Mathematical Reasoning Rules of Inference & Mathematical Induction. 1. Assign propositional variables to the component propositional argument.

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1 Mathematical Reasoning Rules of Inference & Mathematical Induction Example. If I take the day off it either rains or snows 2. When It rains, my basement floods 3. When the basement floods or it snows, I call for help 4. I didn t make any calls today Conclusion: I didn t take the day off. Assign propositional variables to the component propositional argument. 2. Represent the formal argument using variables-in other word write down the hypothese. 3. Use the hypotheses, and the rules of inference( Table-page 69) and any logical equivalences to construct the proof. 4. And finally please use the following format to write your proof! Step Reason. Why 2. you 3. think 4. this 5. step 6. is 7. true... Back to our Example: 02/3/0

2 .Propositions: d: take the day off. r: rains. s: it snows. f: basement floods. c: call for help. 2.Hypotheses: d r V s r f f V s c c Note: We should get to the conclusion: d. Construct the proof in a table format please! Step Reason. c Hypothesis 2. f v s c Hypothesis 3. (f v s) Modus tollens-step 4. f s De Morgans law 5. f Simplification on 4 6. r f Hypothesis 7. r Modus tollens on 5&6 8. s Simplification on 4 9. r s Conjunction on 7&8 0. (r v s) De Morgan s law on 9. d (r v s) Hypothesis 2. d Modus Tollens on 0 & 2 02/3/0

3 Mathematical Induction:A very special rule of Inference! I. Introduction and Definitions. Many theorems in discrete math claim that some propositional function P(n) is true for all n=,2,3 2. P(n) could take on many forms; it could take : An inequality, n < 2 n, A formula A property about a graph a treee of n vertices and e edges has n= e+. A statement about algorithmic complexity, algorithm is O(n^2). A method called mathematical induction is well suited to proving such theorems. Mathematical induction consists of two steps:. Basis step: show that P() is true 2. Induction step: show that for every n P(n) P(n+) P(n) is called the induction hypothesis. In logic form, induction amounts to the statement: why do these two steps imply P(n) for all n? {P() [P(n) P(n+)]} P(n). P() is true P() P(2).then P(2) P(3)..etc so P(n) is true for all n. II. Examples: Please write them down! Remarks: We are not assuming what we want to prove, we are trying to show that if P(n) is true, then P(n+) must also be true. ( for fixed n). Note that induction is NOT a constructive proof technique, it is useful only for verifying a formula, not for producing one. Therefore it works only in conjunction with other constructive methods. 3 02/3/0

4 In both of these examples, establishing the induction step was straightforward, often the hardest part in an induction proof is figuring out how to proceed in the induction step(see examples below). Ex. Harmonic series: H n = 2. Ex2. Show that 5/(7 n -2 n ) n k= k 3. Ex3. Using the product rule show that the power rule for all positive integers n=,2,3, d dx x n = nx n holds III. 2 nd principal of Mathematical Induction The following variant of the principal of induction is sometimes more useful: To show P(n) is true for all n=,2,. Basis step (same as before) show P() Induction Now assume P(k) is true for all k=,2,,n And show that P(n+) is true. Still valid? P() P(2) P() P(2) P(3) P() P(2) P(3) P(4) etc Example: show that if n is an integer >, n is the product of prime factors. (--will do in lecture) Example2: show that any amount of postage 2 cents can be formed using only 4 and 5-cent stamps.( use both regular Induction and 2 nd principal) (--will do in lecture) 4 02/3/0

5 Back to Rules of Inference More definitions! I. Other rules of inference. There are similar rules of inference for quantified statements. We will discuss one in the lecture, make sure you understand the others. Universal Instantiation Let U be the universe of discourse, and P(x) a propositional function: x P( x) c U P(c) Example: Solution: Step Prove the following: No man is an island. Manhattan is an island. Therefore, Manhattan is not a man. P(x): x is a man. Q(x): x is an island. M: Manhattan. Reason. x ( P( x) Q( x) Hypothesis 2. P ( M ) Q( M ) univ. Instantiation 3. Q(M) Hypothesis 4. P(M ) Modus tollens on 2 & 3 II. Techniques of proof We now know how rules of inference can be used to obtain conclusions from hypotheses. In addition, different overall approaches can be taken to prove theorems. We will demonstrate several. 5 02/3/0

6 . Direct Proof: of p q: Assume p true; show that q is true. Example: Show that 3n + 2 is odd if n is odd. Answer: Assume n is odd, so that n = 2k + for some k Z Then 3n + 2 = 3(2k + ) = 6k + 5 =2(3k + 2) + Let k = 3k + 2 then 3n + 2 = 2k + Thus 3n + 2 is odd. 2. Indirect Proof of p q. Assume q true; show p true. (direct proof of contrapositive, which is equivalent to p q) Example: Show that 3n + 2 is odd if n is odd. (---will do in lecture) Remark: These tow together show that 3n + 2 is odd if and only if n is odd. Remark: A variant of indirect proof is proof by contradiction. Here if we are trying to prove p is true, we assume p is false and show some contradiction results; i.e that: p r ( r) for some proposition r Example: show 2 is irrational.(--will do in class) Lemma: p 2 even p even.(--will do the proof in class) Remark: In proving an implication p q Contradiction suppose p q is F, i.e p is T & q is F, i.e p q If we can then show that q p, we will have contradiction p q. This is really the same as proving contrapositive. 3. Theorems with Quantifiers Many mathematical theorems involve either or, i.e 6 02/3/0

7 x P(x) x P(x) ex ex x (x Z x R) x C (x 2 = ) 3. Existence Proofs One approach to showing x P(x) is to actually produce a c such that P(c) is true. This is a constructive existence proof. Example: Show for any integer n, there exists an integer greater than n, i.e n Z m Z (m f n). Proof: n+ > n ( that s all!) A more subtle approach is non-constructive. We indirectly show that a c for which P(c) is T must exist without actually producing it. A famous example is to show that for any integer n there is a prime number > n. Remark: this would imply there are many primes. Logically given n, define P(x): x prime & > n. P(x) Proof: Consider x = n! +. Like all integers, x must have some prime divisor > But notice: no integer in [,n] divides x Thus, x must have a prime divisor > n (possible x itself, if x is prime) Non-constructive proof, since we did not actually produce x just argued it must exist. 3.2 proofs As for proofs, a couple of important points... To prove a statement of the form : ( x P(x)) x ( P(x)) we need only produce a counterexample, i.e a c for which P(c) is false. This is usually much easier than establishing P(x). Example: show that this statement is false. All odd # s are prime. 7 02/3/0

8 Answer: since 9 is odd and 3/9, this is a conterexample & statement is false. 2. On the other hand, no number of examples are sufficient to prove a statement of the form P(x).P(x) may be true for many examples, and yet not all. Example: P(n): n 2 - n + is prime n P(n): is it a theorem? P() = T P(2) = 3 T P(3) = 7 T P(4) = 23 T P(5) = 3 T Must be true for n.right? The answer is No! try P() P() = = 2 which is composite! Theorem : dx x r Proof: let p: r> converges if & only if r>. x r q: dx we wish to show that p q converges Note that [p q ((p q) ( p q)) ] Proof. P q Suppose r> r- -r x dx = x = 0 converges dx x r - r p q Proof 2. Consider two cases: C: r = C2: r < In other words p C C2 8 02/3/0

9 Thus if we show (C q) (C2 q) same as showing p q Tautology [( C C2) s] [(C s) (C2 s)] proof by cases 2a. C q r = dx = lnx diverges x 2b. C2 q r x r < dx = diverges. r x r End of the lecture! Test # is on Thursday- February 22, /3/0

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