CSE 20
DISCRETE MATH
Prof. Shachar Lovett
http://cseweb.ucsd.edu/classes/wi15/cse20-a/
Clicker
frequency:
CA
Todays topics
• Equivalence relations
• Section 6.2 in Jenkyns, Stephenson
(Binary) Relations
• Model a relation between two families of objects
• Formally:
• U universe set
• Relation: π
⊆ π × π
• For π₯, π¦ ∈ π we write: π₯π
π¦ as a shorthand for "(π₯, π¦) ∈ π
“
• Example 1: x<y, formally…
• U=N
• “π₯ < 𦔠represents the relation π
=
π₯, π¦ : π₯, π¦ ∈ π, π₯ < π¦
• Example 1: x<y, formally…
• U = P(N)
• “π₯ ⊂ 𦔠represents the relation π
= { π₯, π¦ : π₯, π¦ ∈ π(π), π₯ ⊂ π¦}
Three important properties of relations
• There are several important properties of relations
• We will focus on the most important three
• A relation R is symmetric if ∀π₯, π¦, π₯π
π¦ ⇔ π¦π
π₯
• A relation R is reflexive if ∀π₯, π₯π
π₯
• A relation R is transitive if ∀π₯, π¦, π§,
π₯π
π¦ ∧ π¦π
π§ → π₯π
π§
• An equivalence relation is a relation which satisfies all of
them
Examples of equivalence relations
• U=Z (integers)
• xRy = “x=y”
Prove that R is an equivalence relation
Examples of equivalence relations
• U=Z (integers)
• xRy = “|x|=|y|” (absolute value of x = absolute value of y)
Prove that R is an equivalence relation
Examples of equivalence relations
• U=“all students in this class”
• xRy = “x and y have the same birthday”
Prove that R is an equivalence relation
Examples of equivalence relations
• U=Z (integers)
• xRy = “x+y is even”
Prove that R is an equivalence relation
What is so special about equivalence
relations?
• Equivalence relations describe a partition of the universe
U to equivalence classes
Equivalence classes
• Let π1 , π2 , … , ππ be a partition of U
• For π₯ ∈ π, define πΆπππ π π₯ to be the (unique!) part that x
belongs to
• Equivalently, define πΆπππ π : π → {1, … , π} and equivalence
classes are defined by all elements with same image
ππ = {π₯ ∈ π: πΆπππ π π₯ = π}
• Theorem: let π
⊆ π × π be an equivalence relation. Then
there is a mapping πΆπππ π : π → πΆ such that
∀π₯, π¦ ∈ π, π₯π
π¦ ⇔ πΆπππ π π₯ = πΆπππ π (π¦)
Equivalence classes: example 1
• U=Z (integers)
• xRy = “x=y”
Class(x)=
A. x
B. y
C. U
D. “x=y”
E. Other
Equivalence classes: example 2
• U=Z (integers)
• xRy = “|x|=|y|”
Class(x)=
A. x
B. |x|
C. “x=y”
D. “|x|=|y|”
E. Other
Equivalence classes: example 3
• U=“all students in this class”
• xRy = “x and y have the same birthday”
Class(x)=
A. x
B. Birthday of x
C. “x=y”
D. “|x|=|y|”
E. Other
Equivalence classes: example 4
• U=Z (integers)
• xRy = “x+y is even”
Class(x)=???
Figure this out in your groups
Equivalence classes: proof
• Theorem: let π
⊆ π × π be an equivalence relation. Then
there is a mapping πΆπππ π : π → πΆ such that
∀π₯, π¦ ∈ π, π₯π
π¦ ⇔ πΆπππ π π₯ = πΆπππ π (π¦)
Proof:
For each π₯ ∈ π, define: Class π₯ = {π¦ ∈ π: π₯π
π¦}
(and so πΆ = πΆπππ π π₯ : π₯ ∈ π ⊂ π(π) )
We need to show that
∀π₯, π¦ ∈ π, π₯π
π¦ ⇔ πΆπππ π π₯ = πΆπππ π (π¦)
Equivalence classes: proof (contd)
For each π₯ ∈ π, define: Class π₯ = {π¦ ∈ π: π₯π
π¦}
(⇐) Assume Class π₯ = πΆπππ π (π¦). We will show xRy.
By definition: we need to show π¦ ∈ πΆπππ π (π₯).
But π¦ ∈ πΆπππ π π¦ since R is reflexive (and so π¦π
π¦)) and
since we assume Class π₯ = πΆπππ π (π¦) then also π¦ ∈
πΆπππ π (π₯).
Equivalence classes: proof (contd)
For each π₯ ∈ π, define: Class π₯ = {π¦ ∈ π: π₯π
π¦}
(⇒) Assume xRy. We need to show Class π₯ = πΆπππ π (π¦).
We will first show πΆπππ π π¦ ⊆ πΆπππ π (π₯). Let π§ ∈ πΆπππ π π¦ .
Then yπ
π§. Since R is transitive, π₯π
π¦ ∧ π¦π
π§ imply that xπ
π§.
So π§ ∈ πΆπππ π (π₯).
Since R is symmetric, we also have yRx. The same
argument as above (with the roles of x,y reversed) then
implies that πΆπππ π π₯ ⊆ πΆπππ π (π¦). So: πΆπππ π π₯ = πΆπππ π (π¦).
Next class
• Modular arithmetic
• Section 6.2 in Jenkyns, Stephenson
• Review session for midterm 2