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.
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Axol x Alex Skrindo - You [NCS Release]
#DiscreteMathematicsByNeso #DiscreteMaths #Relations #EquivalenceRelation

[@sigmapiebonds3191]

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

[@kainaatmakhani6550]

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

[@user-eh3oo8iw2r]

Thank you for this amazing explanation!

[@seradfb345]

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.

[@hannahferrera5591]

This is brilliant explanation. Thank u

[@legcramp4475]

So are all reflexive a transitive?

[@kevinebedi5257]

yeah nice breakdown thanks!!

[@E_Hooligan]

Amazing explanation, thanks!

[@avirajbhandari9811]

Thank you! This is very helpful.

[@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.

[@CebelleWhiteFide]

thank you!!!

[@jadibamaniya9948]

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

[@kosumeghanameghana3360]

Excellent explanation sir,tq soo much

[@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.

[@salimelghersse4822]

it is not transitive i guess it's a mistake

[@okay........]

OH MY GOD SAME HERE I HATE TRANSITIVE

[@naveedahmedsoomro4354]

If something is reflexive it's not mean that it's also a transitive because in some situations reflexive isn't a transitive

[@ramanjaneyuluv479]

It is transitive because there are no different elements to compare by default we consider it as transitive only even it is same with the symmetric also

[@axelsanchez5849]

No, they are 3 different properties

[@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!!

[@faizazam4256]

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...🙏🏻

[@gamer-zy1uj]

Thankyou sir❤️❤️❤️🙏🙏

[@14tajmohammadansari33]

nice explanation sir

[@JuliaGugulski]

Best video on KZbin.

[@AbhishekThakur-fk7px]

Thank you so much sir.

[@lexymrtgt9047]

Well explain sir.....

[@karmaxscience3632]

Upload more videos on data structures and algorithms sir.

[@ahmetkarakartal9563]

thank you so much

[@kailashnayak7748]

Thank you sir😊

[@goodnightvids]

This guy is great

[@lubnaansari3948]

Thank you so much sir

[@codex7299]

Thank you❤

[@masteradil]

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

[@tim-duncan2137]

Thank you sir

[@navdeepkaur2131]

Amazing

[@addybanerjee1767]

Thanks

[@swarnaganesh5693]

thank you :)))

[@samratbarui.]

Thanku sir

[@nikhilnaidu7899]

excellent

[@varzhenenithiyananthan860]

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).

[@damirkoblev4333]

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?

[@nabilwalidrafi5679]

Transitive Definition: if all pairs consist from the same elements and do not connect with other ones

[@zubaerahammed]

The condition is if (a,b) and (b,c) belongs to R, then (a,c) belongs to R. But in that relation, there is no such (a,b) and (b,c). So, there is nothing to check against for its transitivity. It will not be transitive only if there are (a,b) and (b,c) ordered pairs present and there is no (a,c) present.

[@amnakhawaja1548]

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

[@scob16jaybansod44]

Create one discrete mathematics play list

[@ritchmondjamestajarros4439]

In the first equivalence, why is it symmetric?

[@animeshhazra6061]

Nice

[@soummossj2624]

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.

[@anupamagarwal3976]

Sir, How R1 a transitive relation ?

[@alibekbalkybekov4809]

If there aren't different pairs of elements like (1,2) or (2,3) we don't need to check them them for transitive

[@madihadrawingacademy4823]

how R1 is symetric

[@laibahameed4212]

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

[@hannahferrera5591]

Is the relation number 4 really not a symmetric? I am confused. What I have learned in class is that if (a, b) element of R, then (b, a) must be an element of R too. And this property does not need to be true for all elements.

[@hannahferrera5591]

What I am trying to imply is there should be at least one pair of element that is symmetric. In the Video, there is (2,0) in R as well as (0,2). Hence, the relation is symmetric.

[@harshitasharma7665]

Property need to be true for all elements.

[@rahelina7176]

All of this is good but I do not get how R6 is transitive. Thank you anyways.

[@gmoney_swag1274]

I love you

[@adithyadinesh9832]

Approved by Hanena

[@himanshumaurya428]

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

[@Tenaciousplays01]

Its not transitive,he made mistake

[@himanshumaurya428]

@@Tenaciousplays01 yeah

[@nmm6167]

I asked about that and someone told me if there isn't condition then its transitive

[@trusttheprocess4775]

No, he is right. If there are simply no conditions in the relation to check if the relation is transitive or not, then it is transitive. In the question, there was no (a,b) and neither (b,c) hence there were no conditions to check (a,c) whether it belonged to R or not, hence it is still transitive.

[@himanshumaurya428]

@@trusttheprocess4775 oh is that so. thanks mate

[@user-es3pq9lz4r]

Can Anyone tell me how can I understand AXA ?? I will appreciate if any one can help me in this plz

[@akarusardarbekova5927]

A x A={(a,a), (a,b), (b,a), (b,b)} which has 4 elements.

[@sumrelachaunria8776]

If A={a,b} where a and b are elements then AXA={(a,a),(a,b),(b,a),(b,b)}

[@siddheshpandey7905]

R3 is not transitive

[@hhhhhhha954]

sorry why AxA is surely to be equivilant

[@seradfb345]

AXA contains all possible pairs. The matrix for AXA in this case would look like this: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 If the diagonal line from top left to bottom right contains 1s, then it is reflexive. If you put a mirror along the same line and see reflection for all the other entries then it is symmetric. And hopefully you can also see how it can be transitive also. The point is AXA contains all possible pairs of the set

[@tanveersingh3196]

That relation in first seconds of video wasnt transitive

[@hrithikIITR25]

😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍

[@umgsbed3922]

ما عم افهمكسمس

[@shauny4596]

R1 is not transitive

[@weinerblut6869]

What a crummy explanation. Have you shown people here what reflexive, symmetric and transitive mean? Have you done anything else other than say it's obvious yes or obvious no.

[@oziomaodonoekuma1427]

He did that on the previous videos

[@tibetatakan]

what a dumb comment

[@anesumukombachoto8525]

This a lie ,he is a liar

[@shreedevis8019]

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

[@amarsgamer7752]

Thank u sir 🙏🙏 ❤️😘

[@peterchimbuto4095]

Thanks