Am I stressed?

Notice that Alex (a lame friend of mine) wrote something like ... s-t-r-e-s-s-e-d ... Observe that this is the inverse of d-e-s-s-e-r-t-s...

It is very obvious that one is bad and one is sweet, or I termed it good... It seem to be that there is a clearly a bijective correspondence from {good,bad} to {stressed, desserts} via the function f: feelings ->metaphors. Please do not take this mathematically seriously. It is just a bored action of mine to pass time.

On a side note, I am left with one final exam here to do before going back to Singapore. I am rather feeling bad... not sure if its stressed...

Sorry for this post... that is of not much value.

Longing to go home

It certainly seems that there are a lot of criticisms going on regarding the content of my post. :) Ok, I shall not post the proof of the last bit of the theorem then. I shall leave it as a puzzle. I just finished my number theory exam yesterday. I think it should be all right. Anyway, I really lack the momentum to study for the final exam because of the following reasons (not in order of preference):

1) I want to go home
2) My stiff neck and shoulder aches are bothering me quite a lot and I am worried.
3) I think MVA is not going to be as smooth as number theory
4) I miss my gf
5) I miss my family

I hope my upcoming Norway trip will be a breather for me! I hope to get some fresh air and nice scenery. And also, I really quite worried about my chronic shoulder ache... I will see a doctor as soon as I reach Singapore. Can't wait for 14 June to come and I will be home... With that, I want to end with part of the lyrics from a well known National Day song.

Singapore, my homeland, this is where i belong...

Proof of theorem 65 in lecture notes

I shall state the full glory of Theorem 65 here:

Suppose that d is a positive integer that is not a perfect square, and the continued fraction expansion of sqrt(d) has convergents p_n/q_n and is ultimately periodic with period m. Then the only positive solutions (x,y) are given by (x,y)=(p_n,q_n) where n = lm - 1 where l is some interger corresponding to n being odd.

Proof:

Consider the equation x^2 - d*y^2 = 1. This can be transformed to (x+sqrt(d)*y)*(x-sqrt(d)*y)=1 => (x-sqrt(d)*y)=1/[(x+sqrt(d)*y)=1] =>x>sqrt(d)*y.
Whence, this yields 0
Thus we deduce that |sqrt(d) - (x/y)|<1/2*y^2.

We claim that if $theta in R$ and x/y is a rational number with (x.y)=1 satisfying |$theta$ - (x/y)|<1/2*y^2. then x/y is necessary a convergent to the continued fraction expansion of $theta$.

consider the equations

u*p_n + v*p_(n+1) = x
u*q_n + v*q_(n+1) = y

This set of equations do have an integral solution for (u,v) since
p_n*q_(n+1) + q_n*p_(n+1) = (-1)^(n+1).

If u or v =0, we got (x/y) = (p_(n+1)/q_(n+1)) and the coprimailty of x and y yields that v = $+ or -$ 1. Then the claim holds.

So if u and v are both not equal to 0 and that q_n -> infinity as n-> infinity, so we can choose an n such that q_nBut since u*q_n + v*q_(n+1) = y, u and v are of opposite signs and thus

|y*$theta$-x|=|u(q_n*$theta$-p_n) + v(q_(n+1)*$theta$-p_(n+1)| > |(q_n*$theta$-p_n)|.

Since |$theta$ - (x/y)|<1/2*y^2, it follows that |(x/y)-(p_n/q_n)|
This proves our claim.

Now let p_n/q_n be the convergent to sqrt(d).
Then we got the following:

sqrt(d) = (p_n*$theta$_(n+1)+p_(n-1))/(q_n*$theta$_(n+1)+q_(n-1))

whence,
q_n*sqrt(d) = (-1)^n/(q_n*$theta$_(n+1)+q_(n-1))

This leaves that n must be odd in order to have that Pell's equation.
(Check if n is even then there will be a contradiction)

Continue the proof some other day...

Feeling lazy...

Today I had my last lecture for number theory. But after that I feel so lazy. I am not sure what is the reason! Yesterday, I took some photos. However, my hands are shaky so the photos turned out to be very ugly. Nevertheless, I will post some of them here for criticising! (especially by ALEX).

Okay, thats about for photos. I should talk something about number theory today. In all, for this course, I have 65 theorems, lemmas, corollary in total. So obviously, to end the course, it will defintely be a theorem!!! So whats so 'big' about theorem 65? Basically, we learnt continued fractions and Pell's equation. Theorem 65 is the link between them. Consider the equation x^2 - dy^2 = 1, where d is a non square integer. Basically, this theorem says that if p_n and q_n are the convergents of the sqrt(d) then (x,y)=(p_n,q_n) are the solutions of the equation given certain conditions. First, n=lm-1 where l is a natural number and m is the period of the continued fraction of sqrt(d) and n must be an odd number. Seems complicated? Defintely not. Its fun! If you are feeling bored or nothing to do... maybe this can replace your sudoku puzzle when its so common nowadays. I am assure you that its more enjoyable than sudoku. To give yourself the fullest satisfaction, I will recommend you to prove this statement.

Anyway I got this friend who says that 'Life is short, make a fool of it while you can'. I will rephrase it... ' Life is short, make yourself and your family happy while you can and learn more mathematics too'!