(CSC 102) Lecture 12 Discrete Structures

Previous Lecture Summary Floor and Ceiling Functions Definition of Proof Methods of Proof Direct Proof Disproving by Counterexample. Indirect Proof: Proof by Contradiction

Methods of Proof and Number Theory

Today's Lecture Mod Functions Divisibility and Floor Mod Congruence Indirect Proofs Proof by Contra-positive Relation between Contradiction and Contra-positive methods of Proof

Compute following mod mod mod 24: 113 div mod 7: -29 div 7 Mod Functions

Mod and div Definitions

Theorem Divisibility and Floor

Cont…

Use the floor notation to compute 3850 div 17 and 3850 mod 17. Sol: Computing div and mod

Let a, b be integers and n be a positive integer. We say that a is congruent to b modulo n (i.e. a  b (mod n) ) iff n | (b-a), implies that there exist some integer k such that b-a = n·k. Note: a mod n = b mod n Which of the following are true? 1. 3  3 (mod 17) 2. 3  -3 (mod 17)  177 (mod 5)  13 (mod 26) Mod Congruence's

Cont… 1.3  3 (mod 17) True: any number is congruent to itself (3-3 = 0, divisible by all) 2.3  -3 (mod 17) False: (-3-3) = 6 isn ’ t divisible by  177 (mod 5) True: = 5 is a multiple of  13 (mod 26) True: 13-(-13) = 26 divisible by 26.

Congruence's Identities Let n > 1 be fixed and a, b, c, d be arbitrary integers. Then the following properties holds: a)(Reflexive Property ) a  a (mod n). b)(Symmetric Property) If a  b(mod n) then b  a(mod n). c)( Transitive Property) If a  b(mod n) and b  c (mod n) then a  c(mod n). d)If a  b(mod n) and c  d (mod n) then a + c  (b + d ) (mod n) and a·c  b·d(mod n). e)If a  b(mod n) then a + c  b+c(mod n) and a·c  b·c(mod n). f)If a  b(mod n) then a k  b k (mod n) for any positive integer k.

Theorem If k is any integer such that k  1 (mod 3), then k 3  1 (mod 9). Proof:  k  Z, k  1(mod 3)  k 3  1(mod 9) k  1(mod 3)   n, k-1 = 3n   n, k = 3n + 1   n, k 3 = (3n + 1) 3   n, k 3 = 27n n 2 + 9n + 1   n, k 3 -1 = 27n n 2 + 9n   n, k 3 -1 = (3n 3 + 3n 2 + n)·9   m, k 3 -1 = m·9 where m = 3n 3 + 3n 2 + n  k 3  1(mod 9)

1. Express the statement to be proved in the form ∀ x in D, if P(x) then Q(x). 2. Rewrite this statement in the contra positive form ∀ x in D, if Q(x) is false then P(x) is false. 3. Prove the contra-positive by a direct proof. a. Suppose x is a (particular but arbitrarily chosen) element of D such that Q(x) is false (or ¬Q(x) is true). b. Show that P(x) is false (or ¬P(x) is true). Indirect Proofs Method of Proof by Contra-Positive

Proof by Contra-positive Proposition: For all integers n, if n 2 is even then n is even. Contra positive: For all integers n, if n is not even then n 2 is not even. Proof: Suppose n is any odd integer. [We must show that n 2 is odd.] By definition of odd, n = 2k + 1 for some integer k. By substitution and algebra, n 2 = (2k+1) 2 = 4 k 2 + 4k + 1 = 2(2 k 2 + 2k) + 1. But 2k 2 + 2k is an integer because products and sums of integers are integers. So n 2 = 2·(an integer) + 1, and thus, by definition of odd, n 2 is odd.

Relation ship between Contra-positive and Contradiction Proofs In a proof by contraposition, the statement ∀ x in D, if P(x) then Q(x) is proved by giving a direct proof of the equivalent statement ∀ x in D, if ∼ Q(x) then ∼ P(x). To do this, you suppose you are given an arbitrary element x of D such that ∼ Q(x). You then show that ∼ P(x). This is illustrated in Figure

To rewrite the proof as a proof by contradiction, you suppose there is an x in D such that P(x) and ¬Q(x). You then follow the steps of the proof by contraposition to deduce the statement ¬P(x). But ¬P(x) is a contradiction to the supposition that P(x) and ¬Q(x). (Because to contradict a conjunction of two statements, it is only necessary to contradict one of them.) This process is illustrated in Figure Cont….

Proof by Contradiction Proposition: For all integers n, if n 2 is even then n is even. Proof: Suppose n is not even integer. Then n is odd integer. By definition of odd, n = 2k + 1 for some integer k. By substitution and algebra, n 2 = (2k+1) 2 = 4 k 2 + 4k + 1 = 2(2 k 2 + 2k) + 1. But 2k 2 + 2k is an integer because products and sums of integers are integers. So n 2 = 2·(an integer) + 1, and thus, by definition of odd, n 2 is odd. But n 2 is even in hypothesis. Which is a contradiction because any integer cannot be both even and odd. Thus our supposition was wrong. Hence n is even.

When to use which method…??? In the absence of obvious clues suggesting indirect argument, Try first to prove a statement directly. Then, if that does not succeed, look for a counterexample. If the search for a counterexample is unsuccessful, look for a proof by contradiction or contraposition.

Two Classical Theorems Theorem 1: is an irrational number. Proof:

Cont… m = 2k for some integer k. m 2 = (2k) 2 = 4k 2 = 2n 2. n 2 =2k 2 Consequently, n 2 is even, and so n is even. But we also know that m is even. Hence both m and n have a common factor of 2. But this contradicts the supposition that m and n have no common factors. [Hence the supposition is false and so the theorem is true.]

Theorem 2

Lecture Summary Mod Functions Divisibility and Floor Mod Congruence Indirect Proofs Proof by Contra-positive Relation between Contradiction and Contra-positive methods of Proof