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Logic: Connectives AND OR NOT P Q (P ^ Q) P Q (P v Q) T T T T T T T F F T F T F T F F T T F F F F F F P ~P T F F T IF AND ONLY IF IMPLIES Q ( P Q ) T T T F T F F T T F T F F T F F T P Q (P =>Q) T T T T F F F P Converse and contrapositive • For the proposition P => Q, the proposition Q => P is called its converse, and the proposition ~Q => ~P is called its contrapositive. • For example for the proposition "If it rains, then I get wet", Converse: If I get wet, then it rains. Contrapositive: If I don't get wet, then it does not rain. • The converse of a proposition is not necessarily logically equivalent to it, that is they may or may not take the same truth value at the same time. • On the other hand, the contrapositive of a proposition is always logically equivalent to the proposition. That is, they take the same truth value regardless of the values of their constituent variables. • Therefore, "If it rains, then I get wet." and "If I don't get wet, then it does not rain." are logically equivalent. If one is true then the other is also true, and vice versa. Methods of Proof • Direct Proof, Indirect Proof, and proof by contradiction • Direct Proof: The implication p => q can be proved by showing that if p is true, then q must also be true. • Example: Give a direct proof of the theorem “If n is an odd integer, then n 2 is an odd integer”. • Solution: Assume that the hypothesis of this implication is true. Suppose that n is odd. Then n = 2k +1 where n is an integer. It follows that n 2 = (2k + 1) 2 =4k2+4k+1 = 2 (2 k 2 + 2k) + 1 Therefore, n is an odd integer ( It is one more than twice an integer) Methods of Proof • Indirect Proof: We know that an implication p => q is equivalent to its contrapositive ~q => ~p. Thus to prove p => q indirectly, we assume q is false (~q) and show that p is then false (~p) • Example: Give an indirect proof of the theorem “If 3n + 2 is an odd integer, then n is an odd integer”. • Solution: Assume that the conclusion of this implication is false. Suppose that n is even. Then n = 2k where n is an integer. It follows that 3 n + 2 = 3 (2 k) + 2 = 6k + 2 = 2 (3k + 1) Therefore, 3n + 2 is an even integer ( It is an integral multiple of 2) Thus, we show that if n is even, then 3n + 2 is even, which is the contrapositive of the given statement. Hence the given statement has been proved. Mistakes in Proof What is wrong with the following proof “1 = 2’? a≠0, b ≠ 0. Step Reason 1. a = b Given 2. a 2 = ab Multiply both sides by a 3. a2 – b2 = ab – b2 Subtract b2 from both sides 4. (a – b)(a + b)= b(a – b) Factorize both sides 5. a + b = b Divide both sides by (a – b) 6. b + b = b From 1. (Given) 7. 2b = b Addition 8. 2 = 1 Divide both sides by b Mistakes in Proof What is wrong with the proof “1 = 2’? Solution: Every step is valid except for one, step 5, where we divided both sides by (a – b). We cannot divide by (a – b) since a – b = 0 ( a = b Given) Division by zero is not a valid operation. Methods of Proof Proof by contradiction: This method is based on the tautology (p => q) ^ (~q)) => (~p). • Thus the rule of inference p => q ~q Therefore, ~p • Informally, if a statement p implies a false statement q, then p must be false. • We use this technique where q is an absurdity or contradiction. Methods of Proof • Prove there is no rational number p / q whose square is 2. • Solution: Assume (p / q) 2 = 2 where p and q are integers having no common factors. Then, p 2 = 2 q2 Therefore, p 2 is even. This implies p is even, since the square of an odd number is odd. Thus, p = 2n where n is an integer. 2 q 2 = p2 = (2 n) 2 = 4n 2 or, q 2 = 2n 2 Thus q 2 is even, and so q is even. p and q are even, therefore they have a common factor 2. This is a contradiction to the assumption. Thus the assumption must be false. Mathematical Induction • Another proof technique. • In this technique, we need to prove two steps: • Basis step: First we need to prove that the basis element, that is 0, has the property in question. • Inductive step: Then we need to prove that if an arbitrary natural number, denoted by n, has the property in question, then the next element, that is n + 1, has that property. • When these two are proven, then it follows that all the natural numbers have that property. Mathematical Induction • When these two are proven, then it follows that all the natural numbers have that property- Explanation. • Since 0 has the property by the basis step, the element next to it, which is 1, has the same property by the inductive step. Then since 1 has the property, the element next to it, which is 2, has the same property again by the inductive step. • Proceeding likewise, any natural number can be shown to have the property. • This process is somewhat analogous to the knocking over a row of dominos with knocking over the first domino corresponding to the basis step. Mathematical Induction • Example: Prove that for any natural number n, 0 + 1 + ... + n = n( n + 1 )/2 . Proof: Basis Step: If n = 0, then LHS = 0, and RHS = 0 * (0 + 1) = 0 . Hence LHS = RHS. Induction: Assume that for an arbitrary natural number n, 0 + 1 + ... + n = n( n + 1 )/2 . -------- Induction Hypothesis To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis. LHS for n + 1 = 0 + 1 + ... + n + (n + 1) = (0 + 1 + ... + n) + (n + 1) . Using the induction hypothesis, the last expression can be rewritten as n( n + 1 )/2 + (n + 1) . Factoring (n + 1) out, we get (n + 1)(n + 2) / 2 , which is equal to the RHS for n+1. Thus LHS = RHS for n+1. Mathematical Induction • Problem: For any natural number n , 2 + 4 + ... + 2n = n( n + 1 ). Proof: Basis Step: If n = 0, then LHS = 0, and RHS = 0 * (0 + 1) = 0 . Hence LHS = RHS. Induction: Assume that for an arbitrary natural number n, 0 + 2 + ... + 2n = n( n + 1 ) . ---- Induction Hypothesis To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis. LHS for n + 1 = 0 + 2 + ... + 2n + 2(n + 1) = (0 + 2 + ... + 2n) + 2(n + 1) . Using the induction hypothesis, the last expression can be rewritten as n( n + 1 ) + 2(n + 1) . = (n + 1)(n + 2) , which is equal to the RHS for n+1. Thus LHS = RHS for n+1. Mathematical Induction Problem: If r is a real number not equal to 1, then for every n ≥0, r 0 + r 1 + …+r n = (1 – r n + 1)/(1 – r) Proof: Basis Step: If n = 0, then LHS = r0 = 1, and RHS = (1 - r) / (1 - r) = 1, since r ≠ 1. Hence LHS = RHS. Induction: Assume that r 0 + r 1 + …+r n = (1 – r n + 1)/(1 – r) ------ Induction Hypothesis To prove this for n+1, first try to express LHS for n+1 in terms of LHS for n, and use the induction hypothesis. LHS for n + 1 = = r 0 + r 1 + …+r n + r n+1 = (r 0 + r 1 + …+r n ) + r n+1 Using the induction hypothesis, the last expression can be rewritten as: (1 – r n + 1)/(1 – r) + r n + 1 Taking the common denominator, it is equal to ((1 – r n + 1) + (r n + 1 - r n + 2 )) / (1 – r) = (1 - r n + 2 )/(1 – r) which is equal to the RHS for n+1. Thus LHS = RHS for n+1.