(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