quite easily done

Long time no post!

So, I've been [ busy | stressed | sleep deprived ] for the past few weeks. Especially during this past few days, 'coz I was preparing/procastinating for my Math 211 (Abstract Linear Algebra) Midterm. I'm an MS student so I still have to deal with exams and stuff :(.

As expected, the exam was nothing short of suicide due to my very crude understanding of the subject matter. In our defense (this includes Wilson Tan, Lew Tria and me) it's really really difficult. Much of the difficulty stems from the fact that we are trained as engineers and not as mathematicians. The approach to teaching and exams is indeed very different from what we've been used to in our undergrad. Very theoretical instead of practical. Theoretical and filled with rigorous (look ma no loopholes) kinds of proofs.

Majority of the questions in the exam were proving problems. They were really tricky ones too. I do hope we manage to scrape a passing grade for this exam so we don't have too much to worry about later.

Interesting Observations in Abstract Linear Algebra:

• In this subject, you get to prove the most basic things you know in math like:
• (-1)*(-1) = 1
• (0)*x = 0
• f(x) + f(y) = f(x + y)
• We're the only non-math majors in that subject so they can all relate to each other and make fun of the stupid engineers.
  
• In this subject you will learn that almost everything in undergraduate math can be explained in a single course (linear algeb)
• This subject reeks of rigorous proofs in every corner including defining things that seem obvious already. For example, after defining linearly independent, the teacher then defines linearly dependent as something that isn't linearly independent. :D
  

All in all, it's a very interesting subject and definitely a learning experience. I just hope it's not one of those things you learn by failing.

For the curious folks, here is the proof of (0)*x = 0:

1. 0*x = (0 + 0)*x -> existence of additive identity
2. 0*x = 0*x + 0*x -> distributivity
3. 0*x + -(0*x)= 0*x + 0*x + -(0*x) -> existence of additive inverse
4. 0 = 0*x -> definition of additive inverse

QED! (Amusingly my highschool teacher made me believe that this actually meant, Quite Easily Done)

QED = Quod Erat Demonstrandum

yunlang.