# Logic: Connectives AND OR NOT P Q (P ^ Q) T F P Q (P v Q) T F P ~P T F

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Logic: Connectives AND OR NOT P Q (P ^ Q) T F P Q (P v Q) T F P ~P T F
IMPLIES IF AND ONLY IF P   Q   ( P Q )   T T   F F P Q (P =>Q) T F

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 = 4 k k + 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 a = b Given a 2 = ab Multiply both sides by a a2 – b2 = ab – b2 Subtract b2 from both sides (a – b)(a + b)= b(a – b) Factorize both sides a + b = b Divide both sides by (a – b) b + b = b From 1. (Given) 2b = b Addition 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,   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, n = 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 = n + (n + 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 , n = 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, n = 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 = n + 2(n + 1) = ( n) + 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 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.

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