Let S be a finite set. A binary relation on S is simply a...

Let $S$ be a finite set. A binary relation on $S$ is simply a collection $R$ of ordered pairs $(x, y) \in S \times S$. For instance, $S$ might be a set of people, and each such pair $(x, y) \in R$ might mean " $x$ knows $y$." An equivalence relation is a binary relation which satisfies three properties: $\bullet$Reflexivity: $(x, x) \in R$ for all $x \in S$ $\bullet$symmetry: if $(x, y) \in R$ then $(y, x) \in R$ $\bullet$ Transitivity: if $(x, y) \in R$ and $(y, z) \in R$ then $(x, z) \in R$ For instance, the binary relation "has the same birthday as" is an equivalence relation, whereas "is the father of" is not, since it violates all three properties. Show that an equivalence relation partitions set $S$ into disjoint groups $S_{1}, S_{2}, \ldots, S_{k}$ (in other words, $S=S_{1} \cup S_{2} \cup \cdots \cup S_{k}$ and $S_{i} \cap S_{j}=\emptyset$ for all $i \neq j$ ) such that: Any two members of a group are related, that is, $(x, y) \in R$ for any $x, y \in S_{i},$ for any $i$ Members of different groups are not related, that is, for all $i \neq j,$ for all $x \in S_{i}$ and $y \in S_{j}$ we have $(x, y) \notin R$

Let $S$ be a finite set. A binary relation on $S$ is simply a collection $R$ of ordered pairs $(x, y) \in S \times S$. For instance, $S$ might be a set of people, and each such pair $(x, y) \in R$ might mean " $x$ knows $y$."
An equivalence relation is a binary relation which satisfies three properties:
$\bullet$Reflexivity: $(x, x) \in R$ for all $x \in S$
$\bullet$symmetry: if $(x, y) \in R$ then $(y, x) \in R$
$\bullet$ Transitivity: if $(x, y) \in R$ and $(y, z) \in R$ then $(x, z) \in R$
For instance, the binary relation "has the same birthday as" is an equivalence relation, whereas "is the father of" is not, since it violates all three properties.

Show that an equivalence relation partitions set $S$ into disjoint groups $S_{1}, S_{2}, \ldots, S_{k}$ (in other words, $S=S_{1} \cup S_{2} \cup \cdots \cup S_{k}$ and $S_{i} \cap S_{j}=\emptyset$ for all $i \neq j$ ) such that:
Any two members of a group are related, that is, $(x, y) \in R$ for any $x, y \in S_{i},$ for any $i$
Members of different groups are not related, that is, for all $i \neq j,$ for all $x \in S_{i}$ and $y \in S_{j}$ we have $(x, y) \notin R$

Key Concepts

Partition

A partition of a set is a division of the set into non-overlapping, non-empty subsets such that the union of all these subsets is the original set. When an equivalence relation is defined on a set, the resulting equivalence classes form a partition, meaning each element belongs to exactly one equivalence class and no element is left out.

Equivalence Relation

An equivalence relation is a specific type of binary relation that satisfies three critical properties: reflexivity, symmetry, and transitivity. These properties ensure that the relation groups elements in a way that every element is related to itself, the relation is bidirectional, and the relation can be passed along chain-wise between elements.

Binary Relation

A binary relation on a set is a collection of ordered pairs that describes a relationship between elements of the set. It forms the foundational concept for understanding how elements can be connected by various properties and is the starting point for more complex relations such as equivalence relations.

Equivalence Classes

Equivalence classes are subsets of the original set where every pair of elements within a class is related by the equivalence relation. They are defined as the set of all elements that are equivalent to a particular element under the relation, and they capture the idea of grouping similar or 'equivalent' elements together.

Transcript

00:01 Hi, we're giving a relation here we are all the set natural numbers, cross natural numbers.

00:05 A, b, migration with b, b is given as a plus d, as equal to b.

00:08 You can find out how there is reflexive, symmetric, prongative, or equivalent relation.

00:13 So let's go for them.

00:15 First thing we have a plus t equals b plus c.

00:20 So that means we have a b in relation with c.

00:24 Now for symmetric, interchange here we have.

00:27 So in place of a, we put b, so b.

00:32 Plus in place of d we put c equals in place of b we put a plus g b plus e equals a plus g so we have uh i'm sorry we have in place of a so see here we have a correspond to c right so in place of a we just put c here plus we have b correspond to d so in place of d we put b equals in place of b we will have our d here plus we have in place of c is a so c plus b equals d plus a right, we just interchange here, x and y values.

01:10 So, c plus b equals d plus a.

01:12 I mean, in dividing, so a correspond to c, so in place of a we put c, in place of d we put b, in place of d we put b.

01:21 That's also equal to, so it's coming up to be symmetric.

01:23 So we can say that cd is in relation with ab.

01:27 It is symmetric.

01:31 For the black si we just take here, a plus b equals b plus c, the relation is a b, the relation is a, b.

01:37 With td right so what we can do here so here we have it is a plus g so it's come out to be a as it is plus now d corresponds to b so a plus b equals we have b as it is plus c corresponds to a plus a so a plus b is equal to b plus a that's what we do we put x as x and y as y in reflexive uh check reflexivity so it's going to be equal so it is reflex so we can say ab is in relation with ab.

02:08 It is the next for pramptit is so here we have a plus g equals b plus t a b plus b is in relation with c b so what we can do here now we have a plus d equals b plus c and next we have in place of a it is coming out to be c so c plus we will have in place of d it's coming up to be f right so we're checking here a, b relation with cd and c b relation with ef, that corresponds to a .b in relation with ef...

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