The Foundations: Logic and Proofs

1. The Foundations: Logic and Proofs Introduction to Proofs 1

2. 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 2

3. 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. 3

4. 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 4

5. Direct proof • Example 1: • Give direct proof that : Theorem : “If n is an odd integer, then n2 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 n2 is odd , • thus n2 = (2k+1)2 • n2 = 4k2 + 4k + 1 • n2 = 2(2k2 + 2k) + 1 • Therefore n2 is of the form 2j + 1 • (with j the integer 2k2 + 2k), thus n2 is odd 5

6. 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= b2 • Proof • We assume “m and n are both perfect squares” • m = s2 and n=t2 • We want to show that mn is also perfect squares • mn= s2 t2 • mn= (s t)2 • Thereforemn = x2 which x=st • thusmn is also perfect squares from the definition 6

7. 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.) 7

8. 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 8

9. 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. 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. 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. 11

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 12

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 rationalis called irrational 13

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 14

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 ) 15