of the following holds:
(1) it has finite number of decimal places, or
(2) it has infinite number of decimal places but is repetitive with a finite period.

Proof:

If (1) holds, clearly the number is rational because it can be written as n / 10^m, where n is an integer and m is the length of decimals. If (2) holds, let say the decimals are repeating with a period m. Then the number can be written down as
( n / 10^m) ( 1 + 10^-m + 10^-2m + 10^-3m + ... )
= ( n / 10^m ) ( 1 / (1 - 10^-m) )
= ( n / 10^m ) ( 10^m / (10^m - 1) )
= n / (10^m - 1)
where n is an integer, namely the first 10^m decimals. So the number is rational.

Now given any rational number r = q / p, we want to show that it either has a finite number of decimals or has a finite repetitive decimals period. It suffices to show for r = 1 / p, for p a prime number. For p = 2 or 5, we know 1 / 2 = 0.5 and 1 / 5 = 0.2 have finite decimals. Now let assume p is not 2 nor 5, so that p and 10 is relatively prime. Fermat little theorem tells us
10^(p-1) - 1 is divisible by p
Let n be an integer such that
n p = 10^m - 1, where m = p - 1
and therefore the rational number has the form
r = 1 / p = n / (10^(p-1) - 1)
= ( n / 10^m) ( 1 + 10^-m + 10^-2m + 10^-3m + ... )
which has repetitive decimals of period m = p - 1.

2009-12-06 02:01:04 補充：
As a corollary, the theorem implies all irrational numbers have infinite decimals and are not repetitive.

2009-12-06 02:01:27 補充：
More definitions:
(iii) A number is algebraic if it is a root of a polynomial equation with integer coefficient.
(iv) A number is transcendental if it is not algebraic.

e.g. sqrt(3) is algebraic but irrational. pi and e are transcendental.

「循環節為無限長」的意思是一個無限接一個無限的循環??，這是「無理」的說法。但不能因此說這是「無理數」歸為循環小數類的一個條件。

2009-12-16 19:56:05 補充：
004 eodyam 的回答是完全引述舊有資料，並非暢所欲言去答「你能否這樣說 :『無理數可歸為循環小數類,只不過其循環節為無限長。』」。為何成為最佳答案？

我覺得如果你既用意係想令到數字既特性比較統一既話，咁你既講法可以成立。
就好似對於p-adic number咁，R可以當做∞-adic number field。

2009-12-07 12:53:11 補充：
用返極限既講法，其實即係講緊無理數既循環節比一切正整數都大，所以叫做∞。

題目是有語病的，要有限個數才能談循環。

在某程度上你可以這樣說
因為我們的確不可能百分百肯定無理數於某一小數位會開始循環

例如[派]就是一例子
直到現在小數點後過億位仍然沒有開始循環

無理數應該定義為
在可計算範圍內沒有循環節

而可計算範圍則可再相確及定義

我認為不能，因為「丌」是無理數，但不能歸為循環小數類(丌是無限，而且沒有循環節)，所以我認為不能說 :

『無理數可歸為循環小數類,只不過其循環節為無限長。』

Is it ok?

If it is OK?

Tell me.

2009-12-05 23:26:17 補充：
丌約3.14......
我怕你不明~

只作為參考:

先回答你的問題: 對

有盡小數→有理數

無盡小數→無理數


額外資料:
有理數: 可寫成分數
無理數: 不可寫成分數
(所有無理數化為小數均為無盡小數)

希望幫到你