Equivalence Relation

Discrete Mathematics: Equivalence Relation
Topics discussed:
1) The definition of discrete mathematics.
2) Example problems to find out if the given relation is an equivalence relation.

Music:
Axol x Alex Skrindo - You [NCS Release]

#DiscreteMathematicsByNeso #DiscreteMaths #Relations #EquivalenceRelation

[TammanaSultanaFood]

for the first example, if something is reflexive, is it automatically transitive and symmetric? you didn't go over that but i know its symmetric but i am struggling with transitive.

[sigmapiebonds]

Literally in love with the content. It proved so helpful for me as its my exam tomorrow and found this gem!!! So well explained; our clg teachers dont even gives so precisely

[seradfb]

For those who said it is not transitive, I think a possible misconception of transitivity means that a given number can relate to any other number in the set, but this is not what transitivity means:

Definition:
A relation is transitive if xRy and yRz, then xRz.

Example:
To clarify, the definition of x here is the first element of a pair from R1 and so on.
For example looking at R1 we can decide x to be any first value from all of the pairs in R1. Let's say we decide that x = 2
The only pair which contains 2 is (2, 2).
Now, referring to the definition of transitive above, y in this case must be the second item of the chosen pair which is 2.
xRy = (2, 2)
As we know that y = 2, to find z we look for the second item of a pair with 2 in. Again (2, 2) is the only pair, so it follows that z = 2.
yRz = (2, 2)
As x = 2 and z = 2, we can see that it holds true that xRz = (2, 2)
So the definition holds true: if (2, 2) and (2, 2) then (2, 2)

x and y would always be the same number though. They could not be different because there is no such pair.


Proof:
An easy way to check if a matrix is in fact transitive is by the theorem which states:
A relation is transitive if and only if R² ⊆ R
(In this example it means basically the squared matrix does not contain a 1 anywhere the original matrix does not)
In example one the matrix (M) for R1 is just the identity matrix:

M=
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

M² =

1 0 0 0

0 1 0 0
0 0 1 0
0 0 0 1

so M² ⊆ M

Therefore R1 is transitive.

[masteradil]

Can you clear R1 please?
How is it symmetric and transitive

[danichef]

For everyone wondering about R1:

A relation is transitive if
aRb and bRc imply aRc.
R={(1, 1), (2, 2), (3, 3)}
Let’s consider all the cases where aRb and bRc, There are only three ways this can happen:

a=1; b=1; c=1

a=2; b=2; c=2

a=3; b=3; c=3

In every single case, we have aRc. aRb and bRc imply aRc.

[kainaatmakhani]

You are brilliant sir. Thank you so much for making us understand hard concepts easily.

[damirkoblev]

Thanks a lot! But I still can't understand how can the first relation be transitive if all pairs consist from the same elements and do not connect with other ones?

[himanshumaurya]

How is the first relation equivalence? How is it transitive?

[varzhenenithiyananthan]

Can you solve the question ?? Let R be an equivalence relation on a set A, and let a€A and b€A. Prove that aRb if and only if R(a)=R(b).

[faizazam]

For Equivalance Relation all 3 has to be satisfied or one is enough.

As you say R2 in not reflexive and yes they are not but they are symmetric and transitive

Are they not equivalence?... Please Reply...🙏🏻

[soummossj]

On example one R2 you said it is not reflexive because there is no (1, 1) ...there is not a single element of 1 here.. so I think its safe to say 1 doesn't exist in this relation...so its reflexive, symmetric but not transitive. Please take the time to provide correction in the desc.

[jaybansod]

Create one discrete mathematics play list

[amnakhawaja]

how is R1 transitive? there's no c but (a, a)

[دانيهالعيد]

Thank you for this amazing explanation!

[jadibamaniya]

Very well explained.. Up to point...
❤

[MathCuriousity]

Hi may I pose a question: let’s say we have an equivalence relation aRb. Why can’t I represent this within set theory as set T comprising subset of Cartesian product of a and b, mapped to a set U which contains true or false? Thanks so much!!

[jaysonjamesalvarez]

its like an entire semester in 6 mins thank you man, gotta take this exam in a few hours

[laibahameed]

No, R1 is not an equivalent relation, for equivalence relation a function must hold the properties of reflexive, symmetric, transitive, it doesn't hold transitive relation

[ritchmondjamestajarros]

In the first equivalence, why is it symmetric?

[karmaxscience]

Upload more videos on data structures and algorithms sir.