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
We construct out of this a number in (0, 1) as follows: start with 0x 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, ) = 0x 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 = 0y 1 y 2 y 3, and then unwind the process from above to create a sequence of numbers: x 1 := 0y 1 y 3 y 6 y 10 x 2 := 0y 2 y 5 y 9 x 3 := 0y 4 y 8 x 4 := 0y 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 619 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 2
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 619 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 3
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 23 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, 4
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 523 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 5
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 6