The first and the third questions seemed okay. They were straightforward direct proofs which I luckily navigated well and I think I found my way to the correct algebra and methods. The second question, on the other hand, gave me much more trouble. I think I might have panicked over it, and the whole process of proving it, which I completed, felt very loose and messy. I had to cross out many things.
Looking back on it, the second statement seems fairly intuitive. Basically, it states that if you know w is greater than some number d, then you know that floor(w) is greater than some number e. This makes intuitive sense, as the size of w and floor(w) are of course related in a very predictable way. I believe what I ended up doing is using d = e+1, though I cannot quite remember. It was something like this, anyway. This seemed obvious, but it led to me having also to show that floor(e+1) > e, which, though simple, was enough of a distraction that I think it might have effected my test.
I will re-examine this problem, then, for my SLOG problem solving post which is required. Of course, it is not mind bending, per se; rather, it is a simple problem and the real trick is being able to proof it clearly and precisely.
1. Understand the problem
This step is not difficult. I have to prove that floor(e+1) > e, which means that the largest possible integer which is less than or equal to (e+1) is itself greater than e.
It intuitively makes sense. if e is not an integer, then floor(e) will be the integer just to the left of e on the number line. Floor(e+1) will be the integer to the right of floor(e) on the number line, so e itself will sit between them and we can picture pretty clearly that it is less than floor(e+1).
2. Devising a plan
To show this, I will use the same definition of floor(x) that we used on the test and the assignment:
floor(x) [belongs to] Z ^ floor(x) [less than or equal to] x ^ ( [for all] z [belongs to] Z, z [less than or equal to] x [implies] z [less than or equal to] floor(x) )
Now I have to lay out the foundations of my proof, which will be:
- floor(e+1) is an integer
- floor(e) is an integer.
- floor(e) is less than or equal to e.
- floor(e+1) is less than or equal to e+1.
- e+1 is greater than e.
Now, my first thoughts:
- in CASE 1, e is an integer. In that case, it is simple to say that since e is an integer and it is less than e+1, by the definition of floor(x), e is an integer and e is less than or equal to e+1 so e is less than or equal to floor(e+1). e+1 is an integer if e is an integer, so floor(e+1) is equal to e+1. Then since e is less than or equal to floor(e+1) and floor(e+1) = e+1, we know that e is less than, rather than equal to, floor(e+1) because obviously e is not equal to e+1.
- in CASE 2, e is not an integer. In this case, it is the same logic but there are more steps applied which makes it a bit more complicated. e is less than e+1, and there then exists some integer between e and e+1; we will call it k. It is safe to say that k is equal to floor(e+1), because k is an integer and it is less than e+1 but no other integer less than e+1 is greater than k. Then we see that floor(e+1) is less than e+1 and greater than e.
3. Carrying out the plan
First, the formal statement: For all numbers e belonging to the naturals, floor(e+1) > e.
Assume e belong to the natural numbers. #assume domain
Case 1: e is an integer.
Then floor(e) [belongs to] Z ^ floor(e) [less than or equal to] e ^ ( [for all] z [belongs to] Z, z [less than or equal to] e [implies] e [less than or equal to] floor(e) ) #definition of floor(e)
Then since e belongs to the integers and is less than or equal to z, by the definition of floor(e), this implies that e is less than or equal to floor(e).
This in turn implies that floor(e) = e.
If e is an integer, then e + 1 is an integer.
The same logic can be used to arrive at the conclusion that floor(e+1) = e + 1 #since an integer plus an integer is equal to an integer.
Taking this all into account, we have that e is an integer so e + 1 is an integer, and floor e is
equal to e and floor(e+1) = e+1. Therefore: floor(e+1) = e + 1 > e = floor(e) so floor(e+1) > e.
Case 2: e is not an integer.
Then e +1 is not an integer,
Then there exists a k between e+1 and e which is an integer.
Then we have the inequality e+1 > k > e.
Then k [belongs to] Z ^ k [less than or equal to] e+1 ^ ( [for all] z [belongs to] Z, z [less than or equal to] e+1 [implies] e [less than or equal to] floor(e+1) )
So, taking this definition of floor(e), we can say that k is an integer and it is less than k.
Then because k is less than e+1, by the implication, we can say that it is less than or equal to floor(e+1). Because k is the integer between e and e+1, then, we can say that k = floor(e+1) #since it is between e and e+1 and it is less than or equal to e+1.
Then since e+1 > k > e and k = floor(e+1), we have floor(e+1) > e
Then for all e in the integers, floor(e+1) > e
4. Looking Back
The proof seems to hold up, reviewing it after a few minutes. Of course, it is not a complex problem, but sometimes the simplest ones are hard to prove, because, in theory, they "go without saying". But that is what proof is all about; nothing goes without saying. My proof, then, was quite wordy and I think it partially has to do with typing/word processing. It is a bad process to type out mathematical concepts, so that led me to typing it in words, which comes with its own problems. Overall, though, it seems okay. With another definition of floor(e), I definitely can see how I might go about proving this again.
Looking over some other slogs, I see that people had similar trouble with the second question. Here's one that I found : http://gonzzang.blogspot.ca/ (look at the post for week 9) and it says, just like me, that the poster found the second question to be tricky. That is somewhat comforting. It seems like mixing multiple concepts--like floors and limits--is tough, at least when it is given to us with such a time constraint.
No comments:
Post a Comment