# Quod Erat Demonstrandum

## 2008/01/29

### 小報告：濫藥

Filed under: Report — johnmayhk @ 7:44 下午

## 2008/01/28

### Hilbert’s work

Filed under: Family — johnmayhk @ 5:53 下午

## 2008/01/27

### Music appreciation: Vitas

Filed under: Fun — johnmayhk @ 1:37 上午

## 2008/01/25

Filed under: Uncategorized — johnmayhk @ 11:07 下午

### [CE][Math] Statistics: More dispersed?

Filed under: HKCEE — johnmayhk @ 2:52 下午

In CE mathematics, we know that there are at least 3 measurements for dispersion, namely, range, inter-quartile range and standard derivation.

Describing or comparing the dispersion of different sets may be a very important application.

Given two numerical data sets A and B.

If we come across the following situations

range of A > range of B and
inter-quartile range of A < inter-quartile range of B

How to compare which set is more dispersed?

Well, you may say, just look at the standard derivation.

OK. If we have the following

range of A > range of B and
inter-quartile range of A > inter-quartile range of B and
standard derivation of A < standard derivation of B

Is it true to draw the conclusion that set B is more dispersed?

Further, how about when (set A $\ne$ set B) such that

range of A = range of B and
inter-quartile range of A = inter-quartile range of B and
standard derivation of A = standard derivation of B

Urm, let me give the following as an example

Set A ={1,10,10,10}
Set B = (1,1,1,10}

Is it true to draw the conclusion that both sets have the same dispersion?

The problem may be something about the vagueness of the concept of “dispersion". How to resolve it?

## 2008/01/24

### Pure Mathematics 補底

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 3:08 下午

1.
If f(x) is a polynomial of degree 2, and f(1) = f(2) = 0, then it is NOT true to say
f(x) = (x-1)(x-2).
f(x) $\equiv$ k(x-1)(x-2) for some non-zero constant k.

2.
Let f(x) be a polynomial of degree not less than 3.
If f(x) is divided by (x-1)(x-2)(x-3), then it is NOT true to say
f(x) = (x-1)(x-2)(x-3)Q(x) + R
(where R is a constant, the remainder)
f(x) $\equiv$ (x-1)(x-2)(x-3)Q(x) + $ax^2 + bx + c$
that is,
we should set the remainder as a polynomial of degree 2
(just less than the degree of the divisor (x-1)(x-2)(x-3) by 1)

3.
Resolve $\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2}$ into partial fraction.
Don’t write the following
$\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} = \frac{A}{x} + \frac{Bx + C}{x^2} + \frac{D}{x - 1} + \frac{Ex + F}{(x - 1)^2}$.
$\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} \equiv \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$

4.
If $x > k$, it is NOT true to draw the following conclusion
$x^2 - x + 1 > k^2 - k + 1$.
But, if we know $x + k - 1 > 0$, then it is true to have $x^2 - x + 1 > k^2 - k + 1$.
Proof: Multiply $x + k - 1 > 0$ and $x - k > 0$, we have $(x^2 - k^2) - (x - k) > 0$ and result follows.

5.
Let {$x_n$} be a sequence of real numbers.
Knowing that $x_n > n$.
It is NOT true to say
{$x_n$} is increasing.
Just give an example.
$x_1 = 10, x_2 = 9, x_3 = 8, x_4 = 7, x_5 = 6, ...$
Obviously, $x_n > n$ (for n = 1,2,3,4,5), but {$x_n$} is not increasing.

6.
Let {$x_n$} be a sequence of real numbers.
Knowing that $x_n < 1 - \frac{1}{n}$.
We cannot say {$x_n$} is bounded from above by $1 - \frac{1}{n}$.
$1 - \frac{1}{n}$ is NOT an upper bound of the sequence {$x_n$}.
$x_n < 1 - \frac{1}{n} < 1$ for all positive integers n.
And now, it is clear that, 1 is an upper bound of {$x_n$}.

7.
(a) Solve $|x| < 3$
Solution: $-3 < x < 3$
(b) Solve $|x| > 3$
Solution: $x > 3$ or $x < -3$

8.
$ax^2 + bx + c = 0$
Given that the discriminant $\Delta < 0$.
Then it is NOT true to say that the quadratic equation has no real solution.
Just consider a quadratic equation with roots 0 and i.
That is
$x(x - i) = 0$
$x^2 - ix = 0$
Then $\Delta = (-i)^2 - 4(1)(0) = -1 < 0$.
But there is a real root (x = 0) for the quadratic equation.

9.
“Vector" is out of the syllabus now.
“Linearly independent", “linearly dependent" are out-of-syllabus concepts.

10.
Techniques in evaluating limits
(a) l’hôpital rule
(b) Special limit : $\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$
(c) Special limit : $\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = e$
(d) Special limit : $\lim_{a \rightarrow 0}(1 + a)^{\frac{1}{a}} = e$
(e) Boundedness : Knowing that {$a_n$} is bounded and $\lim_{n \rightarrow \infty}b_n = 0$, then $\lim_{n \rightarrow \infty}a_nb_n = 0$
(f) Squeezing (Sandwich) principle
(g) Taking limit on the recurrence relation : $a_{n + 1} = 2a_{n}$ and suppose {$a_n$} converges $\Rightarrow \lim_{n \rightarrow \infty}a_{n + 1} = \lim_{n \rightarrow \infty}(2a_{n}) \Rightarrow \lim_{n \rightarrow \infty}a_n = 2\lim_{n \rightarrow \infty}a_n \Rightarrow \lim_{n \rightarrow \infty}a_n$ = 0

11.
Symbols about left and right hand limits
$x \rightarrow 3^+$" does NOT mean “$x \rightarrow +3$"
$x \rightarrow 3^-$" does NOT mean “$x \rightarrow -3$"
See the difference? Try to obtain the following by sight.
(a) $\lim_{x \rightarrow 3^+}\sqrt{x - 3} = 0$
(b) $\lim_{x \rightarrow 3^-}\sqrt{x - 3}$ does not exist.
(c) $\lim_{x \rightarrow +3}\sqrt{x - 3}$ does not exist (since the left and right lmits are not the same)
(d) $\lim_{x \rightarrow -3}\sqrt[3]{x + 3} = 0$

12.
To evaluate $\lim_{x \rightarrow \infty}[f(2004 + x) - f(x)]$, please do NOT split it as
$\lim_{x \rightarrow \infty}f(2004 + x) - \lim_{x \rightarrow \infty}f(x)$
No! Both limits may NOT exist!
You may try “rationalization", “sum to product", “MVT" etc.

13.
How to evaluate $\lim_{x \rightarrow \infty}[f(\sqrt{2004 + x}) - f(\sqrt{x})]$ = ?
Try the Mean Value Theorem (of course, $f$ should be differentiable at certain interval), that is
$\lim_{x \rightarrow \infty}[f(\sqrt{2004 + x}) - f(\sqrt{x})]$
= $\lim_{x \rightarrow \infty}f'(c)(\sqrt{2004 + x} - \sqrt{x})$ for some $c \in [\sqrt{x} , \sqrt{2004 + x}]$
= $\lim_{x \rightarrow \infty}f'(c)[\frac{2004 + x - x}{\sqrt{2004 + x} + \sqrt{x}}]$ for some $c \in [\sqrt{x} , \sqrt{2004 + x}]$
= $\lim_{x \rightarrow \infty}f'(c)[\frac{2004}{\sqrt{2004 + x} + \sqrt{x}}]$ for some $c \in [\sqrt{x} , \sqrt{2004 + x}]$
(if $\lim_{x \rightarrow \infty}f'(c)$ is bounded, then the limit is zero, read 10.(e))

14.
If $\lim_{x \rightarrow 0}\frac{g(x)}{x}$ exists, then $\lim_{x \rightarrow 0}\frac{g(x)} = 0$.
e.g. Given $\lim_{x \rightarrow 0}\frac{f(x) - 1}{x} = 1$, evaluate $\lim_{x \rightarrow 0}f(x)$. (ANS: 1)

15.
Be careful, the following may NOT be true!
$\lim_{x \rightarrow a}(f(x) + g(x)) = \lim_{x \rightarrow a}(f(x) + \lim_{x \rightarrow a}g(x))$
e.g. Given $\lim_{x \rightarrow 0}\frac{f(x) - 1}{x} = 1$, evaluate $\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}$
It is WRONG to write
$\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}$
= $\lim_{x \rightarrow 0}\frac{f(x) - 1}{x \times 1}$ (for $\lim_{x \rightarrow 0}e^x = 1$)
= 1
You should give somthing like
$\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}$
= $\lim_{x \rightarrow 0}\frac{f(x) - 1 + 1 - e^x}{xe^x}$
= $\lim_{x \rightarrow 0}(\frac{f(x) - 1}{xe^x} + \frac{1 - e^x}{xe^x})$
= $\lim_{x \rightarrow 0}(\frac{f(x) - 1}{xe^x} + \frac{-e^x}{xe^x + e^x})$ (l’hôpital rule)
= 0 (see, NOT 1)

16.
(a) $\int\tan^n(x)dx \neq \frac{tan^{n + 1}}{n + 1} + C$
(b) $\int e^{x^2}dx \neq e^{x^2} + C$
For (a), try the reduction formula.
For (b), there is no closed form, just stop there.

17.
The following definitions MUST be memorized!
(a) “$f(x)$ is well-defined at $x = a$" means “$f(a)$ can be found".
(b) “$f(x)$ is continuous at $x = a$" means “$\lim_{x \rightarrow a}f(x) = f(a)$“.
(c) “$f(x)$ is differentiable at $x = a$" means “$\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$ exists".

18.
Q: At a look at the definitions above, why there is no mention of the left and right hand limits? When should we consider the left and right hand limits?
A: When $f(x)$ has DIFFERENT expressions for $x \ge a$ and $x < a$ (say)
e.g.
(a) Let $f(x) = x + 1$ for $x \ge 3$ and $f(x) = 2(x - 1)$ for $x < 3$. Then $f(x)$ is continuous at $x = 3$. Proof: $\lim_{x \rightarrow 3^+}f(x) = \lim_{x \rightarrow 3^+}(x + 1) = 4$ while $\lim_{x \rightarrow 3^-}f(x) = \lim_{x \rightarrow 3^-}2(x - 1) = 4 = f(3)$ Hence $f(x)$ is continuous at $x = 3$.
(b) Let $f(x) = x + 1$ for $x > 3$ and $f(x) = 2(x - 1)$ for $x < 3$ with f(3) = 5 Then $f(x)$ is NOT continuous at $x = 3$. (why?)
(c) Let $f(x) = x|x|$. To prove $f(x)$ is continuous at $x = 0$, it is NOT necessary to “divide cases", since both left and right hand limits are zero. i.e. $\lim_{x \rightarrow 0^+}x = \lim_{x \rightarrow 0^-}x = 0$, hence $\lim_{x \rightarrow 0}x = 0$ and therefore, $\lim_{x \rightarrow 0}x|x| = 0 = f(0)$. Done.

19.
Given

$f(x) = f_1(x)$ for $x \ge a$
$f(x) = f_2(x)$ for $x < a$

(both $f_1(x)$ and $f_2(x)$ are ‘nice’ functions)
Then, to find $f'(x)$, the problem only appears at $x = a$, hence DON’T write

$f'(x) = f_1'(x)$ for $x \ge a$
$f'(x) = f_2'(x)$ for $x < a$

$f'(x) = f_1'(x)$ for $x > a$
$f'(x) = f_2'(x)$ for $x < a$

And try to check the value of $f'(x)$ at $x = a$ by finding

$f_{+}'(a) = \lim_{h \rightarrow 0^+}\frac{f_1(a + h) - f(a)}{h}$
$f_{-}'(a) = \lim_{h \rightarrow 0^-}\frac{f_2(a + h) - f(a)}{h}$

and if $f_{+}'(a) = f_{-}'(a)$, then $f'(a)$ is the common limit.

20.
Techniques in integrations: substitution
(a) Involves $\sqrt{a^2 - x^2}$ , substitute $x = a\sin(\theta)$ or $x = a\cos(\theta)$
(b) Involves $\sqrt{x^2 - a^2}$ , substitute $x = a\sec(\theta)$
(c) Involves $x^2 + a^2$ , substitute $x = a\tan(\theta)$

21.
Techniques in integrations: partial fractions
(a) Involves $x^2 - a^2$ , partial fractions may help
(b) Involves $x^4 + a^4$ , don’t forget
$x^4 + a^4$
= $x^4 + 2x^2a^2 + a^4 - 2x^2a^2$
= $(x^2 + a^2)^2 - (\sqrt{2}ax)^2$
= $(x^2 + \sqrt{2}ax + a^2)(x^2 - \sqrt{2}ax + a^2)$
and then partial fraction resolving

22.
Techniques in integrations: others
(a) Integration by parts
(b) Taking limits
e.g. $\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx$ = ?
Let $I_n = \int_{0}^{1}\frac{x^{n}}{1 + x^2}dx$, then
$0 \le I_n = \int_{0}^{1}\frac{x^{n}}{1 + x^2}dx \ge \int_{0}^{1}\frac{x^{n + 1}}{1 + x^2}dx = I_{n + 1}$
That is {$I_n$} is decreasing and bounded from below by zero.
Hence $\lim_{n \rightarrow \infty}I_n$ exists, says L.
For sufficiently large $n$, write
$\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx$ = $\int_{0}^{1}\frac{x^{n - 2}(x^2 + 2x + 1 - 2x - 1)}{1 + x^2}dx$ = $\int_{0}^{1}(x^{n - 2} -\frac{2x^{n - 1}}{1 + x^2} - \frac{x^n}{1 + x^2})dx$ = $\frac{1}{n - 1} -2\int_{0}^{1}\frac{x^{n - 1}}{1 + x^2}dx - \int_{0}^{1}\frac{x^n}{1 + x^2}dx$
Taking $n \rightarrow \infty$ from both sides,
$L = 0 - 2L - L$
Hence, $\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx = L = 0$

23.
Many questions involve the Mean Value Theorem in Cauchy form, that is
both $f(x)$ and $g(x)$ are continuous on [a , b] and differentiable on (a , b) with $g(a) \neq g(b)$, then
then $\frac{f(a) - f(b)}{g(a) - g(b)} = \frac{f'(c)}{g'(c)}$ for some $c \in$ (a , b).
DON’T write the following as a so-called proof

$f(a) - f(b) = f'(c)(a - b)$
$g(a) - g(b) = g'(c)(a - b)$
$\Rightarrow \frac{f(a) - f(b)}{g(a) - g(b)} = \frac{f'(c)}{g'(c)}$ for some $c \in$ (a , b)

NO! We CANNOT ensure both $c$ in $f'(c)$ and $g'(c)$ are the same!!! That is, all we have may be the following

$f(a) - f(b) = f'(c_1)(a - b)$ for some $c_1 \in$ (a , b)
$g(a) - g(b) = g'(c_2)(a - b)$ for some $c_2 \in$ (a , b)

and then we can do nothing at all…How to prove? Try the read

24.
Get tired to give details, just give topics in major techniques:
(a) Integration by parts
(b) Fundamental theorem of integral calculus
(c) Inequalities
(d) Riemann sum

25.
Partial fractions : there is a short-cut for distinct linear factors, just in case you may forget…

$\frac{6x^2 - 18x + 6}{x(x - 2)(x - 3)}$
= $\frac{6(0)^2 - 18(0) + 6}{x(0 - 2)(0 - 3)} + \frac{6(2)^2 - 18(2) + 6}{(2)(x - 2)(2 - 3)} + \frac{6(3)^2 - 18(3) + 6}{3(3 - 2)(x - 3)}$
= $\frac{1}{x} + \frac{3}{x - 2} + \frac{2}{x - 3}$

## 2008/01/21

### [報告] 某次崇拜的經驗

Filed under: Life,Teaching — johnmayhk @ 4:16 下午

## 2008/01/16

### [Fun] 濟記狂想 gag : 檢查頭髮

Filed under: Fun — johnmayhk @ 9:58 下午

### [初中] Factorization, useful or not?

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 9:44 下午
Tags: ,

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.

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

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.

– – – – – –

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.

## 2008/01/14

### [AL][PM] Hyperpower function

Filed under: HKALE,Pure Mathematics,University Mathematics — johnmayhk @ 4:39 下午

Justin asked me again about the hyperpower function. Sorry, I just re-post my old message in my old forum, which was prohibited by the adminstrators in the hkedcity. It is known that the following

converges if and only if

We may create an AL pure mathematics question concerning hyperpower function like

Let
(a) Prove that {$x_n$} is increasing.
(b) By M.I. or otherwise, show that {$x_n$} is bounded above by 2.
(c) Hence solve the outdated problem:

## 2008/01/12

### [軟件][教學] Scilab 在中學數學的應用初探

Filed under: HKALE,HKCEE,Junior Form Mathematics,Teaching — johnmayhk @ 6:01 下午

http://johnng.inscyber.net/mathgif2/Scilab-johnmayhk.txt

(1) 直接輸入

(2) 透過系數（coefficients）

(3) 透過根 (roots)

1. 基本運算

2. 有關除法

3. 解多項式方程 (solve polynomial equations)

4. 因式分解（factorization）

5. 求最大公因式（GCD）及解丟番圖方程（Diophant equation）

6. 補充一下：有關整數求 HCF 及 LCM 的方法

7. 有理函數化成部分分式（resolving into partial fractions）
Partial fractions 這類熱門的 pure mathematics 題目，也可以 Scilab 輕易處理。

Scilab 最強大的功能是處理矩陣，諸君請看！開始時，當然要定義矩陣。方法是相當簡單。

Scilab 可做的遠超上述，除了有關統計數據資料之處理，還有下列一些。但因為別的，常用的軟件也有類似功能，我略舉以下數例作結，圖收拋磚引玉之效。

## 2008/01/09

### yummy…

Filed under: Life — johnmayhk @ 7:33 下午

### [CE][AM] 錯在哪裡 – general solutions to trigonometric equations

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 12:01 下午

Given that $a,b$ are non-zero constants such that $|a| < 1$ and $|b| < 1$.

Solve the following simultaneous equations

$\sin(x) = a$ – – – (1)
$\cos(x) = b$ – – – (2)

Solution (so-called)

divide (1) by (2), yield

$\tan(x) = \frac{a}{b}$

Hence

$x = 180^{o}n + \tan^{-1}(\frac{a}{b})$ where $n$ is any integer

## 2008/01/06

### anniversary ceremony of Y.U.81

Filed under: Report,School Activities — johnmayhk @ 8:36 下午

Last night was the 30th anniversary ceremony of the establishment of Y.U.81, we had games and hot-potting, leaving school at 10 p.m. something. Although the activities were just similar to previous years, I think the change should be the quality of members. Senior members took care of the junior like brothers. (Esp. stayed with a crying member) I just recalled the days when chun tak was the HSL (or earlier), the dinner hours were a bit embarrassing for quite a lot of ‘men’s talk’ and ‘foul languages’ there. However, in recent years, members were becoming much more polite and respectful to teachers. As Mr. Ching always claims that the boys of SFXC are of very high quality in conduct. You know, my mother school was a band 5 school and I knew how trouble will be when students acting madly. One old boy, Mak, was there, he was good at solving 3*3*3 Rubik’s Cube and he is playing 5*5*5 Rubik’s Cube now. He said the principles were quite similar. How interesting to tell that two handsome young teachers, Mr. Rock and Mr. Zinc were enthusiastic about playing the 3*3*3 Rubik’s Cube these days in staff room. I think that it may be a good show if there are competitions or performances about playing Rubik’s Cube. A Talentine show? Apart from singing and musical performance, we may have other alternatives. Magic? Yoyo? One F.2 student in my class had a mini-show in the class party, just click the following and have a look!

Occasionally, I found the following movie clip: a Japanese TV program.

Quite surprising that mathematics can be ‘used’ in a popular TV show! I’m afraid that we cannot watch similar programs in Hong Kong in the near future, even after the introduction of digital television broadcasting system with high resolution.

## 2008/01/05

### [報告] 批判思考工作坊 (critical thinking workshop)

Filed under: Report,School Activities — johnmayhk @ 12:32 下午

『今有兩人，阿甲只會在早上說謊，阿乙只會在晚上說謊，如何問兩個問題，可分辨誰是誰？』

（趙同學即問：如何定義早上和晚上？Oops）