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Angie Rangel DISCRETE MATHEMATICS MATH 170 PROJECT PART II Jose De Jesus Melendez Carlos Aguilar Drake Jain PROJECT PART II Section 4.6: # 22, Section 9.2: #33 Section 9.5: #10 Section 10.2: #9, 18 Section 4.6 INDIRECT ARGUMENT: CONTRADICTION AND CONTRAPOSITION WHAT IS ARGUMENT BY CONTRADICTION? Indirect proof It is a statement that is either true or false but not both Reduces assumption by reasoning to a contradiction METHOD OF PROOF BY CONTRADICTION 1. Suppose that the statement that has to be proved is false. 2. Logically, show that it leads to a contradiction 3. Conclude that the statement that has to be proved is true EXAMPLE OF CONTRADICTION Prove: For all integers x and y, x² ≠8y+2 So we must suppose that what we went to prove is false . Suppose there are integer x and y such that x² =8y+2 Then x²= 2 4y + 1 . So x² is even. X is also even, so x=2k for some integer k. Then x²=4x² We have 4x²=2 4y + 1 ≡ 2x²= 4y + 1 2x² is even and 4y + 1 is odd, so they can’t be equal. Thus we have a contradiction, so there must not be any integers x and y such that x²=8y+2 WHAT IS ARGUMENT BY CONTRAPOSITION? Form of indirect argument Logical equivalence between statement & its contrapositive STEPS TO PROVE BY CONTRAPOSITION Take contrapositive of statement Prove contrapositive by direct proof Conclude that original statement is true EXAMPLE OF CONTRAPOSITION ∀x in D, if P(x) Q(x). Take contrapositive: ∀x in D, if ~Q(x) ~P(x). QUESTION 22 Consider the statement: “For all real numbers r, if r² is irrational then r is irrational.” a) Write what you would suppose and what you would need to show to prove this statement by contradiction. b) Write what you would suppose and what you would need to show to prove this statement by contraposition. a) “For all real numbers r, if r² is irrational then r is irrational.” ∀x ∈ ℝ, 𝑟 2 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑟 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 . Proof by contradiction: Suppose not. That is, suppose that there is a real number r such that r² is irrational and r is rational. Show that this supposition leads logically to a contradiction. Contradiction: ∃𝑥𝜖ℝ, 𝑟 2 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑟 𝑖𝑠 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 Let r²= 2 So, if r²=√2, then r= √(√2) Statement is false. Original statement is true done by contradiction. SUMMARY ∀x ∈ ℝ, 𝑟 2 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑟 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 . To prove by contradiction, we must assume that what we went to prove is false. So we suppose that there exist a real number that if 𝑟 2 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 then 𝑟 𝑖𝑠 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 So when we let Let r²=√2, r= √(√2) was not rational. Hence, we have a contradiction. b) “For all real numbers r, if r² is irrational then r is irrational.” ∀x ∈ ℝ, 𝑟 2 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑟 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 Proof by contraposition: Suppose that r is a real number such that r is irrational. Show that r² is not irrational. Contraposition: ∃𝑥𝜖ℝ, 𝑟 2 𝑖𝑠 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑟 𝑖𝑠 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 Let r= a/b, a= 2 and b= 3 So, if r= 2/3 then r²= 4/9 Since r is rational, r² is rational. Original statement is true done by contraposition. Section 9.2 MULTIPLICATION RULE THE MAN BEHIND THE COUNTING Pierre-Simon Laplace Mathematician and astronomer His work led to the development of mathematical astronomy and statistics There was no early biography of Laplace Starring mathematical physicist between Newton and Maxwell MULTIPLICATION RULE If the operation consists of k steps and - The first step can be performed in 𝑛1 ways - The second step can be performed in 𝑛 2 ways . . . - The kth step can be performed in 𝑛 𝑘 ways Then the entire operation can be performed in 𝑛 1 𝑛 2 𝑛 𝑘 ways. - Discrete Mathematics (Susanna S. Epp) HOW TO DO IT Example: How many 4 -digit PIN (Personal Identification Number) numbers are there to create? 1 . Look for how many spaces you have ____ ____ ____ ____ In this case, you have 4. 2. Look at how many digits you can choose from. 0 1 2 3 4 5 6 7 8 9 Overall, you have 10 digits to choose from (including the endpoints). So each space gets to have a choice of the 10 digits. HOW TO DO IT (CONT’D) 3. Plug it in. _10_ _10_ _10_ _10_ 4. Multiply the spaces. _10_ x _10_ x _10_ x _10_ 5. There your number of possible PIN numbers 10 4 = 10000 6. DONE! TRY IT Section 9.2 #33 - Six people attend the theater together and sit in a row with exactly six seats. a. How many ways can they be seated together in the row? b. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat? c. Suppose six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row? SOLUTION a. How many ways can they be seated together in the row? 1 . How many spaces? -6 2. How many to choose from? -6 3. Plug it in. _6_ _5_ _4_ _3_ _2_ _1_ The reason why each space decrease because as each seat is taken, the person who sat down is pulled from the situation. 4. Multiply. 6x5x4x 3x 2x 1 5. Answer: 720 ways (or you could leave it as “6 x 5 x 4 x 3 x 2 x 1”) SOLUTION b. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat? 1 . How many spaces? - 5, because the doctor already occupies a seat 2. How many to choose from? -5 3. Plug it in. _1_ _5_ _4_ _3_ _2_ _1_ 4. Multiply. 1x5x4x 3x 2x 1 5. Answer: 120 ways SOLUTION c. Suppose six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row? 1 . How many spaces? - 3, because a couple can occupy two seats the number is reduced to half. 2. How many to choose from? - 3, because a couple counts as 1 person now 3. Plug it in. _3_ _2_ _1_ 4. Multiply. 3x2x1 5. Answer: 6 ways Section 9.5 COUNTING SUBSETS OF A SET: COMBINATIONS INFORMATION TO KNOW: Formula: C (n, k ) = 𝑛 𝑘 = 𝑛! 𝑘! 𝑛−𝑘 ! 𝑛 𝑘 , means “n chose k” = number of subset of size k that can be chosen from n elements. n = total number of elements with-in a set k = total that are chosen from the set Where n and k are both nonnegative integers with k ≤ n. When C(n, 0) : By definition 0! =1. Therefore, if k is zero and n 𝑛 0 is any nonnegative integer, then is the number of elements with-in the set of n. C(n, 0) = = 𝑛! (1)∗𝑛! 𝑛 0 =1 = 𝑛! 0! 𝑛−0 ! EXAMPLE: CSUMB has created an indoor -soccer league where teams of 6 (or 6-combination) members must be formed. If there are 31 people who want to play how many teams can be formed? n = 31 people who want to play k = 6 people per team C(31 , 6) = 31 6 = = 31! 6! 25 ! = 31∗30∗29∗28∗27∗26 6∗5∗4∗3∗2∗1 = 31! 6! 31−6 ! 31∗30∗29∗28∗27∗26∗25! 6 ∗5∗4∗3∗2∗1 !25! = 736,281 dif ferent teams can be formed. QUESTION 10 Two new drugs are to be tested using a group of 60 laboratory mice, each tagged with a number for identification purposes. Drug A is to be given to 22 mice, drug B is to be given to another 22 mice, and the remaining 16 mice are to be used as controls. How many ways can the assignment of treatments to mice be made? (A single assignment involves specifying the treatment for each mouse—whether drug A, drug B, or no drug.) Solution: Total Mice = 60 = n Drug A = 22 mice = k (60¦22) Drug B = 22 mice = k (38¦22) Control = 16 = k (16¦16) Drug A + Drug B + Control 60 22 + = 38 22 + 60! 22! 38! 16 16 + 60! 22! 60−22 ! 38! 16! + 22!16! 16!0! = + 38! 22! 38−22 ! + 16! 16! 16−16 ! = 60∗59∗59∗58∗57∗57∗56∗56∗54∗53∗52∗51∗50∗49∗48∗47∗46∗45∗44∗43∗42∗41∗40∗39∗38! 22∗21∗20∗19∗18∗17∗16∗15∗14∗13∗12∗11∗10∗9∗8∗7∗6∗5∗4∗3∗2∗1! 38! + 38∗37∗36∗35∗34∗33∗32∗31∗30∗29∗28∗27∗26∗25∗24∗23∗22! 22!16∗15∗14∗13∗12∗10∗11∗9∗8∗7∗6∗5∗4∗3∗2∗1! = 5,848,876,094,823,595 + 22,239,974,430 + 1 = 5,848,898,334,807,026 + 16! 16!0! SUMMARY This question can be solved as an inclusion/exclusion problem. That is because in the problem is states “A single assignment involves specifying the treatment for each mouse”, therefore no mouse can be given more than one drug type. Thus leading it to be and inclusion/exclusion problem, in which this case it is an exclusion problem where the number of elements is reduced for each subset (here the subsets are Drug A, Drug B, and Control). Therefore, you start with 60 mice total (n), and 22 mice (k) are to be chosen to be in the first subset (Drug A). So, Drug A= C(60, 22) Since the first combination has taken away 22 mice from the entire set (n), the second subset (Drug B) only has 38 mice (n) in which 22 mice (k) can be chosen. So, Drug B= C(38, 22) Finally for the last subset, since both first subset (Drug A) and the second subset (Drug B) have used up 44 mice out of the 60 total. The last subset (Control) is left with 16 mice (n) in which they can chose 16 (k) to have in the Control group. So, Control= C(16, 16) therefore there is only 1 combination for this subset. Section 10.2 TRIALS, PATHS, CIRCUITS INFORMATION TO KNOW: Walk: An edge can be repeated, as well as vertices, and also the graph does not have not be connected. Trial: Edges cannot be repeated, but vertices can be repeated. Path: Edges, and vertices cannot be repeated, and so one cannot end at the same point they started with. Closed Walk: Edges and vertices can be repeated and you can end at the same point you started with. Circuit: Edges cannot be repeated, but vertices can, also you can end at the same point you started with. INFORMATION TO KNOW: Simple Circuit: Edges cannot be repeated, and only the first and last vertex can be repeated, therefore you can start and end at the same vertex. Euler’s Circuit: A Euler Circuit can be made when a graph is connected (all vertices are connected to one another by an edge) and the degree of every vertex of the graph has a positive even degree. Hamiltonian Circuit: Is an simple circuit that contains every vertex in the graph, in which every vertex appears only once excluding the first and last vertex which are resulting to be the same. EXAMPLE Does the following graph have a Euler’s circuit? SOLUTION Yes it does because the graph is connected and also each vertex has a degree of positive even integer. Therefore, the Euler circuit is 𝑉2 𝐸1 𝑉1 𝐸8 𝑉4 𝐸7 𝑉5 𝐸6 𝑉4 𝐸5 𝑉3 𝐸4 𝑉2 9. DOES THE GRAPH HAVE AN EULER CIRCUIT? JUSTIFY YOUR ANSWERS. A) G is a connected graph with five vertices of degrees2,2,3,3, and 4. B) B) G is a connected graph with five vertices of degrees 2,2,4,4, and 6. C) C) G is a graph with five vertices of degrees 2,2,4,4, and 6. SOLUTION: A) No this graph G isn’t a Euler circuit, however it is a Euler path. B) Yes by Euler’s theorem 10.2.3, the connected graph G is a Euler circuit. C) Not necessary, because we do not know if the graph is connected or not. SUMMARY: In Euler’s theorem 10.2.3 it states hat a connected graph G has a Euler circuit if all the vertices have a even positive degree. with this being said the graph mush be connected as well with all even degrees. also it is possible to have a Euler path if there is two degrees of odd integers from Euler’s theorem 10.2.4 is states that a connected graph G can be an Euler path if and only if there are exactly two vertices that are odd positive integers. A) This is not a Euler circuit, but a Euler path.10.2.4 Euler’s theorem states that a connected graph G contains a Euler path if and only if there are exactly two vertices with odd degrees. Graph G with degrees of 2,2,3,3,and 4 contains exactly two degrees of odd integers. B) Yes by Euler’s theorem this is a circuit because all the degrees are even integers and the graph is connected. The theorem states that the graph must be connected and contain vertices with all even degrees. the statement states that the graph G is connected and has five vertices of all even integers 2,2,4,4, and 6. C) The reason why the answer to the statement is not necessarily is because the Euler theorem 10.2.3 states that the graph is connected. and the statement does not state that the graph is connected but it does state that there are five vertices with degrees of 2,2,4,4, and 6 even though the theorem states “if and only if the vertices are all even degrees” it still states that the graph G is connected. This statement has nothing about the graph being connected or not so with that being said we can say not necessarily. There is not enough information provided with it besides that all vertices are even. QUESTION 18 Is it possible to take a walk around the city whose map is shown below, starting and ending at the same point and crossing each bridge exactly once? If so, how can this be done? SOLUTION The way that this can be answered is by the following circuit: B →D→C→A→D→E→A→B SUMMARY: This question is looking for a Euler’s circuit that is because which can be applied by Theorem 10.2.3, there are five edges who’s each degree is of even amount. Therefore, since with-in an Euler’s circuit every vertex has a degree of even amount. Which by the map vertex B(2), D(4), C(2), A(4), D(4), E(2). Which means a Euler’s circuit can be made. •Therefore, you start with B and the use the first bridge to D. So, B→D •Since D has an even amount of bridges connecting to it (degree amount) you can connect to either A, C or E. So, D→C •Since, C also has an even amount of bridges connecting to it (degree amount) you can connect to A, you cannot go back to D because the problem states that you cannot repeat the bridge. So, C→A •Then, From A since you have to get to E, you can either go to D then E or just E. So, A→D •Finally, from D all that is left is to take the bridge to E and then to A, and one more bridge up to B. So, D→E→A→B •Therefore, the answer is: B →D→C→A→D→E→A→B BIBLIOGRAPHY Epp, Susanna S. Discrete Mathematics With Applications 4 th Edition. Boston, MA: Brooks/Cole Publishing Company, 2004. Print. Rowlinson, J. S. "Laplace: The Man." Notes and Records of the Royal Society 60.2 (2006): 221-223.