(CSC 102) Lecture 12 Discrete Structures. Previous Lecture Summary Floor and Ceiling Functions Definition of Proof Methods of Proof Direct Proof Disproving.

(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 following1. 113 mod 242. -29 mod 7

1. 113 mod 24: 113 div 24

2. -29 mod 7: -29 div 7

Mod FunctionsMod Functions

411324

17

96

5297

35

6

Mod and divMod and div

Definitions

Theorem

Divisibility and FloorDivisibility and Floor

Cont…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)3. 172 177 (mod 5)4. -13 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

17.

3. 172 177 (mod 5) True: 177-172 = 5 is a multiple

of 5

4. -13 13 (mod 26) True: 13-(-13) = 26 divisible by

26.

Congruence's IdentitiesCongruence'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 k3 1 (mod 9).

Proof: k Z, k 1(mod 3) k 31(mod 9)

k 1(mod 3)

n, k-1 = 3n n, k = 3n + 1

n, k 3 = (3n + 1)3

n, k 3 = 27n 3 + 27n 2 + 9n + 1

n, k 3-1 = 27n 3 + 27n 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 31(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-positiveProof by Contra-positive

Proposition: For all integers n, if n2 is even then n is even.

Contra positive: For all integers n, if n is not even then n2 is not even.

Proof: Suppose n is any odd integer. [We must show that n2 is odd.] By definition of odd, n = 2k + 1 for some integer k. By substitution and algebra,

n2 = (2k+1)2 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.

But 2k2 + 2k is an integer because products and sums of integers are integers. So n2 = 2·(an integer) + 1, and thus, by definition of odd, n2 is odd.

Relation ship between Contra-positive and Relation ship between Contra-positive and Contradiction Proofs 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…. Cont….

Proof by Contradiction

Proposition: For all integers n, if n2 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,

n2 = (2k+1)2 = 4 k2 + 4k + 1 = 2(2 k2 + 2k) + 1.

But 2k2 + 2k is an integer because products and sums of integers are integers. So n2 = 2·(an integer) + 1, and thus, by definition of odd, n2 is odd. But n2 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 TheoremsTheorem 1: is an irrational number.

Proof:

Cont…

m = 2k for some integer k.

m2 = (2k)2 = 4k2 = 2n2.

n2=2k2

Consequently, n2 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 Theorem 2 Theorem 2: Prove by contra dictions 1+3 2 is irrational.

Lecture Summary Lecture Summary

• Mod Functions

• Divisibility and Floor

• Mod Congruence

• Indirect Proofs

• Proof by Contra-positive

• Relation between Contradiction and Contra-positive

methods of Proof