How to Prove: Bezout's Identity

[@bobmarley9905]

That was so elegant, short, and intuitive compared to the other popular videos on youtube. Criminally Underrated 😊 ! Keep making such number theory videos and other such math videos!

[@wishizuk]

Thanks for your kind words! I hope that you will enjoy my other videos, too!

[@markborz7000]

Short, elegant and crystall clear. Best proof!

[@wishizuk]

Thanks for your kind words! I hope that you will like my other videos!

[@vafkamat]

Partial Converse: If a and b are greater 0, and if there exist integers x and y such that ax + by = 1 then gcd(a,b) = 1 Prf: Let d = gcd(a,b) If 1 = bx + ay then 1 = d( b/d)x + d ( a/d)y or 1 = d( xb/d + ay/d) The quantity in brackets is an integer for b/d and a/d are integer; this tells us that d|1 or d=1 so that gcd(a,b) = 1

[@wishizuk]

Thank you for posting this comment!

[@Anonymous_MC]

i like how this guy writes everything in logical statements

[@wishizuk]

Thanks for your kind words!

[@lolo-dg3um]

bookmark this ,hope to figure it out. thank you brother

[@wishizuk]

Thanks for your comment! Let me know how it goes. You are always welcome!

[@NurbolBekbossin]

I couldn't understand Theorem 1.6 from Apostol's book, I believe this is a lot more clear. Not sure why Apostol chose such a convoluted way of proving the theorem, only way I could think of is that he assumed reader/student doesn't know two assisting theorems on the video that would help advance further more efficiently.

[@wishizuk]

I am glad to hear! I hope that you will enjoy my other videos!

[@krumpy8259]

There we go, that’s fantastic thank you.

[@wishizuk]

My pleasure! I appreciate your kind words!

[@leonardopaes2494]

Great and neat proof! I learned a lot, specially regarding the well-ordering principle!

[@wishizuk]

Glad to hear that! I hope that you will enjoy my other videos!

[@qxty]

i have no clue what this is or why its on my recommended but 👍

[@wishizuk]

Thanks for your thumb-up, anyway!

[@Naoseinaosei213]

you earned a sub for this amazing proof

[@wishizuk]

Thank you! I have not made any video during the summer, but I will make videos soon!

[@Naoseinaosei213]

Such a Nice proof. Like

[@wishizuk]

Thanks for your kind words! I hope that you will enjoy my other videos as well!

[@janusium]

Where did all the numbers go

[@wishizuk]

Who needs numbers?

[@epic_editz_x]

The best explanation so far ❤

[@wishizuk]

Thanks for your kind words!

[@Mathcentury626]

Informative 👍🏻

[@wishizuk]

Thanks for your kind word!

[@tetchii]

keep it up man

[@wishizuk]

Will do!

[@77YuvrajSingh77]

Thank you sir !

[@wishizuk]

My pleasure! I hope that you will enjoy my other videos!

[@Linhkinhbrods]

isn't it that when gcd(a,b)|(ax+by) then gcd(a,b) = ax + by already because of well ordering principle and the gcd itself let's say when gcd(4,6) = 2 then we have 2|ax+by hence ax + by must be equal to 2.

[@wishizuk]

Being divisible by 2 does not mean being equal to 2.

[@Linhkinhbrods]

@@wishizuk what?

[@wishizuk]

@@Linhkinhbrods 2|(ax+by) does not imply ax+by=2.

[@Linhkinhbrods]

@@wishizuk why not

[@wishizuk]

@@Linhkinhbrods 2 divides 4, but 4 is not 2.

[@Shams-256]

it's so beautiful :>

[@wishizuk]

Thanks for your kind words!

[@abhigyan2346]

Beautiful sir! thanks a lot

[@wishizuk]

My pleasure!

[@roberttomi8800]

thanks

[@wishizuk]

My pleasure!