Amazing explanation. Been cracking my head trying to understand transitivity and you have explained it concisely with excellent examples. Thank you SIR!
Thank you. my teacher spent two hours to discuss this topic, and I still don't understand what he is talking. However, I spent 10 mins on this video, I understand the logic of transitive relation clearly.
Thank you so much for your amazing explanation about different knowledge points in discrete mathematics! I will take my first discrete mathematics exam tomorrow.😀
Thanks a lot for such clear & concise explanation!! U've made it easy to understand how to check for transitivity in a relation. I'm surely will share this videos with my students. thanks!!
Sir, excellent explanation. Useful examples to highlight transitive relations. Thank you. When testing for transitivity, if the pair of elements is the same then this pair can be paired with itself in the test. For example, if (2,2) is a member of R then it can be paired with itself.
The relation must be defined before it can be declared transitive. In the first example consider the relation between the two numbers in each ordered pair to be "is the brother of...". That relation is transitive. But consider the relation "shook the hand of ...". That relation is not transitive. So, even though the ordered pairs (2, 4) and (4, 7) are connected by the 4, we can't tell if the relation is transitive until we know the nature of the relation we are working with. E.g. are they shaking hands, are they brothers, or what?
Hi Robert, Thanks for the comment. I would certainly agree. It would be better to define the relation to have the property P (transitivity) from the offset and then to proceed, in which case we could then find its transitive closure. If the relation from the offset did not have the property P (transitivity) then the relation resulting from the transitive closure would not make sense. You can probably see that the examples provided are a first step along the learning journey and, in particular; the actual relation resulting has not been given a specific name. Your point is well taken thou. Regards. Jonathan.
HI Thank you so much for this explanation. I have been going through all your videos for my assignments. I have a mock test question though as follows - could you help me? thank you in advance. Let A be the set of members in a family and let R be a relation defined on A as if “ is a child of ”. Explain if R is: (i) Reflexive (ii) Symmetric (iii) Transitive
Interesting. Well let's start with (i) first: A = {m1, m2, m3, ... } is R Reflexive? R is reflexive iff for all m ϵ A we have mRm. If this was true what it is saying is that everyone is related to themselves - in this case, it means that everyone is a child of themselves. This certainly is not true. (ii) second: is R Symmetric? R is symmetric iff for all mRn ϵ R (all relations in R) we have nRm also in R. The implications of this would be that if 'm is a child of n' then 'n is a child of m' this is also absurd. Although with that said, we could have the vacuous case. (iii) finally: is R Transitive? A similar argument to (ii) above - if mRn and nRo in R then for R to be transitive we would require mRo - but look at the chain: 'm is a child of n' and 'n is a child of o' then this would require that 'm is a child of o' which is also absurd I believe. So R is neither reflexive, or symmetric, or transitive. I hope this helps. Regards. Please share my page with your friends. Jonathan
I have a question if you can spare a second. But first some context In the Relation R = { (a, {b} ) , ({b},{2}), (1,{2}) } If you apply the transitive " process" you explain in the video. The order pair (a, {b} ) can be paired with ({b},{2}) and itself. Then if you want to pair the order pair ({b},{2}) to the relation the only pair is (a, {b} ). Question : Can you use (a, {b} ) to pair with ({b},{2}) even tough you have already paired the two off ? Hope this is not a terrible question. Thanks
Hi Andries, Considering the relation R as you defined it, yes you are correct you can pair (a, {b}) with ({b}, {2}) and so if r was transitive you would expect to find the order pair (a, {2}) in R, which clearly is not the case and so R is not transitive. Another nice way to think of transitivity is by considering bus journeys. In your case above you can get on a bus at stop 'a' and take a journey to stop '{b}'; then you can get on another bus at stop '{b}' and ride the bus to stop '{2}'. You have effectively taken two bus journeys. If your bus route was transitive you should be able to take a single bus from stop 'a' through to stop '{2}', hence the ordered pair (a, {2}) would need to be part of your possible journeys 'Relation.' I hope this helps it would be great if you could share with your friends. Any other questions feel free to message. Kindest Regards Jonathan
Amazing explanation. Been cracking my head trying to understand transitivity and you have explained it concisely with excellent examples. Thank you SIR!
Thank you. my teacher spent two hours to discuss this topic, and I still don't understand what he is talking. However, I spent 10 mins on this video, I understand the logic of transitive relation clearly.
The best explanation for this concept on the net
Teach explained it poorly and my test is tonight. Finally understand. You're doing Gods work man. Thanks.
Magnificent , you made it so easy . Thanks I had been really struggling with other youtubers who was were just confusing me even more
At last I found what I was searching for
Legend
Absolute.
Legend.
Extremely thorough explanation, thank you very much sir!
well understood thank you sir
Sir, you are the best
Thank you so much for your amazing explanation about different knowledge points in discrete mathematics! I will take my first discrete mathematics exam tomorrow.😀
Excellent way of explaining the process
Trillions thanks sir, i learned transitive, only for this i watched lot of videos, but you done Great greatest job
perfect explanation
Thanks a lot for such clear & concise explanation!! U've made it easy to understand how to check for transitivity in a relation. I'm surely will share this videos with my students. thanks!!
Sir, excellent explanation. Useful examples to highlight transitive relations. Thank you. When testing for transitivity, if the pair of elements is the same then this pair can be paired with itself in the test. For example, if (2,2) is a member of R then it can be paired with itself.
But he said that
YOU'RE THE BEST!!!
it wasn't informative but super-duper informative, loved the explaination... thanks cleared all my doubts on relations.
Good stuff Mr Maths!
SIR I'm from 2023, Your video still helps struggling students in discrete maths until this day
You are awesome! You are an amazing teacher! Thank you!
Hi Peace. Thank you for the kind words. It would be great if you could share the channel with your friends. Kindest regards. Jonathan
Amazing Explanation. All the best, sir.
From Egypt
Thank you sir
Nice lecture
The relation must be defined before it can be declared transitive. In the first example consider the relation between the two numbers in each ordered pair to be "is the brother of...". That relation is transitive. But consider the relation "shook the hand of ...". That relation is not transitive. So, even though the ordered pairs (2, 4) and (4, 7) are connected by the 4, we can't tell if the relation is transitive until we know the nature of the relation we are working with. E.g. are they shaking hands, are they brothers, or what?
Hi Robert, Thanks for the comment. I would certainly agree. It would be better to define the relation to have the property P (transitivity) from the offset and then to proceed, in which case we could then find its transitive closure. If the relation from the offset did not have the property P (transitivity) then the relation resulting from the transitive closure would not make sense. You can probably see that the examples provided are a first step along the learning journey and, in particular; the actual relation resulting has not been given a specific name. Your point is well taken thou. Regards. Jonathan.
thank you
HI Thank you so much for this explanation. I have been going through all your videos for my assignments. I have a mock test question though as follows - could you help me? thank you in advance.
Let A be the set of members in a family and let R be a relation defined on A as if “ is a child of ”. Explain if R is:
(i) Reflexive
(ii) Symmetric
(iii) Transitive
Interesting. Well let's start with
(i) first: A = {m1, m2, m3, ... } is R Reflexive? R is reflexive iff for all m ϵ A we have mRm. If this was true what it is saying is that everyone is related to themselves - in this case, it means that everyone is a child of themselves. This certainly is not true.
(ii) second: is R Symmetric? R is symmetric iff for all mRn ϵ R (all relations in R) we have nRm also in R. The implications of this would be that if 'm is a child of n' then 'n is a child of m' this is also absurd. Although with that said, we could have the vacuous case.
(iii) finally: is R Transitive? A similar argument to (ii) above - if mRn and nRo in R then for R to be transitive we would require mRo - but look at the chain: 'm is a child of n' and 'n is a child of o' then this would require that 'm is a child of o' which is also absurd I believe.
So R is neither reflexive, or symmetric, or transitive.
I hope this helps. Regards. Please share my page with your friends.
Jonathan
Sir please also make a video on co-reflexive and quasi-reflexive relations
I have a question if you can spare a second. But first some context
In the Relation R = { (a, {b} ) , ({b},{2}), (1,{2}) }
If you apply the transitive " process" you explain in the video. The order pair (a, {b} ) can be paired with ({b},{2}) and itself. Then if you want to pair the order pair ({b},{2}) to the relation the only pair is (a, {b} ).
Question : Can you use (a, {b} ) to pair with ({b},{2}) even tough you have already paired the two off ?
Hope this is not a terrible question.
Thanks
Hi Andries,
Considering the relation R as you defined it, yes you are correct you can pair (a, {b}) with ({b}, {2}) and so if r was transitive you would expect to find the order pair (a, {2}) in R, which clearly is not the case and so R is not transitive.
Another nice way to think of transitivity is by considering bus journeys. In your case above you can get on a bus at stop 'a' and take a journey to stop '{b}'; then you can get on another bus at stop '{b}' and ride the bus to stop '{2}'. You have effectively taken two bus journeys. If your bus route was transitive you should be able to take a single bus from stop 'a' through to stop '{2}', hence the ordered pair (a, {2}) would need to be part of your possible journeys 'Relation.'
I hope this helps it would be great if you could share with your friends. Any other questions feel free to message.
Kindest Regards
Jonathan