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Discussion 4 Spring 2015 Week of Feb. 9 Students are currently working on homework 2 and homework 3 will cover only proofs and will be due on the day of the test. Quiz 1 will be given on Thursday of this week and Monday and Tuesday of next week. There will be no proofs on the quiz. (I sent out an email about this to the class.) So, for the Thursday discussion give the quiz and give them a copy of the worksheet so they know the kinds of problems I ask on proofs. We’re about halfway through proofs in class and should start induction on Tuesday. Discussion Notes: 1. Go over any questions from last week’s worksheet that caused more than a few students difficulty. 2. The topics this week are the remainder of chapter 1 in the Ensley and Crawley text and material on proofs from the beginning of Chapter 2. You probably won’t be able to get through all of this, so save whatever you have left over for next week’s discussion and let me know where you ended up. a. Look at the following problems from the text. Even though the answers are given in the book they need more discussion of how they are done. You don’t have to do truth tables on the problems from 1.6 but be make sure to clearly justify the steps in the argument and the validity of the reasoning. Section 1.5 # 14 c, 16 c Section 1.6 # 10 a, 11 a b. Validity of arguments. If the argument below is valid state whether it is modus ponens or modus tollens. If it is not valid determine what fallacy was used. All honest people pay their taxes Mr. Brown pays his taxes Therefore, Mr. Brown is honest. Change the first sentence to “Only honest people pay their taxes” and see how the logic changes. c. Use a propositional argument to determine where the pencil is. Be sure to specify which type of logical argument is being used to draw conclusions (i.e. modus ponens, modus tollens, etc.) Define propositions and show how to reason through the statements. 1) 2) 3) 4) 5) If I left the pencil on the kitchen table, then I saw it at breakfast. I was doing the crossword puzzle in the living room or in the kitchen If I did the crossword puzzle in the living room then the pencil is on the coffee table. I didn’t see the pencil at breakfast. If I did the crossword puzzle in the kitchen then it is on the kitchen table. (You should be able to conclude the pencil is on the coffee table.) 3. Examples of proofs. Avoid using properties that are very similar to the statement they are to prove. The hardest part for students is usually trying to use properties of numbers such as the product of two odd numbers is odd when that is in effect what they are proving. In general, try to use just definitions and algebra and avoid using number properties. Reasons must be shown for all steps in a proof. Example 1: Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd. Do the “if” and “only if” as separate proofs. Use only the definition of odd and algebra. Example 2: If m + n is odd, then m is even or n is even. Have them work through this one for a few minutes and go around and see how they are doing as they are working on it. If they need a hint suggest using a contrapositive proof. Rather than giving a solution to them point them to the answer on page 632 in the text--problem 13 c.) Example 3: Use a contrapositive proof to show that “if i is irrational and r is rational, then s = i + r is irrational”. Form the negation of the hypothesis so they know where they are headed. Here’s the proof. Let r be a rational number and let i be an irrational number. (given) Assume that r + i = s and that s is a rational number. Then, a, b, c, d Z, b 0 and d 0 such that r = a/b and s = c/d defn. rational number Thus, a/b + i = c/d (substitution) i = c/d - a/b = (bc - ad)/bd (algebra) Since the closure property tells us the product and difference of integers are integers, this implies i is rational, a contradiction of the hypothesis. Example 4: Prove both parts. Be sure to remind them of the definition of divides. a. Prove that if a|b and a|c then a|(b+c). b. Prove that if a|b and c|d then ac|bd (These are exercises 7a and 7 c in Section 2.2—answers are in the back of the book.) Things to watch out for: Be very sure to use only the definitions i.e. a|b means b = ak for k Z. Saying that b/a = k is incorrect; Remind them that divides is a relation not an operation so no numerical “answer” is produced and using the operation of division in the proof is incorrect. Proof of part b. b = ak for k Z d = cm for m Z bd = ak•d = ak•cm = (ac)(km) Then, ac | bd Definition of divides Definition of divides Algebra Definition of divides Example 5: Prove there does not exist a smallest positive real number. Example 6: Prove the square of an odd integer equals 8k + 1 for some integer k. Proof: Let n be an odd integer. (given) Then, n = 2r + 1 for some r Z (defn. of odd) Then, n2 = (2r + 1)2 = 4r2 + 4r + 1 = 4(r)(r + 1) + 1 (algebra) Either r or r + 1 must be even since they are consecutive integers If r is even, then n2 = 4(2j)(r + 1) + 1 = 8j(r + 1) + 1 (defn. even, substitution, algebra) If r+1 is even then n2 = 4(r)(2s) + 1 = 8rs + 1 (defn. even, substitution, algebra) 2 In either case, n has the form 8k + 1 Example 7: Using only the definitions of odd and even and algebra, prove that if x and y are odd then x2 + y is even. 4. Induction Proofs—Satya, skip this on Monday since I haven’t started it in class yet. By Tuesday they will have seen some of these examples in class so if you have time do it then. On induction proofs there are three parts. Explicitly prove the basis. State the induction hypothesis (just saying assume P(k) is not sufficient. The induction step is the hardest part for them. Do not let them work from P(k+1) back to P(k) since that shows the converse of what they must prove. Do not let them algebraically manipulate both sides of an equation or inequality and conclude they must be the same since everything cancels out or they can both be reduced to the same thing. Use the proof in Example 4 below as a model of how the proof should be done. Hold off any extras until next week. Example 1: Use mathematical induction to prove the following formula: 12 + 32 + 52 + … + (2n + 1)2 = (n+1)(2n + 1)(2n + 3)/3 for n 0 Example 2: Prove the following statement in two ways—directly and by induction. n3 + 2n is divisible by 3 for all nonnegative integers n. Example 3: Use mathematical induction to prove 1•1! + 2•2! + … + n•n! = (n + 1)! –1 for n 1 Example 4: Use mathematical induction to prove that 1/(1•2) + 1/(2•3) + … + 1/[n(n+1)] = n/(n + 1) (Substitution of the hypothesis is shown in dark red.) Basis: n = 1 lhs = 1/1•2 = ½ rhs = 1/(1+1) = ½ Hypothesis: Assume 1/(1•2) + 1/(2•3) + … + 1/[k(k+1)] = k/(k + 1) for k > 1 Induction step: 1/(1•2) + 1/(2•3) + … + 1/[k(k+1)] + 1/[(k+1)(k+2] = k/(k + 1) + 1/[(k+1)(k+2] = [k(k+2) + 1]/[(k+1)(k+2)] = [k2 + 2k + 1]/ [(k+1)(k+2)] = (k+1)2/[(k+1)(k+2)] = (k+1)/(k+2) Therefore, by the principle of mathematical induction this holds for all n. Note: Inequalities give the students lots of trouble. You may want to save these for next week. Example 5: Use mathematical induction to prove that 2n+1 < 3n for all n > 1. Example 6: Prove that (1 + x)n > 1 + xn for n > 1, x > 0 Basis: n = 2 (1 + x)2 = 1 + 2x + x2 > 1 + x2 since x > 0. Hypothesis: Assume (1 + x)k > 1 + xk for k > 1, x > 0 Induction step: (1 + x)k+1 = (1 + x)(1 + x)k > (1 + x)(1 + xk) = 1 + xk + x(1 + xk) = 1 + xk + x + xk+1 > 1 + xk+1 (removing x + xk makes the right side smaller) Therefore, by the principle of mathematical induction, this holds for all n 1.