CSCE 235
Homework
Assignment 5 Solution
Assigned: February 16, 2009
Due: 12:30 p.m. February 25,
2009
(Homework 5 minutes late will not be
accepted)
1. (15 points) For each of the following sets, determine
whether there is an one-to-one function from S to T, whether
there is a function mapping S onto T, and whether there is an
one-to-one correspondence between S and T. If the answer is YES give an example; if the
answer is NO explain briefly.
(a) S = { 1, 2 , 3 } and T = { d, e, f }
(b) S = { a, b } and T = { 1, 2, 3, 4 }
(c) S = { 1, 2, 3, 4 } and T = { a, b }
Solution:
First,
remember that a function from a set A to a set B is a binary relation f
from A to B with the property that for every 
, there is
exactly one 
such that 
. That means that every single element in the
domain must have a link to one of the elements in the target.
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Problem |
1-to-1
function? |
Onto? |
1-to-1
correspondence? |
|
(a) S = { 1, 2 , 3 } and T = { d,
e, f } |
Yes. For example, f = { (1,d), (2,e),
(3,f) }. |
Yes. For example, f = { (1,d), (2,e),
(3,f) }. |
Yes. For example, f = { (1,d), (2,e),
(3,f) }. |
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(b) S = { a, b } and T = { 1,
2, 3, 4 } |
Yes. For example, f = { (a,1), (b,2)
}. (It is important to understand that
it is okay that some of the elements in the target set are not mapped onto!) |
No. Since the number of elements in S is
smaller than the number of elements in T, it is impossible to map onto
all elements in T without having an element in S mapping to more than
one element in Tand that is not allowed based on the definition of a
function. |
No. The numbers of elements in the two sets are
different. So, a 1-to-1 correspondence
is impossible. |
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(c) S = { 1, 2, 3, 4 } and T = { a,
b } |
No. This is because the number of elements in S
is greater than the number of elements in T. Thus, a function f mapping S to
T would have to map at least two elements in S to the same
element in T. |
Yes. For example, f = { (1,a), (2,b),
(3,a), (4,a) }. (It is
important to understand that it is okay to map different elements in the
domain to the same element in the target!) |
No. The numbers of elements in the two sets are
different. So, a 1-to-1 correspondence
is impossible. |
2. (24 points) Determine whether each of these functions is
onto and one-to-one (where Z is the set of integers, and R is the set of real numbers).
Explain or show an example to support your answers. (Hint:
For (e)-(g), think carefully about the domain, target, and range
involved.)
(a)
f: Z → Z , 
(b)
f: Z → Z , 
(c)
f: R → R , 
(d)
f: R → R , 
(e)
f: Z → Z , 
(f)
f: R → R , 
(g)
f: R → Z ,
(h)
f: Z+ → R , 
Solution:
First,
remember that a function from a set A to a set B is a binary relation f
from A to B with the property that for every , there is
exactly one such that . That means that every single element in the
domain must have a link to one of the elements in the target. Also, for a function to be onto, then all the
elements in the target, B, must have a link from the domain.
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Problems |
One-to-one? |
Onto? |
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(a) f: Z → Z , |
Yes. Since this function maps from an integer to
another integer (smaller by 1), all elements in the domain set will have a
unique corresponding element in the target.
For example, {
, (-2,-1), (-1, 0), (0, -1), (1,0), (2,1), (3,2),
} |
Yes. All elements in the target will have a
corresponding element in the domain. |
|
(b) f: Z → Z , |
No. For example, |
No. The target set is the set of integers. However, the function only maps to positive
integers since |
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© f: R → R , |
No. For example, |
No. The target set is the set of real
numbers. However, the function only
maps to real numbers smaller than 7 since |
|
(d) f: R → R , |
Yes. Each real number will be mapped to a unique real number. In fact, the function resembles a graph like this:
|
Yes. The graph shows that |
|
(e) f: Z → Z , |
Yes. Each integer will be mapped to itself. (Note that the ceiling function actually
does not do anything when the domain is the set of integers.) |
Yes. Each integer will be mapped to itself and
all integers in the target set will have corresponding elements in the
domain. (Note that the ceiling
function actually does not do anything when the domain is the set of
integers.) |
|
(f) f: R → R , |
No. Now we consider the set of real numbers. So, it is possible to have two elements
from the domain set to map to the same element in the target set. For example, |
No. Now we consider the set of real numbers
in the target set. Since we know that
the ceiling function always return an integer, that means many real numbers
in the target set will not have corresponding elements in the domain. For example, an element such as 6.7 in the
target set does not have a corresponding element in the domain. Thus this function is no longer onto. |
|
(g) f: R → Z , |
No. Now we consider the set of real numbers. So, it is possible to have two elements
from the domain set to map to the same element in the target set. For example, |
Yes. See, here, the target is the set of
integers!!!! So, all the elements of
the target set will have corresponding elements in the domain! |
|
(h) f: Z+
→ R , |
Yes. Since each positive
numbers logarithm is unique. |
No. It is not onto since
there are many other numbers in the target that would not be mapped to. Note that the domain is the set of positive
integers, not the set of positive real numbers. |
.
.
.3. (15 points)
We define functions mapping R to R (where R is the
set of real numbers) as follows: 
, 
, 
. Find
(a) 
(b) 
(c) 
(d) 
(e) 
Solution:
(a) = 
.
(b) 
= 
.
(c) 
= 
.
(d) = 
.
(e) = 
.
4. (16 points) For each of the following functions mapping R to R (where R is the set of real numbers), if it has an inverse, then find the inverse function. If it does not have an inverse, prove that it is not one-to-one correspondence. Make sure that your steps are correct.
(a) 
(b) 
(c) 
(d)
where 
is a constant.
Solution:
(a)
We have . Let 
, and 
. Solving for 
, we have
Thus, 
is the inverse of 
.
(b)
We have 
. Let 
, and 
. Solving for , we have
Thus, 
is the inverse of 
.
(c)
Thus function is not one-to-one-correspondence.
Specifically, it is not one-to-one.
Here is the proof. Suppose that 
. Then we have 
. Expanding the
equation:
And
since 
are both real numbers,
there are many combinations for both to add up to 2 while not being the same
number. One example is 
. Thus, we have shown
that does not guarantee 
.
(d)
We have where 
is a constant. Let 
, and 
. Solving for 
, we have
Thus, 
is the inverse of 
.
5. (50 points total) Let A denote the set
R\{ 0, 1 } (where R is the set of real numbers). Let 
denote the identify
function on A and define the
functions f, g, h, s, r: 
by
, 
, 
, 
, 
(a) (40
points) Show that 
and 
. Complete the table,
thereby showing that the composition of any two of the given functions is one
of the given five or the identity. (The
table you will construct in this exercise is the multiplication table for an
important mathematical object known as a group.
This particular group is the smallest one which is not commutative.) Show all steps clearly.
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First,
we want to show that 
. That is, we want to
show that 
. Given the function
definitions, we have 
. Thus, we have shown
that ; and therefore .
Second, we want to show that 
. That is, we want to
show that 
. Given the function
definitions, we have
.
So,
we have shown that ; and therefore .
Now,
to complete the table, we consider the following:
Let us first consider the identity
functions. The composition of an
identity function with another function always results in a value which is that
another function. Thus, we 
, 
, 
etc. Thus we only have to worry about
compositions that do not involve the identity function.
(1) 
: 
. Thus, 
.
(2) 
: we have shown above
that .
(3) 
: 
. Thus, 
.
(4) 
: 
. Thus, 
.
(5) 
: 
. Thus, 
.
(6) 
: 
. Thus, 
.
(7)
: 
. Thus, 
.
(8)
: 
. Thus, 
.
(9)
: We have shown above
that .
(10)
: 
. Thus, 
.
(11)
: 
. Thus, 
.
(12)
: 
. Thus, 
.
(13)
: 
. Thus, 
.
(14)
: 
. Thus, 
.
(15)
: 
. Thus, 
.
(16) 
: 
. Thus, 
.
(17) 
: 
. Thus, 
.
(18) 
: 
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(19) 
: 
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(20) 
: 
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(21) 
: 
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(22) 
: 
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(23) 
: 
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(24) 
: 
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(25) 
: 
. Thus, 
.
Now we can complete the table:
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(b) (10 points) Which of the given six functions have inverses? Find (and identify) any inverses which exist.
We
know that the composition of a function and its inverse results in an identity
function. Thus, if we see an identify
function in the above column for a particular function, we know that the other
function that causes the composition to yield an identity function is the
inverse of the function. For example, has an inverse
which is simply . Thus, each of
the six inverses has an inverse. The
inverse of f is g. The
inverse of g is f. The
inverse of h is itself. The
inverse of r is itself. The
inverse of s is also itself.
Problems
based on (Rosen 2003), (Ross and Wright 1988) and (Goodaire and Parmenter
2002).