Ah it's been a while since I posted. I hope everyone had a nice Thanksgiving. It was my birthday last week so that was a bonus too. Before Thanksgiving we had a midterm so we only had 2 lectures that week and 2 lectures this week. In both weeks, we mainly covered proofs and methods.

The first method that was covered was proving the contrapositive. If for example we were told to prove that any natural number n, n­² odd implies that n is odd. Then by contrapositive I would instead prove that any natural number n, n is even implies n² is even. The usefulness of this is that proving the contrapositive is sometimes easier, and since the contrapositive is logically the same as the original statement, proving the contrapositive will prove the original statement.

The next method is proving existence where it is sufficient to show that the set is non-empty and there is one element. For example, prove

∃ x ∈ R, x³ + 3x² – 4x = 12

For this proof, we would simply pick an x and show that it works. In this case, if we pick x = 2 which is a typical real number and plug it into the equation, the result is 12. Thus we exhibited one element and the proof is done.

Next is proving a contradiction. If we were to prove the given statement:

∀ n ∈ N, |P| > n

Where P is a prime number, it would be difficult to prove that a prime number is greater than every natural number, in fact impossible. Instead if we prove the contradiction of the statement

∃ n ∈ N, |P| <= n

Then it would be a lot easier to prove that a prime number is less than or equal to some natural number.

Assume e ∈ R

          Pick d = _______ Then d ∈ R⁺ #We are going to leave this blank for now and

                                                            # do some work to find d

          Assume x ∈ R # generic x

                   Assume |x-3| < d

                             Then |x² – 3²| = (|x-3| * |x+3|) = |x-3| * |x+3|

                             Then let d <= 1

                             Then |x-3| <= 1 ⇒ (2 <= x <= 4) #bounded d to be less than or

                                                                                 #equal to 1 and found the interval

                             Then x+3 < 4+3 = 7 #We bounded d to get rid of the (x+3)

                             Then |x² – 3²| < 7 * |x-3| < e

                             Then |x-3| < e/7 #Which now we will choose d = min(1, e/7)

                             So |x² – 3²| < 7 * |x-3| < 7 * d < 7 * e/7 = e

                   So |x-3| < d ⇒ |x² – 3²| < e #Assumed antecedent, got consequent

So ∃ d ∈ R⁺, ∀ x ∈ R, |x-3| < d ⇒ |x² – 3²| < e

Then ∀ e ∈ R⁺, ∃ d ∈ R⁺, ∀ x ∈ R, |x-3| < d ⇒ |x² – 3²| < e #Since assumed e ∈ R

This is my version of this proof which shows that for every e, there exists a d for every x such that the antecedent implies the consequent. 

That's it for this week! Thanks for reading.