我不这样认为咧，欢迎反驳 送交者: Mushy 2008月09月17日13:17:56 于 [灵机一动] 发送悄悄话 回 答:神秘的无理数 - 挑战学院派 由 西线晨雾 于2008-06-24 10:16:03 [quote] 现在讨论戴狄金有理分划定义无理数。我所知道的最常用的定义是，当有理分划的 下组没有最大数，上组没有最小数时，约定分划夹一个数。这个数就是无理数。其 实这个定义也是有缺陷的。首先，所谓夹就是比较大小。在这个数没定义之前，怎 样参与和有理数的比较？如果是逐位比较，那么说明这个数已有定义，不必定义两 次。还有，没有无理数的定义，怎样证明下组没有最大数，上组没有最小数的有理 分划的存在？就算这种分划存在，如何证明这种分划存在空隙而能容纳一个数？如 果数中只有整数被定义，1和2之间还有空隙吗？换句话说，如果没有无理数的事先 定义，就不能证明，下组没有最大数，上组没有最小数的有理分划夹一个数。如果 只定义有理数域，没有更广泛的数域，有理数域的非只能是空集。 [quote] 我只想说，楼主的疑问很fundenmental,我也想了很久，最后还是借助英文资料。我以下用英文，因为参考文献主要是英文，我华文水平有限。 1. 我所知道的最常用的定义是，当有理分划的 下组没有最大数，上组没有最小数时，约定分划夹一个数。这个数就是无理数。 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. 其实这个定义也是有缺陷的。首先，所谓夹就是比较大小。在这个数没定义之前，怎 样参与和有理数的比较？ we can 比较大小, use inclusion of set: let x,y be cuts and a,b be rationals, then xa, if a belongs to x. 如果是逐位比较，那么说明这个数已有定义，不必定义两次。 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. 没有无理数的定义，怎样证明下组没有最大数，上组没有最小数的有理分划的存在？ I dun know if I answers you question, but I can give an instance of 下组没有最大数，上组没有最小数的有理分划的存在. 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]. 就算这种分划存在，如何证明这种分划存在空隙而能容纳一个数？ I dun know to restate this:这种分划存在空隙而能容纳一个数. the dedekind cut 这种分划 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. 如果数中只有整数被定义，1和2之间还有空隙吗？ 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之间没有空隙. 如果没有无理数的事先定义，就不能证明，下组没有最大数，上组没有最小数的有理分划夹一个数。 下组没有最大数，上组没有最小数的有理分划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. 欢迎指教和讨论。 谢谢你的耐心