Methods of Proof

1. Methods of Proof • Proof techniques in this handout • Direct proof • Division into cases • Proof by contradiction • In this handout, the proof techniques will be used to prove properties in number theory.

2. Even and Odd Integers • Definition: An integer n ● is even iff  an integer k such that n=2k; ● is odd iff  an integer k such that n=2k+1. Ex: If x and y are integers, is even or odd?

3. Method of Direct Proof • To prove a statement: xD if P(x) then Q(x). • Suppose x is a particular but arbitrarily chosen element of D for which P(x) is true; • Show the conclusion Q(x) is true by using ♦ definitions; ♦ previously established results; ♦ rules of logical inference.

4. Method of Direct Proof (Ex.) • Show xZifx is odd then3x+9 is even. Proof: Suppose x is an arbitrarily chosen odd integer. Then x=2k+1 for some integer k. (by definition) So 3x+9 = 3(2k+1)+9 (by substitution) = 6k+3+9 (by distributive law) = 2(3k+6) (by factoring out a 2)(*) 3k+6 is an integer. (**) Hence 3x+9 is even based on (*), (**) and definition of even integers. ▀ (this is what we needed to show)

5. Directions for writing proofs • Write the theorem to be proved. • Clearly mark the beginning of your proof with the word Proof. 3) Make your proof self-contained. (Identify all variables used in the proof; state the sources of outside facts). 4) Write proofs in complete English sentences.

6. Common mistakes in proofs • Arguing from examples • Using same letter to mean two different things • Jumping to a conclusion (without adequate reasons)

7. Types of Mathematical Statements • Theorems: Very important statements that have many and varied consequences. • Propositions: Less important and consequential. • Corollaries: The truth can be deduced almost immediately from other statements. • Lemmas: Don’t have much intrinsic interest but help to prove other theorems.

8. Divisibility • Definition: Forn,d Z and d≠0 we say that n is divisible by d iffn=d·k for some k Z . • Alternative ways to say: n is a multiple of d , d is a factor of n , d is a divisor of n , d divides n . • Notation: d | n . • Examples: 6|48, 5|5, -4|8, 7|0, 1|9 .

9. Properties of Divisibility • For xZ, 1|x . • For xZ s.t. x≠0, x|0 . • For a,b,cZ, if a|b and a|c then a|(b+c) . • Transitivity: For a,b,cZ, if a|b and b|c then a|c .

10. Quotient-Remainder Theorem • Theorem:For nZ anddZ+ ! q,rZsuch that n=d·q+r and 0≤r<d. • q is called quotient; r is called remainder. • Notation:q = n div d; r = n mod d. • Examples: 1) 53 = 8·6+5. Hence 53 div 8 = 6; 53 mod 8 = 5. 2)-29 = 7·(-5)+6.Hence -29 div 7 = -5; -29 mod 7 = 6.

11. Example of using div and mod • Last year Halloween was on Thursday. Q.: What day is Halloween this year? Solution: There are 365 days between 10/31/13 and 10/31/14. 365 mod 7 = 1. Thus, if 10/31/13 was Thursday then 10/31/14 is Friday.

12. Proof Technique: Division into Cases • Suppose at some stage of a proof ● we know that A1 or A2 or A3 or … or An is true; ● want to deduce a conclusion C. • Use division into cases: Show A1→C, A2→C, …, An→C. Conclude that C is true.

13. Division into Cases: Example • Proposition:If nZ such that neither of 2 or 3 divide n, (1) thenn2 mod 12 = 1. (2) • Proof: Suppose nZ s.t. neither of 2 or 3 divide n. By quotient-remainder theorem, exactly one of the following is true: a)n=6k, b)n=6k+1, c)n=6k+2, d)n=6k+3, e)n=6k+4, f)n=6k+5 for some integer k. (3) n can’t be 6k, 6k+2 or 6k+4 because in that case 2 | n (which contradicts (1) ). (4) n can’t be 6k+3 because in that case 3 | n (which contradicts (1) ). (5)

14. Division into Cases: Example(cont.) • Proof(cont.):Based on (3), (4) and (5), either n=6k+1 or n=6k+5. Let’s show (2) for each of these two cases. Case 1: Suppose n=6k+1. Then n2 = (6k+1)2=36k2+12k+1 (by basic algebra) = 12(3k2+k)+1 (6) Let p=3k2+k. Then p is an integer. n2 = 12p+1 . ( by substitution in (6) ) Hencen2 mod 12 = 1by quotient-remainder th-m. Case 2: Suppose n=6k+5. (exercise)■

15. Method of Proof by Contradiction • Suppose the statement to be proved is false. • Show that this supposition logically leads to a contradiction. • Conclude that the statement to be proved is true. Example of proof by contradiction. Theorem: There is no greatest integer. The proof on the board. We will see several contradiction proofs in graph theory.