EECS 1019 Lecture 4: Proof by contrapositive + contradiction

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Proof by cases: case (i): x = 0. Chapter 1: the foundation logic + proof. Disproving existence + proof by cases e. g: let x be from d = z. If x = 0, then (cid:1876)(cid:2870)=(cid:884) (cid:884: case (ii): (cid:1876)= (cid:883) If x = 1 or x = -1, then (cid:1876)(cid:2870)=(cid:883) (cid:884) If |x| > 1, then |x| (cid:3410) 2 and |(cid:1876)(cid:2870)| (cid:3410) 2 so (cid:1876)(cid:2870) (cid:884) (cid:882)(cid:2870)=(cid:882)(cid:3409)(cid:883)(cid:882)=(cid:882)+(cid:883)(cid:882) (cid:883)(cid:2870)=(cid:883)(cid:3409)(cid:883)(cid:883)=(cid:883)+(cid:883)(cid:882) (cid:884)(cid:2870)=(cid:886)(cid:3409)(cid:883)(cid:884)=(cid:884)+(cid:883)(cid:882) (cid:885)(cid:2870)=(cid:891)(cid:3409)(cid:883)(cid:885)=(cid:885)+(cid:883)(cid:882) e. g: theorem: if n is an integer and (cid:882)(cid:3409)(cid:1866)(cid:3409)(cid:885) then (cid:1866)(cid:2870)(cid:3409)(cid:1866)+(cid:883)(cid:882) Proof by exhaustion: case (iii): |x| > 1. An exhaustive proof can be used when we need to check finitely many cases: solve the equation. Proof: assume n is even => n = 2k. = (p q) ( p q) p q = (p q) (q p) Suppose we want to prove p true. We can find contration q such that p q. P q p q p f p.

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