Introduction

Routine problems usually require us to have one or many answers . If we are asked to find the smallest of the three consecutive integers whose sum is 18, then our answer would is 5. If we are asked to find the equation of a line passing through (2,3), we can have many answers.

Proofs, however, is different. It requires us to think more and to reason with sound arguments. It requires us to be explicit and to be logical. It requires us to convince our readers and most of all ourselves.

Unless a proof problem is already given, finding mathematical statements to prove requires us to see patterns, generalize them and make conjectures about them. The problem stated above about consecutive integers does not require us to reason much or to generalize at all.

The success of proof writing requires intuition, mathematical maturity and experience. Contrary to proofs written in books, the ideas behind arriving at a proof are not “well-pressed” and elegant.  Mathematicians do not reveal the process they go through, or the ideas behind their proofs. This is also a skill that mathematicians and persons who are good at mathematics possess: they are able to read proofs. The skills of reading proofs may be achieved by learning how to write them.

Proving in higher mathematics, on the other hand, requires formal training. For instance, we have to know how to use logical connectives like and, or, not, and must understand how conditional and biconditional connectives work. Basic set theory concepts are also important. Moreover, we also have to learn proof strategies like direct proof and proof by contradiction to name some.

For now, we will not be discussing these things . Most of the proofs in basic mathematics only require a little intuition and good reasoning.

In the tutorial below, I tried to recreate (amateurishly) the process on how mathematicians see patterns, arrive at a conjecture and how they prove their conjectures.  Of course, in reality, problems mathematicians encounter are a lot harder. In fact, some of the hardest problems take hundreds of years to be solved. For example, no mathematician has proved the Fermat’s Last Theorem for more than 300 years, and the mathematician who proved it solved it for eight years.

The proof that we are about to do below is very elementary. For now, we will highlight the process and not the difficulty. The titles of the processes below are not necessarily in order.

Recognizing Patterns and Making Conjectures

Before mathematicians prove theorems, they usually first see patterns. This happens when they read books, solve problems or prove other theorems. For example, what do we see when we add two even integers? Let’s add some: 2 + 8 = 10, – 24 + 6 = -18, and – 4 + – 8 = -12. We can easily see that if we add two even integers, then their sum is always even.

From here, we might be tempted to say that if we add two integers, then their sum would always be even.  In mathematics, this kind of statement or hypothesis is called a conjecture: an educated and reasonable guess based on patterns observed. Rewriting our guess, we have

Conjecture: The sum of two even integers is always even.

If we want to disprove a conjecture, we only need one counterexample (Can you think of one?). If we can find one counterexample for the conjecture, then it must be false.  If we want to prove it, however, we might be tempted to pair a few more integers and say that “oh, their sum is even, so it must be true”. Note that no matter how many integers we pair, if we can’t exhaust all the pairs, then it cannot be considered a proof.  When we say the sum of two even integers above, we mean ALL even integers.  Of course, there is no way that we can list all pairs of even integers since there are infinitely many of them.

Generalizing Patterns

Since it is impossible to enumerate all pairs of even integers, we need an algebraic expression that will represent any even integer. If we can find this algebraic expression, then all the even integers would be represented. This process is called generalizing.  If you generalize, you represent all members of the set by a single expression. In our case, the members of our set are all even integers.

From the definition, we know that all even integers are divisible by 2. That means that if m is an even integer, then, when we divide m by 2, we can find a quotient which is also an integer. For instance, since 18 is an integer, we are sure that there exists an integer such that 18 divided by 2 is equal to that integer.

In general, suppose that quotient of m/2 is q, then it follows that m/2 = q, for any even integer m. Multiplying both sides of the equation by 2, we have m = 2q. That means that if m is an even integer, then there exists an integer q such that m = 2q. Hence, we may represent any even integer m with 2q for some integer q*. Note that q here is a generalized number, which means that an even integer can also be represented by 2x, 2y, 2z or any variable with the condition that the variables are integers.

Connecting the ideas

Proving is making logical and relevant statements from definitions, facts, assumptions and other theorems to come to a desired conclusion. Before coming up with an elegant proof, mathematicians usually have their scratch work, connecting their ideas to arrive at what they want to prove.

Scratch work:

From the statement above, we have shown that any even integer m, there exists an integer q, such that m = 2q. That means, that if we can show that the sum of two even integers is in the form 2q (or that the sum is divisible by 2), then we can be sure that it is always an even integer.

Since we need two integers, we might let m and n are the two integers that we will add. Since both of them are even integers, then we can represent them as 2q and 2r respectively for some integers q and r. Adding both of them, we have m + n = 2q + 2r = 2(q + r). Now, q + r is an integer since q is an integer and r is an integer from our definition above. This means that 2(q + r) is of the form 2x for some integer x. This means that m + n is of the form 2x for some integer x. Therefore, m + n is even.

Writing (elegantly) the final proof

Here, we write our proof in a shorter and more elegant way. Conjectures that are proven are called theorems. So let us write the proof of our first theorem.

Theorem 1: The sum of two even integers is always even.

Proof. Let m,  n be even integers.  Then m = 2q for some integer q and n = 2r for some integer r. Now, m + n = 2q + 2r = 2(q + r). Since q + r is an integer, clearly, 2(r + s) = m + n is divisible by 2. Therefore, the sum of two even integers is even.

Most proofs are written in a concise way, leaving some details for the reader to fill in. For example, the statement “Since q + r is an integer” did not really state the reason why this is so. This is stated in our scratch work, but not in the proof.

Going Further

If m is an even integer, then m – 1 and m + 1 are odd integers. Since m = 2r, then 2r – 1 and 2r + 1 are also odd integers. In our example below, we will use 2r + 1, to prove that the sum of two odd integers is always even. As an exercise, use 2r – 1 in your proof.

Theorem 2: The sum of two odd integers is always even.

Let p, q be odd integers. Then p = 2a + 1 and q = 2b + 1 for some integers a and b. Now, adding we have p + q = 2r + 1 + 2s + 1 = 2r + 2s + 2 = 2(r + s + 1). Since r + s + 1 is an integer, then 2(r + s + 1) is divisible by 2. Hence, p + q is divisible by 2. Therefore, the sum of two odd integers is even.

Math and Hardwork

Being good in math requires hard work. Andrew Wiles worked on the Fermat’s Last Theorem for seven years, have given up several times thinking that it was impossible. In 1995, he finally thought he had proved it, and presented it in a conference.  A month later, his reviewer thought that there is a part of the proof which was vague (or wrong), so he had to review his work and found out that there was a part which was actually wrong.  He almost gave up. He worked more than a year to correct the error.

Now, he has carved his place in history.

*In technical language, the word “for some” is equivalent to the word “there exists”.

Exercises:

1. Prove that the sum of an even number and an odd number is always odd.
2. Prove that the difference of two odd integers is always even.
3. Prove that the product of two even integers is always even.
4. Prove that the product of two odd integers is always odd.
5. Prove that the product of an even number and an odd number is always even.

From Facebook:

Subroto Roy thinks the core of the *Bhagavad Gita* is captured in Grigori Perelman’s statement declining the Fields Medal after proving Poincaré’s conjecture: “[The prize] was completely irrelevant for me. Everybody understood that if the proof is correct then no other recognition is needed.”

Remember junior high?

Remember quadratic and cubic equations?

Well, turns out that Dan Christensen, a researcher in the University of California, took these mundane objects a little more seriously than most of us, and drew a picture of all the roots of all the polynomials of degree at most 5 with integer coefficients ranging from -4 to 4. Surprisingly, marvelous patterns in the roots of the polynomials. The results can be found here.

While it’s not that easy for us laymen to understand the implications of such a discovery, but even so, I think that no one disagrees that the images are marvelous! Personally, it gives me goose pumps, as it indicates that there’s beauty beyond our imagination even in the simplest of artifacts – equations of the form a*x^2 + b*x + c = 0.

Let be a point on the curve Let be the tangent line at Given intersects the curve again at Prove that

Here is good problem for my IMO students.

Given a real number . Find the greatest area of triangle ABC with circum-radius

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：

（甲）二次方程的情況

若 的複根是 ，則

…………………… (*)

由 (*) 可見

的圖像，其頂點（vertex）的 -坐標是 ，而 就是複根的實部（real part）。

另外，由 (*) 可知 的圖像，其頂點的 -坐標是 （即片中的 ）；

考慮在二次曲線上的一點 ，其 -坐標為 ，那麼 的 -坐標 是什麼？

只要把 代入 ，得

即是說，複根虛部中的 ，就是 的 -坐標 和 的距離。

（乙）三次方程的情況

這個要中六同學才明白，高中一同學要忍耐一下。

為簡化，只考慮 係數為 1。

即設 。

首先，實係數三次方程必有實根；若它只有一個實根 ，則設另外兩個複根為 。

那麼，片中曲線的方程可寫成

……………………… (1)

現在，設經過 並相切於曲線的直線方程為

……………………… (2)

假設曲線和直線的切點（point of contact）為 ，如何證明 的 -坐標 (say) 就是 ？

我們（想像一下）解 (1)，(2)；我們得

易知上式的解為 或 ，即

是下式的根，且是重根（repeated root）：

……………………… (2)

既是重根，把上述等式求導後， 仍是根（Why？），即 滿足：

故 真的是 。

由此可見， 滿足 (2)，故

而片中的 不過是切線的斜率，即 。

這裡有一個問題，如果切線的斜率 是負數，又會怎樣？即出現例如以下情況：

豈不是不能計出 ？

同學，試花一點時間先想想。

開估：當我們考慮 的係數為 1 （或其他正數）時， 一定是正數。

無他，

因為我們考慮的方程是 ；

故此，

代入大於 的 值，對應的 值為正。（即上圖是沒有能的）
代入小於 的 值，對應的 值為負。

這就保證了 。

Let Suppose , where is a rational root.

Then we have .

Multiplying both sides by yields:

Since , we must have divides

Now, from multiplying both sides by and argue similarly yields divides Hence the theorem.

函數 連續，並不保證逆函數 也連續。

(在定義兩個拓樸空間同胚（homeomorphic）時，就是要求他們之間存在一一對應的連續函數 ，並 也要連續。)

舉例，設 並 （即 不過是 中的 unit circle）定義 使 ，一看下圖，立即知道 不連續。

Just finished my CB class tonight and I am impressed by two students. They both expressed interests in the following problem:

Suppose is a polynomial of degree . Suppose further that

(i) for all and

(ii)

Find the leading coefficient of and the degree of

I won’t discuss the solution I presented in class here. Here are two innovative ways suggested by the two students.

1) Using Taylor series. One of the students based his solution using the Taylor’s formula. The ingenious thought is that he centered the expansion at Namely, he writes the expansion as:

This method yields the correct answer and I am very impressed.

2) Using the Mean Value Theorem. Another student suggested using the Mean-Value theorem, but he could not finish the proof. I gave this idea some thought and eventually made it to work. Here is my solution whose idea is inspired by my student.

From Differentiating both sides times gives:

for all

Now applying the mean value theorem to yields:

, where

This gives:

Now note that is a polynomial with infinitely many distinct roots, namely, . [Note that they are indeed distinct because the intervals are disjoint. Therefore, by the identity theorem.

Hence, we have for all This immediately yields the degree of is , and that the leading coefficient can be found by integrating times, which yields the answer to be

For example, integrating once yields: , integrating twice yields

Likewise, integrating 101 times yields:

Book 2, pg 14. (00IQ12)(c)
(i) Let y = x^2, then we have ay^2 – by + a.
Considering the discriminant yields: b^2 – 4a^2 > 0. Therefore, it has two distinct roots: If they are both positive, then we are done. Otherwise, they must be both negative since (Note since .)
Suppose Then we have

By Viete's theorem, we have (after simplification) ), and since

(ii) This part is actually easy, just applying the previous parts. However, one must pay attention to note (and you have to explicitly state this fact) that are not equal to or Since dividing by something potentially zero is a fatal error in HKAL.

Book 4, 02IQ11.
(a) (ii), (Method 1) Here is a better way to present the solution. (to avoid adding the extra line explaining why might not be good enough.)
for all This would imply is strictly increasing for all since f has no local max or local min.

(method 2): Suppose it has more than one real root, then all roots are real since complex roots come in pairs. Let the roots be Then by Viete’s theorem, we have , and Now, This means , which is absurd.

The rest are as discussed in class.

Just take a rest from work, type something boring here…

Students, you may regard the above as a formula or derive it by using trigonometric substitution every time.

As you may know that the expression of a primitive is not unique, we may have other forms being a primitive of , say

where

To show the validity of the above, simply by differentiation (try, esp. the last one). But, apart from the above, any other ‘format’ of a primitive? How to obtain primitives of different ‘formats’ (esp. the last one)?

老題：利用面值 $10，$20 及 $50 紙幣若干張，要拼出 $100，問有多少組合方式？

可以拿 2 張 $50 紙幣，這是一種組合；
可以拿 3 張 $10，1 張 $20 及 1 張 $50，這是另一種組合；
……

那麼，共有多少種可能的組合？

這篇為承接昨天的發帖而寫的例。是舊技巧，高手見諒。

對 $10 紙幣，我們考慮冪級數

我們觀察 的指數，便可對應 $10 紙幣的數目，即

代表 2 張 $10 紙幣；
代表 3 張 $10 紙幣；
……

類似地，

對 $20 及 $50 紙幣，我們分別考慮冪級數

把上面提及的三個級數相乘，得

– - – - – - (*)

同學，想像（是，想像而已）把上式拆開（或曰「爆開」expand），每項皆由三個括弧，分別抽出一項相乘而得。比如，

第一個括弧抽出 ；
第二個括弧抽出 ；
第三個括弧抽出 ；

相乘之，得

– - – - – - (**)

看看當中的意義：

第一個括弧抽出 ，相當於拿出 3 張 $10 紙幣；
第二個括弧抽出 ，相當於拿出 1 張 $20 紙幣；
第三個括弧抽出 ，相當於拿出 1 張 $50 紙幣；

總面值是 3*$10 + 1*$20 + 1*$50 = $100；而 100，正是 (**) 中 的指數。

這樣，(**) 代表著一種拼出 $100 的組合方式（即 3 張 $10，1 張 $20 及 1 張 $50）。

而 (**) 不過是 (*) 拆開後的某一項。

即 (*) 拆出一項 ，代表著一種拼出 $100 的組合方式。

那麼，若 (*) 可以拆出 項 ，即代表著 種拼出 $100 的組合方式。

但若 (*) 可以拆出 項 ，把它們加起來，得 。

也就是說，(*) 拆開後， 的係數就是 ，亦即是說

「有多少種拼出 $100 的組合方式，就等於 的係數」。

但如何求 (*) 中 的係數？昨天也談過。

– - – - – - (*)

可變為

之後就是拆部份分式，但直接考慮上式的部份分式，太艱鉅，我們考慮簡單一點的情況：

不難得到

運用 sum of G.S.，得知

於是

那麼，

故此，(*) 變成

不難尋求上式中， 的係數，即

即是說有 10 種拼出 $100 的組合方式。

驗證一下：

數一數，共 10 個組合方式。

同學你或在取笑這個方法：「就咁數」仲快。不錯，這個特例是，但上述的方法是值得留意的：因為它是有系統的方法。

It is extremely easy to set up questions on number pattern, like

Finally, students, you may find the method of using generating function is a bit clumsy, especially for this particular question. You may have smarter methods to solve for . However, all I want is to introduce the method of using generating function and it is known that some questions can ONLY be solved by using generating function. Hope to share more next, bye bye!

Let

Solve the following matrix equation where

Dear Students,

Please avoid as much calculations as possible. In ideal situation, you should not do any calculation at all.

Problem 1
Let be a complex number.
(a) Find
(b) Find
(c) Find
(d) Find
(e) Find
(f) What is
(g) Suppose , write in terms of

Problem 2
(a) Write in the form
(b) Find the two square roots of and represent the two square roots on the Gaussian Plane.

Problem 3
(a) State Euler’s Formula
(b) State a friend of Euler’s Formula
(c) Write and in terms of powers of

Problem 4
(a) Write done the n-th roots of unity in terms of
(b) What is as a power of ?
(c) Write out one nth root of as a power of
(d) Write out all the nth roots of in terms of your answer in (c) and

Problem 5
(a) Describe geometrically the set of complex numbers satisfying
(b) Describe geometrically the set of complex numbers satisfying
(c) Sketch on the Argand diagram, the set of points satisfying both and
(d) Without doing any calculation, determine the number of points satisfying both and
(e) Consider the circle C: , let be the center of any circle that touches with radius 2. Describe the locus of

Problem 6
(a) Suppose is a non-zero complex number. Let be roots of the equation $z^5 – 1 = 0.$ Describe geometrically, the shape formed by complex numbers
(b) Let be a cube root of unity. What is
How about

Problem 7
Let .
(a) What is
(b) What is
(c) What is

Problem 8
Suppose you are asked to evaluate the two sums:
(i) and
(ii) .
What is your first step? And you next step? Do you see the entire solution lay out in your head?

Problem 9
Let be a polynomial of degree 4 with real coefficients. Given that and are roots of
(a) What is the sum of roots?
(b) What is the product of the roots?
(c) Do you know how to do the problem if the condition “ has real coefficients is removed?

Problem 10
Expand the following in your head:
(a) (1 + x)(1 + y)(1 + z)
(b) (1 – x)(1 – y)(1 – z)

Problem 11
How can you tell whether is rational?

Problem 12
Let C be the set of points satisfying
(a) Describe C geometrically.
(b) Let P = 4 + 3i. What is the shortest distance from P to C?
(c) Find the complex number w on C such that the distance between P and w is the answer you found in (b). [Remember, no calculation, you can work this out in your brain.]

Lesson 1:
Page 26
where the last line follows from part a(ii).

Page 27
(i)
Hence

Note that we have implicitly used the condition to avoid being

(ii) The sum equals

Again note that we have implicitly used the condition to avoid being

Exercise: try to do the problem on page 27 without using the previous parts. This time, use the theory of roots of unity.

Lesson 2

Page 50

(a) (i) , where . Hence the roots are simply , for

(ii) Next, we have

Thus

Now,

Finally, since dividing both sides by yields the desired factorization for at once.

(b)(i)

(b)(ii) Let , we have and

Adding the subtracting the two equations yield

, and

Plugging this in (a)(ii) yields

Dividing both sides by 2i yields the result at once.

(b)(iii) Dividing both sides by and taking limit as yields

by b(ii).

Now, using yields

Finally, for since for We have

(c) From (a)(ii), setting z = -1 yields

Using yields

Since for , we have

Let (E) be the following system:

Let be the augmented matrix.

Determine whether the following statements are True or False.

1) (E) can have exactly 3 solutions.

2) (E) has a unique solution iff

3) A is invertible iff

4) The system: is consistent if or

5) If , then

6) If , then

7) If (E) is consistent and , then (E) has infinitely many solutions.

If (E) has infinitely many solutions, then

9) If , then

10) If , then (E) is consistent.

More at
http://www.youtube.com/view_play_list?p=ED4CEC2CC684DDDD

初中時教解二次方程，我通常順便說一個無聊的例：

= ?

要求出「答案」，我們可設

等號左右兩邊取平方，則

故此

解出 或 （不合），亦即

當然，

的表達有點兒「唔清唔楚」，清楚一點的表達，比如是

即

中六修純數的同學知道，我們「計」極限時，要先證明極限存在，然後才進行上述「代 」 的計算。

同學會質疑：「明明已經計到極限出來（像低年級時做的），豈不就說明它存在嗎？為何還要先（比方說，用 monotonic sequence theorem 來）證明極限存在？」

其實所謂「計到」，不一定代表極限存在，舉個例：

即

如果我們跟隨低年級學的方法，設

取平方，

即

「計出」

不管哪一個值，總之取其中一個作極限，豈不是說極限存在嗎？

錯！{} 根本是不收斂的，且看

可見 {} 是發散的。此乃「計到，不一定存在」也。

1. 高階導數的符號

同學問，為何 D 兩次（即求二階導數）的符號是

而不是

　或　？

記得 CJ 老師曾經出上面的來考同學。

我不知其來源，只是靠估兼無聊地說：

比如求 的導數（w.r.t. x），我們可以表達成 ，這個 可視為運算子（operator）。

求 的二階導數，即 ，我們可以想像為：運算子作用在 兩次。循運算子的看法， 可表為 ，於是，它看起來「好像」「二次方」的運算（當然不是真的二次方啦！），借用一下「二次方」的記法，即 ，於是自然地，寫它為 ，或曰 是也。一般地，由記法 得 。

2. 不定義的函數

有沒有聽過以下的句子？

「證明函數 f 是有定義（well-defined）」
「這個函數 f 沒有定義」
…

現在中一已經引入函數的概念，同學對函數相信也能了解一二，但上述句子，有沒有問題？

純數課告訴我們，如果關係（relation）f 被稱為函數，那麼 f 必然是有定義的。那麼「證明函數 f 是有定義」這說法好像有點問題，「這個函數 f 沒有定義」也有點奇怪。

比如設 ，其中 為正整數。

這個 f 的運算，就是小學生發明的神奇分數加法！這是學分數加法時的一個「難點」。

看看 f 這個東西，明顯地，它不是函數，因為它不符合：「一個輸入，一個輸出」，例如

同樣輸入 及 ，輸出迥異。

f 不是函數（function），那麼我們如何稱呼這些 f？Gower 提議用 “gunction” 這個字，有興趣有耐性的同學，不況看看以下文章，特別是讀者回應部分：

http://gowers.wordpress.com/2009/06/08/why-arent-all-functions-well-defined/

3. 解釋和簡略地解釋

在數學考題，著學生 Explain 和 Explain briefly 有沒有分別？如何界定一個「解釋」，真的是一個有效的解釋？肯定不是字數。

比如，試簡略解釋：「設 ~ ，當區間 的長度（width）固定，則當 時， 的值最大。」

如果單單說一句：從常態分佈的圖像得之。那麼，可否視它為「簡略地解釋」？

雖然這不是百分百錯誤的解釋，但這似乎不算有效的解釋，起碼，究竟從常態分佈的圖像的什麼（特性）得之？

設區間 長度固定為 （）。把圖像畫出如下

當 置中，即 ，，把 右移一些，如圖所示，則在新的位置，，由常態分佈的圖像，可見 ，即區間在新位置對應的新概率（面積） 比區間置中時對應的概率 少。

大家覺得這是不是一個有效解釋？是簡略解釋嗎？有沒有漏洞？

我也頗喜歡擬要學生解釋的題目，但當中涉及一些似乎不純為數學的問題，比如界定解釋的有效與否，除了運算能力，更難的可能是溝通能力。

數學人喜歡的，相信是冰冷無誤語言：

設

因 ，故當 ， 達到最大值。

相信這肯定較之前的解釋有效，但簡略嗎？

前

考中二數學前夕，梁同學致電問數。我都「好野」，一邊行街赴父親節宴會，一邊做數講數。

問題太多，晚上回家，梁同學再問下半場。

他問什麼中二數學問題？列幾個：

1. Factorize .
2. 把 8 cm * 10 cm 長方形一對角（opposite angles）摺疊，求摺痕長度。
3. 二進制轉和十六進制的直接互換方法。
…

等等。

後

考試最後一天，學生考完純數。

鄧同學：「阿 sir，份卷好長呀！」
我：「都唔係呀，都係 F4 size 渣o麻」
鄧同學：「…」

我不是出卷高手，題目往往又長又悶。同學可以下載看看：

f6-pure-mathematics-final-examination-200906022

Q.2 和 Q.5 總共用不到半小時 hea 出。稍為要想一想的，是最後一道短題（Q.6），幾個同學問 有冇打錯（其實沒有問題的）。

從同學的解題，才看到一些題目的「白痴」，例如

Q.6 如果以 為目的，(a) 是極之無聊的，因為上限是極易找出：

由夾逼原理，完工。又例如

Q.7 如果以解 為目標，已知它有 triple root，其實直接去解不難，但用之前的結果，似乎簡單複雜化。

我心想長題 Q.7, Q.8 可以送分，試後，從同學的評語看來，未敢樂觀。

Q.7 在 (b) 原先沒有給同學 ，最後我也給了，望減低傷亡率。
Q.8 明顯是 hea 出的 curve sketching，我是想，這些樣版題目，何來費周章，找幾個可以的數字照改 past paper 已是。

至於另外兩道長題，了無新意：

Q.9 我不過是把已往 MVT （Cauchy form）的問法濃縮，推出 Jensen，再找個可以的函數 做個不等式。
Q.10 是 ratio test，亦容易找一些「樣好」的級數（series）來擬題。

事實上今次擬這純數卷是我有生以來最快的，不過，相信我又令不少同學陣亡，他們或希望打我一頓…

Let’s consider a question from International Mathematical Olympiad 1986…

IMO 1986/3 – To each vertex P_i (i = 1, . . . , 5) of a pentagon an integer x_i is assigned, the sum s = Σx_i being positive. The following operation is allowed, provided at least one of the x_i’s is negative: Choose a negative x_i, replace it by −x_i, and add the former value of x_i to the integers assigned to the two neighboring vertices of P_i (the remaining two integers are left unchanged). This operation is to be performed repeatedly until all negative integers disappear. Decide whether this procedure must eventually terminate. [Proposed by Elias Wegert, East Germany]

Now the answer to this magnificent question is even more magnificent than what the question seems itself.

The key to the answer lies in finding an integer-valued parameter of each situation (after each step), after which the answer is simple with Fermat’s Method of Infinite Descent.

Also it is important to note that the sum of integers (= s) is invariant, that is it doesn’t change with our legal transformations.

Now for the cyclic-5-tuple ((x₁, x₂, x₃, x₄, x₅)), which is unique at each stage S_i,
Define a cyclic summation

as a parameter of the situation.
Now calculating the effect of each legal transformation on our parameter-
Let the negative integer, on which we apply our transformation be x₃
Hence x₁ & x₅ don’t change
ƒ(S_i+1) – ƒ(S_i) = (x₁ – x₃)² + (x₂ – x₄)² + (x₃ – x₅)² + (x₄ – x₁)² + (x₅ – x₂)² – (x₁ + x₃)² – (x₂ + x₃ – x₄ – x₃)² – (-x₃ – x₅)² – (x₄ + x₃ – x₁)² – (x₅ – x₂ – x₃)²
= 2s * x₃
That would imply that as long as x₃ is negative, our parameter is monovariant which will continuously decrease which is not possible as there does not exist any ever-decreasing integer sequence of squares, thus our Transformation Loop will terminate.
Q.E.D.

But this magnificent game doesn’t stop at solution
of the IMO problem.

To be continued…

有人強調數學在生活中的應用，有人指出所謂「數學在生活中的應用」是牽強的。經驗使我比較贊成後者。

無論如何，以趣味的角度向學生介紹（所謂的）數學在生活中應用的例子，或許可以達到某些教學的成效。

讓我在這裡介紹一個「偽應用」吧，起碼三本數普書籍有記載此例。

修純數的同學定會接觸介值定理：

設 f 為定義在 [a,b] 上的連續函數，已知 f(a)f(b) < 0，則在 (a,b) 內存在 c 使 f(c) = 0。

這個定理有沒有生活應用例子？嗯，或許有，見下：

今有方桌子一張，四條腿等長。若把桌子放於凹凸不平但平滑的地面上，證明一定存在某個位置，使四條腿同時著地。

想像從上面看桌面（top view），設 A, B, C, D 為四腿子在地面的投影，見下

連 AC 及 BD，分別視之為直角坐標系中的 x 軸和 y 軸。

想像桌子自由地繞中心轉動，以 代表對角線 AC 轉動後與 x 軸的夾角。

命
= 對應 A, C 兩腿與地面距離之和
= 對應 B, D 兩腿與地面距離之和

因地面光滑，知 為連續函數。

可以想像，我們總可使桌子三腿著地，比如先使對應 A, C 的腿著地，即 for some ，再使對應 B 的腿著地，這時

無論如何，我們有

　 – - – - – - (*)

現在，不況設為對應 A, B, C 的三腿著地在 之處著地，即

命

則

想像把桌子轉動 ，即 AC 和 BD 的位置互換，易知

則

由介值定理，在 中必存在 ，使

即 （第二個等式由 (*) 而得）亦即四腿在 之位置著地。

……這例子讓我憶起兒時和弟妹在同一張小桌子上做家課的景象……

下面的例不是純數而是應數的例，純粹順便打一打，有點 Out-C 的，同學見諒。

設有一表面光滑的橄欖球，其表面形狀是由長半軸為 6，短半軸為 3 的橢圓繞其長軸旋轉所得的旋轉橢圓球面。在無風的細雨天，將該球放於室外草坪上，使長軸在水平位置。求雨水從球面上流下的路線方程。

取 y 軸為長軸，橢圓面方程為

由於雨水會沿 z 下降最快的方向向下流，此方向就是使 z 的方向導數取得最大值的方向，即

設雨水流下的曲線為 , 在 面上的投影曲線的方程為

則 的切向量 應與 平行，故

橢圓面方程兩邊取全微分，得

,

因此

解出

（ 由雨滴初始位置決定），因此所求的曲線為

……趁我還未忘記記下……

早前的一分鐘閱讀介紹了《當我們變成一堆數字》一書

當中出現了一個名詞：Numerati，數字搜客。有時間不況看看書摘：
http://www.ithome.com.tw/itadm/article.php?c=55377

……無聊聯想到：一分鐘「驗毒」、中學生「驗毒」獎勵計劃……

分享一下學生傳給我的圖：

……我做數也是如此，很人性的感覺吧……

不知有沒有授課員用過橢圓規這個教具？（實情我不知這名稱是否正確，網上找到 Ellipsograph 這個字，不知是否橢圓規的正確英文名稱。）

我「靜靜雞」用科組錢買了一個，操作見下：

（是，我好似有點 handicapped，因為我又是單手拍片，單手做野。）
趁中四最後一天上課，在堂上試試（其時，課室包括我，共有三名教師，和四十多名學生，各自進行各自的「活動」），原來，第一：插粉筆的洞洞太小，粉筆插不進去。第二：雖然它有四個「吸盤」，但總是粘不著「綠黑板」，不一會又丟下來，要同學出來按著，麻煩非常。

o拿，又是 IT 出場：花一點吹灰之力，便可在網上模擬：

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877&current_page=&i=1247858&t=1247858

（橢圓規是「單腳」的，但因為我的 Geogebra 功力麻麻，畫了「雙腳」，見諒。）

當然，這些舊東西，網上俯拾，找一個不錯的 gif：

更多的 gif 可看看：

http://bellsouthpwp.net/e/d/edwin222/enter.htm

如果真的要用橢圓規，要找質量好一些的。是，現行的課程沒有研究橢圓，我只是作為軌跡的舊例子而已。

為何動點 D （或 E）點走出來的軌跡是橢圓？

設 , 及 ，已知 及 長度固定，不況設 ，由 section formula，得

得

常數，可見 D 走出的軌跡是橢圓也。太代數了，同學或可找找更幾何的證明。

=================================================================

舊檔重貼：答張同學有關拋物線的問題@濟濟一堂學術討論區 2003-04-11 02:04:03

參考上圖

球體和切面相切點記為 F，此 F 就時所謂的「焦點」（focus）。
球體和錐體相切處是一圓，過這圓的平面是水平面（平行底部的面）。
此面和切面相交之處是直線，此直線是所謂的「準線」（directrix）。

要證明切面和錐體相交之處是拋物線，
即證該相交處上任何一點 P 與焦點距離等於它與準線的距離，
如圖所示，欲證 PF = PQ。

由 tangent from external point，易知 PF = PR。
把 PR 沿水平圓（橙色）旋轉至 P’R’，易知 PR = P’R’。
因切面是平行錐體斜邊切出（平行切才得出拋物線），即 P’R’// PQ，
知 P’R’QP 成平行四邊形，
即 P’R’= PQ。
得出結論 PF = PQ。

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝
P.S.
有關正生事件，你可以罵我「以偏概全」，但聽到以下的 “sound bite”：

居民陳先生怒斥政府無視民意，質問政府︰「若梅窩設置戒毒學校，樓價下跌，誰負責任？」

心中的火又徐徐上升…

Not many students and people are fond of mathematics.  But Dr. Marasigan has liked math most his life.  He has in fact made it a personal mission to raise the level of mathematics education in the country.  He also works to help create a climate of public opinion that values excellence and appreciates the importance of intellectual work especially in mathematics.

Dr. Marasigan earned his BS Mathematics degree from Ateneo de Manila University in 1962, Diplom-Mathematic in 1067, and Dr.rer.nat (Mathematics) in 1971 both from Technische Hochschule Daarmstadt in former West Germany.  He also received fellowship from Deutcher Akademischer Austauschdienst (1962-67, 1969-710), DAAD post doctoral fellow (March-June 1979), and Japan Society for the Promotion of Science exchange scientist at Sophia University (1982), Tokyo Institute of Technology (1982), Osaka University (1990 and 1995), Osaka Kyoiku University (1991), and Hirosaki University (1993).  HE ALSO WENT TO Australian National University (March-June 1984) as fellow of Australian Universities’ International Development Program.

Dr. Marasigan has taught mathematics at AdeMu since 1962.  He was also deeply involved in faculty development program in pure and applied mathematics among teachers on AdeMu, Dela Salle University, and University of the Philippines.  He led a group that drafted Policies and Standards for Mathematics, a joint project of Department of Education, Culture, and Sports, and Science Education Institute of the Department of Science and Technology.

Photo courtesy:  math.admu.edu