So, now we have only 2 last pieces of work to go. One problem set and one assignment. Semester is going to end very soon. Actually, the time passed so fast...
"Time is a great teacher, but unfortunately it kills all its pupils." Louis Hector
This block is dedicated to CSC236 course, which I am taking this year. I will post here my thoughts about the course material and problems.
So, now we have only 2 last pieces of work to go. One problem set and one assignment. Semester is going to end very soon. Actually, the time passed so fast...
"Time is a great teacher, but unfortunately it kills all its pupils." Louis Hector
You might consider this post kind of irrelevant, since it is not related to the course material directly. I accidentally happened to know that the Well-ordering Principle and Well-ordering theorem are different things, even though they are often taken to be synonymous.
The Well-ordering Principle states that that every set of natural numbers has its smallest element, while the Well-ordering Theorem states that every set can be well-ordered.
Well, to begin with, problem set # 1 was easy... I thought that I would need to use one of two famous techniques: wishful thinking or thoughtful wishing. However, it turned out that eaxmples we did in class were in many ways similar to problems included in the problem set. Anyway, since deadline passed long time ago I want to publish my solutions in case if you want to go through.
Q#1
No need to go into details. After going through several cases for n=1, 2, 3, 4 I found that there is a pattern. Each time when we multiply a number by 4, we multiply each digit of that number. For example, if we multiply 45 by 4 we know that the last digit of resulting number must be 0, since 5*4=20. Next step is to write out a proof based on simple induction. So, piece of cake, I would say.
Q#2
Again, I did some scratch work, so see for yourselves:
case1: n=1, {1}, there are 0 pairs
case2: n=2, {1, 2}, there is one pair, 2*(2-1)/2=1
case3: n=3, {1, 2, 3}, possible pairs {1, 2}, {2, 3}, {3, 1}, 3*(3-1)/2=3 pairs
And again, pattern! The great news, sounds like music to my ears, meaning that scratch work had not been done vainly! Any set {1, 2, ..., n-1, n} contains all pairs of different numbers from set {1, 2, ..., n-1}. The rest of the pairs of different numbers are formed just by taking each element of {1, 2, ..., n-1} and adding to it the last element of {1, 2, ..., n-1, n}. So each time when n increases by one, the number of pairs of different numbers increases by n.
n*(n-1)/2+n=n((n-1)/n+1)=n(n+1)/2, simple algebra and simple induction.
Christopher Reeve
Once I asked Prof. Heap the reason why we often try to solve problems which seem to have obvious solutions. His respond is worthy of quoting, I think.
"It's not the result that matters, but the path that you use to solve the problem. It's only path that makes you better... "