Reflection of CSC165 Lec. 6 Oct. 14th

Reflection of CSC165 Lec. 6 Oct. 14th

Today's lecture lasted two hours, so the content would be a little bit shorter, again.

6.1  proof about non-boolean functions

Non-boolean functions are another type of function. Compare to Boolean, the result is not simply True or False. Thus, we cannot make a judge (of True of False) by looking at a non-boolean function. 

The most important non-boolean function introduced in class is called "floor". ⌊X⌋  means the largest integer ≤ x.

Although ⌊X⌋  and x looks quiet similar, but they have different property. X is just a variable,  ⌊X⌋  is a function, which in the same as f(x). Thus, we cannot apply ∀⌊X⌋∈R (a quantifier) to a function.

6.2 prove something false disprove something

To prove a statement S equals to proving ¬S.

6.3 proof about limits

Let's try to solve a typical question of proof about limits which mentioned in class.

∀ϵ>0, ∃δ>0, ∀x∈R, |x−2|<δ→|x^2−4|<ϵ

Here the thinking is, since ϵ could be every value > 0, we must change δ according to ϵ in order to satisfy the requirment. Thus, we need to find a constant relationship between ϵ and δ so that no matter what the value of ϵ is, we  can always find a valid value of δ follows the pattern of such relationship.

Since we know that |x^2−4|<ϵ and  |x−2|<δ and we need to build a relation between ϵ and δ, we can try to find the relation between  |x^2−4|  and |x−2|.

We find out |x^2−4|= |x+2|×|x−2|

Since |x−2|<δ 

 |x^2−4|<δ ×|x+2|

Let x + 2 be (x - 2) + 4 = δ + 4 

 |x^2−4|<δ ×(δ+4) 

Now comes to the tricky part:

 make δ≤1,δ ×(δ+4)<5δ<ϵ  so, δ≤ϵ/5  

We also now  δ≤1(as we assume):

δ=min⁡(1,ϵ/5)

However, all the works above is just rough thought and cannot be used as to prove, but the conclusion would be used in the prove.

Prove:

Assume ϵ>0                  # Assume the prerequisite

    pick δ=min⁡(1,ϵ/5)      # from the conclusion of the rough thought above

    then δ>0                   #we can conclude this from the line above.

    assume x ∈R               #Assume prerequisite

         assume |x−2|<δ  #Assume the antecedent, antecedent must be true if we want to prove the implication is true

             then  |x^2−4|= |x+2|×|x−2|<δ ×(δ+4)<5δ<ϵ  #The same process as the rough thought

Then |x−2|<δ→ |x^2−4|<ϵ