Exam 1 Definitions Flashcards
even
divisible by two
divisible
if b|a then bc = a
odd
a is odd if a = 2x+1
prime
p > 1 and divisors of p are 1 and p
composite
a is composite if b|a such that 1 < b < a
subset
A and B are sets
every element of A is in B
A ⊆ B means A is a subset of B.
power set
set of all subsets of A
union
A or B
A ∪ B
intersection
A and B
A ∩ B
theorem about sets
- A ∪ B = B ∪ A and A ∩ B = B ∩ A (commutative properties)
- A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C
- A ∪ Ø = A and A ∩ Ø = Ø
- A ∪ (B ∩ C) = ( A ∪ B) ∩ (A ∪ C) and A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). (distributive properties)
disjoint
A ∩ B = Ø
there is nothing in both A and B
pairwise disjoint
a collection of sets where Ai ∩ Aj = Ø and i ≠ j so that there are no two of the same elements in common
set difference
A - B = x
x is in A but not B, denoted A ∆ B = (A - B) ∪ (B - A)
cartesian product
A x B, set of ordered pairs by taking a from A and b from B to make (a,b)
relation
set of ordered pairs
relation on, between sets
R is a relation on A provided R ⊆ A x A
R is a relation between A and B provided R ⊆ A x B
inverse relation
reversing the order for all ordered pairs in R
properties of relations
- x ∈ A then xRx (reflexive)
- x ∈ A then x ≠ x (irreflexive)
- x, y ∈ A then xRy ⇒ yRx (symmetric)
- x, y ∈ A then (xRy ^ yRx) x = y (antisymmetric)
- x, y, z ∈ A then (xRy ^ yRz) ⇒ xRz (transitive)
equivalence relation
the relation is reflexive, symmetric, and transitive
congruence modulo n
x ≡ y (mod n) provided n|(x - y)
equivalence class
a ∈ A where [a] is the set of all elements of A related by R
[a] = {x ∈ A : xRa}
partition
the partition of A has to be nonempty, pairwise disjoint and their union is A
function
f = (a, b) and f = (a, c) implying b = c
function notation
f(a) exist provided that (a, b) ∈ f
f(a) = b
if there is no ordered pair than f(a) is undefined
domain
set of all possible first elements in the ordered pair in f
dom f
image
set of all possible second elements in the ordered pair of f
im f
function from A to B
dom f = A and im f ⊆ B
(f : A → B)
proposition inverse function
if f^-1 then dom f = im f^-1 and f = dom f^-1
onto
f is onto B for f(a) = b or im f = B
bijection
function is both one-to-one and onto