Kopitiam! [Archive]

chiunlin

15-09-2004, 10:17 AM

ok, let me have a shot at it (I'm only a freshman, though)

Let's start by assuming that log-base2 of 5 is rational,i.e
log-base2 5=a/b
5=2^(a/b)
and 5=5^1
Now, let c=2^(a/b)
For all x<=b, 2^(1/x) is irrational and an irrational number can't be made rational by multiplying it with an integer. Therefore, if c is to be an integer, it must have 2 as a factor.
However, by unique prime facorization theorem, any integer can only have one set of prime factors and all others are just a rearrangement of the order of the primes.
Since 5 does not have 2 as a prime factor, therefore, 5=2^(a/b) is false. By contradiction, log-base2 5 must be irrational.

luke

15-09-2004, 10:33 AM

Now, let c=2^(a/b)
For all x<=b, 2^(1/x) is irrational and ....
how does this come about?

chiunlin

15-09-2004, 10:50 AM

Now, let c=2^(a/b)
For all x<=b, 2^(1/x) is irrational and ....
how does this come about?
Want me to prove it?
Ok, here it is.
assume that for all 2<=x, 2^(1/x) is rational. Hence
2^(1/x) = c/d and c and d is in the simplest form, i.e, no common factor.
2=(c/d)^x
2d^x=c^x
Since LHS is even, RHS must also be even. By unique prime factorization, RHS must have a prime factor of 2 with multiplicity x.
Hence,
Let c=2e, e is an integer, therefore, c^x=(2^x)(e^x)
Hence 2d^x=(2^x)(e^x)
d^x=(2^(x-1))(e^x)
Since RHS is even, LHS must also be even. Hence d can be expressed in the form of d=2f where f is an integer. However, this means that c/d is not in the simplest form since they share a common factor, which is 2. By contradiction, 2^(1/x) is irrational for all 2<=x.

luke

15-09-2004, 11:13 AM

Owh ok :)

but there is a flaw in your original conclusion
Since 5 does not have 2 as a prime factor, therefore, 5=2^(a/b) is false. By contradiction, log-base2 5 must be irrational.
you say 5=2^(a/b), where a and b are integers, is false. what if (a/b) is not integer, for example a/b=5/2 ... then you can't use unique prime factorization theorem since the theorem basically says that if g = 2x2x2x2x...x2 (n times where n is an integer), you can't prime-factorize g using any other combination of primes ... it doesn't support for n not an integer ;)

you are near there already .. just have to play around with the equation a little bit

digimushu

15-09-2004, 11:16 AM

I feel like i'm watching a cat playing with a mouse...
hahahaha
Good try tho!

chiunlin

15-09-2004, 11:30 AM

Ok, ok, I wrote the previous two arguments without even really thinking. Now, after looking hard at the question for one minute, I finally realise that how dumb i was.

I'll write the whole argument again.
Now, assume that log-base2 5 is rational and thus can be expressed as log-base2 5=a/b
5=2^(a/b)
5^b=2^a
LHS has prime factor 5 with multiplicity b and LHS has prime factor 2 with multiplicity a. Since unique prime factorization states that a number can't have two unique sets of prime factors, LHS can never be equal to RHS. Therefore, log-base2 5=a/b is false and thus it is irrational. #

luke

15-09-2004, 11:34 AM

you got it :D

but to give it a punch you should say:

since 5^b=2^a is not possible for a and b are both integers, (a/b) is not rational .. therefore log base-2 of 5 is irrational ...

but you still get some cookies from me

http://anatilmizun.homeip.net/directlink/cookie.jpg

chiunlin

15-09-2004, 11:35 AM

Well, let me post a few questions also.
1) using purely vector techniques, find the dihedral angle of a
regular tetrahedron.

2)Find the shortest distance to go from point A(4,5) to y-axis,
then x-axis and finally point B(8,4).

luke

15-09-2004, 02:17 PM

2)Find the shortest distance to go from point A(4,5) to y-axis, then x-axis and finally point B(8,4).

I managed to come up with a method but got stuck at finding the solutions. I post it here if others want to continue ;)

first, let say we can do that in the way shown in the picture below:

http://anatilmizun.homeip.net/directlink/kopitiam01.gif

A and B are arbitrary numbers we have yet to find .. to find them we have to find the values that gives the absolute minimum value for the following expression:

distance=sqrt(16+(5-A)^2)+sqrt(A^2+B^2)+sqrt((8-B)^2+16)

for sure A=B=0 is not the minimum (i.e going straight from (5,4) to the origin and then to (8,4) is not the shortest distance)

anyone (but chiunlin of course :P) has any idea how?

yekban81

15-09-2004, 10:35 PM

2) From A(4,5), go to the origin (since at the origin, you have arrived at the y and x axis at the same time) and then straight to B loh.. any other method?

Are this kopitiam for maths wizards hah?

digimushu

15-09-2004, 10:41 PM

if i were to solve this, i would just use luke's distance equation as a cost function and use optimization routines such as Hooke and Jeeves to find the values of A and B that minimizes the cost function. Brute force approach but I'm just too lazy to think of any other way...hehehe

digimushu

15-09-2004, 10:42 PM

hahaha
nah, kopitiam is meant to be the place where engineers can talk about naything under the sun. just that these two are uhmm......exercising their intellectual prowess

chiunlin

16-09-2004, 12:07 AM

hehe, hei, keep in mind that I'm just a freshman, how complicated can the solution be? Cost function and optimization routines like blahblahblah are not required(you may use it if you want though). Think simple.
Hint: Look at the mirror. Hehe...

Anyway,kopitiam is also famous for people to play a game of chess or chinese chess there right? The situation is analogous here.

digimushu

16-09-2004, 12:42 AM

You are right, this is like a chess game.

It does not matter how complicated the problem could be. I believe as an engineer, there is always a better way to do things. yes, the simplest solution may be the best, but as we all know in math, there is often multiple solutions to the same problem.

chenchow

16-09-2004, 01:16 AM

Hey, everyone, spread the words on this SIG and perhaps there could be an announcement for a day or two at the Updates above, to get everyone's attention about Engineers SIG~!

digimushu

16-09-2004, 01:25 AM

Uhmm...would luke or masterof_none mind to do as chenchow suggested? Thank you very much!

luke

16-09-2004, 02:39 AM

2) From A(4,5), go to the origin (since at the origin, you have arrived at the y and x axis at the same time) and then straight to B loh.. any other method?

Are this kopitiam for maths wizards hah?
As I already mentioned, going straight to the origin and then to (8,4) is not the shortest distance .. try going to (0,2.5) and (4,0) first and you'll see that the distance taken is shorter than going thru the origin ;) but of course that is not the shortest .. by brute force I found the solution of A and B to be near 2.5 and 3.1 ... maybe an approach using derivatives will give us a more precise answer ...

owh btw, engineers talk math :P :P :P

p/s: mamak, give me 45 degree celcius 200 milliliter hot coffee with 15 gram sukrose ... make it in between 5 minutes to 10 minutes ... uh, tempat biasa :P

digimushu

16-09-2004, 03:34 AM

Good one luke,
try this, got this off from somewhere:

Pick-Up Lines to use on Engineering Chicks

I won't stop bugging you until I get the address of your home page.

Let's convert our potential energy to kinetic energy.

Wanna come back to my room and see my 166mhz Pentium?

How about you and I go back to my place and form a covalent bond?

You're sweeter than glucose.

We're as compatible as two similar Power Macintoshes.

Wanna see the programs in my HP-48GX?

Your body has the nicest arc length I've ever seen.

You're hotter than a bunsen burner set to full power!

My love for you is like a concave up function because it is always increasing.

luke

16-09-2004, 12:56 PM

Wanna come back to my room and see my 166mhz Pentium?
Don't want!! I have 2.4GHz Pentium IV :lol:

mamak! roti canai radius 10cm satu!

chiunlin

18-09-2004, 09:02 AM

Dah lama lah masih tak ada jawapan? Need another hint? Alright, alright, here's another one.
Hint: The shortest distance between two points is a STRAIGHT line.
Still scartching your head and wondering what's happening? Keep on scratching then.

luke

18-09-2004, 11:14 AM

yekban81

19-09-2004, 02:04 PM

Straight line is the shortest path between two point?

Then the shortest path shall be a straight line from A to B, of which its both ends is extended futher to both y and x-axis. I mean by following this straight line, we go from A to y-axis and surpass A again and B too to X-axis, finally reverse back to B again. Possibly the shortest path. Answer please?

Where did you get sych question?
And digimushu, are you giving us a riddle to guess? Not quite understand..haha. Any hint like Chiun Lin?

digimushu

19-09-2004, 02:08 PM

hahahah
hrmmm
sorry sorry, i was off topic then(does this thread even have a topic?). It was just some pick up lines we are all supposed to avoid using when talking to chicks i guess...

luke

19-09-2004, 02:36 PM

Then the shortest path shall be a straight line from A to B, of which its both ends is extended futher to both y and x-axis. I mean by following this straight line, we go from A to y-axis and surpass A again and B too to X-axis, finally reverse back to B again. Possibly the shortest path. Answer please?
Nawh .. I don't think that will be the shortest distance. If we extend the straight line from Y-axis to X-axis while passing thru point A and B, it's very easy to see that the portion of the line which connects A to Y-axis is not the shortest line for that purpose. You should know that the shortest distance from a line L to a point P is a straight line passing thru P and normal to L. Therefore the shortest line connecting point A and Y-axis is a horizontal line from A to Y-axis. Applying the same idea, the shortest line from point B to X-axis is a vertical line thru B.

And then we can clearly see that from Y-axis to X-axis, instead of passing thru point A and B, we can just go straight to X-axis for a shorter distance, and then only do we go to point B. In the end we get a trajectory similar to what I've drawn. And even that, it's still not the shortest. We have to find the points on Y-axis and X-axis that will give us the shortest distance if we are to pass thru them in our 'journey' from A to B.

yekban81

19-09-2004, 03:06 PM

Yalor, Luke.

How about using mathematical equation to solve it?
We have point at y-axis C(0,y), point at -x-axis D(x,0), A( 4,5) and B(8,4).
We need to find shortest path for A,C,D and finally B.

Using trigonometry rule, we may derive a equation to represent the distance travelled within the path:
Distance, L=.......which has parameter x and y.

Differentiate equation L , for x when y=constant value and y when x=constant value, to get
dL/dx = 0
dL/dy= 0

then you will get two equation to solve simultaneously.
What do you think?

luke

19-09-2004, 11:28 PM

Refer to the first page of this discussion. I have posted a drawing and an equation which illlustrate exactly the approach you are suggesting. The problem is I suck at derivatives and integrations and thus unable to find the solution :P

chiunlin

19-09-2004, 11:59 PM

ok, final hint:(I'll tell the answer if the question remains unsolved for another 24 hours) My first hint is mirror which refers to symmetry and reflection.

chiunlin

20-09-2004, 09:37 PM

2)Find the shortest distance to go from point A(4,5) to y-axis, then x-axis and finally point B(8,4).

http://anatilmizun.homeip.net/directlink/kopitiam01.gif

Hehe, well, since yekban81 is so curious to know the answer, I'll post it a few hours earlier(if you happen to be on the verge of Eureka, I'm sorry).

Now, first of all, using y-axis as the line of reflection, reflect point B to B'(-8,4). Then using x-axis as the line of reflection, reflect it to point B''(-8,-4). Now, draw a straight line connecting A to B''. Since I ask for the distance only, d=sqrt(12^2+9^2)=15 units.

yekban81

20-09-2004, 10:08 PM

Frankly said, I don't understand the solution that you have provided above.

Anyone know how does the word ENGINEER come out?

littlebigone

20-09-2004, 10:25 PM

the tip about the shortest distance between two points being a straight line would help a lot.

what chiunlin did was simply reflect the point (8,4) with respect to y = -x. then if you draw a straight line between the two points, (4,5) and (-8,-4) you should get the shortest line between these two points.

this line will cut the y-axis and the x-axis, such that we can actually have 3 segments; [(4,5):y-axis], [y-axis:x-axis] and [x-axis:(-8-4)].

by reflecting the [y-axis:x-axis] segment with respect to the y-axis and also doing a 180degree rotation on the [x-axis:(-8,-4)] segment with respect to the origin, you would now "bend" the line such that it would hit the y-axis, the x-axis and end up at (8,4).

draw it out, you'll see that you could also reverse the steps and see how some of the other proposed solutions would measure up to the optimal.

yekban81

20-09-2004, 11:18 PM

Got it liao. Smart approach.

chiunlin

20-09-2004, 11:29 PM

hehe, sorry just now for the overbrief solution, I was in a hurry to attend the class.

Engineer?
The word engineer has its roots in the Latin word ingeniare, which means to devise in the sense of construct, or craftsmanship. Several other words are related to ingeniare, including ingenuity.

chiunlin

21-09-2004, 10:15 AM

1) using purely vector techniques, find the dihedral angle of a
regular tetrahedron.
Vectors anyone? FYI, I take the idea of the second question from Paul Zeitz's "The Art and Craft of Problem Solving." I don't have the book with me now but that question should be one of the past AIME(American Invitational Mathematics Examination) questions.

digimushu

23-09-2004, 12:30 PM

Any knows about how the phrase 'debugging' the code comes about?

gonjeng

27-10-2004, 04:29 AM

aaa tidaaaaakkkkkkkk...

the post about the Grand Robo-Tapper is gone...

hehehe...

digimushu

27-10-2004, 08:31 AM

huarghhhhh
noooooooooooooooooooo....

yekban81

12-01-2005, 05:08 PM

Anyone dare to solve the integral of (1/In x) ?

gonjeng

17-02-2005, 03:43 PM

aiseh... what happen to this SIG? 2 months in malaysia without unlimited access to internet really suxx. anyway, how is everybody doing hah? digimushu, whats up??? senyap aje hehehe...

digimushu

17-02-2005, 08:03 PM

wassup dude?

nothing much, just busy with research and teaching this semester. lots of lots of stuff to do. what in the world were u doing in m'sia? dah abis?

gonjeng

18-02-2005, 11:19 AM

heh not much, same old me :)

i'm taking a semester break. saje je... also, i want to test the job market here before i actually done with school, make some connections, etc. right now, i'm working as a contract engineer in a consulting firm doing some cfd job :) and heh, will be back to US, insya-Allah, around end of may :) rase cam malas aje nak balik pulak, now that i am in my comfort zone hehehe...

Benster

19-03-2005, 06:33 PM

hey, i'm actually wondering which engineering course shud i apply 4. Anyone out there can help me a lil'?
1st of all, i would love to do aeronautical engineering but it seems tht there wouldn't be any job opportunities in malaysia.
2nd choice will be mechatronics. but the inti counselor said tht besides sony there wouldn't be any other job fields.
3rd, E n E engineering. Feel tht this course is so so common. Although the job fields are large but the it's a common course where a lot of students apply for n currently studying it. Which means it's very competitive n will naturally decrease it's demand.
lastly, mechanical engineering. another common course like E n E.
By the way, can someone elaborate the job scope of a chemical engineer?

mizi28

11-06-2007, 03:10 PM

i feel kind interested with telecommunication,marine and chemical engineering...

white2020

11-06-2007, 03:12 PM

Studying MBA after completing of Mechanical engineering degree, is it a wise choice?

Normally how many years do MBA take to complete?

jojyi

12-06-2007, 10:53 PM

Studying MBA after completing of Mechanical engineering degree, is it a wise choice?

Normally how many years do MBA take to complete?

What's MBA...mind explain?thanks... :wink:

sAmurAi-X

13-06-2007, 01:54 AM

Chemical engineering could cater to a few other specialisations when you upon graduation. IMO, chemical engineering does has a good career prospect in the future as its market demand is increasing tremendously recently. Somehow, my senior did enlighten me that chemical engineering is one of the toughest field among all engineering fields. Yet i still don't know whether it's true or merely hearsay?? :roll:

sAmurAi-X

13-06-2007, 01:58 AM

Studying MBA after completing of Mechanical engineering degree, is it a wise choice?

Normally how many years do MBA take to complete?

What's MBA...mind explain?thanks... :wink:

MBA = Master of Business Administration

for futher information, please refer to:

http://en.wikipedia.org/wiki/MBA

hopefully this will help to clarify your queries.. :wink:

white2020

13-06-2007, 12:10 PM

Studying MBA after completing of Mechanical engineering degree, is it a wise choice?

Normally how many years do MBA take to complete?

What's MBA...mind explain?thanks... :wink:

Master of Business Administration (MBA)

well, haha, it seems that I am late as our dear sAmurAi-X has explained it.

Anyway, let me elaborate more. Normally for enginnering students, other field I am not sure, they will take up MBA upon their graduation of degree of certain field in engineering, as their ambition is strive for the business world.

Engineering and business world are somehow related, well, to be exact, for every other field also, they are actually got some link to business field. We can skip the degree of the business(whatevver) and straight go for the Master in Business (MBA), as long as we possess the degree certificate.

well, rectify me if I am wrong in any infomation.

Thanks