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
@seradfb345 Жыл бұрын
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 Жыл бұрын
This is brilliant explanation. Thank u
@legcramp4475 Жыл бұрын
So are all reflexive a transitive?
@kevinebedi525711 ай бұрын
yeah nice breakdown thanks!!
@TammanaSultanaFood2 жыл бұрын
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 Жыл бұрын
it is not transitive i guess it's a mistake
@naveedahmedsoomro4354 Жыл бұрын
If something is reflexive it's not mean that it's also a transitive because in some situations reflexive isn't a transitive
@ramanjaneyuluv479 Жыл бұрын
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 Жыл бұрын
No, they are 3 different properties
@sahilchaudhary2412 Жыл бұрын
@@salimelghersse4822yes it is transitive bcz property says if there is (a,b) belongs to R and (b,c) belongs to R then there should be exists (a,c)......soo here (a,b) and (b,c) is not present soo there is no need to have (a,c) therefore it is transitive.....hope it helps
@masteradil10 ай бұрын
Can you clear R1 please? How is it symmetric and transitive
@danichef Жыл бұрын
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 Жыл бұрын
thank you!!!
@kainaatmakhani65502 жыл бұрын
You are brilliant sir. Thank you so much for making us understand hard concepts easily.
@damirkoblev4333 Жыл бұрын
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 Жыл бұрын
Transitive Definition: if all pairs consist from the same elements and do not connect with other ones
@zubaerahammed Жыл бұрын
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.
@himanshumaurya4283 жыл бұрын
How is the first relation equivalence? How is it transitive?
@Tenaciousplays012 жыл бұрын
Its not transitive,he made mistake
@himanshumaurya4282 жыл бұрын
@@Tenaciousplays01 yeah
@nmm61672 жыл бұрын
I asked about that and someone told me if there isn't condition then its transitive
@trusttheprocess47752 жыл бұрын
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.
@himanshumaurya4282 жыл бұрын
@@trusttheprocess4775 oh is that so. thanks mate
@varzhenenithiyananthan8602 жыл бұрын
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).
@JuliaGugulski Жыл бұрын
Best video on KZbin.
@SHUBHHAM26 күн бұрын
Hii Julia
@anupamagarwal3976 Жыл бұрын
Sir, How R1 a transitive relation ?
@alibekbalkybekov4809 Жыл бұрын
If there aren't different pairs of elements like (1,2) or (2,3) we don't need to check them them for transitive
@soummossj26242 жыл бұрын
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.
@ronakop23573 ай бұрын
it is transitive as well, it is an equivalence relation
@faizazam4256 Жыл бұрын
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-zy1uj3 жыл бұрын
Thankyou sir❤️❤️❤️🙏🙏
@jaybansod442 жыл бұрын
Create one discrete mathematics play list
@E_Hooligan2 жыл бұрын
Amazing explanation, thanks!
@LexyMrLee9111 Жыл бұрын
Well explain sir.....
@amarsgamer77523 жыл бұрын
Thank u sir 🙏🙏 ❤️😘
@Bnbnayak5 ай бұрын
Thank you sir😊
@AbhishekThakur-fk7px Жыл бұрын
Thank you so much sir.
@sachingaming61693 ай бұрын
Thank u so much sir😊
@دانيهالعيد2 жыл бұрын
Thank you for this amazing explanation!
@MathCuriousity9 ай бұрын
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!!
@tim-duncan2137 Жыл бұрын
Thank you sir
@goodnightvids Жыл бұрын
This guy is great
@codex7299 Жыл бұрын
Thank you❤
@14tajmohammadansari332 жыл бұрын
nice explanation sir
@peterchimbuto40952 жыл бұрын
Thanks
@swarnaganesh56932 жыл бұрын
thank you :)))
@karmaxscience36323 жыл бұрын
Upload more videos on data structures and algorithms sir.
@laibahameed42122 жыл бұрын
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
@navdeepkaur21312 жыл бұрын
Amazing
@samratbarui. Жыл бұрын
Thanku sir
@nikhilnaidu78992 жыл бұрын
excellent
@amnakhawaja15482 жыл бұрын
how is R1 transitive? there's no c but (a,a)
@siddheshpandey79054 ай бұрын
R3 is not transitive
@jadibamaniya99486 ай бұрын
Very well explained.. Up to point... ❤
@hannahferrera5591 Жыл бұрын
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 Жыл бұрын
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.
@harshitasharma76658 ай бұрын
Property need to be true for all elements.
@ritchmondjamestajarros44392 жыл бұрын
In the first equivalence, why is it symmetric?
@gmoney_swag12745 ай бұрын
I love you
@avirajbhandari981110 ай бұрын
Thank you! This is very helpful.
@animeshhazra6061 Жыл бұрын
Nice
@madihadrawingacademy4823 Жыл бұрын
how R1 is symetric
@kosumeghanameghana33602 жыл бұрын
Excellent explanation sir,tq soo much
@jaysonjamesalvarez15723 ай бұрын
its like an entire semester in 6 mins thank you man, gotta take this exam in a few hours
@rahelina7176 Жыл бұрын
All of this is good but I do not get how R6 is transitive. Thank you anyways.
@adithyadinesh98322 жыл бұрын
Approved by Hanena
@ههخخ-ع3ت2 жыл бұрын
Can Anyone tell me how can I understand AXA ?? I will appreciate if any one can help me in this plz
@akarusardarbekova59272 жыл бұрын
A x A={(a,a), (a,b), (b,a), (b,b)} which has 4 elements.
@sumrelachaunria87762 жыл бұрын
If A={a,b} where a and b are elements then AXA={(a,a),(a,b),(b,a),(b,b)}
@SKP21024 ай бұрын
AXA simply means the Cartesian product from a Non-empty Set A to the same set B which means mapping of every element of A with each element of A. So create one by one all the ordered pairs from setA to set A
@hhhhhhha954 Жыл бұрын
sorry why AxA is surely to be equivilant
@seradfb345 Жыл бұрын
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
@Rocker123ukАй бұрын
Correction: R3 is not transitive
@tanveersingh31962 жыл бұрын
That relation in first seconds of video wasnt transitive
@vipinyadav66612 ай бұрын
Example 1 - Transitive 😂😂😂. But How If (aRb) & (bRc) then must be (aRc) but there is (cRa) . So not Transitive 🤫.
@hrithikIITR25 Жыл бұрын
😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍😍
@umgsbed39224 ай бұрын
ما عم افهمكسمس
@anesumukombachoto852511 ай бұрын
This a lie ,he is a liar
@weinerblut6869 Жыл бұрын
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 Жыл бұрын
He did that on the previous videos
@tibetatakan11 ай бұрын
what a dumb comment
@shreedevis80193 жыл бұрын
You are brilliant sir.Thank you so much for making us understand hard concepts easily.