Cool fact: Bezout's lemma (that's how I learned it) is actually applcable in any group that's similar enough to Z. I learned a general form of the lemma in a course on group theory. Of course in group theory the notions of gcs and lcm are defined more generally in terms of subgroups and their generating sets. Interesting stuff if you're into that sort of thing.
for people unfamiliar with the Euclid's Algorithm, it's actually based upon the lemma: Suppose b = aq + r, then gcd(a, b) = gcd(a, r). You can prove this lemma by contradiction in ~8 lines
I have a few suggestions for videos. Here’s just one: Two circles of radius R intersect each at exactly two points. Lines are drawn from each of those points to the center of one of the circles. Those lines and the inner arc of the other circle define a region. What is its maximum area?
Can you please make a video about why Euclid’s algorithm finds the GCD? We were taught how to do this in the 6th grade I think, but I never thought about why it works until now.
I haven't watched this video yet but just judging from the very first statement you wrote at the beginning isn't this just the extended euclidean algorithm
Hei I really like your posts!! but i have a question for you bout 3D topic. ABCD.EFGH Cube, its side is a. Point O is intersection between AC and BD. Determine the distance between line EO and line HB..please answer thiss
I am sorry, but you did not show that ALL solutions are in the form x=-2+7m and y=7-24m. In theory there can be other solutions then the ones you showed.
Steven Gottlieb I agree with you. But I was just showing an example on how this works. Notice the video is 18 minutes already. Max or I will work out the proof for that in the future on another video.
Dear prof. Blackpenredpen...you are very good speaker,,...your problems are very interesting, but..please...your carioca is writing sometimes, hard for me to read symbols. Can you improve this symbol's visibility?
I have an issue here: After Bézout's identity, ax+by=gcd(a,b), My problem is such as "ax" is known N and "by" is the unknown from the type [-(x-1)*gcd(a,b)] and of course gcd(a,b) is the unknown I'm looking for.
No it is you have to check both sides. 436 = 2 x 2 x 2 x 3 x 3 x 3 x 2 126 = 2 x 3 x 3 x 7 Both sides have a common number 2 x 3 = 6 So, GCD (436,126) = 6
Did my 'Calc 3' paper today. Was nervous and made a couple stupid errors and I didn't get a couple of things but on the whole it went ok. One mistake was 4r^4 x 0 x 0 = 4r^4
Don't the possible values for Y have to be in the range 0 to (n-1) in our case 0 to (432-1)? ie 7 +432 = 439 but that's greater than 432. so only one solution?
One thing is peculiar. Upto 1960 agebra books, there was no word Euclidean algorithm,a term for determining G.C. D. All are forced to swallow this coinage.Again, Bezout' s identity is renamed as Extended Euclidean Algorithm . This is cultural imperialism.as if no civilisation did not think GCD or HCF
It's not an identity, I don't know why he calls it that. It's an existence claim so clearly not an identity. As for the name, I've learned it as "Bezout's lemma", there's a version of it for integers, a version of it for polynomials and a generalized version of it for groups
The extended Euclidean ALGORITHM is how you find the integers x and y in Euclidean domains (such as the integers). The Bezout identity is the resulting equation. There are mathematical structures where such x and y exist, but even the normal Euclidean algorithm doesn't work. As a general rule, the result isn't the algorithm.
Question: A dealer bought a number of horses at $344.00 each, and a number of bullocks at $265.00 each. He then discovered that the horses had cost him in all $33.00 more than the bullocks. Now, what is the smallest number of each that he must have bought? [Source: 536 Puzzles by Dudeney #20]
0:19 "Okay in this video I'm going to demonstrate one of the most useful facts in number theory." but you didn't prove anything, you just showed an example. Honestly I must say I'm a bit disappointed with your videos lately. Anyway, it's just what I think so don't let it bother you.
You explain this so well and the only example I have found that I can actually follow, thank you! 😊
I was lazy and didn't want to watch an 18 minute video!
Now, I am more than grateful I clicked and watched. Thank you, best explanation ever!
Thank you for your upbeat atittude that made me feel a lot better!
Thank you for your explanation, it helps my report so much. And I really like your smile. Thank you
😃
BIG Thanks man your videos helps a lot as a collage student
Nice!! My favorite part is quantifiers "forall" and "exists":D
Eightc yay!
instaBlaster...
This was the best explanation ever! Thank you! You saved my maths exam. :)
Thank you, you explained it so well that I am confident in passing my discrete maths exam
Thank you for a very nice explaination. Keep it up 🙂
Excellent explaination
It feels so weird to have done calculus without having learned this stuff.. thanks!
Woah!! Wonderful explination Love from indiaa💗
Cool fact: Bezout's lemma (that's how I learned it) is actually applcable in any group that's similar enough to Z. I learned a general form of the lemma in a course on group theory.
Of course in group theory the notions of gcs and lcm are defined more generally in terms of subgroups and their generating sets. Interesting stuff if you're into that sort of thing.
That's cool! I have taken a course on group theory but have not heard of Bezout's Lemma. I'll have to look more into that.
Thank You vey much. Sir.
Yay!
Damn, this is flipping awesome explanation! ♥
Thank you man, this is so useful
thank you black pen red pen
nice, i didn't know about the lcm trick. Thanks
Amazing explanation 👌👌
this is better than the resources my uni has thanks
for people unfamiliar with the Euclid's Algorithm, it's actually based upon the lemma: Suppose b = aq + r, then gcd(a, b) = gcd(a, r). You can prove this lemma by contradiction in ~8 lines
8:57 this reminds me of finding the multiplicative inverse in ciphering class
I have a few suggestions for videos. Here’s just one:
Two circles of radius R intersect each at exactly two points. Lines are drawn from each of those points to the center of one of the circles. Those lines and the inner arc of the other circle define a region.
What is its maximum area?
Shree Ganesh I don’t know the answer with 100% certainty, but…
…I think (going off of memory here) it’s (-πR^2 + R√3)/6.
damn this was a really clever solution, props to bezout!
Thank u. U may arrange a video on Bezout identity and Uclid Algorithm.
EUCLID please!
Nice as usual ! thanks!
Please do a videos on how to find inverse of a function
If you can't solve the equation y=f(x) for x then use:
en.wikipedia.org/wiki/Lagrange_inversion_theorem
11:45 "I will call this x naught and y naught because why not" is what you should've said! you missed a pun opportunity. I'm disappointed.
I was so hoping at 11:43 for "I will call this x-naught and y-naught because... why not?"
mjones207 oh... I should have done that. :)
Very helpful!!
Can you please make a video about why Euclid’s algorithm finds the GCD? We were taught how to do this in the 6th grade I think, but I never thought about why it works until now.
This might help medium.com/i-math/why-does-the-euclidean-algorithm-work-aaf43bd3288e
Thanks!
I haven't watched this video yet but just judging from the very first statement you wrote at the beginning isn't this just the extended euclidean algorithm
Awesome!!!
Hei I really like your posts!! but i have a question for you bout 3D topic. ABCD.EFGH Cube, its side is a. Point O is intersection between AC and BD. Determine the distance between line EO and line HB..please answer thiss
Amazing!
How can we prove the validity of a theorem with a single example?
let:
1/(a-b)(a+b)=A/(a+b)+B/(a-b)
and form it
■A(a-b)+B(a+b)=1
■a(B+A)+b(B-A)=1
Here are two cases of a Bezout's Lemma.
say some thing about that.
I am sorry, but you did not show that ALL solutions are in the form x=-2+7m and y=7-24m. In theory there can be other solutions then the ones you showed.
Steven Gottlieb I agree with you. But I was just showing an example on how this works. Notice the video is 18 minutes already. Max or I will work out the proof for that in the future on another video.
blackpenredpen Can you plz prove that Bézout Coefficients are not unique? Are there any numbers were there *_are_* unique?
Steven Gottlieb It does give every solution, unless m is restricted to integers.
Thank you for responding!
Yes, every solution was given. The problem is that was not proven. Even BPRP agrees with that.
Dear prof. Blackpenredpen...you are very good speaker,,...your problems are very interesting, but..please...your carioca is writing sometimes, hard for me to read symbols. Can you improve this symbol's visibility?
I have an issue here: After Bézout's identity, ax+by=gcd(a,b), My problem is such as "ax" is known N and "by" is the unknown from the type [-(x-1)*gcd(a,b)] and of course gcd(a,b) is the unknown I'm looking for.
What do the pen colours represent? Is the black colour to do with the original question while red is an insertion or something?
Do the numbers on the side of Euclid’s algorithm always multiply to also give you the gcd or is that a coincidence
ikr i noticed that also
Hey Blackpenredpen, do a video on the (complex valued) infinite series Sum(i^(n-1)/n) from n = 1 to infinity.
Why do x and y have to be integers? If you fill in complex numbers for m they cancel out as well, right?
Great!
proof is more difficult than application
@4:41 you had a mistake the gcd(436, 126) = 2
wtf
No it is you have to check both sides.
436 = 2 x 2 x 2 x 3 x 3 x 3 x 2
126 = 2 x 3 x 3 x 7
Both sides have a common number 2 x 3 = 6
So, GCD (436,126) = 6
@@itsalencraft1717 divide 436%6!=0 126%6=0 how is 6 the gcd
The Number Theory playlist in the description contains a private video at #4: I have watched all the others so what am I missing?
your way of finding gcd is better than text book
It's technically the same way, but I agree that writing it out in that equation style is confusing as hell
Superb..
Did my 'Calc 3' paper today. Was nervous and made a couple stupid errors and I didn't get a couple of things but on the whole it went ok. One mistake was 4r^4 x 0 x 0 = 4r^4
Don't the possible values for Y have to be in the range 0 to (n-1) in our case 0 to (432-1)? ie 7 +432 = 439 but that's greater than 432. so only one solution?
whats so important about bezouts indentity
why he holding that black ball ?
One thing is peculiar. Upto 1960 agebra books, there was no word Euclidean algorithm,a term for determining G.C. D. All are forced to swallow this coinage.Again, Bezout' s identity is renamed as Extended Euclidean Algorithm . This is cultural imperialism.as if no civilisation did not think GCD or HCF
so the final x is -2+7m and final y is 7-24 m?? :)
Why does he have ball in his hands?
Change your lighting system, board can’t be seen easily
What does the upside-down A and the backwards E mean?
AddQ he literally said it in the video
For All and There Exists
AddQ Quantifiers. You will encounter it in a discrete math or logic course.
there is also a blue pen LOL : )
How to use it if a+b is the exponent and thier gcd(a,b)=1 ??
Bezout Identity? I thought the name is Extended Euclidean
It's not an identity, I don't know why he calls it that. It's an existence claim so clearly not an identity. As for the name, I've learned it as "Bezout's lemma", there's a version of it for integers, a version of it for polynomials and a generalized version of it for groups
The extended Euclidean ALGORITHM is how you find the integers x and y in Euclidean domains (such as the integers). The Bezout identity is the resulting equation. There are mathematical structures where such x and y exist, but even the normal Euclidean algorithm doesn't work. As a general rule, the result isn't the algorithm.
Shachar H Wikipedia calls it the Bezout identity and I think I've heard that term used in Bezout domians (things where the x and y always exist).
Both Bezout's Identity and Bezout's Lemma are correct.
We want more IMO Problem
Question: A dealer bought a number of horses at $344.00 each, and a number of bullocks at $265.00 each. He then discovered that the horses had cost him in all $33.00 more than the bullocks. Now, what is the smallest number of each that he must have bought? [Source: 536 Puzzles by Dudeney #20]
fucking genius
Should I multiply it out?
You are calling it y naught because why not xD
❤❤❤
9:02 something magic happened
Hi!
I love you
Can you do a video about 1+2+3+4....=-1/12
mathologer
“Because we are know we are smart, don’t do that “ 😂😂😂
whohoo solved
Hello
WHY A.L. AGAIN!!!!!!!!!!
Ιωάννης - Αθανάσιος Χαραλάμπους
What's AL?
Ensemble theory d(^^)
0:19 "Okay in this video I'm going to demonstrate one of the most useful facts in number theory." but you didn't prove anything, you just showed an example. Honestly I must say I'm a bit disappointed with your videos lately. Anyway, it's just what I think so don't let it bother you.
He 'demonstrated'
I did demonstrate. Usually this is easier when we first see an example then see the proof.
Okay, then sorry for what I said. But I still prefer proofs than examples.
Max does lots of proofs : )
I will do so too later.
Yeah I know, I subscribed when you uploaded "Practice your Trig".