x*****d 发帖数: 427 | 1 【 以下文字转载自 LosAngeles 讨论区 】
【 原文由 yyshape 所发表 】
我的数学基础大致相当于初二初三学生的水平,
所以可能的话请用通俗的语言来帮我看看这个.
为什么可以用有限的线段表达无限的意义呢?
比如,根号2是无线不循环小数吧,
用几何的形式表现出来是一个直角三角形的斜边,
边长都是1.
我就是不太懂,为什么可以用一条有限长度的线段,
来表示一个无线不循环小数呢??
初中的时候问过我们数学老师这个问题,
当时就北捧杀了,所以我数学一直不好.. | x*****d 发帖数: 427 | 2 呵呵,好像我要皈依比大哥拉丝学派了。。。。
是这样,有人要我举一个生活中的例子来说明无理数,就象三个人分两个苹果
可以用有理数表示一样,我说来说去都是直角三角形的斜边。。。。。可几何
也不能说是生活不是数学。。。 | g****e 发帖数: 5 | 3
This is a broad question. I think what really matters is
how much one wants to give up in order to seek different
levels of satisfying answers to questions of this sort.
Either way, we are better off than not asking questions
at all.
Maybe the questions is about whether a real number is "real".
By definition, rational numbers form a subset of the set of
real numbers. The rest are irrational numbers, and there are
uncontablly many of them.
Quoting Rudin's undergraduate analysis text for some rel
【在 x*****d 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 【 以下文字转载自 LosAngeles 讨论区 】 : 【 原文由 yyshape 所发表 】 : 我的数学基础大致相当于初二初三学生的水平, : 所以可能的话请用通俗的语言来帮我看看这个. : 为什么可以用有限的线段表达无限的意义呢? : 比如,根号2是无线不循环小数吧, : 用几何的形式表现出来是一个直角三角形的斜边, : 边长都是1. : 我就是不太懂,为什么可以用一条有限长度的线段, : 来表示一个无线不循环小数呢??
| x******g 发帖数: 318 | 4 2)如何证pi是无理数?
这个证明属于Ivan Niven。假设pi=a/b,我们定义(对某个n):
f(x) = x^n(a-bx)^n/n!
F(x) = f(x) + ... + (-1)^jf^(2j)(x) + ... + (-1)^nf^(2n)(x)
这里f^(2j)是f的2j次导数.
于是f和F有如下性质(都很容易验证):
1)f(x)是一个整系数多项式除以n!。
2)f(x) = f(pi-x)
3)f在(0,pi)区间上严格递增,并且x趋于0时f(x)趋于0,
x趋于pi时f(x)趋于pi^na^n/n!
4)对于0 <= j < n, f的j次导数在0和pi处的值是0。
5)对于j >= n, f的j次导数在0和pi处是整数(由1)可知)。
6)F(0)和F(pi)是整数(由4),5)可知)。
7)F + F'' = f
8)(F'·sin - F·cos)' = f·sin (由7)可知)。
这样,对f·sin从0到pi进行定积分,就是
(F'(pi)sin(pi)-F(pi)cos(pi)) - (F'(0)sin(0)-
【在 x*****d 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 呵呵,好像我要皈依比大哥拉丝学派了。。。。 : 是这样,有人要我举一个生活中的例子来说明无理数,就象三个人分两个苹果 : 可以用有理数表示一样,我说来说去都是直角三角形的斜边。。。。。可几何 : 也不能说是生活不是数学。。。
| s*****c 发帖数: 753 | 5 Finally a good post.
Actually, the difference between 无理 and 有理 is that, according to
definition, any number which can be represent by the division of two INTEGER
number is 有理. If not, then it is 无理.
So to prove sqrt(2) is 无理, you have to first assume that it can be represent
in the form of a/b and then find a conflict. So we can try
assume:
sqrt(2)=a/b
where a/b is already reduce to most simplified forms, namely, they don't have
common factor other than 1, therefore they can't be further reduce
【在 x******g 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 2)如何证pi是无理数? : 这个证明属于Ivan Niven。假设pi=a/b,我们定义(对某个n): : f(x) = x^n(a-bx)^n/n! : F(x) = f(x) + ... + (-1)^jf^(2j)(x) + ... + (-1)^nf^(2n)(x) : 这里f^(2j)是f的2j次导数. : 于是f和F有如下性质(都很容易验证): : 1)f(x)是一个整系数多项式除以n!。 : 2)f(x) = f(pi-x) : 3)f在(0,pi)区间上严格递增,并且x趋于0时f(x)趋于0, : x趋于pi时f(x)趋于pi^na^n/n!
| S*********g 发帖数: 5298 | 6 You meant the Planck Length, right?
How about if that distance itself is not rational?
To my understand, quantum theory itself doesnot say the distance
can not be irrational.
Nevertheless, using the fundamental constants,
you can write down a quantity with the unit of length.
When one tries to reconcile general relativity, it must involves with
the constats: hbar, c, and G(the Gravitational constant). This gives
the unit of length called Planck Length
L=(hbar G/c^3)^(1/2)
When the scale approach | d****z 发帖数: 9503 | 7 Firstly, there is NO minimal length in QM. A minimal length would mean a
maximum energy. There is clearly no up limit on energy in current QM. Why? For
any object you observe, you can always choose a reference frame which has a
relative speed of 0.9999999999999999999999c with respect to the object.
Current QM assumes the principle of relativity, which means it should describe
the physics in that frame the same way as in any other inertia frame. In that
frame, the energy of the object could be at |
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