Types of Relations (Solved Problem) 

I promise to contribute the day I will get a job because right now my condition is not so good. This channel is the best I have ever found. Thank you so much neso academy

Homework Answer: NOT Transitive For example, (2, 1) and (1, 3) belongs to R, but (2, 3) does not belong to R

Very high quality. Thank you!

Hello Neso Academy, First of all thank you so much you guys for these awesome videos. Your content really helps me. But at some point, it feels like videos are not coming much frequently. No offense, I know it's not easy to make these videos and you guys make videos on many different topic too, and it's so much time consuming. But you guys started Discrete Mathematics series 3 years ago and it's still not half a way. I know it's a very huge area, it's going to take time, but it's just a genuine request from your student (your fan), that could you guys please upload videos much frequently. You guys said anyone in college who have this subject in syllabus or anyone preparing for GATE could watch it, but since the series is not fully completed, we can't understand the whole subject. Kitno ki to exam nikal jayegi jab tak yeh series khatam hogi. Just a genuine appeal, btw I love you guys and your content. ❤️

Thank q so much neso acadamy.....❤❤❤

Hi, great video. Please cover equivalence relation and equivalence classes in the next episode.

The last relation {x,y| x=1 or y=1 } is NOT TRANSITIVE because When we have a set A={1,2,3} (1,2),(2,3),(3,1) in this not all the all the pairs include one hence there’ll be a case where y is not = 1 or x is not =1 ((2,3)).

Good job

thanks

the answer is :not transitive see why if a=1 the answer will be transitive (a,b)=(1,2) and (b,c)=(2,3) then (a,c)=(1,3) but if b=2 the answer will be not transitive : (a,b)=(2,1) and (b,c)=(1,3) then (a,c)=(2,3) and it isn’t satisfied the condition of a=1 or b=1 . i hope it’s clear

In my class, we just end number theory... Maybe I will review discrete using your videos lol. ps. most i love is your c programming and data structure

Hi @4:01 why did you take x and y as 0 for reflexive and symmetric and why did you take 1 and -1 for antisymmetric and transitive? If we take same 0 for antisymmetric and transitive they will be accepted right?

H/W Answer ; Relation is not transitive since (a,1) belongs to R and (1,b) belongs to R but (a,b) does not belong to R

it is transitive. considering a=1, b=1 and c=2, then result will still be 1,2. x=1 or y=1 doesn't mean that both can not be equal to 1

Answer to H.W : R is Not Transitive. For Ex - a=2, b=1, c=2.

Isn't relation b) an equivalence relation on the real numbers ?

❤❤

Homework Answer: Not Transitive. For being transitive, it's not sufficient that X or Y be 1, the union of ordered pairs (a,b) and (b,c) must implies in the ordered pair (a,c). However, it's possible to not reach this answer. For example: a = 3 b = 1 c = 2 (a,b) ^ (b,c) ^ (a,c) (3,1) ^ (1,2) ^ (3,2) Although "b" fits the criteria (Y = 1), the ordered pair (a,c) isn't a implication of the union of the ordered pairs (a,b) and (b,c). Of course if a=1, b=2 and c=1, the implication of the union of the same ordered pair would fit the criteria. However, it isn't the only possibility. Therefore, "X=1 or Y=1" is not a Transitive relation.

ig it is transitive(homework problem)

My answer to homework is not transitive. Why? ⬇️ For it to be transitive: (a,b) belongs to R AND (b,c) belongs to R implies (a,c) belongs to R and that is not true. For example: (0,1) belongs to R AND (1,3) belongs to R but (0,3) doesn't belong to R.

Not transitive

x * y != 0

💀

Not transitive

Not transitive

Not transitive