1. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 11 The Foundations: Logic and Proofs Introduction to Proofs

2. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 22 Some terminology Theorem: is a statement that can be shown to be true. Axioms (postulates): Statements that used in a proof and are assumed to be true. Proof Methods Proving p  q Direct proof: Assume p is true, and prove q. Indirect proof: Assume  q, and prove  p. Trivial proof: Prove q true. Vacuous proof: Prove  p is true

3. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 33 Direct proof Proving p  q Direct proof: Assume p is true, and prove q  Direct proofs lead from the hypothesis of a theorem to the conclusion.  They begin with the premises; continue with a sequence of deductions, and ends with the conclusion.  Direct proof often reaches dead ends.

4. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 44 Direct proof Definition 1: The integer n is even if there exists an integer k such that n=2k, and n is odd if there exists an integer k such that n=2k+1. Axiom: Every integer is either odd or even

5. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 55 Direct proof Example 1: Give direct proof that : Theorem : “If n is an odd integer, then n 2 is an odd integer”. Proof We assume that the hypothesis of this condition is true “ n is odd” n = 2k+1 for some integer k We want to show that n 2 is odd, thus n 2 = (2k+1) 2 n 2 = 4k 2 + 4k + 1 n 2 = 2(2k 2 + 2k) + 1 Therefore n 2 is of the form 2j + 1 (with j the integer 2k 2 + 2k), thus n2 is odd

6. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 66 Direct proof Example 2: Give direct proof that : Theorem : “if m and n are both perfect squares, then mn is also perfect squares”. a is a perfect square when a= b 2 Proof We assume “m and n are both perfect squares” m = s 2 and n=t 2 We want to show that mn is also perfect squares mn= s 2 t 2 mn= (s t) 2 Therefore mn = x 2 which x=st thus mn is also perfect squares from the definition

7. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 77 Direct proof Example Prove that if n is an integer and 3n+2 is odd, then n is odd.  We assume that 3n+2 is an odd integer  This mean that 3n+2=2k+1  There is no direct way to proof that n is odd integer (Direct proof often reaches dead ends.)

8. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 88 Indirect proof  We need other method of proving theorem of p  q, which is not direct  which don’t start with the hypothesis and end with the conclusion ( we call it indirect proof)  Indirect proof (proof by contraposition): Assume  q, and prove  p.  Contraposition (p  q   q   p)  We take  q as hypothesis, and using axioms, definitions any proven theorem to follow  p

9. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity Indirect proof Example 3: Prove that if n is an integer and 3n+2 is odd, then n is odd. Proof Suppose that the conclusion is false, i.e., that n is even (  q) Then n=2k for some integer k. Then 3n+2 = 3(2k)+2 = 6k+2 = 2(3k+1). Thus 3n+2 is even, because it equals 2j for integer j = 3k+1. So 3n+2 is not odd  p We have shown that ¬(n is odd)→¬(3n+2 is odd), thus its contraposition (3n+2 is odd) → (n is odd) is also true.

10. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity Indirect proof Example 4: Prove that if n=ab, where a and b are positive integers, then a  √ n or b  √ n Proof We assume that (a  √n or b  √n ) is false (  q)  (a  √ n  b  √n ) (a > √n  b > √n ) we multiply these to obtain ab > √n √n. ab > n which ab ≠ n (  p) The theorem is proved.

11. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 11 Trivial &Vacuous proofs Vacuous proof Proving p  q Vacuous proof: Prove  p is true Examples Theorem: (For all n) If n is odd and even, then n2 = n + n. Proof: The statement “n is both odd and even” is necessarily false, since no number can be both odd and even. So, the theorem is vacuously true.

12. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 12 Trivial &Vacuous proofs Trivial proof Proving p  q Trivial proof: Prove q true. Example Theorem: (For integers n) If n is the sum of two prime numbers, then either n is odd or n is even. Proof: Any integer n is either odd or even. So the conclusion of the implication is true regardless of the truth of the hypothesis. Thus the implication is true trivially. This kind will be discuss in 4.1

13. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 13 Examples of Proof Methods Definition 2: The real number r is rational if there exist integers p and q with q≠0 such that r =p/q. A real number that is not rational is called irrational

14. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 14 Examples Proof Methods Example 7: Theorem: Prove that the sum of two rational numbers is rational. Proof assume that r and s are rational numbers r=p/q and s=t/u where p,q,t,u are integers and p≠0, u≠0 r+s=(p/q)+(t/u) = (pu+qt)/(qu) Because p≠0 and u≠0, then qu≠0 Both (pu+qt) and (qu) are integers Then the theorem is proved Note that :Our attempt to find direct proof succeeded. Do example 8

15. Fall 2008/2009 I. Arwa Linjawi & I. Asma’a Ashenkity 15 Indirect Proof (Proof by Contradiction) we want to prove that if the negation of p ( which is true ) conclude false, this means that we prove the theorem. Examples:  Prove that √ 2 is irrational by giving a proof of contradiction p = “ √ 2 is irrational “, suppose that  p is true  Give a proof by contradiction that “ if 3n +2 is odd, then n is odd” Let p be “3n+2 is odd“ and q be “n is odd“ We assume that  (p  q ) is true which means p   q is true ( as indirect example )