"Slog", "Blog", such ugly guttural words. Anyways.
I'm glad to be back. I've been looking forward to CSC 236, because I so enjoyed 165 last year. It's strange - I did pretty awfully in MAT137 last year, but when I'm doing logic and proofs I just feel right at home. Maybe it's because every math class I've taken in my life has focused on "calculations of the grocery-bill variety" (a phrase cribbed from Uncle Petros and Goldbach's Conjecture which I reread over the summer), whereas 165 and 236 have felt like they required a lot more independent thought, and not just memorization and repeated application of formulas and procedures. For that reason, I found solving proofs in 165, and now 236, a lot more satisfying. But maybe that's just me explaining away my incompetence at calculus.
I feel like I've done pretty well with the material during these first two weeks. It's taken me a while to warm up to induction since last year, but at this point I feel quite comfortable with it. The explanation the book gives in section 1.2, reconciling the principles of simple and complete induction with induction as proof technique by expressing the natural numbers for which a predicate is true as a subset of the natural numbers, and then using the principle of induction to show that that set is a superset of the natural numbers (and hence, exactly equal to them).
My only difficulty in solving the first problem set came in the second question, trying to decide the degree of rigour with which I would need to justify a certain statement. Specifically, I had partitioned the set of pairs for {0, 1... n+1}, Dn+1 = {{1, 2}, {1, 3}, {2,3}... {n, n+1}} into Y: {s \in Dn+1, n+1 \in s} and N: {s \in Dn+1, n+1 ~\in s}, and was trying to prove that the cardinality of Y was equal to n, which seemed obvious to me, but somewhat messy to prove. My solution was just to add an appendix to the end of the proof, with what ended up being about a page of writing, elucidating the reasoning behind those two lines, with a note to the marker that they could refer to it if they found the lines alone insufficient.
Wasn't that an interesting story? There are more where that came from. Stay tuned for next week.
1 comment:
The gold standard for showing cardinalities are equal is to exhibit a bijection. But, that felt like overkill in this case.
I find expressing a proof much harder than finding a solution I believe. My ideal is few or no symbols (other than the Latin alphabet and punctuation), but I rarely get there.
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