Incoming Finals and Assignments

Week 11

Sunday

I was reviewing the previous tutorial problem sets, and I reversed one of them (assume - implement) and disproved the statement.

The original one is proof 5n^4 - 3n^2 + 1∈ O (6n^5 - 4n^3 + 2n)

Therefore, the reverse one is 6n^5 - 4n^3 + 2n ∈/ 5n^4 - 3n^2 + 1

To disproof the statement

∃ c' ∈ R, ∃ B ∈ N, ∀ n ∈ N, n >=  B => 6n^5 - 4n^3 + 2n =< c (5n^4 - 3n^2 + 1)

∀ c' ∈ R, ∀ B ∈ N, ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)

assume c' ∈ R
   assume B' ∈ N
       let n' = 6, then n ∈ N
       assume n >= B
          then 5n^4 - 3n^2 + 1 < 5n^4 + 1
                                            < 5n^4 + n^4
                                            =< 6n^4
                                            =< n^5
                                            =< 2n^5
                                            =< 6n^5 - 4n^5
                                            =< 6n^5 - 4n^3
                                            =< 6n^5 - 4n^3 + 2n
          then 6n^5 - 4n^3 + 2n > (5n^4 - 3n^2 + 1) 
       then ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)
   then ∀ B ∈ N, ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)
then ∀ c' ∈ R, ∀ B ∈ N, ∃ n ∈ N, n >=  B ^ 6n^5 - 4n^3 + 2n > c (5n^4 - 3n^2 + 1)          

So, there is less than a week left for all the lectures. I will sure work slight harder since it's almost the finals. Beside that, I still got the CSC165 assignment 3 due on next Wednesday, CSC148 project II due on Friday. Well, gotta continue get my hands working on them now. MARCH madness is not over yet!!!

Good luck everyone!

Thanks for reading!

Roland Qiyang Liu

The End of Midterms

Week 10

Wednesday

for last week's CSC165 lecture and tutorial,I did not really follow along well. I was still confused while I was trying to remember all the definitions a.nd rules. Substution is also a big part what I am trying to figure out because I was having a hard time to decide when to use a substation or just assume and assign numbers to unknown values.

I did actually pretty well on the second midterm. Although I haven't took my test paper back yet, I checked online and the result was I got perfect for the first one, lost one mark on the last one. Unfortunatly, I completly made a wrong decision on the second question. I knew to disprove the statement, but just like the first midterm, I once again failed on negation: I was not sure about if I should have changed 'equal' to 'not equal' in the implenment part. For that, although I proved part of the negated statement, I did not disprove all of the original statment. Therefore, I guess I need to find more pratices to fix this problem

Thanks for reading!

Roland Qiyang Liu

Double Trouble

Week 9

Sunday

For this week, we had the second 165 midterm test and I also had the 148 term test as well.

There is one question I wasn't sure if I i did correct or not.

For the last question, my solution is that

Assume x ∈ R

             then [x] is the largest integer =< x

             so [x] =< x

             since x + 7 > x + 6

             then [x] + 1 + 6 > x + 6

             then [x] + 1 > x

then ∀ x ∈ R, [x] + 7 > x + 6

I compared this to the sample solution and I found out that mine is really close to it, but not exactly the same. 

And I also saw the solution for question 1 which has the same idea with mine, just not the style. I bet I am gonna lose marks on that. This is really tough since there are so many places to just make a little mistake and lose marks for it.

Thanks you for reading!

Roland Qiyang Liu

Midterms Round 2

Week 8

Sunday

There will be two midterms for me next week: CSC165 and CSC148. And I was too busy to wrote any blog for the past two weeks due to the amount of assignments I had. But in this blog, I want to take chance to talk about Assignment 2 which I had last week.
Overall, the assignment was not that hard and it took me about 2.5 hours to finished it and checked for a second time. However, for question 5 in the assignment:

4. Prove or disprove: for all quadruples of positive real number w, x, y, z, if w / x < y / z, then:

( w / x < ( w + y ) / ( x + z ) ) ^ ( ( w + y ) / ( x + z ) < y / z )

A: Prove/ true

assume ∀ w, x, y, z ∈ R, w > 0, x > 0, y > 0, z > 0

and w / x < y / z

=> ( w / x < ( w + y ) / ( x + z ) ) ^ ( ( w + y ) / ( x + z ) < y / z )

assume ( w / x < ( w + y ) / ( x + z ) ) 

=> ( ( w / x ) * ( x ( x + z ) ) ) < ( ( w + y ) / ( x + z ) * ( x ( x + z ) )

then    = ( x + 2 ) w < x ( w + y )

             = x w + z w < x w + x y

             = z w < x y

             = z w / x < y

             = w / x < y / z

therefore, it is true

assume ( ( w + y ) / ( x + z ) < y / z )

=> ( ( w + y ) / ( x + z ) * z ( x + z ) ) < ( y / z * ( z ( x + z ) ) )

then    = ( w + y ) z < y ( x + z )

             = w z + z y < y x + z y

             = w z < y x

             = w / x < y /z

therefore, it is true

therefore, ∀ w, x, y, z ∈ R, w > 0, x > 0, y > 0, z > 0, w / x < y / z =>

                   ( w / x < ( w + y ) / ( x + z ) ) ^ ( ( w + y ) / ( x + z ) < y / z )

therefore, it is true

The above was my solution. I also checked the sample solution. But it is much different from mine: I separate the two parts and then solved each one of them. And it also seems that I have so much more steps than the sample solution because I thought just in case that I would lose marks for not having enough steps and explain my answer clearly.

Therefore, my question is that does my solution to the question look right? And I also have noticed that my writing style in the problem-solving is not exactly like the sample solutions nor notes, does it matter a whole lot and might causing me to lose marks in an exam?

Thank you very much for your time!

Roland Qiyang Liu