For the second lecture, we talked about sets and had a little exercise about the relationships between two sets. The "any/not any" and "all/not all" terms sort of confused me, but after breaking down each return statement, the relationships made a lot more sense. The third lecture, on the other hand, stumped me quite a bit. I did not really understand the Venn diagrams and the placements for the 'X' and check mark. I expect get a better understanding of this before the next lecture and tutorial. Besides all of that, the quantifiers section was pretty basic.
I heard many interesting and delightful things about this course - mainly about the way it improves your critical thinking capabilities, but also that this course is pretty challenging. I hope that I find this course straightforward, but still intriguing and thought-provoking. Wish me luck!
We also did a weird, but interesting, problem in class:
A: haven’t seen you in a long time! How old are your three kids now?
B: The product of their ages is 36.
A: That doesn’t really answer my question…
B: Well, the sum of their ages (in years) is —[at this point a fire engine goes by and obscures the rest of the answer.]
A: That still doesn’t really tell me how old they are.
B: Well, the eldest plays piano.
A: Okay, I see, so their ages are—[at this point you have to get off, and you miss the answer]
(Using Polya's approach, as suggest in class)
Understand the Problem:
- B has 3 kids whose ages, when multiplied, is 36
- B has an eldest child, so one of the ages is greater than the others
- since A was not sure about the ages of each child by the sum, there must be set of numbers whose product equals 36 and have the same sum
Devise a Plan:
- list all possible combinations of numbers whose product is 36
- with that list, find the sum of each set
Carry Out:
Child 1 | Child 2 | Child 3| Sum
1 | 1 | 36 | 38
1 | 2 | 18 | 21
1 | 3 | 12 | 16
1 | 4 | 9 | 14
1 | 6 | 6 | 13
2 | 2 | 9 | 13
2 | 3 | 3 | 8
3 | 3 | 4 | 10
Looking Back:
- by realizing that A could figure out the ages by the sum, sets of numbers must have the same sum, in this case, (1, 6, 6) and (2, 2, 9)
- although that these sets have the same sum, the only possible answer is 2, 2, 9 since one of the ages must be the largest in order to satisfy the "eldest" characeristic
Therefore the ages of the children must be 2, 2, and 9.
Commented on:
http://celinasopiniononcsc165.blogspot.ca/2014/09/my-struggle-on-week-1-materials.html#comment-form
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