Quod Erat Demonstrandum

2007/11/30

感想?我不敢想。

Filed under: Life — johnmayhk @ 6:33 下午

政策出籠,勢必作有利己方的宣傳。教育政策無異。要學生『以多角度看事情』,弔詭地,宣傳政策時卻只談『政策是好』這單一角度。

世事就是弔詭。今早的講者談通識的 IES。但私下對我組說:他期望下一次的教改:把通識科刪除。

下午是羅氏中學的同工之實戰經驗分享。是,一年請 24 個代課老師,為其他老師空出課時,為通識作更好的準備,是冒險也是決心吧。陳志強老師給我組帶來歡樂,或是苦中作樂吧。有時做教育的要麻醉自己。

『不要問,只要信』,這年頭已用不著奢侈地去問為什麼,而是要做!如何做才是我們關心。果效比原則重要多。

我姑且稱十多二十來歲為人生的『哲學期』,這個時段很寶貴,對事情的看法沒多大的包袱,偶爾吐出頗具哲理的言談。再年長一點,在工作的巨輪下グルグル,人心還可以容納什麼?就是把『你們有何期望?』說成:『讓我和大家進行期望管理』這種論述的包裝?感謝教師發展日給我有一分數秒的思考時間。

2007/11/25

Hilbert’s space

Filed under: Family — johnmayhk @ 10:06 下午

More than just photos, they are the precious memories of my family. Just share with you my son’s recent photos.

http://www.flickr.com/photos/13918910@N02/sets/72157603291646571/show/

2007/11/24

[初中] Algebraic computation

Filed under: Junior Form Mathematics — johnmayhk @ 1:23 下午

After the uniform test, I’m really sad to see the poor performances made by my F.2 boys. We, teachers, keep on asking WHY, WHY, WHY: for certain kinds of questions, it is a MUST that students will make mistakes. During the discussion of paper, I said, ‘I knew many of you may make mistake here…’, then a student, Ho, replied,’ then why you set up this question?’ Oops.

What is the question in the discussion? It is a M.C. question.

\frac{1}{x - y} - \frac{1}{x + y} =

A. \frac{-2}{x + y}
B. 0
C. \frac{2y}{x^2 - y^2}
D. \frac{2x}{(x - y)(x + y)}

Yes, I intentionally set this question to test whether students could perform algebraic computation correctly. And we, teachers, knew that some students may make mistakes. Hence, I let students practice questions of similar types beforehand. But, according to student Ho’s comment, should I stop setting this kind of question instead?

Students asked what are techniques of doing mathematics? Teachers may reply that “practice more", but students just turn a deaf ear to this and want to have the painless short-cut. I try to clear up the concepts and show common techniques during lessons, but it is insufficent if students have no studying and drilling. Undoubtedly, we have tricks, but the effort takes the lead.

[AL][PM] Class discussions on complex numbers

Filed under: HKALE,Pure Mathematics — johnmayhk @ 1:21 下午
Tags:

We are learning the roots of unity in these F.7 pure math lessons. I randomly gave the following

Solve x^5 - x^4 + x^3 - x^2 + x - 1 = 0

Then I wrote

x^5 - x^4 + x^3 - x^2 + x - 1 = 0
1 - x + x^2 - x^3 + x^4 - x^5 = 0
\frac{1 - (-x)^6}{1 - (-x)} = 0 (for x \neq -1) – – – (1)

then I jumped to

x^6 = 1 – – – (*)

Then I stopped for a while because I had something in my mind:

started with another equation

x^5 + x^4 + x^3 + x^2 + x + 1 = 0

we can jump to the same equation x^6 = 1, why?

At that moment, Harris claimed that he had a quicker method: “just adding a negative sign".

Well, when solving (*), we have

x = cis\frac{2k\pi}{6} (for k = 0,1,2,4,5 < – – – no ‘3’ since x \neq -1) is the roots to the original equation.

Then Harris wrote his method (H’s method) on the board as follows.

From (1), we have

(-x)^6 = 1
-x = cis\frac{2k\pi}{6}
x = -cis\frac{2k\pi}{6} < – – – – this is so-called “adding a negative sign"

Well, it is OK with saying k = 1,2,3,4,5 < – – – – no ‘0’ since x \neq -1.

Is it a quicker way to obtain solutions, as claimed by Harris? Well, it depends.

But the most important thing is how to explain the solution given by Harris is the same as what I gave.

Of course, we can put different values of k and check them one by one. However, it’s time consuming. We should have a general (or easier) way to show it up.

At that moment, some (or few) students tried to suggest something and think it in details. That is a wonderful moment because interaction, active learning began!

The first thing came to me is trying to see the geometric meaning. But how stupid that I was thinking about vertices of a regular pentagon!! I mistake that there are 5 roots so the roots represent the vertices of a regular pentagon!! No, cis\frac{2k\pi}{6} are representing vertices of a regular hexagon!! A student, Lai, reminded me that there are 6 distinct values generated!!

Then the situation is easy now.

Because the vertices of a regular hexagon is symmetrical about O,
that is,
if cis\theta is a root
=> the point representing cis\theta is a vertex P of the regular hexagon
since there is another vertex P’ of the hexagon such that P and P’ are symmetrical about O
and the corresponding complex number represented by P’ is -cis\theta
=> -cis\theta is also a root

Don’t understand geometric meaning? Urm, try to use “pure algebra".

For n being a positive even number,

\cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}
= -\cos(\frac{2k\pi}{n} - \pi) - i\sin(\frac{2k\pi}{n} - \pi)
= -(\cos\frac{(2k - n)\pi}{n} + i\sin\frac{(2k - n)\pi}{n})
= -(\cos\frac{2(k - m)\pi}{n} + i\sin\frac{2(k - m)\pi}{n}) where n = 2m for some natural numbers m.
= -(\cos\frac{2k'\pi}{n} + i\sin\frac{2k'\pi}{n}) where k' = k - m

Back to our question, n = 6, hence m = 3; thus, by above,

cis\frac{2(0)\pi}{6} = -cis\frac{2(0 - 3)\pi}{6} = -cis\frac{2(3)\pi}{6}
cis\frac{2(1)\pi}{6} = -cis\frac{2(1 - 3)\pi}{6} = -cis\frac{2(4)\pi}{6}
cis\frac{2(2)\pi}{6} = -cis\frac{2(2 - 3)\pi}{6} = -cis\frac{2(5)\pi}{6}
cis\frac{2(4)\pi}{6} = -cis\frac{2(4 - 3)\pi}{6} = -cis\frac{2(1)\pi}{6}
cis\frac{2(5)\pi}{6} = -cis\frac{2(5 - 3)\pi}{6} = -cis\frac{2(2)\pi}{6}

For the last round of pure math teaching period, I’d come across other questions from F.7 students, just click here for your further reading.

I still want to say something more including a funny geometric application of “roots of unity" and replies to Justin. Just wait and see.

2007/11/19

[U] Solve sin(z) = 2

Filed under: University Mathematics — johnmayhk @ 3:21 下午
Tags:

A F.5 student, Yan, discussed with me something about “complex angles" in trigonometric formulae, in lunch hour, he asked me the way of solving

\sin(\theta) = 2

I just post it here. Starting with the Euler’s formula

e^{i\theta} = \cos(\theta) + i\sin(\theta)

We can extend the angle \theta to complex number by using the above formula. Replace \theta by -\theta, we have

e^{-i\theta} = \cos(\theta) - i\sin(\theta)

Subtracting, yields

e^{i\theta} - e^{-i\theta} = 2i\sin(\theta), thus

\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}

Now, putting \sin(\theta) = 2, we have

\frac{e^{i\theta} - e^{-i\theta}}{2i} = 2
u^2 - 4iu - 1 = 0 where u = e^{i\theta}

By quadratic formula,

u = \frac{4i \pm \sqrt{(4i)^2 - 4(1)(-1)}}{2} = \frac{4i \pm \sqrt{-16 + 4}}{2} = (2 \pm \sqrt{3})i

Thus,

e^{i\theta} = (2 \pm \sqrt{3})i
e^{i(x + iy)} = (2 \pm \sqrt{3})e^{(2n\pi + \pi/2)i}

Here, we put \theta as a complex number x + iy and try to find out values of x and y. Also, by putting \theta = 2n\pi + \frac{\pi}{2} into the Euler’s formula, we have i = e^{(2n\pi + \pi/2)i}. Continue,

e^{-y}e^{ix} = (2 \pm \sqrt{3})e^{(2n\pi + \pi/2)i}

\therefore x = 2n\pi + \pi/2
e^{-y} = 2 \pm \sqrt{3}, taking logarithm, yields
\therefore y = -\ln(2 \pm \sqrt{3})

Last time, I’d derived the formula of \ln(z) to Yan, hence it is done.

2007/11/16

Filed under: Life — johnmayhk @ 5:00 下午

Just copy and paste (sorry) an article written by my secondary school English teacher Mr. James Hon (韓連山老師) in 教協報. Apart from daily heavy workload in secondary school, he committed himself in Hong Kong Professional Teachers’ Union. The following is a stereotype.

日期: 2007-10-05
Tomorrow is another day

早上七時卅分回到學校,還來不及吃完早點,五丙班俊文來找我,詢問有關課外活動的安排,報了名參加我的網球組卻不見自己名字在名單上。我對他說:「我也不知道,是課外活動組的老師負責統籌和安排的。」答允他替他跟進後,把未吃完的早餐囫圇吞掉,早會鈴聲已響起。

今天早會由外籍老師和我負責「英語一分鐘」時間,向全校學生介紹一些英語慣用語,推介完後便趕回教員室去拿已準備好的教材,然後到課室上課。

今天要上六節課,上午四節、下午兩節,三班不同級別的同學,兩個課程、三個程度。通常要於周末把下周用的素材備好、把要教的東西鋪排妥當,否則於當天才張羅必然倒瀉籮蟹!

聲嘶力竭教完兩課節,小息時剛想倒杯水喝,張Sir過來找我,說很抱歉遲了交朗誦報名表,我也唯有說:「不打緊!明年請早吧。」跟著校務處小陳來找我,遞過來一張教育局查詢學校英文科老師達標情況的表格,心裡在罵教育局的官僚,難道不知道那婦人連烏沙也掉了嗎?還替她延續這荒謬劇?但也對小陳說:「不必緊張,我來應付吧!」正想坐下歇歇,電話來了,是教育局跟進我校的「英語提升計劃」建議書,想跟我安排「專業對話」的日期。

沒有小息的小息過了,跟著兩節要用CD機,趕緊跑到資訊科技室借CD機,然後左手一大疊工作紙,右手一部CD機,跑上四樓上課。幸虧學生也算合作,順利完成授課。若像那天出現學生打架或「問候」老師娘親的情景,不單累壞,氣也氣死了!

滿以為午膳前有一空課節可以清清檯面堆積的來函、通告、海報和學生的習作,還未坐下,副校長過來跟我說:「又有英文科教師辭職啦!」想起要重新請人,登報、讀信、見人、篩選等工作,又是一陣頭痛。

把學生習作塞進背包,是必須帶回家的「宵夜」或「周末美點」;把海報丟在一角,再看通告:替學生訂購報紙的通告、圖書館知會各班有關廣泛閱讀的通告、德育組知會各班壁報比賽的通告、校務處收取校刊費用的通告、課外活動的學生名單……在在都需時處理,心裡還牽掛著要找時間呼籲同工加入教協發動之支持小班教學登廣告行動……。

還未拆閱信件已鈴聲大作,午膳時間到了。Peter仔氣急敗壞跑來,說:「有人踢爛了我的貯物箱!我……我……」然後一輪粗話,要打要殺的氣憤難平。我勸他平靜下來,先去吃飯,回來再尋解決方法,他也答允待會兒再跟我談。

匆匆吃過午飯,一進入教員室已經看到 Peter仔在跟羅Sir對話,羅Sir是紀律組老師,正在訓斥Peter,看樣子是Peter又闖了禍。原來他心有不甘,吃完飯後去踢爆了三個貯物箱!今天放學又要跟他詳談了。

下午要上兩節課,一群大男孩,飯氣攻心、睡眼惺忪,唯有放下教本,遊戲中學習吧!把要教的詞彙變作Bingo,只要留心聽便有機會取得小禮物,玩完Bingo玩hangman,雖然不及「超級無敵掌門人」那樣好玩,大伙兒也頗投入,卻看不到老師已力困筋乏。

放學後,替俊文向王Sir查詢課外活動事宜、輔導Peter仔,然後到圖書館替學生找讀物,突然想起要致電書商催促他們把教師用書送來,看看腕錶,已過了辦公時間,還是明天才跟進吧。

回家途中,背包裡的學生習作,沈甸甸的……拿出日記本看看,尚待處理的雜務,密密麻麻……。

仿效《隨風而逝》的主角,望望灰濛濛的天空,對自己說:
“Well, tomorrow is another day.”

Cheering team

Filed under: School Activities — johnmayhk @ 4:52 下午

[2007-11-15 morning]When taking about “green house”, people may think about “greenhouse effect”, however, the first concept turning up in Xavierians’ minds should be one of the four houses. In particular, the sense of belonging is overwhelming during the period of athletic meets. Cheering dancing on the track is one of the programs. Bystanders may have no idea on the enormous cost (I’m not professional and had no idea of what kind of movements should be good-looking and safe, I just randomly created something seemed to be so-called ‘dance’. Trial and errors, time consuming! Then followed by series of training and adjustments for 2 to 3 weeks) for the 3-miniute-only show. As the teacher-in-charge of the green house cheering team, I think our committee members and F.1 students were trying their best to practice and perform. Yellow house was the champion in the dancing part. Yes, as one of the judges, I voted for them. Although I believed that the dance steps in our team was OK, judges preferred varieties in overall pattern to details in dance steps.
Click to watch the practice
[2007-11-15 night] My mother-in-law’s birthday, my wife spent time on preparing the feast and made many surprises. But, the happiness was struck by one of the family members. Sigh.
[2007-11-16 morning] The pre-walk in Shing Mun Reservoir lasted about 2 hours something. I just remembered that I was attacked by a monkey because of my staring.

2007/11/07

[CE] A typical curve sketching question

Filed under: HKCEE — johnmayhk @ 9:40 下午

Just busy with the cheering team practice. For a break, write something quick…

This is a part of the question of 2000-CE-A.Math-Q.10, a typical question.Let y = f(x) = \frac{7 - 4x}{x^2 + 2}.
Show that the maximum and minimum values of f(x) are 4 and -\frac{1}{2} respectively.

When it is given to F.5C students, some of them puzzled and asked, ‘is that something wrong in the question?’ They added, ‘-\frac{1}{2} is maximum and 4 is minimum.’ How come?

In fact, students’ concept is so vague to give that comment. Saying “-\frac{1}{2} is maximum and 4 is minimum." is not clear! They should say something like:

y attains its maximum value when x = -\frac{1}{2}. Hence,
to find maximum value of y,
we put x = -\frac{1}{2} into y = \frac{7 - 4x}{x^2 + 2},
that is, the maximum value of y = \frac{7 - 4(-\frac{1}{2})}{(-\frac{1}{2})^2 + 2} = 4.

Also, they should say
y attains its minimum value when x = 4. Hence,
the minimum value of y = \frac{7 - 4(4)}{4^2 + 2} = -\frac{1}{2}.

The designer of this question played little tricks on numbers. It may be quite interesting to set up similar question by choosing ‘suitable’ numbers a and b satisfying

a < b;
f(a) = b = maximum value of y;
f(b) = a = minimum value of y; and
It is not too easy (or too difficult) to find derivatives of f(x).

Students try to find such function to help me set up A. Math Examination Paper ^_^.

[初中] Ratio

Filed under: Junior Form Mathematics — johnmayhk @ 9:32 下午

I gave F.2 boys the following simple question in a quiz.

Edmund has x stamps while Kelvin has 12 stamps more. Given that the ratio of the number of Edmund’s stamps to that of Kelvin’s is 2 : 3.
(a) Find the value of x.
(b) Kelvin then takes some stamps from Edmund such that the ratio of the number of Edmund’s stamps to that of Kelvin’s is changed to 7 : 23. How many stamps does Kelvin have finally?

Many students could give correct answer using something like

(a)
x : (x + 12) = 2 : 3
\frac{x}{x + 12} = \frac{2}{3}
x = 24

(b)
Let y be the number of stamps taken by Kelvin.
(24 - y) : (36 + y) = 7 : 23
\frac{24 - y}{36 + y} = \frac{7}{23}
y = 10
Hence, Kelvin has (36 + 10) = 46 stamps in the end.

This method is converting ‘ratio problem’ into ‘solving equation problem’. Natural?

However, some students gave the following.

(a)
x = 2 \times 12 = 24.

(b)
Kelvin has \frac{23}{7 + 23} \times 5(12) = 46 stamps.

If you were a teacher, will you grant full marks to this kind of ‘one-line’ solution. For me, affirmative! That is mere RATIO. That is equation-free! How stupid that we had found y = 10 which is “useless" with respect to the method above. Yes, we emphasize always on the presentation of solution, writing appropriate steps. However, we may need to strike a balance for allowing delicate solution.

2007/11/06

[CE] Exploring on curve sketching

Filed under: HKCEE — johnmayhk @ 3:31 下午

1.

Could you guess that equation of the graph of the curve in red (as shown)? You may say it looks like y = x^3. Well, it is actually y = x|x|. Justin discovered that the graphs of y = x^3 and y = x|x| are quite similar in shape. Further, the following shows the graph of y = x^3|x| which looks similar to the graph of y = x^5.

Can we generalize the above as the graph of y = x^{n-1}|x| looks quite ‘similar’ (quite a vague idea actually) to y = x^{n+1} for n > 1? How to explain that finding? I just gave comment briefly that y = x^{n + 1} = x^{n - 1}\times x^2 ‘behaves similarly’ to y = x^{n - 1}\times |x| in the way that x^{n - 1} are multiplied by a non-negative factor. Could you give further comments and explanations?

2.

Many interesting matters have rooms for further discussions. Um, just give you something to ‘play’ with. Can you draw a square? What a silly question. Um, let me refine, could you use one single equation to plot a square with curve plotting software (e.g. Winplot.exe)? I had done one, see

Could you guess the equation of the graph above? Strictly speaking, the above graph is NOT a square (see the trouble corners). How to make the curve above? Just consider:

We all know that the equation of circle is something like

x^2 + y^2 = 1

Instead of making the power of 2, how about

x^3 + y^3 = 1 or
x^4 + y^4 = 1

You may use curve plotting software to see the difference of the graphs of the above curves (could you explain the difference?). When the power of x and y is getting larger and larger, what will be the changes? Try to use winplot.exe to check. In fact, the figure I posted is the graph of x^{100} + y^{100} = 1. It looks like a square except the situation at corners. Students, can you explain, without using any software, why the shape is something like a square? More, if we consider when the powers of x and y are getter smaller, what will be the changes of the shape of graph? Could you imagine the shape of the graphs of the following

x^{\frac{1}{2}} + y^{\frac{1}{2}} = 1
x^{\frac{1}{3}} + y^{\frac{1}{3}} = 1

Finally, when the powers are different, any further observation you may discover? As for example

x^2 + y^3 = 1

We still have a lot of elementary mathematics stuff for exploring, think more!

Also read
http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1231449&t=1231383&v=t

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=358012&t=358012

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=445770&t=445770

2007/11/03

Singing contest

Filed under: Life,School Activities — johnmayhk @ 5:11 下午

My wife was not at home and I’m happy to take care of my son one whole day. He’s sleeping sweet now. Just do some recalling.

The annual important event of the SFXC music club “sing^3 singing contest” was held yesterday. I’m not professional, but students asked me to be one of the judges (together with FC & SHY) of that event. Just like last year, singing while dancing is always impressive. The champion of the senior group event went to “4 sons” because we teachers enjoyed the happy atmosphere and their effort in preparing the dance.

When it comes to the score of ‘cooperation’, in my school days, my teachers would focus on the harmony of voices, like Simon and Garfunkel (Hello darkness my old friends…), but today, I seldom find that kind of harmonic performance, instead, the dancing or making jokes may be more dominant, part of the reasons should be the culture “fed” by similar Hong Kong TV programs.

It is very happy to see students on the stage and become ‘another one’, like some F.7 boys I know. Richard, Yip, Tsui, H, Wong, they had tried hard. A F.4 student, Chow, he was a naughty boy but I’m quite shocked to listen to his performance, his voice is clear with a bit sorrowful feeling. Actually, many senior students performing good, the major difference may be something about the “lasting power” (i.e. made no mistake even in the last note of the singing)

Actually, I prefer singing to listening, but I know my voice (together with my whole self) is “deteriorated”, but I think it’s enough to sing songs to my son (and wife sometimes :p).

2007/11/02

[Fun] Gag

Filed under: Fun — johnmayhk @ 12:06 下午

店員:「先生, 個Pizza想切開做八塊定六塊呀?」
顧客:「切六塊啦, 太多我驚食唔晒~」
店員:「=.=’」

[AL][PM] Asymptote

Filed under: HKALE,Pure Mathematics — johnmayhk @ 12:05 下午

A livejournal writer, Mr. Ng, felt surprising that many Hong Kong secondary school mathematics teachers have no idea in the definition of asymptote. He quoted that “\lim_{x\rightarrow \infty} f(x) = 0 \Rightarrow x-axis is a horizontal asymptote of y = f(x)" is completely wrong! He then gave a counter-example y = \frac{\sin(x)}{x}. He says that even in MathWorld, the definition that “An asymptote is a line or curve that approaches a given curve arbitrarily closely." is 一塌糊塗。Some times ago, I found the following in the internet.

From a page of an article “The Asymptotes of Plane Curves" written by H. G. Green, published in “The Mathematical Gazette" (Vol. 13, No. 185 (Dec., 1926), pp. 232-235), we see a definition of asymptote as follows.

An asymptote of a curve of the n^{th} degree is the limit of a continuous series of parallel lines which cut the curve in less than n points and whose intersections therewith, as the lines approach the limit, become and remain farther from the origin than any given distance, however great."

As in Mr. Ng’s further discussion, there are MANY definitions for asymptote. I think it is a good topic to study further.

2007/11/01

[CE] Linear programming – system of inequalities

Filed under: HKCEE — johnmayhk @ 3:58 下午

1.

Just start teaching the solving of system of 2-variable linear inequalities. One F.5C student, Yeung, told me that “no solution for parallel lines", Oops, his comment was too brief but I understood him immediately. Let me refine his observation, as follows.

Consider a system of two 2-variable linear inequalities (x and y are real with no extra restriction), the system MAY have no solution only when two different corresponding straght lines are parallel to each other.

As for example,

x + 2y < 1
2x + 4y > 4

The two corresponding lines x + 2y = 1 and 2x + 4y = 4 are parallel, and the system above has no solution.

In other words, the (two 2-variable linear inequalities) system MUST be solvable if the system is ‘corresponding to’ two non-parallel lines. Although it’s not something very striking, I appreciate the discovery, because it is totally orignated from a student himself. Now, could you show me the proof of Yeung’s observation?

[But, most of F.5C students were out of mood of attending the lesson that day, may be they thought that the content is TOO EASY.]

2.

To solve Ax + By + C < 0,
we draw L: Ax + By + C = 0, and
L divides the x-y plane into 2 regions, namely R_1 and R_2.
Knowing that P(x_1,y_1) lies on R_1 and
P satisfies Ax + By + C < 0, that is
Ax_1 + By_1 + C < 0, then
R_1 is the required region.
That is, R_1 represents the solution to the inequality Ax + By + C < 0.

The question is: why we just try one point? Should we try another? Is it possible that there is another point Q lying on R_1 but Q does not satisfy the inequality Ax + By + C < 0? How to prove that “one point is enough for representing the whole region"?

Well, the first thing in my mind is the signed distance (有向距離) from a point to a line.

Additional Mathematics tells us that the distance between a point P(x_1,y_1) and a line L : Ax + By + C = 0 is |\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}|. The absolute sign is for ensuring non-negative distance. How about getting rid of the absolute sign? The numerator Ax_1 + By_1 + C may be positive, zero or negative. What is the significant geometric meaning of the positive and negative values obtained?

Just tell you the following.

If Ax_1 + By_1 + C is positve, then P and O are at the different sides of L (as shown)

If Ax_1 + By_1 + C is negative, then P and O are at the same sides of L (as shown)

Back to our questions, if O does not lie on L, O must lie on one of the regions R_1 and R_2. Suppose O lies on R_1, then all the points on R_1 and O are at the same side of L. Hence any point Q(a,b) lying on R_1 will lead to the result A(a) + B(b) + C < 0. Hence one point is enough! It is impossible to have another point T lying on R_1 such that T does not satisfy Ax + By + C < 0.

Please complete the discussion when O lies on L.

3.

F.5C students asked, ‘how about the case for 3 or more variables?’ To the best of my memory, we may use simplex method.

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