discrete math test 4

Reflexivity ∀x ∈ A. xRx.

(xRx) If every element in a set, can point to itself, meaning it is in both sets ( A -> A) How to prove: xSy if ∃k ∈ N. (x + k = y) Pick any x ∈ A. We need to prove that xRx. [...] Consider some integer x. We need to prove that xSx, meaning that we need to show that there's a natural number k such that x + k = x. [...]

Transitivity ∀x ∈ A. ∀y ∈ A. ∀z ∈ A. (xRy ∧ yRz → xRz).

(xRy, yRz, xRz) if A -> B and B -> C, then A -> C How to prove: xSy if ∃k ∈ N. (x + k = y) Consider some arbitrary x, y, z ∈ A where xRy and yRz. We need to prove that xRz. [...] Proof: Consider some arbitrary integers x, y, z where xSy and ySz. We need to prove that xSz. Since we know that xSy and ySz, we know that there exist natural numbers m and n where x + m = y and y + n = z. [...] Consider some arbitrary x, y, z ∈ A where xRy and yRz. We need to prove that xRz. Since R is cyclic, from xRy and yRz we learn that zRx. [...]

Inverse relation

A -> B inverse is B -> A ( R or R^-1). Bijection goes both ways, but not anything else.

Identity Map

Also known as equality relation where A = B, and each A goes to each B

Binary Relation A -> B (AxB)

Any subset from A to B, where A is domain, B is codomain. If A -> B is a binary relation, then (S,T) ∈ R, then S is a preimage of T under R, and T is an image of S under R. if A is n and B is m, and AxB is nm, then there are 2^nm binary relations.

Bijective

Both one to one and onto, So exactly one element of X maps to one element of Y. If A -> B is bijective, then so is B -> A. Some famous examples Identity Functions: f : A ! A defined by f(a) = a Linear Functions: f : Q ! Q by f(n) = an + b, a 6= 0 Translation Functions: f : Zm ! Zm by f(n) = n + a (mod m) ExponentialFunctions:f:R!R+ byf(x)=ax,a>0• Logarithmic Functions: f : R+ ! R by f(x) = loga(x), a > 0 String-Reversing Functions: f : A ! A, where A is the set of all bitstrings of finite length and f(b) = the bitstring obtained by writing the characters of b in reverse order.

Onto / surjective (∀y∈Y, x∈X, such that f(x) = y)

Every element of Y has a mapped element of X. Basically no element in Y is empty. How to prove: We must show that the inverse of Y (i.e. X = something) so take Y = whatever (x) and solve for X. plugging into F(x) equals Y. I.E. Let X = (inverse of y) F(x) = Y

Bipartite graph

Consists of n isolated vertices, adjacent to every m isolated vertex. K n,m has order n + m and size ___

One to One / injective (∀x₁, x₂ ∈ X, f(x₁) = f(x₂) implies x₁ = x₂)

Each element of X goes to a different element of Y. Basically no 2 arrows point to the same element. How to prove: Let X1 and X2 be elements of your domain. Set functions F(x1) = F(x2) and then simplify the equations until x1 = x2

Complete graph

Every vertex is adjacent to every other vertex, meaning the line has 2 dots on each end. Kn has order of n and size of (n / 2)

Function

F : A -> B is a binary relation such that every element in domain has exactly one image. (A , B) ∈ F, we often write F (A) = B. If F (A1) =/ F (A2), then A1 =/ A2. If |A| = n, |B| = m, then any function F : A -> B must have |F| = n (exactly one image for every member of A), If A = {A1,...,An}, then f = {(A1, f(A1)), . . . , (An, f(An))}

Polynomial Variant example

F : R --> Z Domain = real #'s, codomain = integers F(x) = Ceiling of (x^2 - 4) find x intercept: (x^2 - 4) = 0 --> (-1 < x^2 -4 < 0) --> +4 (3 < x^2 < 4) --> x = (sqtr3 , 3) or x ∈ (2, -sqrt3)

Edge density

Fraction of total possible edges G has. 0 <= |E| / (|V| / 2) <= 1 Sparse graphs have low density, Dense graphs have higher density

Cartesian product A x B = { (a,b) : a ∈ A and b ∈ B }

Given sets A and B, the Cartesian product is AxB is a set whose elements are all the ordered pairs of A and B combined (order matters). AxB doesn't not = BxA

Relational digraph

Graph representing Binary relations by making each element a circle, and point to other elements or itself.

Empty graph / independent set

Has dots but no lines or vertices. nK1 has order of n and size of 0. https://kennesaw.view.usg.edu/d2l/lp/documents/aqBidvvqOCW7Ts6YhmRh27551966/28/1

Symmetry (xRy --> yRx)

If 2 elements have arrows pointing to and from themselves ( A->B and B ->A ) How to prove: x − y Suppose x, y ∈ R and xRy. Then x − y is an integer. Since y − x = − ( x − y ) , y − x is also an integer. Thus ,yRx.

Partitions

If A is a nonempty set, then there is a one-to-one correspondence between the set of all equivalence relations on A and the set of all partitions of A. "Every partition of a nonempty set can be matched with a unique equivalence relation on that set."

Countability

If a set or subset is finite is it countable, and uncountable if it is infinite

Image of a subset

Let F : A -> B be a function and let C⊆A and D⊆B. The image of C under F is F(C) = {F(x) : x∈C}. And the inverse image of D under F is f^-1(D) = {a∈A : F(a)∈D} For every C⊆A, we have f(C)⊆B. F(A) is the range of F. For every D⊆B ,we have f^-1(D) ⊆ A

Equivalence Class proof example ( [a] = [b] ⇔ (a, b) ∈ R )

Let R be an equivalence relation on a nonempty set A and let a, b ∈ A. How to prove: (⇒): As R is reflexive , (a,a) ∈ R. so a∈ [a]. As [a]=[b] we have a ∈ [b] so (a,b) ∈ R. (⇐): Let c ∈ [a]. Then (c,a) ∈ R. As (a,b) ∈ R and R is transitive , (c,b) ∈ R so c ∈ [b]. Thus [a] ⊆ [b]. Conversely, let c ∈ [b]. Then (c, b) ∈ R. As (a, b) ∈ R and R is symmetric (b,a) ∈ R, and as R is transitive, (c,a) ∈ R so c ∈ [a]. Thus [b] ⊆ [a]. Therefore,[a] = [b]. All elements in the same equivalence class are essentially viewed as equal. Any one element in a class can be used to represent the class.

Classes are identical or disjoint ( [a] = [b] or [a] ∩ [b] = ∅)

Let R be an equivalence relation on a nonempty set A and let a, b ∈ A. How to prove: Suppose that [a] ̸= [b]. We must show that [a] ∩ [b] = ∅. Assume to the contrary, that thereexistsc∈[a]∩[b]. Then(c,a)∈Rand(c,b)∈R. Bysymmetry,(a,c)∈R. By transitivity, (a, b) ∈ R. From previous theorem [a] = [b], a contradiction.

Corollary / class partitions on a set

Let R be an equivalence relation on a nonempty set A. The set of equivalence classes P = {[a] : a ∈ A} are a partition of A.

Tree graph

Minimally connected, no cycles. Order is n, size is n-1 How to prove: If n = 1 ,then T ∼= K1 and has m = 0 = 1 − 1 edges. IH: Suppose for an arbitrary but specific integer k ≥ 1, any tree of order k has size k − 1. Consider a tree T of order k + 1. T has an end-vertex, say u, so T′ = T− u is a tree of order(k+1) − 1 = k. By IH ,T′ has size k−1. Since u was an end-vertex, | E(T) | = | E(T′) | + 1, so T has size k − 1 + 1 = k. https://kennesaw.view.usg.edu/d2l/lp/documents/aqBidvvqOCW7Ts6YhmRh27551966/35/1

Cardinality |x|

Number of elements in a set, not to be confused with range. Denoted by |x|. Same cardinality means there is a bijection from A to B. To prove same cardinality, must prove bijection.

Equivalence Relation proof example

Prove R = {(a, b) : 3 | (a − b)} is an equivalence relation on Z. How to prove: Reflexive: Let a be a particular but arbitrarily chosen element of Z. Then 3 | 0 because 0 = 3 · 0, so 3 | (a − a) and thus (a, a) ∈ R. Symmetric: Suppose that (a, b) is a particular but arbitrarily chosen element of R.Then 3 | (a−b), so a−b = 3k for some integer k, thus b−a = 3(−k), so 3 | (b − a) meaning (b, a) ∈ R. Transitive: Suppose that (a, b) and (b, c) are particular but arbitrarily chosen elementsofR. Then3|(a−b)and3|(b−c)soa−b=3kandb−c=3l forsomekandlinZ. Thena−c=(a−b)+(b−c)=3(k−l)so 3|(a−c). Thus(a,c)∈R.

Floor Function

Round down to the nearest integer (floor if 2.14 is 2) Domain = Real #'s , Codomain = Integers. Definition : n ≤ −x < n + 1.

Ceiling Function

Round up to the nearest integer (ceiling of -1.89 is -1) Domain is Real #'s , Codomain = Integers Definition: n−1<x≤n

Range {b ∈ B : ∃ a ∈ A, (a , b) ∈ R}

Set of all values of f in Y, that have a pre image. Set of all the images. Not to be confused with cardinality.

Composition (F: X -> Y and G: W -> Z : g°f)

The range of F is a subset, and Domain of G. The composition of F and G denoted by g°f = X -> Z = (g°f)(x) = g(f(x))

Degree Sum Formula

The sum of the degrees of all vertices equals twice the number of edges. LetX = {(v,e) : v∈V(G), e∈E(G), and v is an end point of e}. 1 Vertex-Perspective: for any fixed vertex v ∈ V(G), it is an endpoint of exactly deg(v) edges. So |X| = v∈V (G) deg(v). 2 Edge-Perspective: for any fixed edge e∈E(G), it has exactly two endpoints. So |X| = 2 · | E(G)|.

Permutation

Type of bijection where X and Y are the exact same

Modular function

Used in coding and cryptography, or finding a day of the week. A ( Mod N ) = R where 0 < R < N EX) What day of the week will it be in 267 days if today is Thursday. 267 = 38 * 7 + 1 = Friday

Graph basics

Vertex (V) = dot, object, node Edge (E) = relationships, lines, links elements of E are subsets of V having a cardinality of 2 Order of G is the cardinality of V = |V| Size of G is the cardinality of E = |E|

Image and Pre image

X is preimage(inverse image) of Y Y is the image of X under F. If A -> B is a binary relation, then (S,T) ∈ R, then S is a preimage of T under R, and T is an image of S under R.

Circle graph

https://kennesaw.view.usg.edu/d2l/lp/documents/aqBidvvqOCW7Ts6YhmRh27551966/34/1 https://kennesaw.view.usg.edu/d2l/lp/documents/aqBidvvqOCW7Ts6YhmRh27551966/33/1

Networks

some examples: transportation, ecological, social media / internet, historical, art, biological /human body. https://kennesaw.view.usg.edu/d2l/lp/documents/aqBidvvqOCW7Ts6YhmRh27551966/60/1

Notation

| or : = such that {(a,b) : 3 | (a - b)} (n / 2) = n pick 2 A -> B = If A, then B ∈ = member or element of the set ( a ∈ A) A ⊆ B = SUBset, A contains some/ all elements of B A ⊇ B = SUPERset, A has same elements as A, or more {} = set of elements Q = rational #'s R = real #'s P = prime #'s (2,3,5,7,11,13,17) A − B = Difference, in A but not in B A ∪ B = Union or both together A ∩ B = Intersection or what's common in both

Universal Quantifier {∀x ∈ D, Q(x)}

∀ = "for all". D = domain "being human" Q(x) = predicate "drinking water". Then this means, All humans, also drink water. Negation of this is an existential quantifier.

General proof format

∀x. P Direct proof: Pick an arbitrary x, then prove P is true for that choice of x. By contradiction: Suppose for the sake of contradiction that there is some x where P is false. Then derive a contradiction. ∃x. P Direct proof: Do some exploring and find a choice of x where P is true. Then, write a proof explaining why P is true in that case. By contradiction: Suppose for the sake of contradiction that P is always false and derive a contradiction. ¬P Direct proof: Simplify your formula by pushing the negation deeper, then apply the appropriate rule. By contradiction: Suppose for the sake of contradiction that P is true, then derive a contradiction. P∧Q Direct proof: Prove each of P and Q independently.By contradiction: Assume ¬P ∨ ¬Q. Then, try to derive a contradiction. P∨Q Direct proof: Prove that ¬P → Q, or prove that ¬Q → P. (Great question to ponder: why does this work?) By contradiction: Assume ¬P ∧ ¬Q. Then, try to derive a contradiction. P→Q Direct proof: Assume P is true, then prove Q.By contradiction: Assume P is true and Q is false, then derive a contradiction.By contrapositive: Assume ¬Q, then prove ¬P. P↔Q Prove both P → Q and Q → P.

Existential Quantifier {∃x ∈ D, Q(x)}

∃ = "for some, there exists at least one". D = "being human" Q(x) = predicate "drinking alcohol". Then this means, There exist some humans, that also drink alcohol. Opposite of this is a Universal Quantifier.