School

CUNY New York City College of Technology **We aren't endorsed by this school

Course

CST 2357

Subject

Mathematics

Date

Nov 20, 2023

Type

Other

Pages

10

Uploaded by MasterLobsterMaster534 on coursehero.com

Question 1:
Solve the following questions from the discrete Math zybook:
A. Exercise 4.1.3, sections b, c
b)
Solution:
f(x) is not a well-defined function from R to R. x = +2 & -2. The function is not defined.
c)
Solution:
f(x) is a function from R to R because for every
, there is
.
Range:
B. Exercise 4.1.5, sections b, d, h, i, l
b)
Let A = {2, 3, 4, 5}.
f: A
Z
such that f(x) = x
2
Solution:
f(x) = {4, 9, 16, 25}
d)
f: {0,1}
5
Z
. For x
∈
{0,1}
5
, f(x) is the number of 1's that occur in x. For example f(01101) = 3,
because there are three 1's in the string "01101".
Solution:
Range of f(x) = {0, 1, 2, 3, 4, 5}
h)
Let A = {1, 2, 3}.
f: A × A
Z
×
Z
, where f(x,y) = (y, x).
Solution:
Range of f(x, y): {(1,1), (1, 2), (1, 3), (1, 4), (2,1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

i)
Let A = {1, 2, 3}.
f: A × A
Z
×
Z
, where f(x,y) = (x,y+1).
Solution:
Range of f(x, y): {(1,2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}
l) Let A = {1, 2, 3}.
f: P(A)
P(A). For X
⊆
A, f(X) = X - {1}.
Solution:
Range of f(x): {{0}, {2}, {3}, {2, 3}}.

Question 2:
Solve the following questions from the discrete Math zybook:
A. Exercise 4.2.2, sections c, g, k
c)
Solution:
One - to -One:
Yes
Onto:
No
Not Onto example:
There is no
such that h(x) = 3. Hence, the function is not onto.
g)
f:
Z
×
Z
Z
×
Z
, f(x, y) = (x+1, 2y)
Solution:
One - to -One:
Yes
Onto:
No
Not Onto example:
There is no
such that 2y = 3. Hence, the function is not onto.
k) f:
Z
+
×
Z
+
Z
+
, f(x, y) = 2
x
+ y.
Solution:
One - to -One:
No
Onto:
No
Not One to One example:
= (1, 5),
&
= (2, 3),
. Since,
=
when
& one-to-one function requires that
when
. Hence, the function is not one-to-one.
Not Onto example:
There is no
&
such that
. Hence, the function is not onto.

B. Exercise 4.2.4, sections b, c, d, g
b)
. The output of f is obtained by taking the input string and replacing the first
bit by 1, regardless of whether the first bit is a 0 or 1. For example, f(001) = 101 and f(110) =
110.
Solution:
One - to -One:
No
Onto:
No
Not One to One example:
f(000) = 100 & f(100) = 100, also f(010) = 110 &f(110) = 110, since multiple input string lead to
the same output string, the function is not one-to-one.
Not Onto example:
Since 010 is in solution space, the range of solutions does not include any spring starting with
0.
Hence, the function is not onto.
c)
. The output of f is obtained by taking the input string and reversing the bits.
For example f(011) = 110.
Solution:
One - to -One:
Yes
Onto:
Yes
d)
. The output of f is obtained by taking the input string and adding an extra
copy of the first bit to the end of the string. For example, f(100) = 1001.
Solution:
One - to -One:
Yes
Onto:
No
Not Onto example:
There is no input string in {
} such that the input string = 1100.
Hence, the function is not
onto.

g) Let A be defined to be the set {1, 2, 3, 4, 5, 6, 7, 8} and let B = {1}. f: P(A)
P(A). For X
⊆
A,
f(X) = X - B. Recall that for a finite set A, P(A) denotes the power set of A which is the set of all
subsets of A.
Solution:
One - to -One:
No
Onto:
No
Not One-to-one example:
For X = {1, 2}, f(X) = X-B = {1, 2} - {1} = {2} & X = {2}, f(x) = X -B = {2} -{1} = {2}. Since both X = {1,
2} & X ={2}, which is = to {2}.
Hence, the function is not one-to-one.
Not Onto example:
There is no
such that f(X) = X-B = X -{1} = {1}. So, f(X) = {1} in P(A).
Hence, the
function is not onto.

Question 3:
Give an example of a function from the set of integers to the set of positive integers that is:
One-to one , but not Onto
Solution:
Onto, but not one-to-one
Solution:
One- to -one and onto
Solution:
Neither one-to-one nor onto
Solution:

Question 4:
Solve the following questions from the Discrete Math zyBook:
A. Exercise 4.3.2, sections c, d, g, i
c)
Solution:
Since the function is a bijection (both one-to-one and onto) , the inverse
is well-defined and
d)
Let
A
be defined to be the set
{1,2,3,4,5,6,7,8}. f:P(A)
{0,1,2,3,4,5,6,7,8}.
For
X
⊆
A
,
f(X)=|X|
. Recall that for a finite set
A
,
P(A)
denotes the power set of
A
which is the set of all
subsets of
A.
Solution:
Since the X = {1,2} & X ={1,3} both lead to f(x) = |{1,2}| = |{1,3}| = 2, the function is not
one-to-one. Since the function is not bijection,
is not well-defined.
g)
. The output of
f
is obtained by taking the input string and reversing the bits.
For example,
f(011)=110.
Solution:
The function is bijection.
. The output of
is obtained by taking the input
string and reversing the bits. Example,
(110) = 011.
i)
Solution:
This function is a bijection.
.

B. Exercise 4.4.2, sections b-d
b)
Evaluate (f ο h)(52)
Solution:
(f ο h)(52) =
= 121.
c)
Evaluate (g ο h ο f)(4)
Solution:
(g ο h ο f)(4) =
= 16.
d) Give a mathematical expression for h ο f.
Solution:
h ο f =

Question 5:
Solve the following questions from the Discrete Math zyBook:
A. Exercise 4.4.6, sections c-e
c)
What is (h ο f)(010)?
Solution:
h ο f(010) = h(110) = 111
d)
What is the range of h ο f?
Solution:
Range of f: {{110}, {110}, {101}, {111}}
Range of
h ο f: {{101}, {111}}
e)
What is the range of g ο f?
Solution:
Range of f: {{100}, {110}, {101}, {111}}
Range of
h ο f: {{001}, {011}, {101}, {111}}

B. Exercise 4.4.8, sections c, d
c) f ο h
Solution:
f ο h =
=
d) h ο f
Solution:
h ο f =
=