# Quod Erat Demonstrandum

## 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

## 2007/12/17

### [初中][玩玩] 幾道幾何題

Filed under: Junior Form Mathematics — johnmayhk @ 3:32 下午

Q.1

Q.2

Q.3

ABCD 是正方形，E 在 BC 上。已知 ∠DCF = 45∘，∠AEF = 90∘，證明 AE = EF。

Q.4

### [AL][AM] 出卷心『程』

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 2:27 下午

## 2007/12/13

### [AL][PM][AM] 以應數技巧解純數中涉及 nCr 的問題

Filed under: Additional / Applied Mathematics,HKALE,Pure Mathematics,Teaching — johnmayhk @ 4:07 下午

1. $C_{r+1}^{n+1} = C_{r}^{n} + C_{r+1}^{n}$
2. $C_{r}^{m + n} = \sum_{k=0}^{r}C_{r-k}^{m}C_{k}^{n}$ ($r \leq \min (m , n))$
3. $C_{n+1}^{m+n+1} = \sum_{k=n}^{n+m}C_{n}^{k}$ (即 1994 年 Paper I Q.7)

Q.1 堂上我已提及，由柏斯卡三角的構作方式，已體現了它。再給一法：

Q.2 $C_{r}^{m + n}$ 就是在 m + n 人中選出 r 個的組合方式之數目。現在考慮 m + n 人內有 m 男 n 女。於是，要選出 r 人，可以是

r 男 0 女，共 $C_{r}^{m} \times C_{0}^{n}$ 種組合，或
r – 1 男 1 女，共 $C_{r - 1}^{m} \times C_{1}^{n}$ 種組合，或
r – 2 男 2 女，共 $C_{r - 2}^{m} \times C_{2}^{n}$ 種組合，或

0 男 r 女，共 $C_{0}^{m} \times C_{r}^{n}$ 種組合。

Q.3 在純數課中，我們可以考慮 $(1+x)^n + (1+x)^{n+1} + ... + (1+x)^{m+n+1}$ 及 G.P. sum 便可解之。但我們也可用應數中的『概率』解之如下：

P(n) + P(n + 1) + P(n + 2) + … + P(n + m) = 1

### 無聊分享

Filed under: Life — johnmayhk @ 12:58 上午

『十七年華』這類題材並不陌生，可能只是相對較少吧。還記得 SARS 及至對政府表達很多不滿那段期間，網上多了一些針對時弊的歌。但『十七年華』以 Rap 形式，從『新聞標題』跳到『死者死前第一身的自述』，再帶出『第三身的新聞內容報導』，頗為特別。尾段『沒人理會這新聞』是寫得很有意思，對這段的感覺，立時讓我聯想到一首我極之喜愛的歌：『十個救火的少年』。潘源良的詞加上黃耀明的演繹，再加上當年的社會背景，令此歌成為不可多得的佳作。『在理論裡沒法滅火跟煙』『葬身於這巨變』『大眾議論到這三位少年，亂說亂說，愈說只有愈遠。』歌詞信息頗沉重，但歌曲的旋律卻是輕快，非常特別。我非專業寫詞人，但相信填 Rap 的詞比較易，因為沒有旋律的『包袱』，自由度大。無論如何，『十七年華』令一個對時下樂壇脫節的人把它聽了又聽，相信已是非常厲害了。

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

$i > 0$, 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 $ac > bc$

by 4) $i^2 > 0$, $-1 > 0$

(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。

(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 $> 0$

1,2,3,…

1 < 2 < 3 < …

1 > 2，2 > 3 等等（參考 order dual）

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。）

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

（大家領略到好像『查字典』的感覺嗎？）

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）

（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。

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

x $\leq$ y iff x $\subseteq$ y

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

（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。

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

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

『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。

（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。

0 ⊿ i 或 i ⊿ 0 或 i = 0（不合）

## 2007/12/03

### [PM] Solving inequality with absolute signs

Filed under: Pure Mathematics,Teaching — johnmayhk @ 3:27 下午

Solve $|x - 1| + |x - 2| \geq 5$

$x \geq 4$ or $x \leq -1$

## 2007/12/02

### [初中] Identities

Filed under: Junior Form Mathematics — johnmayhk @ 4:22 下午

The following is a level zero question in F.2 math textbook.

Determine whether $4(3x + 1) = 2(6x + 2)$ is an identity.

Being a math teacher, I must suggest the following solution.

L.H.S.
= $4(3x + 1)$
= $12x + 4$

R.H.S.
= $2(6x + 2)$
= $12x + 4$

$\because$ L.H.S. = R.H.S.

$\therefore 4(3x + 1) = 2(6x + 2)$ is an identity.

Everything fine.

But we may have alternative ways.

One student, Lam, gave

L.H.S.
= $4(3x + 1)$
= $2\times 2(3x + 1)$
= $2(6x + 2)$
= R.H.S.

Quite good.

For discussion purpose, I gave two more ways.

Method 1

$4(3x + 1) = 2(6x + 2)$
$12x + 4 = 12x + 4$
$0 = 0$

Do you accept this solution? Why or why not?

Method 2

When $x = 0$,
L.H.S. = $4(3(0) + 1) = 4$
R.H.S. = $2(6(0) + 2) = 4$
L.H.S. = R.H.S.

When $x = 1$,
L.H.S. = $4(3(1) + 1) = 16$
R.H.S. = $2(6(1) + 2) = 16$
L.H.S. = R.H.S.

Hence $4(3x + 1) = 2(6x + 2)$ an identity.

What is your comment on Method 2? For me, it is OK!

### [初中] Factorization

Filed under: Junior Form Mathematics — johnmayhk @ 3:50 下午
Tags:

The following is a question in the uniform test.
Factorize $12xy - 6x^2$.
One student, Lai, gave the solution as
$12xy - 6x^2$
=$12x(y - \frac{x}{2})$
If you were a teacher, will you give marks?
Well, you may have the “correct answer" in your mind, it should be $6x(2y - x)$, right? But why the answer given by Lai was not an answer? More, if somebody gives the following, what is your comment?
$12xy - 6x^2$
= $24x^3y(\frac{1}{2x^2} - \frac{1}{3xy})$
Urm, let me further my discussion by giving two more examples.
Many students know how to factorize $x^2 - y^2$, that is
$x^2 - y^2$
= $(x + y)(x - y)$
Urm, can I further my calculation in writing something like
$x^2 - y^2$
= $(x + y)(x - y)$
= $(x + y)(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$
Do you think the above is an answer to the factorization problem?
Some may say we cannot factorize $x^2 + 1$, but after introducing the complex numbers, can we say
$x^2 + 1$
= $(x + i)(x - i)$ where $i^2 = -1$
is a process of factorization?
All in all, the above may force us to think about what do we mean by factorization exactly?

### 小報告

Filed under: Report — johnmayhk @ 2:25 下午

http://www.wretch.cc/blog/giawgwan&category_id=169115