Posted: February 7th, 2015

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