Although a2 reached deadline, the argument within our group members just started. We couldn't come to an agreement on where a comment should be provided. We have different opinions on what should be called 'obvious'. Apparently, that's really subjective.
For example,
x > y
Then 2x > 2y
Is it obvious? For me, it's too obvious.
Also, in terms of the proof structure. Such as
Assume x in N
Do we need a comment for that saying 'to introduce for all'? I don't think so since the intended audience should equip that kind of knowledge, otherwise, we should also comment on the indentation of the proof structure because not everyone would find it obvious.
My partner said he was copying the comments on the slides and prof wrote them for every proof. But I think that's because he was introducing proof to all new learners not experienced scholars and my TA didn't write structure comments in tutorial.
Yes, we might solve the conflict by adding comments on every single line. But that's really annoying and ruin the whole style and format (since sometimes we need to start a new line for comment) and also that's ineffective for both readers and writers.
That really bothers me.
Showing posts with label proof. Show all posts
Showing posts with label proof. Show all posts
Monday, 11 March 2013
Tuesday, 5 March 2013
Last Question of Assignment 2
For the last problem of A2, we have to prove if m >= n, then gcd(m,n) = gcd(n, m-n). I got stuck at first until being inspired by the project of CSC148.
For CSC148, we wrote bunch of methods such as __less_equal__(self, s2), and __equal__(self, s2). For __equal__, the comment says we can just use __less_equal__ by saying (self <= s2 and s2 <= self). That's smart and tricky then I realized a = b is equivalent to a<= b and b <= a.
Based on what I found above, instead of proving the equation directly, which I found impossible, I can prove it indirectly by proving gcd(m,n) >= gcd(n, m-n) and gcd(m,n) <= gcd(n, m-n).
Here are the steps. Hope that make sense.
For CSC148, we wrote bunch of methods such as __less_equal__(self, s2), and __equal__(self, s2). For __equal__, the comment says we can just use __less_equal__ by saying (self <= s2 and s2 <= self). That's smart and tricky then I realized a = b is equivalent to a<= b and b <= a.
Based on what I found above, instead of proving the equation directly, which I found impossible, I can prove it indirectly by proving gcd(m,n) >= gcd(n, m-n) and gcd(m,n) <= gcd(n, m-n).
Here are the steps. Hope that make sense.
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