CSC165 Cases, multiple quantifiers, limits

Week #6

Topic: Cases, multiple quantifiers, limits

I found really interesting both the proof by cases and the proof about limits; therefore, I would like to use this space to practice and enhance the knowledge about these two interesting proofs.

Proof by cases

I realized that in this kind of proofs we are considering all the natural numbers by dividing them in odd and even numbers.

In other words: ℕ = {x: ∃ k ∈ ℕ, x = 2k + 1} U {x: ∃ k ∈ ℕ, x = 2k}

Now, let’s try to proof that ∀ n ∈ ℕ, 3n² + n +15 is odd

Definitions: n is odd when ∃ k ∈ ℕ, n = 2k + 1 and n is even when ∃ k ∈ ℕ, n = 2k

Assume n ∈ ℕ #Typical

            Case1: Assume n is odd, n = 2k + 1

                         Then 3n² + n +15 = 3(2k + 1)² + (2k + 1) + 1

                                                          = [3(4k² + 4k +1)] + 2k + 2

                                                          = 12k² + 12k + 3 +2k + 2

                                                          = 12k² +14k + 5

                                                          = 12k² +14k + 4 + 1 # 5 = 4 + 1

                                                          = 2(6k² +7k + 2) + 1

                        Then ∃ k₀ ∈ ℕ, n = 2 k₀ + 1 # k_{0 =} 6k² +7k + 2

                        Then 3n² + n +15 is odd #by definition

            Case2: Assume n is even, n = 2k

                         Then 3n² + n +15 = 3(2k)² + (2k) + 1

                                                          = [3(4k²)] + 2k + 1

                                                          = 12k² + 2k + 1

                                                          = 2(6k² + k) + 1

                        Then ∃ k₀ ∈ ℕ, n = 2 k₀ + 1 # k_{0 =} 6k² + k

                        Then 3n² + n +15 is odd #by definition

            Then 3n² + n +15 is odd #since in both cases this was true

Conclude ∀ n ∈ ℕ, 3n² + n +15 is odd

Now, let’s try to prove that  the limit of 3x - 1 as x approaches 2 is 5

In this case, we apply a concept very similar to the delta-epsilon proof.

∀ e ∈ ℝ, ∃ d ∈ ℝ, ∀ x ∈ ℝ, |x - |<d ⇒ |(3x - 1) - 5| < e

Assume e ∈ ℝ #Generic x

            Pick  d = e/3

                        Assume |x - 2|<d

                        Then |(3x - 1) - 5| = |3x - 6| = |3 (x - 2)| < 3d = e   # |x - 2|<d

   # |(3x - 1) - 5| < e

                        Then ∀ x ∈ ℝ, |x - 2|<d ⇒ |(3x - 1) - 5| < e

            Then ∃ d ∈ ℝ, ∀ x ∈ ℝ, |x - 2|<d ⇒ |(3x - 1) - 5| < e

Conclude ∀ e ∈ ℝ, ∃ d ∈ ℝ, ∀ x ∈ ℝ, |x - 2|<d ⇒ |(3x - 1) - 5| < e