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送交者: Mushy 2008月09月17日13:17:56 於 [靈機一動] 發送悄悄話
回 答:神秘的無理數 - 挑戰學院派 由 西線晨霧 於2008-06-24 10:16:03
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現在討論戴狄金有理分劃定義無理數。我所知道的最常用的定義是，當有理分劃的
下組沒有最大數，上組沒有最小數時，約定分劃夾一個數。這個數就是無理數。其
實這個定義也是有缺陷的。首先，所謂夾就是比較大小。在這個數沒定義之前，怎
樣參與和有理數的比較？如果是逐位比較，那麼說明這個數已有定義，不必定義兩
次。還有，沒有無理數的定義，怎樣證明下組沒有最大數，上組沒有最小數的有理
分劃的存在？就算這種分劃存在，如何證明這種分劃存在空隙而能容納一個數？如
果數中只有整數被定義，1和2之間還有空隙嗎？換句話說，如果沒有無理數的事先
定義，就不能證明，下組沒有最大數，上組沒有最小數的有理分劃夾一個數。如果
只定義有理數域，沒有更廣泛的數域，有理數域的非只能是空集。
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我只想說，樓主的疑問很fundenmental,我也想了很久，最後還是藉助英文資料。我以下用英文，因為參考文獻主要是英文，我華文水平有限。
1. 我所知道的最常用的定義是，當有理分劃的
下組沒有最大數，上組沒有最小數時，約定分劃夾一個數。這個數就是無理數。
Let me restate the definition:
A Dedekind cut is a set over Q with these properties:
1. a cut is not empty, neither it is Q.
2. if a rational p belongs to a cut, then any rational q smaller than p belong to the cut as well.
3. if a rational p belongs to a cut, then p < r for some r in the cut.
This is from Principle of Mathematical Analysis, by Walter Rudin.
其實這個定義也是有缺陷的。首先，所謂夾就是比較大小。在這個數沒定義之前，怎
樣參與和有理數的比較？
we can 比較大小, use inclusion of set:
let x,y be cuts and a,b be rationals,
then x a, if a belongs to x.
如果是逐位比較，那麼說明這個數已有定義，不必定義兩次。
yeah this is true. Because the definition of a decimal representation depends on the archimedean property of R which needs to be defined first.
archimedean property of R states that for x and y in R and x positive there exists a positive integer n such that nx > y. this may seem obvious,but it has to be deduced from condition 2 of definition of dedekind cut above.
沒有無理數的定義，怎樣證明下組沒有最大數，上組沒有最小數的有理分劃的存在？
I dun know if I answers you question, but I can give an instance of 下組沒有最大數，上組沒有最小數的有理分劃的存在.
quoted from the same book as above:
let A be the set of all positive rationals p such that p^2 > 2 and B be the set of all positive rationals q such that q^2 < 2.
Now I claim that A contains no smallest number and B contains no biggest number.
Proof by Infinite Descent:
For every rational p, we consider
p -(p^2-2)/(p+2)= (2p+2)/(p+2), [1]
then
q^2 -2 = 2(p^2-2)/(p+2)^2. [2]
If p is in A, then q is smaller than p by [1] and belongs to A by [2].
If p is in B, then q is bigger than p by [1] and belongs to B by [2].
就算這種分劃存在，如何證明這種分劃存在空隙而能容納一個數？
I dun know to restate this:這種分劃存在空隙而能容納一個數.
the dedekind cut 這種分劃 itself is the real number.
the cut of all rationals p^2 < 2 defines what we intuitively know as sqrt(2).
It is Q that 存在空隙, so we fill these gaps with sets constructed from Q so that
if a set over the field is defined and is bounded above, the sup of the set exists in this field.
如果數中只有整數被定義，1和2之間還有空隙嗎？
I only know this construction of natural number,
http://en.wikipedia.org/wiki/Natural_number#Constructions_based_on_set_theory
If I guessed correctly, 2 is constructed from 1, so 1和2之間沒有空隙.
如果沒有無理數的事先定義，就不能證明，下組沒有最大數，上組沒有最小數的有理分劃夾一個數。
下組沒有最大數，上組沒有最小數的有理分劃exists and is the number itself.
That is all I have to say. Well, just treat everything(natural numbers, rationals,reals) as sets and leave the rest to set theorists.
歡迎指教和討論。
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