For transitive relation we are checking all the order pairs and if the condition is satisfied for all order pairs then its called transitive. For R8= (2,1) is there but there is not any order pair which starts with 2. I think I am right?
Yes you are right I've got your problem Let me tell you, hamza you think that (2,1) is our first order pair ,but 1st order pair is (3,2) then (2,1)and then (3,1) General form: (a,b)--->(b,c)--->(a,c) That is we have:(3,2)--->(2,1)--->(3,1) I hope you'll understand If you understand then answer me
Let me explain you, Antisymmetric = { (1,1) , (1,2) , (2,3) } Here if 1,2 ( say a,b ) exists then 2,1 ( that is b,a ) is not allowed , however if 1,1 ( say a,a ) exists then 1,1 ( agai a,a ) is allowed Assymetric = { (1,2) , (2,3) , (3,1) } Here if (a,b) exists then (b,a) is not allowed, however (a,a) for a ∈ set A is also strictly not allowed I hope you got it!
The other user explained it well, As an addition I like to also see it this way: Symmetric: If (x,y) is in the relation, then (y,x) is in the relation. Antisymmetric( if (x,y) is in the relation, then (y,x) is in the relation ONLY if x == y Asymmetric: If (x,y) is in the relation, then (y,x) CANNOT be in the relation. Asymmetric is the complete opposide of Symmetric. Antisymmetric is a bit different, it onlly allows is if both numbers are equal to each other
Consider a relation R1 = { (1,2) , (2,3) , (1,3) } R2 = { (1,2) , (1,3) , (2,3) } R3 = { (1,3) , (2,3) , (1,2) } R4 = { (1,1) , (1,2) , (2,3) , (1,3) } R5 = { (1,2) , (2,2) , (1,3) } Here R1,R2,R3 all are transitive As they satisfy the following condition , If a,b and b,c exists then there must exist a,c That is a,b ^ b,c -> a,c Note : from R3 we have , { (1,3) , (2,3) , (1,2) } can be rearranged as { (1,2) , (2,3) , (1,3) } Clearly it satisfies a,b ^ b,c -> a,c R4 is a transitive as 1,1 ^ 1,1 -> 1,1 But not R5 I hope you got it!
Additionally, R11 is also transitive !!! Do you agree?
YEP👍👍exactly what I was gonna type in the comment section
Yuppp 👍🏻👍🏻
Yeahhh totally
these videos are extremely well made, big props to you mad keep doinh great work its very apppreciated
this helped me sooo much THANK YOU
the way of teaching is very excellent and impressive
equivalence relation is left
neso the goat fr
Beautiful teaching Sir!!!
After a full year...it is today because of this video, on the day of my exam....this topic is finally clear to me...I'm so happy😌💕🌟
you are saving my grades thank you so much
Thank u teacher ur explanation really saved me ❤❤
For transitive relation we are checking all the order pairs and if the condition is satisfied for all order pairs then its called transitive. For R8= (2,1) is there but there is not any order pair which starts with 2. I think I am right?
Yes you are right
I've got your problem
Let me tell you, hamza you think that (2,1) is our first order pair ,but
1st order pair is (3,2) then (2,1)and then (3,1)
General form: (a,b)--->(b,c)--->(a,c)
That is we have:(3,2)--->(2,1)--->(3,1)
I hope you'll understand
If you understand then answer me
@@muzamilhussain6459how did you know which one is a, b or c?
By god's grace @@kriistyy.
@neso accademy R11 is also irreflexive relation am i right 🤔🤔
Amazing Class ❤❤
Thank you sir again
As simple as that😊🙂
Very helpful lecture 🎉
Thanks a lot sir .. Really helpful 🙏❤
Hi sir. Thank you for these videos. I have a question! What can we say about the cardinality/size of the transitive relation in R5? Is it 3 or 4?
Thanks
SUCH A HELPFUL VIDEO.
Thank you so much sir.
Helpful~🎉
thanks sir
sir please a video on binomial theroem
What of meaning of r
Nice question dude
at 1:33 what if for all a,b,c in R (a,b) is present but not (b,c) then will it be transitive?
In my opinion (a, b) is considered transitive,, strange but true
sir plz complete this discrete mathematics playlist plz
Could you explain the difference between asymmetric and antisymmetric relation sir?
Let me explain you,
Antisymmetric = { (1,1) , (1,2) , (2,3) }
Here if 1,2 ( say a,b ) exists then 2,1 ( that is b,a ) is not allowed , however if 1,1 ( say a,a ) exists then 1,1 ( agai a,a ) is allowed
Assymetric = { (1,2) , (2,3) , (3,1) }
Here if (a,b) exists then (b,a) is not allowed, however (a,a) for a ∈ set A is also strictly not allowed
I hope you got it!
The other user explained it well, As an addition I like to also see it this way:
Symmetric: If (x,y) is in the relation, then (y,x) is in the relation.
Antisymmetric( if (x,y) is in the relation, then (y,x) is in the relation ONLY if x == y
Asymmetric: If (x,y) is in the relation, then (y,x) CANNOT be in the relation.
Asymmetric is the complete opposide of Symmetric. Antisymmetric is a bit different, it onlly allows is if both numbers are equal to each other
its really helpful
Please complete DSA
Now let me tell you
😂
😂😂
Please upload ppt sir
omg you are literally a math god
such a helpful video for me thankyou sir
thank you so much
In network theory playlist “ generation of alternating EMF”
Please do this
Literally waiting for this topic
You did not explain Identity relations
can u teach us microcontroller and microprocessor for 4th sem Electronics
Wao what a way nice
Sir I don't understand transitive 😢
Consider a relation
R1 = { (1,2) , (2,3) , (1,3) }
R2 = { (1,2) , (1,3) , (2,3) }
R3 = { (1,3) , (2,3) , (1,2) }
R4 = { (1,1) , (1,2) , (2,3) , (1,3) }
R5 = { (1,2) , (2,2) , (1,3) }
Here R1,R2,R3 all are transitive
As they satisfy the following condition ,
If a,b and b,c exists then there must exist a,c
That is a,b ^ b,c -> a,c
Note : from R3 we have ,
{ (1,3) , (2,3) , (1,2) } can be rearranged as { (1,2) , (2,3) , (1,3) }
Clearly it satisfies a,b ^ b,c -> a,c
R4 is a transitive as 1,1 ^ 1,1 -> 1,1
But not R5
I hope you got it!
@@Siva_Shyamwhy R5 not??? (1,2) Is present
I can't understand types of relations🥹😥
the organic chem tutor has nothing on you
For sickness call 108 for mathematics problem and computer science problems call neso academy
Thanks sir