Quod Erat Demonstrandum

2015/09/13

一式過

Filed under: Pure Mathematics — johnmayhk @ 5:19 下午
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比如要寫出以下數列的通項

-1,1,-1,1,\dots

應該不難得出

(-1)^n

吧。(這樣假設上述數列的變化模式不變,一直都是「負 1 正 1」咁去。)

但還有沒有其他可能?

有的,比如 (more…)

2015/08/27

某 monic 多項式

Filed under: NSS,Pure Mathematics — johnmayhk @ 10:09 上午
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偶見高X論壇某題:

證明 x^4+x^3+x^2+x+1 > 0

回應者用了較麻煩的方法處理。

其實,當 x\ne 1(more…)

2014/02/20

平方和

Filed under: Fun — johnmayhk @ 11:44 下午
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隨便考慮 n 個正整數

a_1,a_2,a_3,\dots a_n

b=\sqrt{a_2^2+a_3^2+\dots +a_n^2}

(a_1+bi)^m=P+Qi

其中 m 是正整數,i^2=-1,那麼我們必有 (more…)

2012/05/01

無聊談通項

那天觀課,同事開始教等差數列(arithmetic sequence),(估計是隨便)問學生:

2,1,4,\frac{1}{2},8,\frac{1}{4},\dots

的通項(general term)是甚麼?

關於通項,之前也談過,除非先有「特殊規定」,否則所謂通項是無定義的。

談回上題,假設同事的「特殊規定」是把兩個等比數列(geometric sequence)併在一起。當然,學生在中一時接觸過數型(number pattern),但要寫出上述數列的通項,似乎不是一蹴而就的事。因那是引起動機的其中一例,同事沒有繼續討論,但卻引發我思考一些東西。

只要時間許可,相信學生不難得到: (more…)

2012/02/26

答網友:y=z^2

Filed under: HKALE,Pure Mathematics — johnmayhk @ 6:29 下午
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以下是網友在 2011-03-06 的提問:

By considering

(1+z)^8+(1-z)^8=0

show that

\displaystyle\sum_{k=0}^7 \tan^2\frac{(2k+1)\pi}{16}=56

不難 (more…)

2012/01/20

arctan,pi,complex numbers

Filed under: mathematics,Pure Mathematics — johnmayhk @ 12:02 下午
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式子

\pi=\tan^{-1}1+\tan^{-1}2+\tan^{-1}3

美麗嗎?我比較多見的形式是

\tan^{-1}(\frac{1}{2})+\tan^{-1}(\frac{1}{3})=\frac{\pi}{4}

(易知 \tan^{-1}(\frac{1}{n})+\tan^{-1}(n)=\frac{\pi}{2},故上述兩式等價。)

以圖來證明上式,可以考慮 (more…)

2011/11/26

利用複數尋寶

Filed under: Fun,Pure Mathematics — johnmayhk @ 4:51 下午
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此刻,純數還欠複數(complex numbers)這個課題未教完。當完成後,課題中精采之處,將長埋「新」高中之墓塚。

同事問:複數有沒有有趣的幾何應用?很多,這裡舉一個經典例子。

海盜留下藏寶位置之提示:

島上有一個絞刑架,一棵橡樹及一棵松樹。

由絞刑架出發前往橡樹, (more…)

2011/11/18

polar form

Filed under: Pure Mathematics — johnmayhk @ 11:46 上午
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基礎題:

Convert z=-\cos \theta -i\sin\theta into polar form.

問:「在何象限,cosine 和 sine 值也是負?」 (這是極不精確的說法,但學生又明我說甚麼。)

答:「第三。」 (more…)

2011/03/02

當 x 接近零,(1+1/x)^x 如何?

Filed under: Pure Mathematics — johnmayhk @ 5:07 下午
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課堂上,我通常會談「非例子」。

比如教(講)了

\displaystyle \lim_{x\rightarrow \infty}(1+\frac{1}{x})^{x} = e

隨即問

\displaystyle \lim_{x\rightarrow \infty}(2+\frac{1}{x})^{x} = ?

這個易,再問

\displaystyle \lim_{x\rightarrow 0}(1+\frac{1}{x})^{x} = ?

學生 (more…)

2009/10/26

利用圖像尋找非實根

Filed under: HKALE,NSS,Pure Mathematics — johnmayhk @ 8:24 下午
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在高中一 NSS 數學課,我開始教二次圖像和二次方程之根(roots)的關係。現在課程涉及複數,卻沒有教如何利用圖像尋找複根(complex roots)或非實根(unreal roots),我在此補充一下。

以下是在下用極速粗製濫造的 ETV,同學先看看:

解說: (more…)

2009/09/29

今天雜記

Filed under: Life,mathematics,NSS,School Activities — johnmayhk @ 11:02 下午
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難以轉化

科主任擬了兩班高中一 M2 的聯測卷,其中有一道題目只有三名學生懂得處理:

Given that x and y are purely imaginary numbers. If (x + y) + (2x - y)i = 12 - i, find (the values of) x and y. (more…)

2008/04/21

Complex trigonometric functions

Filed under: University Mathematics — johnmayhk @ 1:03 下午
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F.5 student Hoover found it interesting to know some properties of complex trigonometric functions. He asked if the following is still true?

\sin^2(z) + \cos^2(z) = 1

for any complex number z. (more…)

2008/01/16

[初中] Factorization, useful or not?

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 9:44 下午
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Is the topic “factorization" in junior form useful? Urm, apart from simplification, solving equation etc, will factorization play a critical role in advanced mathematics? No idea, but at least, I could tell you that we may generalize the idea of factorization in set theory. Take it easy. I just wanna say something in my daily teaching. This topic may have no application (?) in further study in mathematics, however, it may be useful as one of the tools of training of students’ mind. When letting my students play with some challenging (at least, not-straight-forward) problems, nealy all of them are willing to do and ask for help. Well, it is a good moment of promoting the beauty of mathematics!

The following are questions asked from one of my students, Lo, in class. Factorize

1. 2x^2 + 6x - 15z - 2xy + 5yz - 5xz
2. x^4 + 2x^3 + 3x^2 + 2x + 1

Questions above are not difficult, but at least, we could not just apply identities directly without doing some grouping beforehand. The art (art? yes, sometimes, it is more than ‘technique’, it may be an ‘art’) is: how?

For Q.1

2x^2 + 6x - 15z - 2xy + 5yz - 5xz
= 2x^2 - 2xy + 6x - 15z + 5yz - 5xz
= 2x(x - y) + 3(2x - 5z) + 5z(y - x)
= 2x(x - y) + 3(2x - 5z) - 5z(x - y)
= (x - y)(2x - 5z) + 3(2x - 5z)
= (2x - 5z)(x - y + 3)

Teachers can easily set up factorization problems by just expanding polynomials, i.e., starting from (2x - 5z)(x - y + 3), we can come up with 2x^2 + 6x - 15z - 2xy + 5yz - 5xz and ask students to factorize it back. However, if we change z (say) into number, the problem may not be easy to solve. That is, considering

(2x - 5)(x - y + 3)
= 2x^2 - 2xy + 6x - 5x + 5y - 15
= 2x^2 - 2xy + x + 5y - 15

Now, may be students find it difficult to factorize 2x^2 - 2xy + x + 5y - 15.

Further, if we consider the product of trinomials (three terms) with 3 variables x,y,z, it may already be a nightmare, say, could you factorize the following

3x^2 - 4y^2 - 3z^2 - 4xy + 8yz - 8zx ?

For Q.2

x^4 + 2x^3 + 3x^2 + 2x + 1
= x^4 + 2x^3 + (x^2 + 2x^2) + 2x + 1
= x^2(x^2 + 2x + 1) + 2x^2 + 2x + 1
= x^2(x + 1)^2 + 2x(x + 1) + 1
= (x(x + 1))^2 + 2x(x + 1) + 1
= (x(x + 1) + 1)^2
= (x^2 + x + 1)^2

Following this idea, we can create many, like

(a) x^8 + 2x^7 + 3x^6 + 4x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1
(b) x^{10} + 2x^9 + 3x^8 + 4x^7 + 5x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1
(c) x^9 + 3x^8 + 6x^7 + 10x^6 + 12x^5 + 12x^4 + 10x^3 + 6x^2 + 3x + 1

It is extremely easy to set up questions above, while, it may not be easy to solve them immediately.

Just help you a bit, observe the following

111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321

Well, it is also an art to strike a balance: give some challenging questions to students without frightening them.

When I was in F.2, I found an old little mathematics book of only one single theme: factorization. I’d lost it long time ago. It taught me many techniques in factorization, as for example, I knew how to factorize something like a^3 + b^3 + c^3 - 3abc (read my old post if you want to). But, it is quite demanding to ask a F.2 student to factorize the following

x^{27} - 1

though it is just a piece of cake for F.6 or F.7 students to use the techniques in complex numbers and obtain

x^{27} - 1
= (x-1)(x^2 - 2x\cos(\frac{2\pi}{27}) + 1)(x^2 - 2x\cos(\frac{4\pi}{27}) + 1)(x^2 - 2x\cos(\frac{6\pi}{27}) + 1)\dots(x^2 - 2x\cos(\frac{26\pi}{27}) + 1)

 There was a funny problem called Tschebotarev problem concerning the factorization of x^n - 1, see if I can find more and share with you next time.

– – – – – –

Just wanna add something below.

When introducing the identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

in class, one student, Choi, puzzled that, ‘why can’t we write something like

a^3 - b^3 = ((a^{1.5})^2 - (b^{1.5})^2) = (a^{1.5} - b^{1.5})(a^{1.5} + b^{1.5})

Good question! When putting concrete numbers, it works fine, e.g.

16^3 - 9^3 = (16 ^{1.5})^2 - (9 ^{1.5})^2 = (16 ^{1.5} - 9 ^{1.5})(16 ^{1.5} + 9 ^{1.5})

But, what exactly is the question asking? Factorization of polynomials.

The mathematics object on the L.H.S., a^3 - b^3 is a polynomial in a, b; however, the thing on the R.H.S., like (a ^{1.5} - b ^{1.5}) is NOT a polynomial in a, b. It is because the indices involved are NOT non-negative integers. Hence, the suggestion by the student should not be acceptable.

This is one of the rules. The other is the kind of coefficients.

Also read

https://johnmayhk.wordpress.com/2007/12/02/%e5%88%9d%e4%b8%ad-factorization/
 

2007/12/18

[AL][PM] Applications of roots of unity

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 11:42 上午
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Oh, it’s the time for switching the channel into English. Actually, I’m afraid to write in English because I’m blamed not to use English to write blog too often, it may do harm to students. (My English is poor ma, right!) But, firstly, there are less than 10 students reading this blog, so it may be harmless relatively. And next, because of the beautiful fonts, I try to use English in this blog, and that is the only reason ^_^!

F.7C students were just given a quiz, here are some old stuff for your (so-called) enrichment.

Q.1 [Remainder theorem?]

Show that x^{1987} + x^{1997} + x^{2007} is divisible by x^{1987} + x^{1988} + x^{1989}.

Q.2 [Binomial theorem?]

Evaluate C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007}.

Q.3 [Geometry problem?]

Let P_1, P_2, \dots P_{2007} be the vertices of a regular 2007-gon inscribed in a unit circle. Evaluate P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2007}.

The questions above may look different to each other, however, by using the roots of unity, we can solve them.

Solution to Q.1

Let \omega be a complex cube root of unity, i.e. \omega^3 = 1 and \omega \neq 1.

If P(x) is a polynomial (over \mathbb{R}) such that P(\omega) = P(\omega^2) = 0, then (x - \omega)(x - \omega^2) is a factor of P(x). Hence P(x) is divisible by 1 + x + x^2.

Now, let P(x) = 1 + x^{10} + x^{20}

It is easy to check

1 + \omega^{10} + \omega^{20} = 1 + \omega^{20} + \omega^{40} = 0 (Why?)

Hence P(\omega) = P(\omega^2) = 0; thus P(x) is divisible by 1 + x + x^2. Or

1 + x^{10} + x^{20} = (1 + x + x^2)Q(x) for some polynomial Q(x)

Now, x^{1987} + x^{1997} + x^{2007}
= x^{1987}(1 + x^{10} + x^{20})
= x^{1987}(1 + x + x^2)Q(x)
= (x^{1987} + x^{1988} + x^{1989})Q(x)

Result follows.

Solution to Q.2

Let \omega be a complex forth root of unity, i.e. \omega^4 = 1 and \omega \neq 1.

The following results are trivial.

1 + \omega^k + \omega^{2k} + \omega^{3k} = 4 if k is a multiple of 4
1 + \omega^k + \omega^{2k} + \omega^{3k} = 0 if k is not a multiple of 4

Now

(1 + x)^{n} = C_{0}^{n} + C_{1}^{n}x + C_{2}^{n}x^2 + \dots + C_{n}^{n}x^n (where n = 2007)

Put x = 1, \omega, \omega^2, \omega^3 into the above accordingly. We have

2^{n} = C_{0}^{n} + C_{1}^{n} + C_{2}^{n} + \dots + C_{n}^{n}
(1 + \omega)^{n} = C_{0}^{n} + C_{1}^{n}\omega + C_{2}^{n}\omega^2 + \dots + C_{n}^{n}\omega^n
(1 + \omega^2)^{n} = C_{0}^{n} + C_{1}^{n}\omega^2 + C_{2}^{n}\omega^4 + \dots + C_{n}^{n}\omega^{2n}
(1 + \omega^3)^{n} = C_{0}^{n} + C_{1}^{n}\omega^3 + C_{2}^{n}\omega^6 + \dots + C_{n}^{n}\omega^{3n}

Sum up the above 4 equations and use the trivial results just mentioned, we have

4(C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007})
= 2^{2007} + (1 + \omega)^{2007} + (1 + \omega^2)^{2007} + (1 + \omega^3)^{2007}

Hey, we can simplify it further because we may take \omega = i, \omega^2 = -1, \omega^3 = -i, hence

C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007}
= \frac{1}{4}(2^{2007} + (1 + i)^{2007} + (1 - 1)^{2007} + (1 - i)^{2007})
= \frac{1}{4}(2^{2007} + 2^{\frac{2007}{2}}(2\cos\frac{2007\pi}{4}))
= \frac{1}{4}(2^{2007} + 2^{\frac{2007}{2}}(2\cos\frac{\pi}{4}))
= 2^{1002}(2^{1003} + 1)

Solution to Q.3

The complex numbers corresponding to the vertices of a regular n-gon inscribed in a unit circle are roots of z^n = 1. Hence, we may let

1, \omega, \omega^2, \dots, \omega^{2006} be the complex numbers corresponding to the vertices of the regular 2007-gon; where \omega = \cos\frac{2\pi}{2007} + i\sin\frac{2\pi}{2007}.

It is easy to have \omega, \omega^2, \dots, \omega^{2006} are roots of the equation

x^{2006} + x^{2005} + \dots + x + 1 = 0 – – – (1)

Suppose we translate the regular 2006-gon to the left by 1 unit, then the vertices will become 0, \omega - 1, \omega^2 - 1, \dots, \omega^{2006} - 1, and hence

P_1P_2 = |\omega - 1| = |z_1| (say)
P_1P_3 = |\omega^2 - 1| = |z_2|
P_1P_4 = |\omega^3 - 1| = |z_3|

P_1P_{2007} = |\omega^{2006} - 1| = |z_{2006}|

Now, if we can find out an equation whose roots are z_1, z_2, \dots , z_{2006}, then P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2006} is simply the modulus of the product of roots.

By (1), just set z = x - 1 then an equation with roots z_1, z_2, \dots , z_{2006} is

(z + 1)^{2006} + (z + 1)^{2005} + \dots + (z + 1) + 1 = 0
z^{2006} + \dots + 2007 = 0

Since the product of roots of the above equation = 2007,
P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2006} = 2007

Not enough? If you want to know how powerful in using complex numbers for solving elementary mathematics problems, I suggest an old Chinese popular mathematics for your own leisure reading: 神奇的複數─如何利用複數解中學數學難題

I do admire Mr. Siu’s concept : “I don’t teach: I share". Read his blog to learn better English!
http://hk.myblog.yahoo.com/siu82english

2007/12/11

[AL][PM][U] 比較複數的大小?

Filed under: HKALE,Pure Mathematics,University Mathematics — johnmayhk @ 6:27 下午
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以下是中七同學 Justin 早前的留言。

John sir早前上堂提到 complex number: i 和 0 大小比較的問題

當時的內容:
i > 0, then i^2 > 0,則 -1 > 0, contradiction。反之設 0 > i, 亦生矛盾。


">"關係能滿足以下四個性質:
1)對任何兩個不同的數 a,b; a > b or b > a 不能同時成立
2)if a > b,b > c then a > c
3)if a > b then a + c > b + c
4)if a > b, c > 0 then ac > bc

現設 i > 0
by 4) i^2 > 0, -1 > 0
書中提到此未能導致矛盾(*)
(因為”>”不一定是實數中規定的含義)←這和阿sir上堂提到的問題有分別。
繼續:
利用剛得到的 -1 > 0,by 4), both sides time -1 then (-1)^2 > 0, 1 > 0
(這裏並未有將inequality sign倒轉),這裡是否和(*)所提到的有關?
we have -1 > 0
by property 3), add 1 into both sides
-1 + 1 > 0 + 1,then 0 > 1
by from above, 0 > 1 and 1 > 0,which contradicts property 1)
then we can conclude that i is not > 0
反之 0 > i 也有同樣結果

然而,我的問題是:inequality sign 的使用層面 or 定義有沒有特定規限?因為 (*)

感謝同學指出我的錯誤。我以『-1 > 0』證明『複數不能比較大小』是太快得到結論。讓我趁機又吹吹水。

小孩初接觸自然數,

1,2,3,…

自然地為數字『排序』, 1 先,之後是 2,之後是 3,如此類推。及後,我們引入『不等式』符號 <,得到

1 < 2 < 3 < …

代表著一種『序』(order),也方便我們把兩個自然數比較大小,諸如 3 比 5 小,記之曰 3 < 5。

數學上,我們有辦法定義其他不同的『序』,比如,我們真的可以有

1 > 2,2 > 3 等等(參考 order dual)

那個符號『>』,就不是我們一般的大於(not in usual sense),而是符合某些條件的一種『序』。

對於 complex numbers,我們可否定義出一種『序』?

可以,最常用的是『字典排序』(lexicographic order),即是把字編入字典的方法,比如编 happy 先於 harass,因為我們由左至右比較字母,頭兩個也一樣,直至發覺 p 先於 r,便規定 happy 先於 harass。不妨記之曰 happy < harass。

對 complex number,我們不妨定

a + bi < c + di 當且僅當 a < c 或 (a = c 及 b < d)

(當心,在 A-level 千萬不可出現 a + bi < c + di 等等的式子,老師會二話不說給你零蛋。我們現在討論的是一種『序』,隨便你用什麼符號,但用『<』,是讓我們有一種熟悉的感覺而已,它不一定是一般的不等式符號,它只用來表達熟先熟後的標記。所以 Justin 可以放心,在中學,『<』這個符號一定是代表『嚴格小於』,只是到了大學,就要看情況了。)

由上述定義,我們有

3 + 100i < 4 + 2i
5 + 6i < 5 + 7i
8 – 2i < 8
等等。

(大家領略到好像『查字典』的感覺嗎?)

那麼,任何兩個 complex numbers,我也可以『知其先後』。
但,我們一般不會稱:任何兩個 complex numbers 都可『比較大小』。

其實,何謂『序』?數學上,『序』是一種關係(relation),起碼包括以下三種:

1. 偏序(partial order):亦可再細分為弱偏序及嚴格偏序
2. 線序(linear order)或稱全序(total order)
3. 良序(well order)

先談偏序。

設 A 是一個集。如果存在一種關係,姑且稱之曰 \leq,滿足以下 3 個特性

(自身性 reflexivity)對所有 a \in A, 恒有 a \leq a。
(反對稱性 antisymmetry)對所有 a , b \in A, 若 a \leq b 及 b \leq a,則 a = b。
(傳遞性 transitivity)對所有 a , b , c \in A, 若 a \leq b 及 b \leq c,則 a \leq c。

我們稱滿足上述 3 個條件的關係 \leq 為偏序。存在這種偏序的集,稱為偏序集(partially ordered set,或稱 poset)。只要考慮一般意義下的不等式 \leq,整數集明顯地是偏序集。

讓我給一個非不等式的例子。

例如


a = {1}
b = {2}
c = {3}
d = {1,2}
e = {1,3}
f = {2,3}
g = {1,2,3}

設 A = {a,b,c,d,e,f,g}

我們很容易為 A 定義一種偏序,就是所謂 set inclusion。即

x \leq y iff x \subseteq y

見上例,因 a \subseteq d,有 a \leq d。又例如,b \subseteq d 而 d \subseteq g,有 b \subseteq g;亦表示 b \leq d,d \leq g,有 b \leq g。

因為 \subseteq 存在自身性,反對稱性和傳遞性,立知 \leq 亦符合這 3 個條件,即 \leq 是偏序。

Justin 給的序和偏序差不多,即是所謂嚴格偏序(strict order)。

設 A 是一個集。如果存在一種關係,姑且稱之曰 <,滿足以下 3 個特性

(非自身性 irreflexivity)對所有 a \in A, 無 a < a。
(非對稱性 asymmetry)對所有 a , b \in A, 若 a < b,則無 b < a。
(傳遞性 transitivity)對所有 a , b , c \in A, 若 a < b 及 b < c,則 a < c。

稍微推論一下,我們可以由『傳遞性』及『非自身性』推出『非對稱性』,那麼嚴格偏序的條件可減少到 2 個。

相對於嚴格偏序,有人稱前一種偏序為弱偏序。

單有偏序,足夠嗎?不,起碼,偏序不能滿足我們習以為常的一種認為:任何兩個數都可以『比較』。

舉例,我們在正整數集上定義這樣的一種偏序:若 a 整除 b,則定義 a \leq b。例如

2 \leq 6 [因 2 整除 6]
5 \leq 20 [因 5 整除 20]

只要稍微檢查,可知這種序滿足偏序的三個要求:自身性,反對稱性和傳遞性。然而,對於某對正整數,諸如 4 和 5,因 4 不整除5 及 5 也不整除 4,故我們既無 4 \leq 5 亦無 5 \leq 4。這就是所謂不能『比較』了。

那麼,我們只要在嚴格偏序條件上加多一個:所謂『三分律』或『三一律』,即集合中任何兩個元素必有『三種關係』的其中一種,即

(三分律 trichotomy)對所有 a , b \in A, 必有 a < b, b < a 或 a = b。

滿足非自身性,非對稱性,傳遞性再加上三分律這四個條件的偏序,稱為線序或全序。

『線』者,取其把所有『元素』都可如『數線』,按序排列。
『全』者,即謂『全部』元素都可知其先後。

稍微檢查一下,複數集上定義的『字典排序』:a + bi < c + di 當且僅當 a < c 或 (a = c 及 b < d)
是全序,故複數集是所謂全序集。

如果複數集只是盛載著複數的集合,沒有任何運算發生的話,這個數學物體可說是沒有價值。但當我們在這個集合賦予了一些運算,諸如加減乘除,我們彷彿把這個死物『有機化』,滿注『生命力』。當複數集定義了『加法』和『乘法』後,這個東西不單是複數集,數學上,我們可稱它為複數域(field)。

如果有個域 F(簡單說,就是可以在 F 當中定義『加法』和『乘法』兩種運算),一旦存在全序 < 滿足額外的兩個要求:

(1)對所有 a , b , c \in F,若 a < b,則 a + c < b + c。
(2)對所有 a , b \in F,若 0 < a 及 0 < b,則 0 < ab。

我們稱 F 為全序域。這個『<』更進一步符合我們慣常意義下的不等式之要求。

那麼,複數域是全序域嗎?Justin 的留言已經說明了,複數域不是全序域。讓我也不厭其煩,重述之。

我們用前述的『字典排序』,有 0 < i。
假如複數域是全序域,由條件(2),得 0 < i^2
得 0 < -1
有違『字典排序』,生矛盾。

好,若我們在複數域上定義另外的全序(不用『字典排序』),表之曰⊿。由全序定義,我們必有
0 ⊿ i 或 i ⊿ 0 或 i = 0(不合)

先考慮 i ⊿ 0
由全序域的條件(1),得 i + (-i) ⊿ 0 + (-i)
即 0 ⊿ -i
由全序域的條件(2),得 0 ⊿ (-i)(-i)
即 0 ⊿ -1 ———————————- (*)
再由全序域的條件(2),得 0 ⊿ (-1)(-1)
即 0 ⊿ 1
由全序域的條件(1),得 0 + (-1) ⊿ 1 + (-1)
即 -1 ⊿ 0 ———————————– (**)
由 (*) 及 (**) 可見,⊿ 已不符合全序的其中一個條件:三一律,所以我們不能在複數域上定義這種滿足全序域的序 ⊿。

類似地,若考慮 0 ⊿ i,我們也可推導出 ⊿ 有違全序的條件,諸君不妨一試。這樣,正是這個原因,我們會(粗疏地)說『兩個複數可排序但不能比較大小』。

總結

複數集是全序集,但
複數域不是全序域。

至於第三種序:良序。對歸納法之所以可行和研究無限集序數等有重要意義,但和本篇主題不太相關,就此在結。

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