Principles of Mathematics 2660-Final Exam

Definition 12.8

If f : A → B is bijective then its inverse is the function f ^−1 : B → A. The functions f and f^ −1 obey the equations f ^−1 ◦ f = i A and f ◦ f ^−1 = iB.

Definition 4.3 Two integers have the same parity if

they are both even or they are both odd. Otherwise they have opposite parity

Difference between induction and strong induction

. Strong induction works just like regular induction, except that in Step (2) instead of assuming Sk is true and showing this forces Sk+1 to be true, we assume that all the statements S1, S2, ..., Sk are true and show this forces Sk+1 to be true. The idea is that if the first k dominoes falling always forces the (k +1)th domino to fall, then all the dominoes must fall.

Definition 12.9 Suppose f : A → B is a function.

1. If X ⊆ A, the image of X is the set f (X) = {f (x) : x ∈ X} ⊆ B. 2. If Y ⊆ B, the preimage of Y is the set f −1 (Y) = {x ∈ A : f (x) ∈ Y} ⊆ A.

The Pigeonhole Principle (function version) Suppose A and B are finite sets and f : A → B is any function.

1. If |A| > |B|, then f is not injective. 2. If |A| < |B|, then f is not surjective.

Definition 11.2 Suppose R is a relation on a set A

1. Relation R is reflexive if xRx for every x ∈ A. That is, R is reflexive if ∀x ∈ A, xRx. 2. Relation R is symmetric if xR y implies yRx for all x, y ∈ A. That is, R is symmetric if ∀x, y ∈ A, xR y ⇒ yRx. 3. Relation R is transitive if whenever xR y and yRz, then also xRz. That is, R is transitive if ∀x, y, z ∈ A, ((xR y)∧(yRz)) ⇒ xRz.

Theorem 12.4 Given f : A → B, let W, X ⊆ A, and Y, Z ⊆ B. Then

1. f (W ∩ X) ⊆ f (W)∩ f (X) 2. f (W ∪ X) = f (W)∪ f (X) 3. X ⊆ f ^−1( f (X) ) 4. f ^−1 (Y ∪ Z) = f ^−1 (Y)∪ f −1 (Z) 5. f ^−1 (Y ∩ Z) = f ^−1 (Y)∩ f −1 (Z) 6. f (f ^−1 (Y) ) ⊆ Y.

Definition 12.4 A function f : A → B is:

1. injective (or one-to-one) if for all a, a' ∈ A, a ≠ a' implies f (a) ≠ f (a' ); 2. surjective (or onto B) if for every b ∈ B there is an a ∈ A with f (a) = b; 3. bijective if f is both injective and surjective.

Definition 11.5

A partition of a set A is a set of non-empty subsets of A, such that the union of all the subsets equals A, and the intersection of any two different subsets is ∅

Definition 6.1

A real number x is rational if x = a b for some a,b ∈ Z. Also, x is irrational if it is not rational, that is if x ≠ a b for every a,b ∈ Z.

Definition 11.3

A relation R on a set A is an equivalence relation if it is reflexive, symmetric and transitive

Definition 11.7

A relation from a set A to a set B is a subset R ⊆ A×B. We often abbreviate the statement (x, y) ∈ R as xR y. The statement (x, y) ∉ R is abbreviated as xR/y

Definition 11.1

A relation on a set A is a subset R ⊆ A × A. We often abbreviate the statement (x, y) ∈ R as xRy. The statement (x, y) ∉ R is abbreviated as xR/y

Theorem 14.3

A set A is countably infinite if and only if its elements can be arranged in an infinite list a1,a2,a3,a4,...

Theorem 14.8

An infinite subset of a countably infinite set is countably infinite.

How to disprove P with contradiction:

Assume P is true, and deduce a contradiction.

Theorem 12.1

Composition of functions is associative. That is if f : A → B, g : B → C and h : C → D, then (h◦ g) ◦ f = h◦ (g ◦ f ).

Example 11.5

Consider the set R = {(x, y) ∈ Z×Z : x − y ∈ N} ⊆ Z×Z. This is the > relation on the set A = Z. It is infinite because there are infinitely many ways to have x > y where x and y are integers.

How to show a function f : A → B is injective:

Direct approach: Suppose a, a' ∈ A and a ≠ a 0 . . . . Therefore f (a) ≠ f (a' ). Contrapositive approach: Suppose a, a' ∈ A and f (a) = f (a' ). . . . Therefore a = a' .

Example of Cantor-Bernstein-Schröder Theorem

Example 14.6 The intervals [0,1) and (0,1) in R have equal cardinalities. Surely this fact is plausible, for the two intervals are identical except for the endpoint 0. Yet concocting a bijection [0,1) → (0,1) is tricky. (Though not particularly difficult: see the solution of Exercise 11 of Section 14.1.) For a simpler approach, note that f (x) = 1 4 + 1 2 x is an injection [0,1) → (0,1). Also, g(x) = x is an injection (0,1) → [0,1). The Cantor-Bernstein-Schröder theorem guarantees a bijection h : [0,1) → (0,1), so |[0,1)| = |(0,1)|.

Definition 12.2

For a function f : A → B, the set A is called the domain of f . (Think of the domain as the set of possible "input values" for f .) The set B is called the codomain of f . The range of f is the set { f (a) : a ∈ A } = { b : (a,b) ∈ f } . (Think of the range as the set of all possible "output values" for f . Think of the codomain as a sort of "target" for the outputs.)

Definition 12.6

For a set A, the identity function on A is the function i A : A → A defined as i A(x) = x for every x ∈ A. For example, if A= {1,2,3} , then iA = {(1,1),(2,2),(3,3)} . Also iZ= {(n,n) : n ∈ Z} . The identity function on a set is the function that sends any element of the set to itself

Definition 12.7

Given a relation R from A to B, the inverse relation of R is the relation from B to A defined as R ^−1 ={ (y, x) : (x, y) ∈ R} . In other words, the inverse of R is the relation R ^−1 obtained by interchanging the elements in every ordered pair in R.

Definition 5.1

Given integers a and b and n ∈ N, we say that a and b are congruent modulo n if n | (a− b). We express this as a ≡ b (mod n). If a and b are not congruent modulo n, we write this as a /≡ b (mod n).

(The Division Algorithm)

Given integers a and b with b > 0, there exist unique integers q and r for which a = qb + r and 0 ≤ r < b.

Corollary 14.1

Given n countably infinite sets A1, A2,..., An, with n ≥ 2, the Cartesian product A1 × A2 ×··· × An is also countably infinite.

Theorem 14.5

If A and B are both countably infinite, then so is A ×B.

Theorem 14.6

If A and B are both countably infinite, then their union A ∪B is countably infinite

Theorem 14.7

If A is any set, then |A| < |P(A)|.

Theorem 14.9

If U ⊆ A, and U is uncountable, then A is uncountable.

Fact 4.1

If a and b are integers, then so are their sum, product and difference. That is, if a,b ∈ Z, then a+ b ∈ Z, a− b ∈ Z and ab ∈ Z.

Theorem 14.10 (The Cantor-Bernstein-Schröder Theorem)

If |A| ≤ |B| and |B| ≤ |A|, then |A| = |B|. In other words, if there are injections f : A → B and g : B → A, then there is a bijection h : A → B.

Example 14.5

Let E = { 2k : k ∈ Z } be the set of even integers. The function f : Z → E defined as f (n) = 2n is easily seen to be a bijection, so we have |Z| = |E|. Thus, as |N| = |Z| = |E|, the set E is countably infinite and |E| = ℵ0.

Theorem 12.3

Let f : A → B be a function. Then f is bijective if and only if the inverse relation f −1 is a function from B to A.

Definition 11.6

Let n ∈ N. The equivalence classes of the equivalence relation ≡ (mod n) are [0],[1],[2],...,[n − 1]. The integers modulo n is the set Zn = {[0],[1],[2],...,[n − 1]} . Elements of Zn can be added by the rule [a]+[b] = [a+ b] and multiplied by the rule [a]·[b] = [ab].

How to disprove P(x) ⇒ Q(x).

Produce an example of an x that makes P(x) true and Q(x) false.

How to disprove ∀x ∈ S, P(x).

Produce an example of an x ∈ S that makes P(x) false.

How to Prove A ⊆ B (Direct approach)

Proof. Suppose a ∈ A. . . . Therefore a ∈ B. Thus a ∈ A implies a ∈ B, so it follows that A ⊆ B.

How to Prove A ⊆ B (Contrapositive approach)

Proof. Suppose a ∉ B. . . . Therefore a ∉ A. Thus a ∉ B implies a ∉ A, so it follows that A ⊆ B.

Proposition If n ∈ N, then 1+(−1)n (2n−1) is a multiple of 4.

Proof. Suppose n ∈ N. Then n is either even or odd. Let's consider these two cases separately. Case 1. Suppose n is even. Then n = 2k for some k ∈ Z, and (−1)n = 1. Thus 1+(−1)n (2n−1) = 1+(1)(2·2k −1) = 4k, which is a multiple of 4. Case 2. Suppose n is odd. Then n = 2k + 1 for some k ∈ Z, and (−1)n = −1. Thus 1+(−1)n (2n−1) = 1−(2(2k +1)−1) = −4k, which is a multiple of 4. These cases show that 1+(−1)n (2n−1) is always a multiple of 4. ■

How to Prove A = B

Proof. [Prove that A ⊆ B.] [Prove that B ⊆ A.] Therefore, since A ⊆ B and B ⊆ A, it follows that A = B. ■

Outline for Proving a Conditional Statement with Contradiction

Proposition If P, then Q. Proof. Suppose P and ∼ Q. . . . Therefore C ∧ ∼ C. ■

Outline for Direct Proof

Proposition If P, then Q. Proof. Suppose P. . . . Therefore Q. ■

Outline for Contrapositive Proof

Proposition If P, then Q. Proof. Suppose ∼ Q. . . . Therefore ∼ P. ■

Outline for If-and-Only-If Proof

Proposition P if and only if Q. Proof. [Prove P ⇒ Q using direct, contrapositive or contradiction proof.] [Prove Q ⇒ P using direct, contrapositive or contradiction proof.]

Outline for Proof by Contradiction

Proposition P. Proof. Suppose ∼ P. . . . Therefore C ∧ ∼ C. ■

Outline for Proof by Induction

Proposition The statements S1, S2, S3, S4, ... are all true. Proof. (Induction) (1) Prove that the first statement S1 is true. (2) Given any integer k ≥ 1, prove that the statement Sk ⇒ Sk+1 is true. It follows by mathematical induction that every Sn is true.

Outline for Proof by Smallest Counterexample

Proposition The statements S1, S2, S3, S4, ... are all true. Proof. (Smallest counterexample) (1) Check that the first statement S1 is true. (2) For the sake of contradiction, suppose not every Sn is true. (3) Let k > 1 be the smallest integer for which Sk is false. (4) Then Sk−1 is true and Sk is false. Use this to get a contradiction. ■

Definition 14.4

Suppose A and B are sets. 1. |A| = |B| means there is a bijection A → B. 2. |A| < |B| means there is an injection A → B, but no bijection A → B. 3. |A| ≤ |B| means there is an injection A → B.

Definition 12.1

Suppose A and B are sets. A function f from A to B (denoted as f : A → B) is a relation f ⊆ A × B from A to B, satisfying the property that for each a ∈ A the relation f contains exactly one ordered pair of form (a,b). The statement (a,b) ∈ f is abbreviated f (a) = b.

Definition 14.2

Suppose A is a set. Then A is countably infinite if |N| = |A|, that is, if there exists a bijection N → A. The set A is countable if it is finite or countably infinite. The set A is uncountable if it is infinite and |N| 6= |A|, that is, if A is infinite and there is no bijection N → A.

Definition 11.4

Suppose R is an equivalence relation on a set A. Given any element a ∈ A, the equivalence class containing a is the subset {x ∈ A : xRa} of A consisting of all the elements of A that relate to a. This set is denoted as [a]. Thus the equivalence class containing a is the set [a] = {x ∈ A : xRa} .

Theorem 11.1

Suppose R is an equivalence relation on a set A. Suppose also that a,b ∈ A. Then [a] = [b] if and only if aRb.

Theorem 11.2

Suppose R is an equivalence relation on a set A. Then the set {[a] : a ∈ A} of equivalence classes of R forms a partition of A.

Prove that {x ∈ Z : 18| x} ⊆ {x ∈ Z : 6|x }

Suppose a ∈ {x ∈ Z : 18| x} . This means that a ∈ Z and 18|a. By definition of divisibility, there is an integer c for which a = 18c. Consequently a = 6(3c), and from this we deduce that 6|a. Therefore a is one of the integers that 6 divides, so a ∈ {x ∈ Z : 6| x} . We've shown a ∈ {x ∈ Z : 18| x} implies a ∈ {x ∈ Z : 6| x}, so it follows that {x ∈ Z : 18| x} ⊆ {x ∈ Z : 6| x}

How to show a function f : A → B is surjective:

Suppose b ∈ B. [Prove there exists a ∈ A for which f (a) = b.]

Definition 12.5

Suppose f : A → B and g : B → C are functions with the property that the codomain of f equals the domain of g. The composition of f with g is another function, denoted as g◦ f and defined as follows: If x ∈ A, then g◦f (x) = g(f (x)). Therefore g◦f sends elements of A to elements of C, so g◦ f : A → C.

Theorem 12.2

Suppose f : A → B and g : B → C. If both f and g are injective, then g ◦ f is injective. If both f and g are surjective, then g ◦ f is surjective.

inductive hypothesis.

The assumption that Sk is true

Definition 14.3

The cardinality of the natural numbers is denoted as ℵ0. That is, |N| = ℵo. Thus any countably infinite set has cardinality ℵo.

Example 11.9 Consider R = {(−1,−1),(1,1),(2,2),(3,3),(4,4)}

The equivalence class containing 2 is the set [2] = {x ∈ A : xR2} . Because in this relation the only element that relates to 2 is 2 itself, we have [2] = {2 } . Other equivalence classes for R are [−1] = {− 1} , [1] = {1} , [3] = {3} and [4] = {4} . Thus this relation has five separate equivalence classes.

Definition 4.6

The greatest common divisor of integers a and b, denoted gcd(a,b), is the largest integer that divides both a and b. The least common multiple of non-zero integers a and b, denoted lcm(a,b), is the smallest integer in N that is a multiple of both a and b.

inductive step

The second step (2)

Theorem 14.4

The set Q of rational numbers is countably infinite.

Theorem 14.1

There exists a bijection f : N → Z. Therefore |N| = |Z|.

Theorem 14.2

There exists no bijection f : N → R. Therefore |N| ≠ |R|.

existence theorems

To prove an existence theorem, all you have to do is provide a particular example that shows it is true.

Definition 12.3

Two functions f : A → B and g : A → D are equal if f = g (as sets). Equivalently, f = g if and only if f (x) = g(x) for every x ∈ A.

Definition 14.1

Two sets A and B have the same cardinality, written |A| = |B|, if there exists a bijective function f : A → B. If no such bijective f exists, then the sets have unequal cardinalities, written |A| ≠ |B|.

Example 12.4 Show that the function f : R− © 0 ª → R defined as f (x) = 1 x +1 is injective but not surjective.

We will use the contrapositive approach to show that f is injective. Suppose a,a' ∈ R− {0} and f (a) = f (a'). This means 1 a +1 = 1 a' +1. Subtracting 1 from both sides and inverting produces a = a' . Therefore f is injective. The function f is not surjective because there exists an element b = 1 ∈ R for which f (x) = 1 x +1 ≠ 1 for every x ∈ R− {0} .

Suppose A is an n × n matrix. The following statements are equivalent:

a) The matrix A is invertible. (b) The equation Ax = b has a unique solution for every b ∈ R^n . (c) The equation Ax = 0 has only the trivial solution. (d) The reduced row echelon form of A is In. (e) det(A) ≠ 0. (f) The matrix A does not have 0 as an eigenvalue.

Definition 4.4 Suppose a and b are integers. We say that a divides b, written a | b, if

b = ac for some c ∈ Z. In this case we also say that a is a divisor of b, and that b is a multiple of a.

Definition 4.5 A number n ∈ N is prime if

it has exactly two positive divisors, 1 and n. If n has more than two positive divisors, it is called composite. (Thus n is composite if and only if n = ab for 1 < a,b < n.)

Definition 4.1 An integer n is even if

n = 2a for some integer a ∈ Z.

Definition 4.2 An integer n is odd if

n = 2a+1 for some integer a ∈ Z.

constructive proof

one that produces (or constructs) two explicit irrational numbers x, y for which x^y is rational.

fundamental theorem of arithmetic

states that any integer greater than 1 has a unique prime factorization Any integer n > 1 has a unique prime factorization. "Unique" means that if n = p1 · p2 · p3 ··· pk and n = a1 · a2 · a3 ···al are two prime factorizations of n, then k = l, and the primes pi and ai are the same, except that they may be in different orders.

basis step

the first step (1)