Notice that (|Atimes B|=mn). Also, (|C_i|=n) for all (i=1,2,3,cdots ,m). hence (sum_{i=1}^m |C_i|=sum_{i=1}^m n=mn). Similarly, we can prove that (sum_{j=1}^n |D_j|=mn).

And hence, we get our desired result.

This is just proving that:

Theorem: 

Given any three non-collinear points (A, B, C)  there exists a unique circle passing through (A, B, C).

Proof: 

Let us consider three vertices of (Delta ABC) i.e. (A, B, C). Suppose the perpendicular bisectors of (BC) and (CA) meet at (S). Then (S) lies on the perpendicular bisector of (BC) implies that (S…Read More

No the statement is false.

Let’s take an example,

if we make a rule and declare that the difference between x and y is  divisible by 3 which makes an equivalence relation

so in this way we can partition many sets like

{0,3,6,9,12……………}

{1,4,7………………}

{2,5,8,11…………..}

by this we cannot partition all the numbers…Read More

No this is not true that from equivalence relations we can get partitions and from partitions we can get equivalence relations.

There is a simple reason why-

If we make a rule in which we declare that the difference between any two numbers X and Y is divisible by 2 then it is an equivalence relation.

By this  Sets can be formed which can be…Read More

EQUIVALENCE TO PARTITONs

We can partition the points/elements according to their equivalence like – suppose we have a set S which has 6 points/elements- {A1,A2,A3,A4,A5,A6}

• SYMMETRIC-If A1~A3(A3 ~ A1 as well) then we can arrange them like {A1,A3} (or like {A3,A1})
• REFLEXIVE – If A2~A2 then we can arrange it like {A2}
• TRANSITIVE– If A4~A5 ,…

Read More

No it is not true that equivalence relation makes a partition and Partition makes an equivalence relation always.

Let’s take a way of changing an equivalence relation to a partition first

And for an example let’s make a rule that take two numbers x and y such that x and y is divisible by 3,then it is an equivalence relation

Like this many Sets…Read More

It is true because:

every equivalence relation is

-symmetric

-reflexive and

-transitive

so when we partition the sets they are reflexive, symmetric and transitive

Equivalence Relation guarantees Partition derivation but reverse is not true.

It is true that every equivalence relation gives a partition

For an equivalence relation , the relation must be reflexive,symmetric and transitive

All partitions are reflexive,symmetric and transitive, so it is an equivalence relation.And the equivalence classes of a partition are the sets.So every equivalence relation partitions its set into…Read More

Yes, since making equivalence relations means that the elements are in the same set, we can say that stating elements equivalent means to break the set and form some new set (partition).

Equivalence relation -> Partition

Constrtuct partitions from the relation as follows,

1) pick an element from set S, take all y in S such that x~y and put them all in a new partition

2) repeat step 1 until all elements of S are exhausted

step 1 guarantees disjointedness and step 2 guarantees exhaustiveness

Partition -> Equivalence…Read More

We can partition the points/elements according to their equivalence like – suppose we have a set S which has 6 points/elements- {A1,A2,A3,A4,A5,A6}

• SYMMETRIC-A1~A3 then we can arrange them like {A1,A3}
• REFLEXIVE – A2~A2 then we can arrange it like {A2}
• TRANSITIVE– A4~A5 , A5~A6 and A4 ~A6 then we can arrange them like {A4,A5,A6}

So,  we…Read More

Proof:1. If there is x in a set, then it will have to be in a partition.(symmetric)

2.If x ~y, they are in the same partition.

=>y~x(reflexive)

3.If x~y,y~z,

Then x~z as all are in the same partition.(transitive)