Discrete Mathematics Assignment Solution - Functions and Relations

This document provides a detailed solution to a discrete mathematics assignment, focusing on the concepts of functions, one-to-one functions, onto functions, domain, range, and codomain. The solution addresses questions related to the properties of functions, including determining whether a function is one-to-one or onto, identifying the domain, range, and codomain of different functions, and analyzing composite functions. The assignment covers specific examples, such as determining the range of a function, and evaluating the existence of inverse functions based on the one-to-one property. Moreover, the solution includes analysis of composite functions, determining their domain, range, and codomain and whether the composite functions exist or not. The solution provides a comprehensive understanding of function properties and their applications within the scope of discrete mathematics.

Question 4
a. The cardinality of set A=n while that of set B=m. This implies that the n elements of the
domain set can map to any of the m elements of the codomain hence the number of
functions
mn
b. The cardinalities in A, n=2, the cardinalities in B, m=3
The size of the
BA =32=9 functions
c. From the table
Domain {1,2,3}
Range {1,2}
Codomain {1,2}
Question 5
One to one functions
g 1: A → B such that g 1= { ( 1, y ) , ( 2, w ) , ( 3 , x ) } isa one ¿ one function
g 3 : A → C such that g 3= { ( 1 , a ) , ( 2 , c ) , ( 3 , b ) } is a one ¿ one function
g 1∧g 3 are one ¿ one functions because no two elements∈the domain of g 1, , g 3 iscoreeposnding
to the same element in the range of g1, g3.

G2 is not one to one function because g2(x)=g2(w)=b
G4 is not one to one because g4(1)=g4(2)=g(3)=b
Onto functions
G1 ,g2 and g3 are onto functions because in the three functions, the range is equivalent to the
codomain
G4 is not an onto function because in this function the codomain is {b} while the range is
{a,b,c}. From the definition of onto functions, if the range and codomain are different then the
functions are not onto functions.
Question 6
a.
Ranges for
p:[1,2] → [0], range {0}, zero
q:Z →Z where q(n)=2n, range is all even integers
r: R-(-1) → R where r(x)= 1
1+ x , range is all real (R) except zero
s: Range is {a,b}
Onto functions
p is an onto function because both range and domain are zero

q is not onto function because the codomain contains all the integers while the range is
only made of even integers
r is an onto function because both the range and codomain are all real numbers
s is not an onto function because the range {a,b} is not the same as the codomain (range
is all real while codomain {x : a ≤ x ≤ b }
b. One to one functions
p is not one to one because p(1) =p(2)=0
q is one by one function because every value of the input we get a unique integer z
r is one-to-one because every value of every value of the input we get a unique integer
s is one to one because for every input, there is a unique value of output.
c. They have inverse functions because they are one-to-one functions and one-to-one
function have true inverses.
n= q(n)
2
Domain: All real
Range: All real
Codomain: all real
For r,
We have
r ( x )= 1
x+1

r (x )(x+ 1)= 1
1
r ( x ) x=1−r ( x )
x= 1−r (x )
r (x)
r−¿ ( y )=r= 1−r (x)
r (x) ¿
Domain: all real except r(x)=0
Range: all real except for r(x)=1
Codomain: all real except for r(x)=1
For s,
We have
s−¿ ( y ) =x= s (x)−a
b−a ¿
Domain {a b}
Range {0,1)
Codomain(0,1)
Solution to Question 7
p o q= p(q(n))=p(2n)=

[1,2] →[0]
Domain=[1,2]
Range=codomain=[0]
The inverse does not exist. The inverse will exist when p o q (1)= p o q (2)
r o q= r(q(n))= r(2n)
The composite cannot exist because q is a function of n while r is function of x
q o s
The composite cannot exist because s is function of x while q is a function of n
r o s= r(s(x))=r((b-a)x-a)= (b-a) 1
x+1
The core domain, range and domain of a composite function are those of the last function
in the composite
Domain
[0,1]
Codomain
{x :a ≤ x ≤ b }
Range is {a,b}