Math 300: Final Exam Practice Solutions

Math 300: Final Exam Practice Solutions
1. Let A be the set of all real numbers which are zeros of polynomials with integer coefficients:
A := {α ∈ R | there exists p(x) = a n x n + · · · + a 1 x + a 0 with all a i ∈ Z such that p(α) = 0}.
Show that A is countable. (Elements of A are called algebraic numbers.) The set of transcendental
numbers is the set R − A. Show that the set of transcendental numbers is uncountable.
Remark. To show that A is countable we must either: construct an explicit bijection Z + → A, a
surjection Z + → A, an injection A → Z + , find a way to “list” the elements of A, show that A
is a subset of a countable set, or (as will be applicable here) show that A is a countable union of
countable sets. For relatively “complicated” sets like A, this last technique is often useful. The
key is to build up: how many polynomials with integer coefficients of a fixed degree are there, then
how many such polynomials are there if we allow any degree, and finally how much roots does each
such polynomial have?
Proof. The first part of this, showing that A is countable, is on the “Worked Examples” handout
so check the proof there.
If R − A were countable, then A ∪ (R − A) = R would be countable since it would be the union
of two countable sets. Since R is uncountable, R − A must be uncountable. (Side question which
has nothing to do with the final: do you know any examples of transcendental numbers?)
2. For a set A, A ∞ denotes the set of sequences of elements of A:
A ∞ = {(a 1 , a 2 , a 3 , . . . , a n , . . .) | a i ∈ A for all i}.
(a) (Tricky) Show that R ∞ has the same cardinality as R.
(b) Show directly, without quoting any theorem in the book, that Z ∞ is uncountable.
Remark. Part (a) requires the construction of a bijection between R ∞ and R. As is often the case
when dealing with R, it might be easier to focus on numbers between 0 and 1 first and then use
the fact that (0, 1) and R have the same cardinality. Constructing a bijection (0, 1) ∞ → (0, 1) is
tricky (too tricky for the final), but uses an idea similar to the “splicing” technique we’ve seen for
showing that R 2 and R have the same cardinality. (This is on the “Worked Examples” handout.)
The splicing in this case though is tougher: given an infinite number of decimal expansions, we
have to construct a single real number which encodes all of those digits. Arranging this all in a
table suggests that we can mimic the technique used to show that Q is countable.
For part (b) we should use Cantor’s diagonalization argument, which is a standard tool for
showing directly that sets are uncountable. The same technique shows up in Problem 5.
Proof. (a) First we define a bijection (0, 1) ∞ → (0, 1). Let (x 1 , x 2 , x 3 , . . .) be an element of (0, 1) ∞ ,
so each x i is a real number between 0 and 1. As such, we can write x i according to its decimal
expansion:
x 1 = x 11 x 12 x 13 · · ·
x 2 = x 21 x 22 x 23 · · ·
x 3 = x 31 x 32 x 33 · · ·
.
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We construct out of this a number in (0, 1) as follows: start with 0.x 11 , then append the digits x 21
and x 12 , then x 31 , x 22 , and x 13 , and so on following the successive diagonals moving from lower-left
to upper-right. Define the function f : (0, 1) ∞ → (0, 1) by sending the sequence (x 1 , x 2 , x 3 , . . .) to
this number:
f(x 1 , x 2 , x 3 , . . .) = 0.x 11 x 21 x 12 x 31 x 22 x 13 · · · .
This function is bijective since it is invertible with inverse f −1 : (0, 1) → (0, 1) ∞ defined as
follows. Take y ∈ (0, 1), write it according to its decimal expansion y = 0.y 1 y 2 y 3 . . ., and then
“unwind” the process from above to create a sequence of numbers:
x 1 := 0.y 1 y 3 y 6 y 10 . . .
x 2 := 0.y 2 y 5 y 9 . . .
x 3 := 0.y 4 y 8 . . .
x 4 := 0.y 7 . . .
where we list y 1 in the upper-left corner, then y 2 and y 3 along the next diagonal down, then y 4 , y 5 , y 6
along the next diagonal down, and so on. The inverse of f is f −1 (y) = (x 1 , x 2 , x 3 , . . .).
Since (0, 1) and R have the same cardinality, there exists a bijection g : (0, 1) → R. Also, as a
result of Exercise 6.1.9 in the book (or a generalization of this with a similar proof), (0, 1) ∞ and
R ∞ have the same cardinality so there exists a bijection h : R ∞ → (0, 1) ∞ . The composition gfh
is then a bijection R ∞ → R, so R ∞ and R have the same cardinality as claimed.
(b) Take any list of elements of Z ∞ :
(n 11 , n 12 , n 13 , . . .)
(n 21 , n 22 , n 23 , . . .)
(n 31 , n 32 , n 33 , . . .)
.
Define an element (m 1 , m 2 , m 3 , . . .) of Z ∞ simply by choosing for m i an integer different from n ii .
This element is then not in the list above since it differs from the kth element of this list in the kth
location. Hence any listing of elements of Z ∞ cannot possibly contain all elements of Z ∞ , so Z ∞ is
uncountable.
3. Consider the relation ∼ between sets defined by: A ∼ B if A and B have the same cardinality.
(a) Show that ∼ is an equivalence relation.
(b) Determine which of the following sets are equivalent to which. No justification is needed.
P(Z), Q × Q, P(P({∅})), F (R), P(Z × Z × Z), (0, 1) ∩ Q, P(P(Z + )), R 2 , F ({2, 3})
Remark. This is a mixture of older material and newer material. In part (a) we must as usual
show that R is reflexive, symmetric, and transitive. This amounts to using properties of bijective
functions. In part (b) we use facts about cardinality we’ve already built up, and is a good example
of how to compare cardinalities.
Solution. (a) For any set A, the identity function i A : A → A sending anything in A to itself is
bijective so A has the same cardinality as A. Thus ARA so R is reflexive. Now suppose that
ARB. Then A and B have the same cardinality, meaning there exists a bijection f : A → B.
Then f −1 : B → A is also bijective, so B and A have the same cardinality. Hence BRA so R is
symmetric.
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Finally, suppose that ARB and BRC. Then A and B have the same cardinality so there exists
a bijection f : A → B, and B and C have the same cardinality so there exists a bijection g : B → C.
The composition gf : A → C is then also bijective, so A and C have the same cardinality. Thus
ARC so R is transitive and we conclude that R is an equivalence relation.
(b) No justification is needed, but we give some anyway. Two sets are in the same equivalence
classes when they have the same cardinality.
• P(P({∅})) and F ({2, 3}) both have 4 elements and thus have the same cardinality. These
are the only finite sets in this list.
• Q × Q and (0, 1) ∩ Q are both countably infinite, and so have the same cardinality.
• Since Z × Z × Z and Z have the same cardinality, their power sets have the same cardinality,
which is the same as the cardinality of R. Since R 2 and R also have the same cardinality, we
get that P(Z), P(Z × Z × Z), and R 2 all have the same cardinality.
• Finally, an example on the “Worked Examples” handout shows that F (R) and P(R) have the
same cardinality. Since R and P(Z + ) have the same cardinality, their power sets do as well
so we get that F (R) and P(P(Z + )) have the same cardinality.
Notice that in none of these did we construct any explicit bijections, but rather used cardinality
facts we’ve previously built up.
4. Show that Z × Z + and E × O have the same cardinality, where E is the set of even integers and
O the set of odd integers, by constructing an explicit bijection between them.
Remark. Since an explicit bijection is what is being asked for, an explicit bijection is what we
should give. This will require giving a candidate for the bijection f : Z → Z + → E × O and then
showing that f is indeed bijective. To do all this we use previous bijections we’ve seen on homework
or in class involving integers, even integers, and odd integers. Also, using the same kind of idea
as in Exercise 6.1.9 in the book, we should really be looking for bijections Z → E and Z + → O
separately.
Proof. Define f : Z × Z + → E × O by
f(m, n) =
{
(2m, n − 1) if n is even
(2m, −n)
if n is odd.
We claim that f is bijective. To check injectivity, suppose that f(m, n) = f(a, b). Since whether
or not the second component of f(m, n) = f(a, b) is positive or negative depends on whether n, b
are even or odd, these can be equal only when n and b are both even or both odd. If they are both
even, then f(m, n) = f(a, b) becomes
(2m, n − 1) = (2a, b − 1), so 2m = 2a and n − 1 = b − 1,
giving m = n and a = b. If n, b are both odd, then f(m, n) = f(a, b) becomes
(2m, −n) = (2a, −b), so 2m = 2a and −n = −b,
again giving m = n and a = b. Thus f(m, n) = f(a, b) implies (m, n) = (a, b) in either case, so f
is injective.
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For surjectivity, suppose that (x, y)E × O, so x is even and y is odd. Since x is even x 2
is an
integer. Hence if y < 0 we get f( x 2 , −y) = (x, y), while if y > 0 we have f(x 2
, y + 1) = (x, y). Thus
either way there is something in Z × Z + mapping to (x, y), so f is surjective. Thus f is bijective
so Z × Z + and E × O have the same cardinality as claimed.
5. (Possibly Tricky) The Cantor set is defined at the end of Section 2.3 in the book, under the
heading “Mathematical Perspective: An Unusual Set”. Show that the Cantor set is uncountable.
Remark. This is tricky, especially if you’ve never seen the Cantor set before. My purpose for
including this here is to show how one question can be converted into another, which you already
know the answer to. This is vague, but the point here is that we will show that an element of the
Cantor set can be uniquely described by a sequence of 0’s and 1’s, and the set of such sequences
can be shown to be uncountable using Cantor’s diagonalization argument. As for preparing for the
final, don’t worry so much about the part of this solution having to do with the Cantor set, but
you should at least be able to show that the set of sequences of 0’s and 1’s is uncountable.
Proof. We will use the notation the book uses when defining the Cantor set C. Let x ∈ C. We
construct an element of {0, 1} ∞ associated to this as follows. Since C = ⋂ A n , x ∈ A n for all n. In
particular, x ∈ A 1 so x is in one of the two intervals making up A 1 ; take the first element in our
sequence to be 0 if x is in the “left” interval [0, 1/3] and take the first element in our sequence to
be 1 if x is in the “right” interval [2/3, 1].
Now, whichever of these intervals x is in will itself split into two smaller intervals in the construction
of A 2 . Since x ∈ A 1 , x will be in one of these smaller intervals; take the next element
in our sequence to be 0 if it is the “left” interval x is in and take it to be 1 if x is in the “right”
interval. For instance, the interval [0, 1/3] splits into [0, 1/9] and [2/9, 1/3]. If x ∈ [0, 1/9] the first
two terms in the sequence we are constructing will be 0, 0, while if x ∈ [2/9, 1/3] we have 0, 1 as
the beginning of our sequence. Continuing in this manner, whichever interval making up A 2 that
x is in will split into two smaller pieces; take 0 as the third term in our sequence if x is in the left
piece and 1 if x is in the right piece, and so on. By keeping track of which interval x is in at each
step in the construction of the Cantor set in this manner we get a sequence of 0’s and 1’s.
For instance, if we get the sequence 0, 1, 1, 1, 0, 0, 0, . . ., x is in the “left” interval of A 1 , then in
the “right” smaller interval which this interval splits into, then in the “right” smaller interval this
splits into, then “right” again, then in the “left” smaller interval that this splits into, and so on.
(This is easier to imagine if you draw a picture of this splitting into smaller and smaller intervals
as we did during the review.)
This assignment of a sequence of 0’s and 1’s to an element x ∈ C defines a function C → {0, 1} ∞ .
It is injective since different elements in the Cantor set produces different sequences (at some point
in the construction, two different numbers in the Cantor set will belong to two different “smaller”
intervals), and it is surjective since given any sequence we can use it to single out an element of
the Cantor set. Thus C and {0, 1} ∞ have the same cardinality.
We claim that {0, 1} ∞ is uncountable, which then shows that the Cantor set is uncountable.
Given a listing of elements of {0, 1} ∞ :
n 11 , n 12 , n 13 , . . .
n 21 , n 22 , n 23 , . . .
n 31 , n 32 , n 33 , . . .
.
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where each n ij is either 0 or 1, the sequence
defined by declaring
m k =
m 1 , m 2 , m 3 , . . .
{
0 if nkk = 1
1 if n kk = 0
is an element of {0, 1} ∞ which is not in the above list. Thus no listing of elements of {0, 1} ∞ can
include every element of {0, 1} ∞ , so {0, 1} ∞ is uncountable.
6. Let S be a set and define the binary operation ∗ on P(S) by: A ∗ B = (A ∪ B) − (A ∩ B).
(a) Show that ∗ is commutative, has an identity, and that everything has an inverse.
(b) Give an example of a subset of P(Z) with four elements which is closed under ∗.
Remark. The point of this problem is to give an example of a binary operation on a set which is
not just a set of numbers, like Z, Q, or R. Note that I had originally asked in part (a) to also show
that ∗ is associative, but this turns out to be a lot of messy work which isn’t very enlightening,
so forget that part. Another good example to try is the following. Define ∗ on F (R) by saying
that f ∗ g : R → R is the function given by (f ∗ g)(x) = f(x) + g(x). Is this operation associative?
commutative? Does it have an identity? What about inverses?
Proof. (a) For any subsets A and B of S, we have
A ∗ B = (A ∪ B) − (A ∩ B) = (B ∪ A) − (B ∩ A) = B ∗ A
simply because A ∪ B = B ∪ A and A ∩ B = B ∩ A. Thus ∗ is commutative. For any A ⊆ S,
A ∗ ∅ = (A ∪ ∅) − (A ∩ ∅) = A − ∅ = A,
so ∅ is an identity for ∗. Finally, for A ⊆ S we have:
A ∗ A = (A ∪ A) − (A ∩ A) = A − A = ∅,
so A is its own inverse. Thus everything in P(S) has an inverse under ∗.
(b) We claim that the subset {∅, E, O, Z} of P(Z) is closed under ∗. We check that A ∗ B is one
of these subsets whenever A and B are one of these subsets. For sure, A ∗ ∅ = A for any of these
subsets since ∅ is the identity of ∗. Also, A ∗ A = ∅ for any of these as shown in part (a). Now:
E ∗ O = Z, E ∗ Z = O, O ∗ Z = E,
all of which are in the given set.
{∅, E, O, Z} is closed under ∗.
Thus A ∗ B ∈ {∅, E, O, Z} whenever A, B ∈ {∅, E, O, Z}, so
7. Define a 1 = 1, a 2 = 1, and a n+2 = a n+1 + a n for n ≥ 1. Using induction prove that for all
positive integers n, a n ≤ φ n−1 where φ = (1 + √ 5)/2. Hint: φ satisfies φ 2 − φ − 1 = 0.
Remark. This is an example of using what the book calls the “Second Principle of Mathematical
Induction”, Theorem 5.2.3. The point is that in the induction step we want to use the given
inequality to replace both a k and a k−1 , so we need to assume that the given inequality holds for
all integers smaller than the one we want to establish it for. Note in the proof why it would not
be enough to only use the usual “if it is true for k, then it is true for k + 1”.
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Proof. First, a 1 = 1 ≤ 1 = φ 0 so the inequality holds for n = 1. For a second base case (we’ll
see why we need two base cases), we have a 2 = 1 ≤ φ 1 so the inequality holds for n = 2 as well.
Suppose now that for some k, a i ≤ φ i−1 for all 1 ≤ i ≤ k. (So, not only for n = k but for all
positive integers smaller than k.) We want to show that a k+1 ≤ φ k . Since we’ve already check the
base cases n = 1 and n = 2, we may assume here that k ≥ 3, which we need to do in order to write
a k+1 = a k + a k−1
according to the recursive definition. Both a k and a k−1 are elements to which the induction hypothesis
applies to, so
a k ≤ φ k−1 and a k−1 ≤ φ k−2 .
Then, noting that φ 2 = φ + 1 as a result of the hint, we have:
a k+1 = a k + a k−1 ≤ φ k−1 + φ k−2 = φ k−2 (φ + 1) = φ k−2 φ 2 = φ k .
Thus a k+1 ≤ φ k as required, so by induction we conclude that a n ≤ φ n−1 for all positive n.
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