As TIME Goes By ............

Time can be seen as instant or interval.

While I was writing this blog, Johnson came to ask me, "What time's it now, dad?" To which, I answered, "At this instant, it's 7 am on Wednesday 6 April, 2011." To be exact, 7 am on Wednesday 6 April, 2011 was the instant when I was looking at my watch before telling Johnson the time. By the time I finished saying that sentence, it was probably 15 seconds past 7, and it was some other instant already. Hence, instant is a snapshot at some point in time where there is no motion of any sort. At 7 am, I was looking at my watch. Johnson was standing beside me. Everything was in a standstill as if captured by a camera. But as I was telling the time (with some motion), TIME continues to advance continuously. There is saying that instant does not exist for if it does, motion will not be possible.
The situation is like real numbers. We cannot have answer to the question, "what is the next number after 1?". Naturally we say 2 is the number next to 1. But very quickly, we know that 1.5 is more 'next' to 1 than 2 is, and we can further get smaller number closer than 1 than 1.5. Even though we come to the number 1.00 ......... 001 to tens of thousands of digits, we can always get yet another number more closer to 1 than it.

The Clock Problem

I recall in my primary school days, we were taught to solve the clock problem in arithmetic, an example of which is as follows:

"Between the hour 8 and 9 o'clock, what time will the hour-hand and the minute-hand overlap with each other?"

We know that at 8 o'clock, the minute-hand is at the 0-minute position while the hour-hand is at the 40-minute position. As the minute-hand moves to catch up with the hour-hand, the hour-hand also moves albeit at a much slower pace. In fact, when the minute-hand moves 60 minutes (1 round), the hour-hand moves 5 minutes (1 hour). So they are moving at different speeds in the ratio of 12 : 1. When the minute-hand reaches the 40-minute position, the hour-hand would have moved a little bit further. We also know that the minute-hand will catch up with the hour-hand somewhere between the 40-minute and 45-minute position (They cannot go beyond the 45-minute position otherwise the hour-hand would have gone beyond 9 o'clock already).

In those school days, we were taught to obtain the answer arithmetically (frankly without a full understanding) as:

"40 x 12/11 or 43.64 minutes after 8 o'clock"

As we advanced into secondary school, we learnt Algebra. With the power of Algebra, the problem could be easily (and logically) understood and solved :

Let m be the position (in minutes) of the minute-hand and h be the position (in minutes) of the hour-hand.
As the minute-hand moves m minutes, the hour-hand moves m/12 minutes.
At 8 o'clock, the hour-hand is at 40 while the minute-hand is at 0 (i.e. the hour-hand has a head start of 40 before the minute-hand tries to catch up).
Hence, the position of the hour-hand (h) at any time is related to the position of the minute-hand (m) as follows:

h = 40 + m/12     ............... (1)

When the position of the hour-hand overlaps with the minute-hand,

h = m

Putting this back into the relation (1),

m = 40 + m/1 2

Solving this equation, we have:

m = 40 x 12/11 or 43.64 minutes

Hence, the time between 8 and 9 o'clock when the hour-hand and minute-hand overlap with each other is 43.64 minutes past 8.

Naturally, the problem can be re-phrased to ask for the same situation in any hour rather than between the hour 8 to 9 o'clock. In that case, we just replace the head start position 40 by the corresponding starting position of the hour-hand accordingly.

In addition, there are other variations of the clock problem: e.g.

A. "Between the hour 8 and 9 o'clock, what time will the hour-hand and the minute-hand form a straight line with each other?"

As from above, when the hour-hand forms a straight line with the minute-hand,

h = m + 30

Putting this back into the relation (1),

m + 30 = 40 + m/12        

Solving this equation, we have:

m = 10 x 12/11 or 10.91 minutes

Hence, the time between 8 and 9 o'clock when the hour-hand and minute-hand form a straight line with each other is 10.91 minutes past 8.

B. "Between the hour 8 and 9 o'clock, what time will the hour-hand and the minute-hand form a right angle  with each other?"

As the hour-hand can be right-angled in front of or behind the minute-hand, so when the hour-hand forms a right angle with the minute-hand,

h = m + 15 or h = m - 15

Putting this back into the relation (1),

m + 15 = 40 + m/12 or m - 15 = 40 + m/12

Solving this equation, we have:

m = 25 x 12/11 or 27.27 minutes

or

m = 55 x 12/11 or 60 minutes

Hence, the time between 8 and 9 o'clock when the hour-hand and minute-hand form a right angle with each other is either 27.27 minutes past 8 or 9 o'clock.

If at any point of the calculation, the number becomes negative, add 60 to the number before going on, reason being the minute-hand runs only between 0 and 60 in the clock face.

0, 1 and Infinity

Quiz - "Can you walk 1 mile south, then walk 1 mile east and finally walk 1 mile north to end up where you started?"

Answers:
1. No. We need to walk 1 mile west to come back to where we started.
    Begin from the starting point A, walk 1 mile south to point P. Then walk 1 mile east to point Q. Finally, walk 1 mile north to point B. We are still 1 mile east of A where we started.

2. Yes, when we are at North Pole.
     At North Pole N, any direction will head south. So just choose one and walk 1 mile to get to point P. Then walk 1 mile east to get to point Q. Finally, walk 1 mile north to get back to North Pole N where we started.

3. Yes, there are infinitely many places we can do so.
    There is a circle with circumference of 1 mile around the South Pole S. From any point Q on this circle, we can walk east for 1 mile encircling the South Pole S once to get back to Q. 1 mile north of Q, we can find a place P.
    Now, starting from place P, walk 1 mile south to get to point Q. Then walk 1 mile east encircling the South Pole S once to get back to Q. Finally, walk 1 mile north to get back to P where we started. Since Q is an arbitrary point on the circle, P is also arbitrary. Hence there are infinitely many places of P we can do so.

   

        Furthermore, we can find a circle with circumference of 1/2 mile around the South Pole S. From any point Q on this circle, we can walk 1 mile east encircling the South Pole twice to get back to Q. From Q, find place P as above. Likewise, we can find circles of circumferences of 1/3 mile, 1/4 mile and 1/5 mile etc. around the South Pole S and from any point Q on the circles, walk 1 mile east encircling the South Pole S 3 times, 4 times and 5 times etc respectively to get back to Q. Thus theoretically, we have infinitely many circles and infinitely many points Q, and hence infinitely many places of P we can do so.

數學不好的好處

Recently I read a following article:

"數學不好的好處"
1. 數學不好的人都比較愛笑, 因為沒有數學就沒有煩惱.
2. 數學不好的人都比較天真浪漫, 比較感性.
3. 數學不好的人都比較幽默, 生活充滿樂趣, 感情和想像力都比較豐富.
4. 數學不好的人都比較直爽, 實在, 不會拐彎抹角.
5. 數學不好的人長得都比較漂亮.
只有一個缺點 ...... 就是數學不好!!

Negating the negatives, we have:

"數學好的壞處"
1. 數學好的人都比較不懂得笑, 因為有了數學就煩惱多多.
2. 數學好的人都比較正經實際, 比較理性.
3. 數學好的人都比較嚴肅, 生活缺乏樂趣, 感情和想像力都比較平淡.
4. 數學好的人都比較保守, 虛偽, 時常拐彎抹角.
5. 數學好的人長得都比較平凡.
只有一個優點 ...... 就是數學好 !!

OMG !!!

Squares

How many squares are there are in this diagram?

The answer is :

http://www.youtube.com/watch?v=RpkBjMams54

Illusion or Reality?

You probably have the following experiences:
1. You were flying on a plane from one destination to another (say Sydney to Hong Kong). The plane was so smooth and quiet that you could hardly feel any movement. You remained seated without moving around in the plane. So you thought you were not in motion. You threw a ball straight up in the air, and the ball fell straight down back into your hand, indicating that you were not in motion at all. But on a second thought, you knew the plane was moving at a ground speed of 500 km/hr away from Sydney. You also knew you yourself were moving (together with the plane) similarly at a very fast ground speed of 500 km/hr away from Sydney.
2. You were traveling on a train from one city to another. You felt tired and fell into a nap. The train was so smooth and quiet that, when you woke up, you didn't know whether the train was moving or not. You thought the train was stationary at a train station. You looked through the window on the left side of your train. You saw another train overtaking yours slowly carriage by carriage from your back to the front at a speed of about 10 km/hr. You then turned your head and looked through the window on the right side of your train. From the views outside, you then realized that your train was in fact moving forward at a very fast speed of 100 km/hr passing non-stop over the station. You then also realized that the other train on your left was in fact moving forward at a even faster speed of 110 km/hr passing non-stop over the station.
3. On another similar occasion, you woke up from your nap on a long train journey. You thought the train was stationary at a train station. You looked through the window on your left side. You saw another train passing by yours, carriage by carriage from your front to the back. When all the carriages had passed over, you could see other views outside the window. You then realized that the other train was in fact stationary at the train station, and your train was in fact moving forward passing non-stop over the station at a speed of 100 km/hr.

Your experiences above indicate that:
1. Your position at any time is determined by some point of reference. After an hour, you were still in the middle of the PLANE, but you were 500 km away from SYDNEY.
2. If your position (to some point of reference) doesn't change over time, you are stationary and not in motion (to the same point of reference). During the entire hour, you remained seated in the PLANE, you were stationary and not in motion (with reference to the PLANE).
3. If your position (to some point of reference) changes over time, you are in motion (to the same point of reference). During the same hour, you were further and further away from SYDNEY. So you were in motion (with reference to SYDNEY). Hence, motion is the change of position (to some point of reference) over time.
4. As motion is the change of position over time and position is determined by some point of reference, so motion is also determined by some point of reference. While you were on the train, your train was moving at 100 km/hr (passing over the STATION), but the other train on your left side was moving at 10 km/hr (passing over YOUR TRAIN).
5. Likewise, non-motion (being stationary and not in motion) is also determined by some point of reference. You were not in motion (in the PLANE) although both YOU and the PLANE were moving at a very fast speed (away from SYDNEY).
6. If you are moving (with reference to some point of reference) and the point of reference is also moving (with reference to some other point of reference), then you are also moving (with reference to that common point of reference). The train on your left side was (overtaking YOUR TRAIN) at 10 km/hr, and your train was moving forward at 100 km/hr (passing non-stop over the STATION), then the train on your left side was also moving at 110km/r (passing non-stop over the STATION).
7. When there is no common point of reference, motion can only be relative. As in your experience 3 above, without any common point of reference (the STATION), it can be regarded that either YOUR TRAIN was in motion and the OTHER TRAIN was stationary, or YOUR TRAIN was stationary and the OTHER TRAIN was in motion. Hence, without any common point of reference, it can only be regarded that YOUR TRAIN and the OTHER TRAIN were moving towards each other at a speed of 100 km/hr.

As motion is the change of position over time, apart from being determined by some point of reference, it is also determined by TIME. We shall leave the discussion of TIME in some later blogs as TIME permits.

The Pythagoras Theorem

My point of departure is the Pythagoras Theorem:
Everybody with secondary education knows the Pythagoras Theorem. But I bet over 90% of people don't know why it is true (i.e. they don't know how to prove it)! While proving the Pythagoras Theorem appears to be difficult, I present below a classical proof by Euclid (circa 330 - 260 BC) which requires no more than elementary Geometry known to every secondary student!

Please feel free to skip the rest if Geometric Proof doesn't appeal to you.
First, draw squares ABFG, BCDE and CAHK. Obviously, our task is to prove the area of square ABFG is the sum of the area of square BCDE plus the area of square CAHK. Usually people are stuck at this point. But what Euclid did was to draw a few lines as follows:

. draw CM parallel to AG, cutting AB at N and FG at M respectively
. join BH and CG

With Euclid's hints, the picture now becomes much clearer as in above figure, we can then proceed as follows:

In ΔABH and ΔAGC,
....................... AB = AG ......................... (sides of square ABFG)
.................. ∠BAH = ∠BAC + ∠CAH = ∠BAC + 90°
.................. ∠GAC = ∠GAB + ∠BAC = 90° + ∠BAC
............... ∴ ∠BAH = ∠GAC
...................... AH = AC .......................... (sides of square CAHK)
............... ∴ ΔABH Ξ ΔAGC ...................... (SAS)

....... Area of ΔABH = ½*AH*HK
........................... = ½* (AH*HK)
........................... = ½*(area of square CAHK)
....... Area of ΔAGC = ½*AG*GM
........................... = ½*(AG*GM)
........................... = ½*(area of rectangle AGMN)

Since ΔABH Ξ ΔAGC as proved above,
................ then area of ΔABH = area of ΔAGC
So, ½*(area of square CAHK) = ½*(area of rectangle AGMN)
......... ∴ area of square CAHK = area of rectangle AGMN ......... (1)

Similarly,
............. area of square BCDE = area of rectangle BFMN ......... (2)

Combining (1) and (2) together,
area of square CAHK + area of square BCDE = area of rect AGMN
................................................................... + area of rect BFMN
............................................................... = area of square ABFG

i.e. .............................................. a² + b² = c² .............. Q. E. D.

The Monty Hall Problem

You are on a TV show. You are given a choice of 3 doors. Behind one door is a car; behind the others, goats. You are to pick a door, winning whatever is behind it. So you pick one, say door 1. Since there are 2 goats behind 3 doors, whatever you choose, there is at least one goat left behind the 2 other doors. The host, Monty Hall, who knows what's behind them, opens one of these 2 other doors, say door 3, to reveal a goat. He then asks you, "Do you want to stick to your choice, or do you want to switch to door 2?" The question really is to ask, "Does switching increase your chance of winning the car?"
When you pick door 1, the chance of winning the car is 1/3 (i.e. the chance of having the car behind door 1 is 1/3 since there is 1 car in 3 possible doors). And then Monty eliminates door 3 for you by revealing a goat behind door 3. Intuitively, you think you then get a better chance of winning the car as the chance of having the car in door 1 is now 1/2 (since there is 1 car in only 2 possible doors) and likewise in door 2 is also 1/2. So you think it makes no difference whether to switch or not. So you choose to stick with door 1. Right? WRONG!

Let's go back and examine the probabilities in more details.

When you pick door 1, your chance of having the car behind door 1 is 1/3. There's no question about it. Let's now group doors 2 and 3 together. Your chance of having the car NOT behind door 1 (i.e. behind either door 2 or 3) is 2/3 since the total probability must equal to 1. By eliminating door 3 from the scene (by Monty), your chance of having the car behind door 1 is NOT increased from 1/3 to 1/2 as intuitively thought. Your chance of having the car in door 1 is still 1/3 as when you make your choice, you choose 1 out of 3 possibilities (doors 1, 2 and 3). As the saying goes, the dice has been cast. The probability will not, and cannot be altered. Only IF you are free to choose at this point, you are then to choose 1 out of 2 possibilities (doors 1 and 2), and the chance of having the car behind door 1 is indeed 1/2. So by eliminating door 3 from the scene, the effect is ONLY simply changing our group consisting of doors 2 and 3 to a group consisting of door 2 only. Hence your chance of having the car NOT behind door 1 (now behind door 2 only) is still 2/3. In other words, AT THIS POINT, the chance of having the car behind door 2 is 2/3, not 1/2 as intuitively thought. So after Monty eliminates door 3 for you by revealing a goat behind door 3, the chance of having a car in door 1 is 1/3, and in door 2 is 2/3.
Hence, YES, you WILL increase your chance of winning by switching. IN THEORY, YOU SHOULD ALWAYS SWITCH.

The Seven Bridges of Konigsberg

Konigsberg was a small town in East Prussia (now Russia), divided by a river into several parts which were connected by seven bridges as shown in the diagram. The citizens of Konigsberg crossed these bridges for leisure walks on Sundays. One day, they wondered, "Can we take a walk in Konigsberg in such a way that we cross each of the seven bridges once and only once?"

At first sight, we seem to be faced with a tedious and daunting task of tracing out all the possible routes with the seven bridges, and showing whether there is a particular route that works. To address such problem more systematically may require techniques of topology and the like. But when Leonhard Euler (1707-1783) looked at the problem, he immediately claimed that it was NOT possible to have such a walk. Despite Euler (pronounced as "oi-ler") was a great mathematician, he was able to prove his claim by a simple and clever way which can be understood by almost anyone. Euler's strategy to tackle the problem was by method of proof by contradiction. Now let's see how the genius was at work:

First note that Konigsberg is divided into FOUR regions, A, B, C and D interconnected by the seven bridges as shown in the diagram. Next assume that it IS possible to have a walk in the town by crossing each of the seven bridges once and only once. The walk may start in any one of the 4 regions, A, B, C or D, and end in any one of them (which may or may not be the starting region). In any case, we must have at least TWO regions which are neither the starting region nor the ending region.
Now consider any one of these regions. Since it is not the starting region nor the ending region, if we go into this region to visit, we must go out from it accordingly. To visit this region once, we have to go into the region through one bridge and out through another since we cannot cross the same bridge more than once. So we have to have 2 bridges connected to this region in order to visit it once. If we visit this region a couple of times, we have to have an even number of bridges so that we don't cross the same bridge more than once. But looking at the diagram, NO region in this town has such a property (i.e. connected with an even number of bridges: the island C has 5 bridges while the other regions A, B and D all have 3 bridges each), let alone there are at least 2 such regions. Hence, our original assumption that it IS possible to have such a walk leads to some contradiction to the given facts, and thus it cannot possibly be true. In other words, we cannot have a walk in Konigsberg by crossing each of the seven bridges once and only once.

Q. E. D.

The Story of the Barber

In a certain remote village, there is a barber who shaves all and only those villagers who do not shave themselves.
The story is simple, but now comes the question, "Who shaves the barber?"
If the barber shaves himself, then he does not (since the barber shaves only those who do not shave themselves). If the barber does not shave himself, then he indeed does (since the barber shaves all those who do not shave themselves). So, the logic shows the barber shaves himself if and only if he does not shave himself, which is totally absurd.

The only possible answer to the question is that there cannot possibly be such a barber, nor such a village. Put simply, the story is impossible!

The Beauty of Mathematics

1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 + 10= 1111111111

9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888
987654321 x 9 - 1 = 8888888888
9876543210 x 9 - 2 = 88888888888

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

The Arrow's Dichotomy

The story of 'Achilles and the Tortoise' as mentioned in my blog on 2 August, 2010 is likened to the story of 'The Arrow's Dichotomy' in which it goes as follows:
For an arrow to be shot from one side of the room to the other, the arrow has first to travel one half of the room. When the arrow reaches one half of the room, it still has to travel one half of the remaining length (i.e. one quarter) of the room. When the arrow reaches that position, it still has to travel yet one half of the remaining length (i.e. one eighth) of the room, and so on and so forth. Thus, the total distance the arrow needs to travel in order to cross the room is an infinite series:

1/2 + 1/4 + 1/8 + ..............

Being an infinite series, there is no finite value for the sum. Rather, the limit of this sum is 1 meaning the more terms we add, the sum will get closer and closer to 1 but will never be 1 and cannot exceed 1. Hence, the total distance the arrow travels will never be equal to the length of the room. In other words, the arrow cannot get to the other side of the room.
If we replace the length of the room by some shorter length, say a metre, with similar arguments, we can say the arrow cannot get to the end of the metre. If we replace the metre by yet a shorter length, say an inch, with similar arguments, we can say the arrow cannot get to the end of an inch. If we go on with shorter and shorter length, we will come to the conclusion that the arrow cannot move at all. Finally we conclude motion is not possible, which obviously is a paradox.

Both 'Achilles and the Tortoise' and 'The Arrow's Dichotomy' are different representations of the Zeno's Paradox, and both stories concern about motion, position and time - i.e. physics. We shall discuss Zeno's Paradox from the physics perspective in some later blogs as time permits.

Counting with Infinities

The set of natural numbers 1, 2, 3, 4, 5, ..................... is infinite meaning that no matter how large a number we go, there are always larger numbers we can go further. Likewise, the set of even numbers 2, 4, 6, 8, 10, ..................... is also infinite. While both are infinite, we want to ask, "Are there more natural numbers than even numbers?"
Our answer tends to be yes because natural numbers consist of both even numbers and odd numbers. Apparently the number of natural numbers is doubled that of even numbers. This answer may be correct IF both of them are finite (i.e. of fixed quantities). But things behave differently when they are infinite. Twice of infinite quantity is still infinite quantity. Seemingly amazingly, the correct answer would be: "No, there are as many even numbers as natural numbers."

Suppose we ask, "Are there as many fingers on the right hand as the left hand?" The simple way to determine the answer is to count the number of fingers on the right hand and the left. If they both come to five, we know the answer is yes. But suppose we haven't developed the number system, and we don't have any knowledge of one, two, three and so on. Then we would not be able to count. But it doesn't mean we cannot answer the question. We can still match the fingers on the right hand to those on the left; thumb against thumb, index finger against index finger, middle finger against middle finger and so on. If all the fingers on the right hand can find a corresponding partner on the left, then we know the answer is yes.
Now suppose we have infinite number of fingers on the right and left hands (a monster!). We would not be able to count as they are infinite. But we can still use the method of matching to find the answer. Although the process of matching goes on indefinitely, if each finger on the right hand is able to find a one-to-one corresponding partner on the left, and each finger on the left hand can find a one-to-one corresponding partner on the right, then we know the answer is yes. However, if we can find some finger on the right hand that doesn't have any partner on the left while each finger on the left hand can find a partner on the right, we know there are more fingers on the right hand than the left.
Alternatively, imagine we have a very very large audience and a very very large cinema hall. In order to determine whether there are as many spectators as seats in the hall, we could ask the audience to seat themselves one by one. If there is no spectator left unseated and if there is no seat left empty, then we know there are as many spectators as seats in the cinema hall without going through the tedious process of counting with very very large numbers.

So, using the same methodology, we list the natural numbers N and the even numbers E as follows:

N: 1 2 3 4 5 ............ n ......
E: 2 4 6 8 10 ........ 2n .....

We can see that for each natural number, there is always a one-to-one corresponding even number by doubling its value. And for each even number, there is always a one-to-one corresponding natural number by halving its value. Although the process goes on indefinitely, there is no natural number nor even number left unmatched. Hence, we conclude there are as many even numbers as natural numbers.

Situation of this type is said to be countably infinite, and is the basis upon which Georg Cantor (1845-1918) developed his theory of infinite sets.

Achilles and the Tortoise

The tortoise challenged the great Greek warrior, Achilles to a race with an allowance of 100-metre head start. Although Achilles was such a strong and fast warrior, he would never be able to catch the tortoise, not to mention to overtake him.
In order for Achilles to overtake the tortoise, he had to run the 100 metres, bringing him to the tortoise's starting point. Since both Achilles and the tortoise were moving at the same time, by the time Achilles reached the tortoise's starting point, the tortoise should have moved forward some distance, say a metre. So the tortoise was a metre ahead of Achilles. In order to overtake the tortoise, Achilles then had to run that one metre, bringing him to the tortoise's new position. Again, since they both were moving at the same time, by the time Achilles reached the tortoise's new position, the tortoise should have moved forward yet some distance, albeit smaller this time. So the tortoise was still at some distance ahead of Achilles. Achilles then had to catch up with the new distance. But the tortoise yet moved forward another distance at the same time, albeit much smaller and smaller. So the tortoise was always at some distance ahead of Achilles, no matter how small it was. Therefore, Achilles would never catch the tortoise.
This is the famous Zeno's Paradox by the Greek philosopher, Zeno (490BC - 430BC). We know from common sense Achilles could overtake the tortoise in matter of seconds. But Zeno's argument was perfectly logical. How can we resolve this paradox?
I shall leave it to some later blog.

Mathematics Achieves Longevity

You probably have been taught in school about the many benefits of Mathematics. I'd just like to mention a less obvious one.

Mathematicians usually enjoy long lives. We can always pick up a few well known figures and illustrate below:

Archimedes : 287B.C.-212B.C. (75)
Isaac Newton : 1643-1727 (84)
Galileo Galilei : 1564-1642 (78)
Albert Einstein : 1879-1955 (76)
Carl Gauss : 1777-1855 (78)
Johann Bernoulli : 1667-1748 (81)
Leonhard Euler : 1707-1783 (76)
Augustin Cauchy : 1789-1857 (68)
Pierre-Simon Laplace : 1749-1827 (78)
David Hilbert : 1862-1943 (81)
Chern Shiing-shen (陳省身) : 1911-2004 (93)
Huo Luogeng (華羅庚) : 1910-1985 (74)
Tsien Hseu-shen (錢學森) : 1911-2009 (98)

In 'The Longevity Bible' (2006) by Gary Small, sharpening the mind is considered as the first essential for extending life expectancy, together with other essentials like physical exercise, healthy diet, positive attitude, harmonious relationship, stress-free environment and appropriate medication etc. As the author put it, "Fix the brain first, the rest will follow." A study published in the New England Journal of Medicine found that frequent participation in mentally stimulating activities such as board games, cross word puzzles, sudoku and book reading lowers the risk for Alzheimer's disease (loss of memory and mental abilities usually associated with old age leading subsequently to death) by nearly one third. Hong Kong people have long been aware that playing mahjong avoids or reduces the chance of Alzheimer, although there is no formal study to substantiate. Mathematics is a much more mentally stimulating game. Frequent usages/exercises can achieve similar results, if not better. Most mathematicians work on Mathematics until their very last days, and enjoyed longevity as shown by observations above. Mathematics is indeed rewarding!

Distance Learning in Mathematics

Back in 2003, I somehow had the desire of taking some distance learning in Mathematics. The idea came about 2 years ago when I was helping my second daughter in her study of 4-unit Mathematics as preparation of the HSC examination (HSC is the joint examination for secondary school students in Australia and is the credential awarded for successful completion of high school study and as basis for entrance to universities). To facilitate the assistance to my daughter, I had to review and refresh my knowledge of my high school mathematics. And in so doing, I came to have rekindled my interest in this subject which was my favourite during my early school days.
I searched the internet for distance learning in university level Mathematics. To my surprise, despite there were lots and lots of distance learning courses everywhere, there were hardly anything in Mathematics. Mathematics is quite a wonderful thing. Everybody learns Mathematics, but few study it. Mathematics is compulsory for everyone in School, but only a handful take it in College. There seemed to be no market for higher level Mathematics. So I searched the universities one by one. Finally I saw in the website of Stanford University, California that they had a program called EPGY (Education Program for Gifted Youths) in which they offered university level Mathematics courses to talented high school students in the form of distance learning. I then wrote to Stanford for possibility of enrollment. I said frankly that I was not sure whether I was gifted, but definitely I was not a youth given my age. Stanford declined insisting their program was aimed for talented high school students only. I expressed that I had gone through the whole world (electronically) and was only able to find this suitable program in Stanford, and that I was very keen to pursue knowledge at this level. After several exchanges of emails, Stanford advised that they could offer the program to teachers of Mathematics. So I suddenly became a teacher and was accepted.
Thence I undertook the program part-time from 2003 to 2006 while I was still working with Qantas. I completed subjects including:

. Multivariable Differential Calculus
. Multivariable Integral Calculus
. Ordinary Differential Equations
. Partial Differential Equations
. Real Analysis
. Complex Analysis
. Logic
. Number Theory
. Linear Algebra
. Modern Algebra

Despite the hard work (it was particularly hard for me being away from school so long), the experience was indeed very enjoyable and satisfying. Although the materials covered were quite involved, some of them are still regarded as 'Introduction' and 'Elementary'. Really there are still so much more lying ahead!

Prisoner's Dilemma

The police had been watching two suspects of bank robbery for some time but they didn't have sufficient evidence for a conviction. One day, these two guys were caught stealing packs of bubble gum in a supermarket. The police put them into separate rooms, visited each of them and offered the same deal: If one testifies (defects from the other) for the prosecution against the other and the other remains silent, the betrayer goes free and the silent suspect receives the full 10-year sentence for bank robbery. If each betrays the other, each receives a 5-year sentence for bank robbery. If both remain silent, both suspects will be sentenced to only a 1-month jail for theft in supermarket. Each suspect was assured that the other would not know about the betrayal before the end of the investigation and each was told that the other was also offered the same deal. Thus, each suspect was facing a dilemma of choosing either to co-operate with the police (betray the other) or not to co-operate with the police (remain silent).
How would the suspect act?
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The possible outcomes (pay-off) can be summarized diagrammatically as follows:

	B remains silent	B betrays
A remains silent	A: 1 month B: 1 month	A: 10 years B: goes free
A betrays	A: goes free B: 10 years	A: 5 years B: 5 years

In 'win-lose' terminology, the table looks like this:

	B remains silent	B betrays
A remains silent	win-win	lose much-win much
A betrays	win much-lose much	lose-lose

Clearly from the diagrams above, the best result (for BOTH suspects) could be achieved IF both co-operated with each other and not with the police by remaining silent. Then each would only be sentenced to a 1-month jail for theft. However, since they were shut up separately, they would not be able to negotiate with each other for a co-operation, and they could only guess what the outcomes would be for different scenarios. Without loss of generality, if suspect A remained silent, he had 50% chance for 1-month jail and 50% chance for 10-year jail. His average pay-off was 5 years and a half month. If A betrayed, he had 50% chance to go free and 50% chance for 5-year jail. His average pay-off was 2 and a half years. A might think that, regardless what B chose, he would always receive a higher pay-off (lesser sentence) by betraying. A could actually say, "No matter what B does, I personally am better off betraying than remaining silent. Therefore, for my own sake, I should betray." All things being equal, suspect B would act similarly. Then they both betrayed each other, and both actually got a lesser pay-off (5 years) than they would get by remaining silent (1 month).

This classical prisoner's dilemma illustrates the different strategies that the player (suspect) would take in the game with different perspectives. If the player could look at the broad picture (which only the police could but the suspects could not), he would opt for remaining silent to achieve the best result. If the player looked at the game only from his own perspective (which the suspects were constrained to do), the strategy to remain silent is clearly dominated by that to betray. So the only possible equilibrium for the game is for all players to betray. No matter what the other player does, one player will always gain a greater benefit by playing betrayal. Since in any situation, playing betrayal is more beneficial than playing remaining silent, all rational players will betray. It also shows how a win-win situation (for the suspects) can be turned into a lose-lose situation (as exactly what the police wanted).

Techniques of Teaching Mathematics

I taught private tuition in Mathematics to students junior than me while I was studying in secondary school. Over time, I came to become aware of some skills in handling some special situations during the lessons. Here are things you may say in class under certain circumstances:

. You don't want to go through all the possible cases. You just show one, and let the student figure out the rest,

"Without loss of generality, .......... "

. You know it's true, but you've lost your notes and you don't know how to prove it,

"It is obvious that .......... "

. You are struggling on how to explain, and your student suddenly reminds you the critical point,

"It now becomes clear that .......... "

. You feel tired after a long day's study yourself, and you don't want to explain in details,

"This is trivial."

. You are desperate to go on Friday afternoon,

"The rest are left as exercises to students."

Quite Easily Done - Is It?

When mathematician (or anyone working with mathematics) finishes a problem, he or she tends to tell the whole world that the work was quite easily done (Q.E.D.).

More often than not, they have already worked on it for days, if not weeks!

Gauss in His Childhood - A Glimpse of Talent

Talent is something you can do easily while others find it difficult to do.
Genius is something you can do easily while others find it impossible to do.

When Gauss (1777-1855) was a child, he attended the primary school in his local town. One day, the teacher found the class was too unsettled. In order to keep the class quiet for a while, he asked the children to sum the numbers from 1 to 100 before they could be dismissed. The class did stay silent while everyone was busy sketching their calculations on papers. Gauss didn't move. He stared at the blackboard for a few seconds, raised his hand and said, " The answer is 5050." The teacher looked at Gauss in disbelief, thinking he was just keen to leave the classroom. Gauss then explained step by step as follows:

"We call the sum of the numbers 1 to 100 S. Then,

S = 1 + 2 + .................. + 99 + 100

If we reverse the order of the numbers and add, the sum is still the same S.

S = 100 + 99 + .................. + 2 + 1

If we take the two together (i.e. add them vertically), we have two times the sum 2S, but the terms become 101 all the way for the 100 terms.

2S = 101 + 101 + .......... + 101 + 101

Since the numbers are the same, we don't need to add 100 times but we can multiply by 100. Simpler still, multiplying by 100 is just to add two zeros at the end of the number.

2S = 101 x 100 = 10100

To get back to the sum S, we just divide by 2.

S = 10100/2 = 5050."

Q.E.D.

Following on Gauss idea, we can develop further to generalize to sum the numbers from 1 to N (no matter what N is):

1 + 2 + ............... + N = N(N+1)/2

or, to sum from the number M to number N:

M + (M+1) + .......... + (N-1) + N = (N-M+1)(N+M)/2

If we still want to go further, we can research more to sum their squares:

1 + 2² + 3² + ………. + N²

or their cubes,

1 + 2³ + 3³ + ………. + N³

or, to push to the extreme,

1 + 2^m + 3^m + ………. + N^m

(no matter what m is).

But these require further mathematical techniques like recursive formula and Bernouilli numbers. I can touch on more as time permits.