Abstract Algebra: Bézout's Identity

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In this video, we learn how to find the gcd of two integers A and B, and then compute integers u and v such that uA + vB = gcd(A,B).

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Thanks you kind sir, took you 8 minutes to do what my teacher could not in 2 class sessions, explain the backwards substitution effectively.

nicholasbraud

Thank you. After about 2 hours of trying to figure this out through a textbook, it only took 7 minutes of your video. Now, understood and done.

rauljg

This video helped SO MUCH! With all classes being online, my prof is no longer doing lectures how he used to and this topic was poorly executed. I'm glad a classmate found this video!

tianilavanchy

6 years later and this man is still a math god

Obligedcartoon

After many videos, this video clarified everything for me. Thank you!

GabrielGonzalez-fqup

i rarely like a video, but you sir deserved it ! Thank you very much

E-Dh

Thank you sir. Was struggling on a online math class where the textbook insufficiently describes this process.

jamesthompson

Thank you so much, this is so much clearer than what i'm learning in university!

danverzhao

I think the general solutions to the problem are u = 7 + 31k and v = -16 - 71k. There are infinitely many solutions to this because after finding the lcm, there's no limit to the amount of factors that lead to a difference of 10.

doodelay

This broke it down perfectly for me, thank you.

logankrall

Thank you so much! This was so much easier than the other videos :)

BrennanBecerra

This really helped me understand that method. Great video! Thank you!

publicNickname

Amazing! First time this was explained properly. Thank you!

ruqayyapervaiz

Thank you very much :) even if i'm french and don't have a good english level, you really helped me. TY !

ZeDanseur

Try to implement it
My implementation is iterative with stack, but stack can be eliminated

holyshit

Super 👍 helped me very much :) thank you

fesix

Sir, can you please tell me how to find more coefficients?

harshjha

It seems a monstrously complicated way of doing it. There are so many operations necessary for even a fairly simple calculation, and of course the more separate operations you have to do the more likely you are (anyway, I am!) to make an error.

mikelindsay

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