Math 1.2.3.4 home work redo

Math 1.2.3.4 home work redo Mathematics: Name Institution: Date Question 1: Answer Let P be the set of people Let Q be the set of Occupation Define function f: P?Q What is true? Everybody cannot have more than one occupation but somebody will have more than one occupation. Question 2: Solution: By the definition of f -1, f -1(y) = x    such that    f(x) = y But; f(x) = y 6x - 5 = y by definition of f y = (y + 5) / 6      simple algebra Hence f -1(y) = (y + 5) / 2 6a – 5 = (6(y + 5) / 6)-5 a = y so a = b Question 3: Solution f: R ?R defined by f(x) = 5x – 9 Proving that f is onto Let y = f(x) y = 5x – 9 Solve for x x = (y + 9) / 5 Then (y + 9) / 5 is real number y = f ((y + 9) / 5) = y y+9=5y y=2.25 x=12.5 Hence f(x) is onto f: R ?R defined by f(x) = 5x – 9 Question 4: Answer Let P be the set of people Let m: p ?P M(x) = the birth mother Y = some person Relationship of (m o m)(y) to y The person (m o m)(y) is the sister of y Question 5: Answer Let V be the set of all vertices in G Let E be the set of all edges in G Let i: E ?V x V Where i (e) = (a, b) i (e1) (a, b) = ve1, w e1 i (e2) (a, b) = ve2, we2 i (e3) (a, b) = ve3, vxe3 i (e4) (a, b) = we4, x e4 i (e5) (a, b) =  xe5, ye5 i (e6) (a, b) = x e6, ze6 i (e7) (a, b) = z e7, ye7 i (e8) (a, b) = ze8, ye8 Question 6: Answer R does not define an equivalent relation on the set (2, 3, 4, 6) R = {(2, 2), (3, 3), (4, 4), (6, 6), (3, 4), (3, 6), (6, 3)} Reflexivity fails fails. For instance 2R3 0r 3R2 but 2?3 Question 7: Answer Let T be the set of all movie actors and actresses For x, y ? T Define x R y The relation R is  reflexive, because every actor in b appears in a movie with a The relation R is symetric because a has been in a movie with b The relation R is not transitive because a could appear with b, b could appear in another movie with c, and that R does not guarantee that a will appear in a movie with c. Question 8: Solution A = {1, 2} Let A = {1, 2}. Subset of A x A are A,{(1 * 1), (1 * 2), (2 * 1), (2 * 2)} A , {1, 2, 2, 4} Question 9: Answer x R y, x R Z but y R` Z and x y Z is not even taking x= 1, y = 2, z= 3 xy = 1 * 2 = 2 (even) yz = 2 * 3 = 6 (even) xz = 1 * 3 = 3 (odd) x, y, z = (2,6,3) Question 10: Answer The = mod 3 relation is an equivalence relation on the set {1, 2, 3, 4, 5, 6, 7} Listing the equivalence classes: (1,4,7), (2,5), (3,6) Section 2: Question 1: Solution If teams = 18, Matches played per team = 18 – 1 = 17 Hence here the case will be 18 * 17 =306 games Question 2: Answer The minimum number of colors = No of Vertices – No of lines per vertex The minimum number of colors = 7 – 3 = 4 At least 4 colours Question 3: Answer Applying Euler’s formula Edges=vertices+faces-2 120+62-2 =180 edges Question 4: Answer The graph has 45 vertices The graph has an Euler circuit No. There are is an odd-degree vertices Question 5: Answer The graph has 45 vertices Yes, there are exactly two odd degree vertices Question 6: Answer The graph has 45 vertices Every vertex has degree 4, so each vertex "owns" half each of 4 different edges = 4 / 2 * 45 = 90 edges Question 7: Answer S = {1, 2, 3, 5, 10, 15, 20} (S, I) is a Poset Hasse Diagram Question 8: Answer Let X = {5, 10, 15, 20, 25, 30, 35, 40} Define a Relation ¦ on X For a, b e X, a ¦ b 35 and 40 are incomparable, because 35 ? 40 and 40 ? 35 Question 9: Answer R on {0, 2, 4, 6} is givem = n by R = {(0, 0), (2, 2), (4, 4), (6, 6), (0, 2), (2, 4), (0, 4), (4, 2)} R is not partially ordered Relation R fails. 2 R 0 and 4 R 2, but 2 ?6 Question 10: Answer Let X = {5, 10, 15, 20, 25, 30, 35, 40} Define a Relation ¦ on X For a, b e X, a ¦ b List all minimal elements of (x, ¦) {{5}, {10}} Section 3 Question 1: Answer L (n) = L (n - 1) + L (n - 2) How the definition of L (n) differ from F (n) More than 2 previous terms are required in the calculation of a new term Question 2: Answer P (n) = [P (n - 1)] 2 – n P(n)=p(n2-3n+1) P (1) = -1 P (2) = -3 P (3) = -2 P (4) = 1 Question 3: Answer Q (n) = Q (n - 1) + Q (n - 2) + Q (n - 3) Q (5) = 4 + 3 + 2 = 9 Question 4: Answer Sequence = 2, 6, 18, 54, 162, 486, 1458…… P (n) = P (n - 1) 0 1 P (n) = 3(n – 1) Question 5: Answer 0 1 R (n) = (n / 4) - 1 R (64) = (64 / 4) – 1 = 16 – 1 = 15 21milligram into 64th square = 21 * 1000 * 10^14 = 21000 * 10^14 Question 6: Answer f (n) = an +b when f(1) is = 4 a (1)+b = 4 or a + b = 4 f (2) = 10 This implies that a (2) + b = 10 By elimination a + b = 4 2a + b = 10(-)-a = -6 Therefore a = 6 Substitute a by 6 a + b = 4 6 + b = 4 b = 4 – 6 b = -2 Hence f (n) = 6n – 2 Question 7: Answer Q (n) = 2.Q (n – 1) – 3 Prove that Q (n) = 2n + 3 First box = 0 Second box = 20 + 3 = 1 + 3 = 4 Third box= 2^ (n - 1) + 3 Fourth box = 2^ (n - 1) + 3 Fifth box = f (n) So Q (n) = F (n) Question 8: Answer G (n) = G (n) = G (n - 1) + G(n - 2) - G(n - 8) Coefficients of G    Years n-1    n-2    n-8 1    0    -1    -7    -8 2    1    0    -6    -5 3    2    1    -5    -2 4    3    2    -4    1 5    4    3    -3    4 6    5    4    -2    7 7    6    5    -1    10 8    7    6    0    13 9    8    7    1    16 10    9    8    2    19 11    10    9    3    22 12    11    10    4    25 Rabbits live 11 + 10 + 4 = 25Years Question 9: Answer H (n) = H (n - 1) + 6n - 6 H (5) = (5 – 1) + (6 * 5) – 6 H (5) = 4 = 30 – 6 H (5) = 28 Question 10: Answer Sequence is S (n) = S (n - 1) First box = 0 Alice’s salary after n years = 35000 + S (1 + 0.04) n + 2500