week 6 -- solving negation problem analysis

This week we get out first midterm back, overall, I did OK on this test since I got 85%. But I found myself keep having the same problem on how to negate an implication without hesitation.

For an implication like:
                       ∀a∈ N, P(a) ⇒ Q(a)

P(a) ⇒ Q(a) equals:
                      ∀a∈ N, ¬P(a) ∨ Q(a)

hence the negation:

                       ∃ a∈ N, P(a) ∧ ¬Q(a)


Let's try another way(this is a tricky False statement):

the only false of it :
                       ∃ a∈ N, P(a) ⇒ ¬Q(a)
P(a) ⇒ ¬Q(a) equals:
                       ∃ a∈ N,  ¬P(a) ∨ ¬Q(a)

Let's use "negation" of implication:(False)
so the negation of ∀a∈ N, P(a) ⇒ Q(a)
                       is ∃a∈ N, ¬P(a) ⇒ ¬Q(a)
equals:
                   ∃a∈ N, P(a) ∨ ¬Q(a)

week 5 proof outline

proof outline:

1. for all x in R, P --> Q:

The method we learned in the lecture is trying to prove a statement from base case or in other words, using P to get to Q by transforming the equation or inequation. (it's like the firse step of mathmatic indaction)

2. there exist a x in R, P --> Q:

Just find a example.

3. if the sttement is hard to prove:

try to prove the contropositive.

week 4 review Symbolic Logic

1. the sentence P --> Q:

    (a) If P, then Q.

    (b)P implies Q.

    (c) P only if Q.

    (d) It is necessary that Q for P.

    (e) It is sufficient that P for Q.

2. the sentence: All x in R, P --> Q:

    (a) equavelent to : All x in R, not P or Q

    (b) negation: Exist x in R,  P and not Q

    (c)convers: All x in R, not P --> not Q
                       All x in R, Q --> P

    (d)contropositive: All x in R, not Q --> not P