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This is one of my favorite proofs and you've explained it beautifully! This is your best video yet! My only suggestion is to slow down your speech just a bit in future videos.
It is a wonderful proof! Thanks a lot and I appreciate the feedback! I always try to monitor the speed of my voice, but sometimes I no doubt lose track of it. I've got to take more time to breathe - which will naturally slow me down a bit!
Hi, thanks for this video. I have a question For the contradiction proof, are we assuming that n - m + r ≠ 2 is true for graph G with a minimum edges e? If that's the case, I don't understand how the G-e graph contradicts the n - m + r ≠ 2 because m-1 edges is already less than the minimum edges e so n - m + r ≠ 2 shouldn't apply to it
because after showing that n - m + r for G equals n - m + r for G-e it contradicts n - m + r ≠ 2 basically, so long as there are cycles you can delete an edge from a cycle while leaving the formula unchanged, I like to think of applying this over and over until you get to a tree (no cycles) where we've already proven it!
let the vars of G-e be n', m' and r' so G-e holds n'-m'+r'=2 now place the vars of G inside it n'-m'+r'=2=n-(m-1)+(r-1) and get n'-m'+r'=2=n-m+r so we did not change anything by deleting the edge regarding the formular we know the formular holds for G-e thus the formular holds for G aswell
Dr, Could you please prove this question? Let G and H be connected graphs different from K1 and K2.Show that both factors are paths or one is a path and the other a cycle.
(Copied from J) Hi, thanks for this video. I have a question For the contradiction proof, are we assuming that n - m + r ≠ 2 is true for graph G with a minimum edge m? If that's the case, I don't understand how the G-e graph contradicts the n - m + r ≠ 2 because m-1 edges is already less than the minimum edges e so n - m + r ≠ 2 shouldn't apply to it
The task at hand involves proving a statement about a cycle graph. To do this, a minimum counterexample approach is being used, wherein the smallest possible instance that does not satisfy the statement is being considered. In order to prove the statement, it is necessary to show a contradiction, in this case we show it by deleting the edge. The goal of this contradiction is to demonstrate that the statement is, in fact, true for all cycle graphs, and that the counterexample is invalid.
Support the production of this course by joining Wrath of Math as a Channel Member for exclusive and early videos, original music, and upcoming lecture notes for the graph theory series! Plus your comments will be highlighted for me so it is more likely I'll answer your questions!
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Great video! your enthusiasm makes a 15m proof feel like a 1 minute cat video
Thanks so much! That is as high praise as I can hope for, thanks for watching and let me know if you ever have any questions!
Perfectly described in easy and simple language . All doubts cleared
Glad it helped!
Your videos help me out so much!! I have exams soon and watching your videos make the work so much easier to understand. Thank you!!
You're very welcome and thanks for watching! So glad the videos help, and good luck on your exams!
This is one of my favorite proofs and you've explained it beautifully! This is your best video yet! My only suggestion is to slow down your speech just a bit in future videos.
It is a wonderful proof! Thanks a lot and I appreciate the feedback! I always try to monitor the speed of my voice, but sometimes I no doubt lose track of it. I've got to take more time to breathe - which will naturally slow me down a bit!
Thank you again.
I didn't get the part on the minimum size condition. If it is minimum, how can we remove one edge?
awesome explanation and a very passionate one as well :)
Thank you!
Beautifully explained better than my lecture notes which contains proof by induction( PS: We deep down all hate proof by induction xd )
Hi 👋🏻
Could you make a video about Kuratowski theorem?
Thank you for your work 🙏🏻
Hi, thanks for this video. I have a question
For the contradiction proof, are we assuming that n - m + r ≠ 2 is true for graph G with a minimum edges e? If that's the case, I don't understand how the G-e graph contradicts the n - m + r ≠ 2 because m-1 edges is already less than the minimum edges e so n - m + r ≠ 2 shouldn't apply to it
because after showing that n - m + r for G equals n - m + r for G-e it contradicts n - m + r ≠ 2
basically, so long as there are cycles you can delete an edge from a cycle while leaving the formula unchanged, I like to think of applying this over and over until you get to a tree (no cycles) where we've already proven it!
let the vars of G-e be n', m' and r'
so G-e holds n'-m'+r'=2
now place the vars of G inside it
n'-m'+r'=2=n-(m-1)+(r-1)
and get n'-m'+r'=2=n-m+r
so we did not change anything by deleting the edge regarding the formular
we know the formular holds for G-e
thus the formular holds for G aswell
Dr, Could you please prove this question?
Let G and H be connected graphs different from K1 and K2.Show that both factors are paths or one is a path and the other a cycle.
Hi why do we apply induction on m edges? Why not we apply induction on n vertices?
Can you prove the Jordan Curve Theorem?
Why do we remove an edge? When you use induction, aren't you supposed to go from k edges to k+1 edges?
Euler's Formula? More like "All these proofs are fantastic; thank ya'!" 👍
Perfect, thank you very much.
My pleasure, thanks for watching!
Thank you very much!
My pleasure, thanks for watching!
(Copied from J)
Hi, thanks for this video. I have a question
For the contradiction proof, are we assuming that n - m + r ≠ 2 is true for graph G with a minimum edge m? If that's the case, I don't understand how the G-e graph contradicts the n - m + r ≠ 2 because m-1 edges is already less than the minimum edges e so n - m + r ≠ 2 shouldn't apply to it
The task at hand involves proving a statement about a cycle graph. To do this, a minimum counterexample approach is being used, wherein the smallest possible instance that does not satisfy the statement is being considered. In order to prove the statement, it is necessary to show a contradiction, in this case we show it by deleting the edge. The goal of this contradiction is to demonstrate that the statement is, in fact, true for all cycle graphs, and that the counterexample is invalid.
Amazing!
Thanks for watching, Vishnu!
great !!!
Thank you! If you're looking for more graph theory check out my graph theory playlist! ruclips.net/p/PLztBpqftvzxXBhbYxoaZJmnZF6AUQr1mH
#Excelent!
I love you
❤️
I wish you would have used , V, E, and F labels instead of n, m and r.
Aha the french