I have this assignment for introduction to discrete structure. Attached are the questions with my answers but there is some answers that I have to fix and I just cant seem to figure it out please help.

What I have to fix? Here it is:

Problem 1b, though the union of the three sets is Z, there is overlap, so it is not a partition. In problem 4, the definition of symmetric is attempted but is not quite right. It should say, for every pair (x,y) found in the relation, the pair (y,x) must also be found in the relation. Consider this pair of pairs: (x,y) is (-3, 0), and (y,z) is (0, 3). Both are in R. Then consider (x,z), which is (-3,3). It is not in the relation. Problem 4b > and the real numbers fails to be a totally ordered set because it is not reflexive. Problem 6 asks for two proofs. Problem 7 asks for a proof by mathematical induction. It would be nice to see a base case and a proof of an inductive step. Problem 9a. For the case where the rational number p/q has p=0, there is no inverse. Problem 9b should express the two quantified variables. There exists x that for all y... Problem 10 e: Please be careful to use definitions that the community uses. See Wolfram math world, for example 

 

 

1. (10 pts) For each the following groups of sets, determine whether they form
a partition for the set of integers. Explain your answer.
a. A1 = {n Z: n > 0}
A2 = {n Z: n < 0}
A1 contains all integers greater than 0.
A2 contains all integers less than 0.
0 is not covered in both cases.
A1 ∩ A2 = ∅ ; however, A 1∪A 2 ≠ Z , as 0 is missing. Therefore, it is not
a partition of the set of all integers.
.
b. B1 = {n Z : n = 2k, for some integer k}
B2 = {n Z : n = 2k + 1, for some integer k}
B3 = {n Z : n = 3k, for some integer k}
B1 is the set of all even integers, B2 is the set of all odd integers, and B3
is the set of all integers divisible evenly by 3. E.g. B3 = {3, 9, 12, 15, 18…}
This is a partition of Z since B1 U B2 U B3 ¿ Z.
2. (10 pts) Define f: Z → Z by the rule f(x) = 6x + 1, for all integers x. a. Is f onto?
No. For a function to be onto the codomain must equal the range. A
counter example would be f(x)=y and solve for x. We find that x = y-1/6.
Let y=0 and x = -1/6. This means that, in order to get 0 (an integer) as the
resulting value for f(x), we have to input -1/6 into the function. -1/6 is not
an integer; therefore, the function is not onto.
b. Is f one-to-one?
Yes, f is one-to-one. For every x there will be a different f(x) and for every
f(x) there will be a different x.
c. Is it a one-to-one correspondence?
For a function to be a one-to-one correspondence it must be both one-to-one and
onto. It must have a codomain equal to the range, and each element of the domain
must map to only a single element in the range so based on exercise a we can see that
is not onto therefore is not one-to-one correspondence. d. Find the range of f
Range = {n ∈ Z | 6n+1 ∈
= {…, -11, -5, 1, 13, 19, …} Z} 3. (10 pts) f: R → R and g: R → R are defined by the rules:
f(x) = x2 + 2 ∀
g(y) = 2y + 3 ∀ x ϵ R y ϵ R Find f ◦ g and g ◦ f
2
2
f â—¦ g = f ( g ( y ) )=(2 y+ 3) +2=4 y +12 y +11
2
f â—¦ g = 4 y +12 y +11 g â—¦ f = g(f(x)) = g(x2 + 2)
=> g(x2 + 2) = 2(x2 + 2) + 3
=> 2x2 + 4 + 3 = 2x2 + 7
g â—¦ f = 2x2 + 7
4. (10 pts) Determine whether the following binary relations are reflexive,
symmetric, antisymmetric and transitive:
a. x R y ⇔ xy ≥ 0 ∀ x, y ϵ R Reflexive - Any relation to be reflexive, (x,x) should belong to R.
If we consider any value of x then x*x will always be an positive value >0. For
example X=2, Y=2 2*2 > = 0 or X= -4 Y= -4, -4*-4> = 0 therefore we can say R is
reflexive. Symmetric - any relation to be symmetric, (x,y) should belong to R and (y,x)
should also belong to R. here for any value of x and y if (x,y) belongs to R i.e,
x*y>=0 then y*x will also be > = 0 thus (y,x) will also belong to R. It is also
symmetric. Not antisymmetric because it is symmetric. Transitive - any relation to be transitive, must hold if (x,y) and (y,z) belongs to R
then (x,z) should belong to R. When x*y>=0 and y*z>=0 the we can say x*z will
also be >=0, thus (x,z) belongs to R. Is∀an
relation.
x equivalence
,
x> 0 [ x ] = {∀ y∨ y> 0 } ,
x< 0 [ x ] = { ∀ y| y< 0 } ,
x=0 [ x ] = R b. x R y ⇔ x > y ∀ x, y ϵ R Not reflexive: A counterexample to prove is not would be x=y, x=4; therefore, x
should be greater than y, but since 4 is not greater than 4, this relationship is not
reflexive. Not Symmetric - any relation to be symmetric, (x,y) should belong to R and (y,x)
should also belong to R. For any value of x and y if (x,y) belongs to R i.e, x>y
then y>x is not possible so we can say that R is not symmetric There is no x, y pairs that relate back to each other. E.g. (x, y) is found, but not
(y, x) for all x, y in R. Therefore, it is antisymmetric. Transitive - x, y, z are related. If x > y, and y > z, then x > z. Is not an equivalence relation, nor partial order.
c. x R y ⇔ |x| = |y| ∀ x, y ϵ R Reflexive - Any relation to be reflexive, (x,x) should belong to R. If we consider
any value of x then |x|=|x| will hold. R is reflexive Symmetric - It is symmetric was for all (x, y) there is a corresponding (y, x) pair.
E.g. (-1, 1), (1, -1). Because it is symmetric cannot be antisymmetric. Not transitive since no number is related to each other. Is not an equivalence relation, nor partial order. 5. (10 pts) Determine whether the following pair of statements are logically
equivalent. Justify your answer using a truth table.
p → (q → r) and p ∧ q → r p ∧ q → r p (q r)
p Q r T
T
T
T
F
F
F
F T
F
T
F
T
F
T
F T
T
F
F
T
T
F
F (q r)
T
F
T
T
T
F
T
T p p (q r)
T
F
T
T
T
T
T
T Statements are logically equivalent. T
T
T
T
F
F
F q T
T
F
F
T
T
F T
F
T
F
T
F
T r T
T
F
F
F
F
F p
∧ p q q ∧ → r
T
F
T
T
T
T
T 6. (10 pts) Prove or disprove the following statement:
∀ n, m ∈ Z, If n is even and m is odd, then n + m is odd
Then write the negation of this statement and prove or disprove it.
n + m is not odd
Given that n is even and m is odd, is always going to be odd; therefore, the negation of
the previous statement is false. 7. (10 pts) Prove the following by induction:
n
n+1
3 n2−n
3
i
–
2=
∑
=> ∑ ( 3 i – 2 ) +(3 n+1)
2
i=1
i=1
2 3 (n+1) −(n+ 1)
3 n 2−n
+(3 ( n+1 ) −2)=
2
2 2 3n^2 – n / 2 + (3n+1) = 3 n +5 n+ 2
2 8. (10 pts) Use the permutation formula to calculate the number permutations of
the set {V, W, X, Y, Z} taken three at a time. Also list these permutations. 5P3 = 5!
5!
= =60
( 5−3 ) ! 2! permutations {V,W,X} {V,W,Y} {V,W,Z} {V,X,W} {V,X,Y} {V,X,Z} {V,Y,W} {V,Y,X} {V,Y,Z} {V,Z,W} {V,Z,X}
{V,Z,Y} {W,V,X} {W,V,Y} {W,V,Z} {W,X,V} {W,X,Y} {W,X,Z} {W,Y,V} {W,Y,X} {W,Y,Z}
{W,Z,V} {W,Z,X} {W,Z,Y} {X,V,W} {X,V,Y} {X,V,Z} {X,W,V} {X,W,Y} {X,W,Z} {X,Y,V}
{X,Y,W} {X,Y,Z} {X,Z,V} {X,Z,W} {X,Z,Y} {Y,V,W} {Y,V,X} {Y,V,Z} {Y,W,V} {Y,W,X} {Y,W,Z}
{Y,X,V} {Y,X,W} {Y,X,Z} {Y,Z,V} {Y,Z,W} {Y,Z,X} {Z,V,W} {Z,V,X} {Z,V,Y} {Z,W,V} {Z,W,X}
{Z,W,Y} {Z,X,V} {Z,X,W} {Z,X,Y} {Z,Y,V} {Z,Y,W} {Z,Y,X} 9. (10 pts) Translate the following English sentences into statements of predicate
calculus that contain double quantifiers and explain whether it is a true
statement.
a. Every rational number is the reciprocal of some other rational number.
P ( x ) ∙ N ( x )=1
( ∀ P ( x ) ϵ θ ) (∃ N ( x ) ϵ θ) ¿
statement is true
Reciprocal is inversion of rational number. Dividing would equal 1.
b. Some real number is bigger than all negative integers.
x is real numbers
( ∃ x ∈ R ) , x> y
y is negative integers
statement is true.
x = 2 and y = -4
2>4
10. (10 pts) Consider the following graph: In each case, answer the question and then write the rationale for your answer.
a. Is this graph connected? Yes, no corners are separated from the rest of the graph
b. Is this a simple graph?
Yes, there are no multi edges. There are no loops nor parallel edges.
c. Does this graph contain any cycles?
It is possible if we start in the middle traveling to the left and then come
back to the starting point.
d. Does this graph contain an Euler cycle?
This cycle requires that all edges be used in a path that starts and stops at
the same vertex.
Is this graph a tree?
No. It does not have any open ends or a root vertex.

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