Agenda Proofs (Konsep Pembuktian) Direct Proofs & Counterexamples Indirect Proofs Proof by Contraposition Proof by Contradiction Exhaustive & Resolution Proof Mathematical Induction Latihan Soal

Proof Pembuktian tentang kebenaran suatu mathematical statement. Formal proof: - utilize rule of inference - all steps were supplied - long and hard to follow - suitable for computer Informal proof: - steps maybe skip - utilize assumption - suitable for human

Proof Example: Suppose you did commit the crime. Then at the time of the crime, you would have had to be at the scene of the crime. In fact, you were in a meeting with 10 people at that time, as they will testify. This contradicts the assumption that you committed the crime. Hence the assumption is false.

Mathematical Systems Axiom: statement we assume to be true. Definition: statement that are used to create new concepts in terms of existing ones Theorem: statement that can be shown to be true. Lemma: theorem that is usually not too interesting in its own right, but useful in proving another theorem Corollary: a less important theorem (follows easily from another theorem. Use proof to demonstrate that a theorem is true.

Methods of Proving Theorem p  q Proof: Demonstrate that q is true if p is true Methods: Direct Proofs Indirect Proofs: Proof by Contraposition Proof by Contradiction Exhaustive Proof

Direct Proof & Counterexample

Direct Proof p  q Direct Proof: Assume that p is true Use axiom, definition, rule of inference etc. Show that q must also be true

Give a direct proof of the theorem: Example Give a direct proof of the theorem: “IF n is an odd integer, THEN n2 is odd” Definition: 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. Solution: Assume that “n is odd” is true Use axiom, definition, rule of inference etc n = 2k + 1 (from definition) n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 Conclusion: it is proved that n2 is an odd integer

Example Give a direct proof of the theorem: “IF m is an odd integer and n is even integer, THEN m + n is odd” Definition: The integer n is even if there exists an integer k1such that n = 2k1, and m is odd if there exists an integer k2 such that n = 2k2 + 1. Solution: Assume that “m is odd and n is even” is true Use axiom, definition, rule of inference etc m = 2k2 + 1 and n = 2k1 (from definition) m + n = 2 (k1 + k2 ) + 1 Conclusion: it is proved that m + n is an odd integer

Give a direct proof that: Example Give a direct proof that: “IF m and n are both perfect squares, THEN nm is also a perfect square” Definition: An integer a is a perfect square if there is an integer b such that a = b2 Solution: Assume that “m and n are both perfect square” is true Use axiom, definition, rule of inference etc There are integers s and t such that m = s2 and n = t2. mn = s2t2, mn = (st)2 Conclusion: it is proved that mn is a perfect square, because it is the square of st, which is an integer.

Counterexample To disprove a universally quantified statement: x P(x) We simply need to find one member x in the domain of discourse that makes P(x) false The value of x is called a counterexample

Example Show that the statement n  Z+ (2n + 1 is prime) Is false A counterexample is n=3, since 23+1 = 9, which is not prime

Existence Proof A proof of x P(x) Is called an existence proof. Show that there exists one number a in the domain of discourse that makes P(a) true.

Example  

Indirect Proof: - Contraposition - Contradiction

Proof by Contraposition Indirect Proof: proof that do not start with the hypothesis and end with the conclusion. Contraposition: p  q is logically equivalent to ~q  ~p. Proof by Contraposition: Assume that ~q is true Use axiom, definition, rule of inference etc. Show that ~p must also be true

3n + 2 is even, and therefore not odd = ~p Example Prove that: “IF n is an integer and 3n + 2 is odd, THEN n is odd” Solution: Assume that “~q = n is even” is true Use axiom, definition, rule of inference etc n = 2k (from definition) 3n + 2 = 6k + 2 = 2 ( 3k + 1) 3n + 2 is even, and therefore not odd = ~p Conclusion: it is proved by contraposition

Example ab > √n . √n = n So ab ≠ n Prove that: “IF n = ab, where a and b are positive integers, THEN a ≤ √n or b ≤ √n ” Solution: ¬q is a > √n and b > √n ¬p is n ≠ ab Assume that ¬q is true Use axiom, definition, rule of inference etc ab > √n . √n = n So ab ≠ n Conclusion: it is proved by contraposition

Proof by Contradiction is a statement form that is always false regardless of the truth values of the individual statements substituted for its statement variables. q Ù ~q Proof by Contradiction: To prove that a statement k is true: Suppose the statement to be proved is false. Means its negation ~k is true. Show that this supposition leads logically to a contradiction, so the assumption is false. Conclude that the statement to be proved is true.

Example Prove that: “IF 3n + 2 is odd, then n is odd” Proof by Contradiction: p = 3n + 2 is odd, q = n is odd. Assume that ~(p  q) is true OR (p Ù ~q) is true. It means that both p AND ~q must be true To show contradiction, show that q is also true, OR Show that ~p is also true. ~q = n is not odd = n is even n = 2k 3n + 2 = 6k + 2 = 2 (3k + 1) = 2t is even ~p is true Thus, p and ~p are both true  contradiction

Example Prove that: “IF n2 is even, then n is even” Proof by Contradiction: p = n2 is even, q = n is even. Assume that ~(p  q) is true OR (p Ù ~q) is true. It means that both p AND ~q must be true To show contradiction, show that q is also true, OR Show that ~p is also true. ~q = n is odd n = 2k + 1 n2 = 2 (2k2 + 2k) + 1 = (2 t + 1) is odd ~p is true p and ~p are true  contradiction

More Methods of Proof: - Exhaustive - Resolution

Exhaustive Proof Some theorems can be proved by examining a relatively small number of examples. Prove that (n+1)3  3n if n is a positive integer with n ≤ 4. Solution: n = 1; (1+1)3  31 n = 2; (2+1)3  32 n = 3; (3+1)3  33 n = 4; (4+1)3  34

Proof Strategies: When to use Exhaustive Proof Analyze what the hypothesis (premises) and conclusion mean. If it is a conditional statement: Try a Direct Proof Try a Proof by Contraposition Try a Proof by Contradiction If it has a relatively small number of domain: Try an Exhaustive Proof

If (p  q) and (~ p  r) are both true, then (q  r) is true Resolution Used to automate the task of reasoning and proving theorems, proposed by JA. Robinson in 1965 that depends on a single rule: If (p  q) and (~ p  r) are both true, then (q  r) is true ((p  q) Ù (~ p  r))  (q  r) (p  q) (~ p  r)  (q  r)

Proof by Resolution In a proof by resolution, the hypothesis and the conclusion are written as clauses. A clause consists of terms separated by or’s. A direct proof by resolution proceeds by repeatedly applying resolution to pairs of statement to derive new statements until the conclusion is derived. If a hypothesis is not a clause, it must be replaced by an equivalent expression that is a clause.

Example Jasmine is skiing or it is not snowing. It is snowing or Bart is playing hockey. Conclusion: Jasmine is skiing or Bart is playing hockey. p: it is snowing q: Jasmine is skiing r: Bart is playing hockey (~p  q) (p  r)  (q  r)

Example (a  b) (~a c) (~c  d) (b  d) From 1 and 2: b  c  4 From 3 and 4: b  d

Mathematical Induction

Proof by Mathematical Induction How to prove n P(n)? Utilize Mathematical Induction Basic Step: Show that the statement is true for the positive integer 1. Inductive Step: Show that if the statement is true for a positive integer (k) then it must also be true for the next positive integer (k+1).

Proof by Mathematical Induction Basic Step: P(1) is true Inductive Step: Assume that P(k) is true for an arbitrary integer k. P(k)  P(k+1) is true for all positive integer k. Notation: [P(1)  k (P(k)  P(k+1))]  n P(n) It is not a tool for discovering formula or theorems.

P(1) is true, because 1 = 1(1+1)/2 Example 1 1 + 2 + … + n = n(n+1)/2 Basic step: P(1) is true, because 1 = 1(1+1)/2 Inductive step: Assume that P (k) is true for an arbitrary integer k. 1 + 2 + … + k = k(k+1)/2 Show that P(k+1) is true.

Example 1 Show that P(k+1) is true 1 + 2 + … + k + (k+1) = (k+1)[(k+1)+1]/2 = (k+1)(k+2)/2 Proof: Add element (k+1) to P(k) 1 + 2 + … + k + (k+1) = k(k+1) / 2 + (k+1) = (k(k+1) + 2(k+1))/2 = (k2 + 3k + 2)/2 = (k+1) (k+2)/2

Assume that P (k) is true for an arbitrary integer k. Example 2 1 + 3 + 5 + … + (2n -1) = n2 Basic step: P(1) is true, because 1 = 12 Inductive step: Assume that P (k) is true for an arbitrary integer k. 1 + 3 + … + (2k – 1) = k2 Show that P(k+1) is true.

Example 2 Show that P(k+1) is true 1 + 3 + … + (2k-1) + (2k+1) = (k+1)2 Proof: Add element (k+1) to P(k) 1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k+1) = k2 + 2k+1 = (k+1)2

Assume that P (k) is true for an arbitrary integer k. Example 3 1 + 4 + 7 + … + (3n -2) = n2 Basic step: P(1) is true, because 1 = 12 Inductive step: Assume that P (k) is true for an arbitrary integer k. 1 + 4 + … + (3k – 2) = k2 Show that P(k+1) is true.

Example 3 Show that P(k+1) is true 1 + 4 + … + (3k-2) + (3k+1) = (k+1)2 Proof: Add element (k+1) to P(k) 1 + 4 + … + (3k-2) + (3k+1) = k2 + (3k+1) = k2 + 3k+1 k2 + 3k + 1 ≠ (k+1)2

Example 4: starting from other 1 + 2 + 22 + … + 2n = 2n+1 -1 Valid for all nonnegative integers (0, 1, 2, …) n = 0, 1, 2,… Basic step: P(0) is true, because 1 = 20 + 1 – 1 Inductive step: Assume that P (k) is true for an arbitrary integer k. 1 + 2 + 22 + … + 2k = 2k+1 -1 Show that P(k+1) is true.

Example 4: starting from other Show that P(k+1) is true 1 + 2 + 22 + … + 2k + 2k+1 = 2(k+1)+1 -1 = 2k+2 -1 Proof: Add element (k+1) to P(k) 1 + 2 + 22 + … + 2k + 2k+1 = 2k+1 -1 + 2k+1 = 2 . 2k+1 -1

Assume that P (k) is true for an arbitrary integer k. Example 5 Show that: 1 + 22 + … +n2 = n(n+1)(2n+1)/6 Basic step: P(1) is true, because 1 = 1(1+1)(2.1+1)/6 Inductive step: Assume that P (k) is true for an arbitrary integer k. 1 + 22 + … + k2 = k(k+1)(2k+1)/6 Show that P(k+1) is true.

Add element (k+1) to P(k) Example 5 Show that P(k+1) is true 1 + 22 + … +(k+1) 2 = (k+1)(k+1+1)(2(k+1)+1)/6 = (k+1)(k+2)(2k+3)/6 Proof: Add element (k+1) to P(k) 1 + 22 + … k2 + (k+1)2 = k(k+1)(2k+1)/6 + (k+1)2 = (k+1)[k(2k+1)/6 + (k+1)] = (k+1)[(k(2k+1)+ (6k+6))/6] = (k+1)[(2k2+k+6k+6)/6] = (k+1)[(2k2+7k+6)/6] = (k+1)[(k+2)(2k+3)/6]

Assume that P (k) is true for an arbitrary integer k. k < 2k Example 6 Proving Inequality: n < 2n Basic step: P(1) is true, because 1 < 21 Inductive step: Assume that P (k) is true for an arbitrary integer k. k < 2k Show that P(k+1) is true.

Example 6 (k+1) < 2k+1 k+1 < 2k + 1  1  2k k+1 < 2k + 2k Show that P(k+1) is true (k+1) < 2k+1 Proof: Add 1 to P(k) k+1 < 2k + 1  1  2k k+1 < 2k + 2k k+1 < 2. 2k k+1 < 2k+1

Strong Form of Induction & The Well-Ordering Property Strong Form of Induction and the Well-Ordering Property

Inductive Step Inductive Step (section 2.4): Assume that statement n is true Then prove that statement n+1 is also true The Strong Form of Mathematical Induction: Inductive Step (section 2.5): To prove the truth of statement n, assume the truth of all of the preceding statements (statement n0 to n-1).

Strong Form of Mathematical Induction Suppose that we have a propositional function S(n) whose domain of discourse is the set of integers greater than or equal to n0. Suppose that: - S(n0) is true; - For all n > n0, if S(k) is true for all k, n0 ≤ k < n, then S(n) is true. Then S(n) is true for every integer n ≥ n0

Example Show that postage of four cents or more can be achieved by using only 2-cent and 5-cent stamps. Basis Step: n0 = 4  S(4) is true, because 4 = 2 x 2 cents stamp n0 = 5  S(5) is true, because 5 = 1 x 5 cents stamp

Example Inductive step: Assume that n ≥ 6 and that postage of k cents or more can be achieved by using only 2-cent and 5-cent stamps for 4 ≤ k ≤ n Show that S(n) is true. Proof: For any k ≥ n0 , we can add 2 cents to make k-cents postage. Then, for k = n, we can add 2 cents to make n-cents postage.

Well-Ordering Property The well-ordering property for nonnegative integers: every nonempty set of nonnegative integers has a least (smallest) element. Let B be a set of integers. An integer m is called a least element of B if m is an element of B, and for every x in B, mx. Example: 3 is a least element of the set {4,3,5,11}. Let A be the set of all positive odd integers. Then 1 is a least element of A. Let U be the set of all odd integers. Then U has no least element.

Quotient-Remainder Theorem If d and n are integers, d > 0, there exist integer q (quotient) and r (remainder) satisfying n = dq + r 0 ≤ r ≤ d Furthermore, q and r are unique; that is, if n = dq1 + r1 0 ≤ r1 ≤ d1 And n = dq2 + r2 0 ≤ r2 ≤ d2 Then q1 = q2 and r1 = r2

Example Divide n = 74 by d = 13 We obtain q = 5 and r = 9 Where: 74 = 13 x 5 + 9 and 0 ≤ 9 ≤ 13 n = dq + r and 0 ≤ r ≤ d If q = 3 then r = 35, which is too large If q = 6 then r = -4, which is negative