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MTH 231 Lecture Notes - Lecture 9: Contraposition, If And Only If

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20 Jul 2018

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Oregon State University

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(3) unpack defs (3) unpack defs (1) proof by contraposition (1) prof by contradiction. Last time proving (cid:1868) (cid:1869) (2) suppose (cid:1869) (2) suppose (cid:1868) (cid:1869) (4) argue that (cid:1868) is true (4) argue that (cid:1870) (cid:1870) are true for some proposition (cid:1870) (5) conclude (cid:1868) (cid:1869) (5) conclude (cid:1868) (cid:1869). To prove (cid:1868) (cid:1869: show (cid:1868) (cid:1869, show (cid:1869) (cid:1868, conclude (cid:1868) (cid:1869) . To prove (cid:1868)(cid:2869) (cid:1868)(cid:2870) (cid:1710) (cid:1868) (1) (cid:1868)(cid:2869) (cid:1868)(cid:2870) (cid:884)(cid:4666)(cid:1866) (cid:883)(cid:4667) (2) (cid:1868)(cid:2870) (cid:1868)(cid:2871) arguments (cid:1709) (cid:4666)(cid:1866) (cid:883)(cid:4667)(cid:1868) (cid:2869) (cid:1868) To prove (cid:1868)(cid:2869) (cid:1868)(cid:2870) (cid:1868) (1) (cid:1868)(cid:2869) (cid:1868)(cid:2870) (2) (cid:1868)(cid:2870) (cid:1868)(cid:2871) (cid:1709) (cid:1866) argument (cid:4666)(cid:1866) (cid:883)(cid:4667)(cid:1868) (cid:2869) (cid:1868) (cid:4666)(cid:1866)(cid:4667)(cid:1868) (cid:1868)(cid:2869) Saving of (cid:4666)(cid:1866)+(cid:883)(cid:4667) conclude (cid:1868)(cid:2869) (cid:1868)(cid:2870) (cid:1710) (cid:1868) (cid:884)(cid:4666)(cid:1866) (cid:883)(cid:4667) (cid:1866)=(cid:1866) (cid:884) arguments for (cid:1866) propositions. Example: (cid:1868)(cid:2869) (cid:1868)(cid:2870) (cid:1868) (cid:2869) (cid:2870) (cid:2871) (1) direct proof. Example: (cid:1868)(cid:2869):(cid:1866) is odd (cid:1868)(cid:2870):(cid:1866)+(cid:883) is even (cid:1868)(cid:2871):(cid:1866)(cid:2870) is odd. Proof (cid:1868)(cid:2869) (cid:1868)(cid:2870) (2) suppose (cid:1866) is odd (3) then (cid:1866)=(cid:884)+(cid:883) for some (4) then (cid:1866)+(cid:883)=(cid:4666)(cid:884)+(cid:883)(cid:4667)+(cid:883)

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Related Questions

Complete problem 3 using the information above.

Math 151 Sections 34, 35, 36 Workshop 11 ted to arn This workshop deals with an addition formula in section 5.1. This formula is rel oft told story about the great mathematician Gauss. Here is one version of the story When Gauss was ten years old, he attended a certain math class with other kids. One fine day, the math teacher discovered that he was running out of time to grade the latest batch of math homeworks. This is why he decided to finish the grading during his math period. To keep the students busy while he graded, he gave them the following task at th of the period: Use your slate (the small personal blackboard that every student had start in class) to add up the numbers 1,2,3,... , 100 Seconds after the teacher gave this in-class assignment, little Gauss pointed to his slate Here is the answer." Then Gauss did nothing for an hour, while the other students labored with the additions and the teacher graded. At the end of the perioc Gauss showed the correct answer 5050 on his slate. Only Gauss got the correct answer. Then Gauss told everybody that the sum 12 +3.+ 100 is (1+100)+(2+99) + (3498)+ Problem 1. Explain carefully what Gauss had done. Include the reason why it is impor tant that 100 happens to be an even number Problem 2. Let N be an even positive integer. Use the method of Problem 1 to prove the formula 1 +2 +3+.. . +N(N. Include the reason why it is important that N +(50+51) = 101 + 101 +101 + +101-101.50-5050 an even number. Problem 3. The formula is still valid when N is an odd integer. Prove this, using a variation on the argument given by Gauss The story continues: Gauss then gave the class a lecture on how his method could be used to sum an arithmeric series (a +1.b) (a +2.b) (a +3.b)+. (a +N b), where a and b are constants Problem 4. Find a formula for the sum (a+1.b)+ (a+2.b) + (a +3.b)+ +(a+ N.b) Prove this formula the way Gauss would have done it. You have to do the case when N is even and the case when N is odd Now let us consider a fictional alternate history where Gauss had not thought of this very clever trick. We can imagine the ten year old Gauss going home and doing the sequential computations 1+2 1+2+3 1+2+3+4 1+2+3+4+5 and so forth, just to see if he could find a pattern in the sequence of answers. He would have discovered somethin like this: once he had the sum 1+2+3+...+ N, he could simply add N +1 to that sum and that would give him the sum 1+2+3+ + (N +1) very quickly Now ass N(N+1 ume that, by trial and error, Gauss discovered that the formula 1+2+3+ is true for several values of N. How could he prove that the formula is r for ALL N? The insight described in the previous paragraph might have led Gauss to other words, he only needed to show: Formula for 1+2+3+ + N plus N +1 equals formula for 1+2+3++N+1). He would have discovered the principle of mathema -.. + N eally true 1)+1) think the following: All he had to do was prove mm) + (N + 1) NN+1 (N+1)(( In ered the principle of mathematical induction Problem 5. Prove N(N+1) (N + 1 of the formula for 1+2+3+.+N. (Nt1MNHHI) Show how this leads to a uCtion a proof

Problem 3

Math 151 - Sections 34, 35, 36 - Workshop 11 This workshop deals with an addition formula in section 5.1. This formula is related to an ld story about the great mathematician Gauss. Here is one version of the story Wh en Gauss was ten years old, he attended a certain math class with other kids. One fine day, the math teacher discovered that he was running out of time to grade the latest batch of math homeworks. This is why he decided to finish the grading during his math lecture period. To keep the students busy while he graded, he gave them the following task at the start of the period: Use your slate (the small personal blackboard that every student had in class) to add up the numbers 1,2,3,... , 100 Seconds after the teacher gave this in-class assignment, little Gauss pointed to his slate and yelled "Here is the answer." Then Gauss did nothing for an hour, while the other students labored with the additions and the teacher graded. At the end of the period Gauss showed the correct answer 5050 on his slate. Only Gauss got the correct answer Then Gauss told everybody that the sum 1+2+3+ + 100 is ( 1 + 100) + (2+99)+ (2+98)+--+ (50+51) = 101 + 101 + 101 +. . . +10 101. 50-5050 Problem 1. Explain carefully what Gauss had done. Include the reason why it is impor tant that 100 happens to be Problem 2. Let N be an even positive integer. Use the method of Problem 1 to prov the formula 1 +2+3+.+N) be an even number Problem 3. The formula is still valid when N is an odd integer. Prove this, using a variation on the argument given by Gauss The story continues: Gauss then gave the class a lecture on how his method could be used to sum an arithmeric series (a+1.b) + (a+2 b) + (a+3.b)+. + (a + N b), where a and b are constants Problem 4. Find a formula for the sum (a+1.b)+(a+2.b)+ (a +3.b)+(a+ N b) Prove this formula the way Gauss would have done it. You have to do the case when N is even and the case when N is odd Now let us consider a fictional alternate history where Gauss had not thought of this ve clever trick. We can imagine the ten year old Gauss going home and doing the sequential computations 1+2 1+2+3 1+2+3+4 1+2+3+4+5 and so forth, just to see if he could find a pattern in the sequence of answers. He would have discovered somethin like this: once he had the sum 1+2+3++N, he could simply add N+1 to that sum, and that would give him the sum 1+2+3 Now assume that, by trial and error, Gauss discovered that the formula 1+2+3+ +N an even numbe +) Include the reason why it is important that +Â·Â·Â· + (N +1) very quickly is true for several values of N. How could he prove that the formula is really true for ALL N? The insight described in the previous paragraph might have led Gauss to think the following: All he had to do was prove N(N+1) Â± (N + 1) other words, he only needed to show: Formula for 1 + 2+3+.+ N plus N +1 equals formula for 1+2+3+(N+1). He would have discovered the principle of mathematical induction Problem 5. Prove of the formula for 1+2 +3+... + N (N +1)(+1X++1) Show how this leads to a proof N(N+1 N+)

I have the answer for the following question. But some of theanswers are not correct.

Please do the following question by yourself and check which arethe wrong answers,please post your answers and compare to it.

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S