Week #4

Week #4

#10

Transitivity
  forall x in X, (P(x)=>Q(x)) and (Q(x)=>R(x)) => (P(x)=>R(x))
  Proof:
  1) P=>Q means, as sets, P in Q.
     Q=>R means, as sets, Q in R.
     so, P in R, i.e. p=>R

  2) By Venn diagram.

  3) neg[(P=>Q) and (Q=>R) => (P=>R)]
  <=>((P=>Q) and (Q=>R) and neg(P=>R)
  <=>(¬P V Q) and (¬Q V R) and (P and ¬R)
  <=>(¬P and ¬Q) V ( Q and ¬Q) V [(¬P and R) V (Q and R)] and P and ¬R
  <=>[(¬P and ¬Q) V (¬P and R V Q and R)] and P and ¬R
  <=>(¬P and P and ¬Q and R) V (¬P and P and R and ¬R) V (Q and R and P and ¬R)
             False                     False                    False

For all, one, ... v.s. One, for all, ...
  ∀x∈S1, Ǝy∈S2, x+y=5 (1)
  Ǝy∈S2, ∀x∈S1, x+y=5 (2)
  S1=S2={1,2,3,4}

  (1) >>> all({any({x+y==5 for y in S2}) for x in S1})
      True
  (2) >>> any({all({x+y==5 for y in S2}) for x in S1})
      False

Pic Here!

#11

  ∀e∈R+, Ǝd∈R+, ∀e∈R, |x-0.6|<d => |x2-0.36|<e
  lim x2 = infinity
  x->infinity
  ∀e∈R+, Ǝd∈R+, ∀e∈R, x>d => x2>e
  Switch ∀,Ǝ do not preserve truth.

  ∀m∈N, ∀n∈N, mn∈N
  It changed nothing when switching ∀ pairs


Proof
  Argument to convince audience something is True/False.
  1) Think about why you believe the statement is true.
  2) Write up your reasoning in a clear way.

#12

Thoughts:
Assignment #1

I'm not doing very well at the first assignment. I messed up using word converting doc file into pdf. And I also didn't get the format of the last question right, so I lost all 40% of the last question. However as a friend of mine saying, what I have to do is to move forward and try my best on my following assignments and tests. The worst assignment is only weighting 6% or less in the total course. So I will learn how to use LaTeX and work hard in the future.