Alaska Pacific University Composition of Functions having Infinite Domains and Codomains Exam I have attached the files ……………………………..

Alaska Pacific University Composition of Functions having Infinite Domains and Codomains Exam I have attached the files ………………………………………………….. Worksheet #3 Sec 7.3 Composition of Functions having Infinite Domains and Codomains
When we compose functions with infinite domains and codomains we usually do not try to draw an
arrow diagram. Instead we use algebra, specifically substitution, to find the composition of two
functions, ?? ? ??. The inner function is the one that happens first and the outer function is the
function that acts last. So in the composition ?? ? ?? we call ?? the inner function and ?? the outer
function.
We construct the composition “?? of ??” as follows: Substitute the entire inner function into the outer
function, replacing every occurrence of the variable in the outer function with the entire inner
function.
Ex 1: Let ??: ? ? ? and ??: ? ? ? be defined as follows: ??(??) = ?? + 1 and ??(??) = ??2 .
Do (i) – (v) for each composed function: (a) ?? ? ??
(i)
(ii)
(iii)
(iv)
(v)
(b) ?? ? ??
(c) ?? ? ??
and (d) ?? ? ??.
Find the expressions for each composed function.
Find the domain and codomain for each composed function.
Is the composed function one-to-one? If yes, try and explain why you think it is.
If you think it is not one-to-one, please give a counterexample to specifically
show that there exist elements ??1 and ??2 in the domain of the composed
function (let’s just call that function ? for the purposes of this discussion) such
that ?(??1 ) = ?(??2 ) but ??1 ? ??2.
IS the composed function, call it ?, an onto function? If so, explain why you think
so. If not, please give a specific element ?? in the codomain of ? that is not in
Range(?).
Is the composed function a bijection? Justify your response.
Solutions to (a) and (b):
(a) (i)
(ii)
(?? ? ??)(??) = ??(??(??)) = ?? (?? + 1) = (?? + 1)2
The domain is ? and the codomain is ?.
(iii)
The composed function ?? ? ??: ? ? ? is not one-to-one. Let ??1 = 0 and ??2 = ?2.
Then (?? ? ??)(0) = (0 + 1)2 = 1 and (?? ? ??)(?2) = (?2 + 1)2 = (?1)2 = 1, but 0 ? ?2.
(iv)
The composed function is not onto. Consider the integer 3 in the codomain of ?? ? ??.
If there was an integer ?? in the domain of ?? ? ?? such that (?? ? ??)(??) = 3 then we would
have (?? + 1)2 = 3, i.e., ?? + 1 = ??3 , i.e., ?? = ?1 ? ?3. But ?1 + ?3 isn’t even a
rational number, much less and integer. Neither is ?1 ? ?3. This contradicts the fact
that the domain of ?? ? ?? is ?. So it is not possible that 3 is in the range of ?? ? ??, which mean
it is not onto ?.
(v)
Since the function is not one-to-one, it is not a bijection. It isn’t onto either, and
needs to be both to be a bijective function.
(b) (i)
(ii)
(?? ? ??)(??) = ??(??(??)) = ??(?? + 1) = (?? + 1) + 1 = ?? + 2
The domain is ? and the codomain is ?.
(iii)
The composed function ?? ? ?? is one-to-one. [We need to show that ?? ? ?? satisfies the
implication “If (?? ? ??)(??1 ) = (?? ? ??)(??2 ), then ??1 = ??2 .] So suppose (?? ? ??)(??1 ) = (?? ? ??)(??2 ) for
integers ??1 and ??2 . By the definition of ?? ? ?? this means ??1 + 2 = ??2 + 2. Subtracting 2 from
both sides of this equation shows ??1 = ??2 .
(iv)
The function ?? ? ?? is onto its codomain ?. [We must show that an arbitrary element
?? in the codomain, ?, of ?? ? ?? has an inverse image (preimage) in the domain, which is also ?.] If we
set ?? + 2 = ?? and solve for ?? we see ?? = ?? ? 2. So the preimage of an arbitrary element ??
is ?? ? 2, because (?? ? ??)(?? ? 2) = (?? ? 2) + 2 = ??.
(v)
The function ?? ? ?? is a bijection because it is both one-to-one and onto.
Now you do (i) – (v) for the composed functions given as (c) ?? ? ?? and (d) ?? ? ??.
Worksheet #4
Section 8.1
BINARY RELATIONS
p.1
Recall: Given sets A and B, the cross product of A and B, which we read as “A cross B” is the set of
all ordered pairs having first coordinates from A and second coordinates from B. In other words
A × B = {(??, ??) | ?? ? A ?????? ?? ? B}
Recall: Given sets A and B, a relation from A to B is any subset of A × B. (See Worksheet #1)
Recall: A function from A to B, denoted by ??: A ? B , is a relation from A to B that satisfies
(i)
(ii)
Every element of A has at least one image in B
Every element in A has at most one image in B
Defn: Given a set A, we define a binary relation on A, often denoted by an upper case script letter
like ?, ?? or ??, is any relation from A to A.
Ex 1: Let A = {??, ??, ??, ??}.
a)
?1 = {(??, ??), (??, ??), (??, ??), (??, ??)} is a relation on set A.
b)
?2 = {(??, ??), (??, ??), (??, ??), (??, ??)} is a relation on A.
c)
?3 = ? is a relation on set A. It is called the empty relation.
d)
?4 = (??, ??) is not a relation on A because all relations are sets, and require
either set brackets or a special name.
e)
?1 = {(??, ??), (??, ?? )} is not a relation on A because the element ?? ? A.
Problem #1: Let A = {??, ??, ??, ??}. Find 4 distinct (different) relations on A. Do not use any of the
relations given in the examples above.
Problem #2: How many relations are there on each of the following sets:
a)
b)
c)
d)
e)
The empty set.
Any set having exactly one element. Such sets are called singleton sets.
Any set having exactly two elements.
Any set having exactly three elements.
Any set having exactly n elements, where ?? ? 0.
Problem #3: Let A be the set we have been using so far in this worksheet.
a) Is every binary relation on A a function from A to A? If so, explain why. If not, give a
counterexample, i.e. find a binary relation on A that is not a function on A and explain
which part of the definition of function the binary relation violates.
b) Is every function from A to A a binary relation on A? If so, explain why. If not, give a
counterexample, i.e., find a function from A to A that is not a relation on A.
Worksheet #4
Section 8. 2
PROPERTIES of BINARY RELATIONS
p.2
Relation Notation
Given set A and a binary relation ? on A, if an ordered pair (??, ??) ? ?? we can write ??????.
Now we will learn and practice the definitions of several special properties that some relations
have. For each of these definitions, let A be a set.
Defn (R): A binary relation ? on set A is reflexive if the ordered pair (??, ?? ) is in ? for every ?? ? A.
Using our new notation, we can say ? is reflexive if and only if ??? ? A, a???.
Defn (S): A binary relation ? on A is symmetric if whenever (??, ??) ? ?, its symmetric twin
(??, ??) ? ? as well. In other words, ???, ?? ? A, ((??, ??) ? ? ? (??, ??) ? ?).
Defn (T):
A binary relation ? on A is transitive if whenever (??, ??) ? ? ?????? (??, ??) ? ?, then the
element (??, ??) ? ? as well. In other words, ???, ??, ?? ? A, (((??, ??) ? ? ? (??, ??) ? ?) ? (??, ??) ? ?).
Defn (A):
A binary relation ? on A is antisymmetric if having (??, ??) ? ? and (??, ??) ? ?, implies
?? and ?? are the same element. In other words, ???, ?? ? A, (((??, ??) ? ? ? (??, ????) ? ?) ? ?? = ??).
Ex 2: Let A = {??, ??, ??, ??}.
a)
?1 = {(??, ??), (??, ??)(??, ??)(??, ??)} is a reflexive relation. It is also symmetric, because the
hypothesis of symmetry is that (??, ??) ? ?. Since ?? = ?? for every element in ?1 , the
required symmetric twin is (??, ??) = (??, ??), which is clearly in ?1 .
b)
?2 = {(??, ??), (??, ??)(??, ??)(??, ??)} is a symmetric relation. But it is not reflexive (it is as
“far away” from being reflexive as any relation on A can be). It is also not transitive
because (??, ??) ? ?2 and (??, ??) ? ?2 , but (??, ??) ? ?2 .
c)
?3 = {(??, ??), (??, ??), (??, ??)} is not reflexive or symmetric. But ?3 is transitive because
the only way the hypothesis of transitivity is satisfied is if we state the following:
(??, ??) ? ?3 ? (??, ??) ? ?3 or a similar hypothesis using (??, ??). The required result,
that (??, ??) ? ?3, clearly holds.
d)
?4 = {(??, ??), (??, ??)} is not reflexive and not symmetric. It is transitive. The hypothesis
of transitivity is satisfied in a couple of ways. One of those is this:
(??, ??) ? ?4 ? (??, ??) ? ?4 . The required result, namely that (??, ??) ? ?4 , is clearly true.
e)
?5 = {(??, ??), (??, ??), (??, ??), (??, ??), (??, ??), (??, ??), (??, ??), (??, ??)}
Problems are on next page.
Worksheet #4
Problem #4:
Section 8. 2
PROPERTIES of BINARY RELATIONS
p.3
Look at ?1 in Ex 2 (a) above. Explain why ?1 is symmetric.
Problem #5: Again, look at ?1 from Ex 2 (a). ?1 is a transitive relation, but not by false
hypothesis. Explain how the hypothesis of transitivity is satisfied by every element in ?1 , and how
each element satisfies the required conclusion in the definition of transitivity. [Hint: The hypothesis
of the transitive property is ((??, ??) ? ? ? (??, ???? ) ? ?), but it does not require ?? ? ?? ? ??.]
Problem #6: Look at Ex 2(b). There are at least 3 counterexamples other than the one given in Ex
2 (b) to show ?2 is not transitive. Find two other counterexamples to the claim “?2 is transitive.”
Problem #7: Look at Ex 2(b) again. Show exactly why ?2 is not antisymmetric by producing an
example to show antisymmetry fails.
Problem #8: Look at ?1 in Ex 2 (a). Show ?1 is transitive using an argument similar to the
argument showing ?3 is transitive in Ex 2 (c).
Problem #9:
Look at ?3 in Ex. 2 (c). Explain why ?3 is not reflexive.
Problem #10:
Look at ?4 in Ex. 2 (d). Explain why it is not symmetric.
Problem #11:
Look at ?4 in Ex. 2 (d). ?4 is antisymmetric. Why?
Problem #12:
Look at ?5 in Ex. 2 (e). Is ?5 reflexive? If so, explain. If not, explain why not.
Problem #13:
Look at ?5 in Ex. 2 (e). Clearly 5 is reflexive.
When we check for symmetry in ?5 it turns out we do not need to worry about any of the
elements of the form (??, ??), ?? ? A, because these “reflexive” elements satisfy symmetry for
the same reason as that given to explain why ?1 is symmetric in Ex 2 (a). We must check the
other elements in ?5 for symmetry.
Is ?5 symmetric? If so, explain why. If not, explain why not.
Problem #14: When checking to see if ?5 is transitive, we do not have to look at any of
these reflexive elements of the form (??, ??), ?? ? A. There are 2 reasons this checking is not
necessary. The first reason is the same reason that was used in Ex 2 (c) to show ?3 is
transitive. The second reason is the same as the reason explained in Ex 2 (d) to show ?4 is
transitive. We must check elements that are not of the “reflexive” form (??, ??) when we are
checking any relation for transitivity, including ?5 .
Is ?5 transitive? If so, explain why. If not, explain why not.

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