Discrete Mathematics

Running head: Discrete Mathematics
DISCRETE MATHEMATICS
Name of the Student
Name of the university
Name of the author

1Mathematics
Part 1
Answer to the question 1:
a) As the domain of the mentioned relation is {a, b, c, d}.
Here c is an element of the domain set which has two possible range d and a, thus
relation is not a function of the set {a, b, c, d}.
b) As the domain of the mentioned relation is {a, b, c, d}.
Here, every element of the domain set has a defined range, thus this relation is a
function of the set {a, b, c, d}.
c) As the domain of the mentioned relation is {a, b, c, d}.
Here, every element of the domain set has a defined range, thus this relation is a
function of the set {a, b, c, d}.
d) As the domain of the mentioned relation is {a, b, c, d}.
Here, every element of the domain set has a defined range, thus this relation is a
function of the set {a, b, c, d}.
e) As the domain of the mentioned relation is {a, b, c, d}.
Here, every element of the domain set has a defined range, thus this relation is a
function of the set {a, b, c, d}.
Answer to the question 2:
a) The inverse of the function is f (A) is f-1(A) = {(e, a), (d, b), (b, c), (a, d), (d, e)}.
b) The inverse of the function is f (A) is f-1(A) = {(a, b), (d, c), (c, a), (a, d), (b, e)}.
c) The inverse of the function is f (A) is f-1(A) = {(a, a), (c, b), (b, c), (d, d), (e, e)}.
Answer to the question 3:
a) f(x) = 1/x
x = all the real numbers. Hence the answer will come in decimal form.

2Mathematics
f(x)= 1/2 = 0.5 where x= 2
Hence it is not a function.
b) f(x)= y such that y2 = x
For x = 2
y = √ 2
Hence, it is not a function for real number
c) f(x) = x2-1
Where x= 2
f(2)= 4 – 1
=3
Hence it is function for all real number.
Answer to the question 4:
a) The domain of the function is the set of all bit strings.
Domain= {0, 1, 00, 01, 10, 11, 000, 001, 0101, 011,....}
Co-domain= {0, 1, 00, 01, 10, 11, 000, 001, 0101, 011,....}
The range of this function is the set of all numbers of zeros in the bit string.
b) The function is f(x) = x2-3, domain of this function is (-∞ < x < ∞), interval notation
is (-∞, ∞). Range is f(x) ≥ -3, [-3, ∞]
Answer to the question 5:
a) One to one
f(x) = -3x+7
f(a) = f(b), a=b
-3a+7 = -3b+7

3Mathematics
a=b (hence it is one to one)
Onto
Let us assume,
f(x) = y belongs to all real numbers,
f(x) = y
-3x+7 = y
-3x = y-7
x = y-7/-3
Putting value of x in the equation, where x = y-7/-3
f(x) = -3(y-7/-3)+7
f(x) = y (proved)
The mentioned function is a bijective function since it satisfies the both one to one and onto.
b) One to one
f(x) = x2-2x+1
f(a) = f(b), if a = b
a2-2a+1 = b2-2b+1
a2-2a+2b-b2=0 (not proved)
hence this equation is not bijective.
c) One to one
f(x) = 3x3-3
f(a) = f(b) if a = b
3a3-3 = 3b3-3

4Mathematics
1
2
3
4
x
y
z
a3 = b3
a=b (Proved)
Onto
f(x) = 3x3-3,
f(x) = y belongs to all real numbers R
f(x) = y
3x3-3 = y
3x3 = y+3
x3 = y+3/3
x = 3
√ y +3/3
Now, putting the value of x in the function where, x = 3
√ y +3/3
f(x) = 3*(y+3/3)-3
f(x) = y (proved)
The mentioned function is a bijection function since it satisfies the both one to one and onto.
Answer to the question 6
a) R = {(1, x), (2, y), (3, z)}

5Mathematics
x
y
z
a
b
c
a
b
c
d
a
b
c
d
Here, in this relation there is no mapping for the element 4. Thus, it is not a function.
b) f(x)= {(x, a),(y, b),(z, a)}
It is a function. Only Surjective as two element of domain set has same value.
c)
It is a function. Bijective function.