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10/27/2006 Set of Rational RepresentativesStill on a set of real numbers, define an equivalent relation between 2 real numbers a and b: a is equivalent to b if and only if a-b is a rational number, written as a~b. such relation is an equivalent relation. Proof. a-a=0, rational, so a~a; if a~b, then a-b=q for some q in Q, thus b-a=-q, which is still a rational, so b~a; if a~b, b~c, that is a-b=q1, and b-c=q2 for some q1 , q2 in Q, then a-c=a-b + b-c = q1+q2 still rational, so a~c. An equivalent relation defined on a set, will have equivalen classes that partition the designated set. Let Ka be such a set that for all x in R: x in Ka if a-x is rational. In other words, Ka is one equivalent class of all the real numbers that are equivalent to a. Thus for Ka, Kb, two equivalent classes, either Ka ^ kb = O, or Ka = Kb, (proof left to the reader). Ok, now here comes our funny set K of all equvivalent classes. Pick any one number from the Ka's, then for all the Ka's, whose union is R, K is formed. Is K Countable? definitely not, although all the rationals collapse into one point, there are still uncountable many irrationals. Is K of measure zero? nah, consider the r_n in Q, then the union of r_n + K convers the whole real number, the union of null sets is still a null set (set of measure zero), and the measure of K is the same as the measure of r_n + K, so if K is a null set, respectively one can have R as a null set, however R is non measurable; henceforce, contradiction. So K is uncountable and non measurable... think of K as points in R collapsing down to certain points and yet remains certain properties of R, such as uncountable and non measurable... such K came arosed because of the discussion of the degree of infinity. It's known (well to math students) that, there are infinitely many degress of infinity, starting from the first one being countable and up. so there are finite (including the empty set), and then countable (N, Z or Q), then R (the first uncountable or the set of all irrationals), and then... more to the end of our universe, (cardinality is one important thing when dealing with infinity of a set. two sets have the same cardinality, if there exists a bijection function from one to the other). so, countable union, countable Cartesian products both preserve the countability of sets. and a set always has less cardinality then its power set ( a power set of some set A is defined as a set of all subsets of A), also a set is infinite if and only if a proper subset of it has the same cardinality as the superset. so questions regarding this infinity degree thing is whether the power set of Q has the same cardinality of R? is there any other uncoutanble subset or R that's "really" smaller than R and "bigger" than Q, as far as cardinality is concerned? I think the power set of Q does have the same cardinality of R (proved somewhere), so the question becomes if there's any set inbetween an infinite set and its power set while having different cardinality? I'm not concentrating in number theory.. but that kinda thing is really fun... side notes, Georg Ferdinand Ludwig Philipp Cantor , the creator of Cantor Set, suffered great depression due to his too deep research into Cantor set and related studies (all those infinity stuffs, he proved and axoimitized.) Brief introduction about Cantor set, which is a set of all the ternary expression left out those with the 1's in any position of their ternary expression. Cantor set is a set of uncountablly many points and yet ..with measure zero, Cantor set is also used to contrsuct a continuous function (think of it as a smooth function, changing gradually) monotonically increasing from 0 to 1 and yet no where differentiable (differential only on a null set). Let C be the set of all continuous functions on [0, 1], the the set of all differential functions on [0, 1] is a proper subeset of C and nowhere dense in C (that's really small). the set of irrationals is dense in R, btw.
now here comes my question: pi, the ratio of the circumference to the diameter of a circle;e, the base of the natural system of logarithms; can pi and e appear in the set K at the same time? K is a set of representatives from Ka's, as defined above; so for K_pi, one can pick either pi, or pi+1. for K_e, one can also pick e or e + 1/1000000. This is equivalent of asking if K_pi = K_e; or if pi - e is a rational? Derived from calculus, e can be described as a limit of a series of 1/n! (n factorial). I forgot the Taylor expansion of pi...so probably some conclusion can be made if one takes the difference of these two numbers as difference of serieses (since two serieses converge)
Luckily.. I don't think I will go kill myself when digging too deep into those number theories >.< 10/23/2006 A Small Test from Yimin ChenWell, I just stole this small personality test from Yimin's blog, I personally feel it's pretty accurate, since... well I will tell you which ones are for which ones when you post your answer as replies.. Let's start the test. I will leave enough space so that you won't go through it too fast. Please answer with what comes first into your mind, don't spend too much time thinking about what exactly fits the questionaries...it's just this immediate feeling that counts..
Pick one animal you like.
Use 3 adjectives to describe it.
Pick one color you like.
Use 3 adjectives to describe it.
Pick one environment you like (any type of Earth surface).
Use 3 adjectives to describe it.
Imagine you're stuck in a completely closed space, there's a door in front of you. you walk up to it and open it, suddenly, nothing can be seen, there's only whiteness ahead, what's your feeling under such circumstance?
So have fun thinking about it thorougly..
Here are my answers to the above 7 questions:
1. Cat
2. Free, Independent, Cute
3. Blue
4. Celestial, oceanic, cold
5. Plain
6. Live, capacious, fresh
7. Divine, absolutely sacred.
10/17/2006 Limit of Integration = Integration of Limit?for a sequence of continuous functions f_n (x) on a closed interval [a, b], converges to another continuous function f(x) on [a, b] ([a, b] is the domain for all); in other words lim f_n(x) = f(x) as n goes to infinity. Now here comes the question will the lim of the integration of f_n(x) over the interval [a, b] equal the integration of f(x) on the same interval? the convergence here is only pointwise; if it was uniform convergence, then the answer is obviously yes. If it was a pointwise convergence of a sequence of continuous function to a limit function, then the answer is not necessarily true. Since the limit function is continuous, and all of the domains are the same closed interval [a, b], the continuinty becomes uniform continuity, would this integration of (f_n(x) - f(x)) approach zero when n gets to infinity instead? It's hard to prove, and hard to come up with counterexamples...Consider the function F(x) = f_n(x) - f(x) is continuous, composition of two continuous functions, and defined on a closed inteval, so it's uniformly continuous. Let x in [a, b] be given, then given e > 0, there exists a q such at for any y when |x-y| < q, |f_n(x) - f(x)| < |f_n(x) - f_n(y)| + |f_n(y) - f(y)| + |f(y) - f(y)| < e/3 + e/3 + e/3 < e. will this f_n(y) - f(y) get too big in a small q-neighborhood of x? since, the convergence is pointwise, it might happen....(finite of infinite = infinite of finite, that's easy to prove as long as linearity exists; however infinite of infinite = infinite of infinite? no fixed answers). After a brief discussion with my professor, actually the head of our department, we came to realize it's actually better to consider the following hypothesis: let {f_n(x)}be a sequence of continuous function on [0, 1] converging pointwisely to a constant function f(x) = 0. would the limit of the integration of f_n(x) over [0, 1] approach to zero when n goes to infinity? the answer is yes, since a continuous function on a closed interval (which is also bounded and campact) is uniform continuous and bounded as well, the upper and lower bounds are both approaching to zero, so the integration of f_n(x), which is trapped between the m_n(the lower bound of f_n(x)) and M_n(the upper bound of f_n(x)), thus the limit of the integration is zero, so problem solved :)
Ok, enough math. gotta try switching motherboard and see if my old harddrive is really broken. then it's time for me to write up a resume... for jobs... >.< enough sitting back and doing nothing these days.. gotta get some extra cash for computers >.< school's going well, as long as I ask questions.....such as the one above... trying my best not to ditch any class so far...boring but meaningful life... I think :) let's hope tomorrow I won't fail myself again hehe
10/7/2006 Emergency IM SwitchSpent like 5 hours finally got this computer back to life... however, I just can't log on to msn messenger ..hehe stupid IE offline setting. anyway.. gonna use yahoo messenger for a while when I'm trying to figure out how to fix this MSN thing.... so if you still wanna get in touch with me, please add my yahoo messenger handler is : mingrzhong@yahoo.com... add that to your windows live messenger, so that we can still talk; though it's through two different types of messengers, it's possible......thanks for your understanding and cooperation
P.S. : with a cheap 160 GB SATA2 samsung hard drive... I finally got everything worked out...haha...now gonna test if I can return the new motherboard... in that case.. the only bad apple part in my computer is my beloved hard drive ... /sigh
I can use msn again :p 10/6/2006 Computers ...Got a desktop built up, however I was having this small trouble of having the harddrive boot the system up all the time. Similar problem has also occured while I was trying to reinstall an OS on my laptop.... I'm suspecting the bios on both motherboards are crippled, since the harddrves are detectable on the first few boot up screens and them, the bios just couldn't hand over the OS to windows.... gonna get a cheap motherboard tomorrow and see if I can get things working again.... /pray...
BTW, Happy Mid Autumn Day, how I miss the ice cream Mooncake back in China >.<..../drooool |
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