You mathematicians are so ingenious. I get the shivers out of seeing this proof. It is just beautiful. Thank you for posting this video.
@devesh3648
4 жыл бұрын
I was just casually surfing Number Theory youtube and came across your tutorial. Stopping at 7:32 (obviously going to resume again) to just let you know how awesome this video is. (consider adding the alternative bit of solving the system through residue mod Pi) but all in all, you teach really really well!
@Don-ev5ov
2 жыл бұрын
I think you are doing a great service in these videos. A long time since school, you make the Olympiad and other similar challenging problems I could never solve understandable to me. The proof of this theorem reminded me of orthogonal polynomials and Gaussian integration, which might make a good topic, if you haven't covered it.
@rizalpurnawan3796
4 жыл бұрын
Great video. Now it gets very clear to me, since I was struggling to understand the part of proof from wikipedia. In wikipedia, it is given 4 sections of proof: Uniqueness, Existence proof 1 (by proving the existence of a map from x mod N to (x mod n1, ... , x mod nk) that defines ring isomorphism), Existence proof 2 (constructive proof only on 2 divisiors; n1 and n2; this part is that I get so confused on how they construct x from x = a1m2n2 + a2m1n1 for ai are the remainders, mi are Bezout's coefficients), Existence proof 3 (the general proof as that of yours). I found yours enlightening. Thanks Professor, keep creating more videos of mathematics, cheers 👍.
@georgesadler7830
2 жыл бұрын
Professor M. Penn, thank you for a mammoth proof of the classical Chinese Remainder Theorem.
@prakashmadaksirashamrao5961
3 жыл бұрын
Hats off to your dedication to explaining mathematics ,to your lung power and not the least to your amazing back flips.God bless you with more and more opportunity to serve the cause of Mathematics. Thank you.
@harrysakata3082
3 жыл бұрын
Great video! Part about d seems a bit inaccurate because d can have non-one factors that are distributed over multiple nj (j n.e. i). Stating in advance that d is either 1 or a prime factor of ni should make it accurate.
@purusharthverma2300
4 жыл бұрын
Fantastic. Short and to the point.
@danielkoynov5627
Жыл бұрын
He finishes with "okay, good". Shivers.
@adenpower249
4 жыл бұрын
Is it true that you are a master at making pizza?
@user-uw1ut4ss2q
3 жыл бұрын
I was amazed that this Chinese remainder theorem can be applied to prove that Euler's totient function is multiplicative under two relatively prime integers.
@Suav58
3 жыл бұрын
I always feel a bit of discomfort when quantifiers are being added at the end of the statement. Years of inculcation made it.
@sebastianocampo2093
2 жыл бұрын
i have a question : in the proof of chinese theorem if n1 = 4 , n2=3, n3 = 5, then 2 divide 4 and 4*3*5 but 2 doesn't divide 3 or 5 . but you then said that d(=2) divide nj for every j different of ni(=2 in this case)
@fredrickelvis7883
Жыл бұрын
Thanks for your video! I am just wondering how why X can be constructed as X=X1N1B1+.....XkNkBk? What's the idea behind this? Thanks!
@nishithbaravkar7549
3 жыл бұрын
Thanks a lot for this proof micheal
@mysteriousperson0845
6 ай бұрын
I don't understand the last part, how does X ≡ Y mod N show uniqueness?
@fonso-ir8xz
9 ай бұрын
I need help at 3:19. We suposed that d|n_i and d|N_i because d is the gcd(n_i,N_i). We know that gcd(n_i,n_j)=1 so if w|n_i it cannot divide n_j for w not equal to 1. But then the prof concludes that d|n_j. Any help pls? Thank you
@akshaygowda259
8 ай бұрын
Very good explanation.
@johnvandenberg8883
2 жыл бұрын
If d divides both n_i and n_j, then d is a common divisor of n_i and n_j. But gcd (n_i, n_j) = 1, so d=1. Simpler.
@lgooch
Жыл бұрын
sorry but why are we considering x to be that sum of products?
@Timmy-zl1gu
2 ай бұрын
my dad gave me one day to understand this i have to watch this video
@escobarlohe6571
4 жыл бұрын
It's good but my request to you is, next time before you proceed to proof please give some examples and please don't wipe out the proof in between
@MichaelPennMath
4 жыл бұрын
Thanks for the feedback! I have some more videos devoted only to examples that you should check out.
@TechToppers
3 жыл бұрын
Can you make things a little more intuitive? I mean in a sense in which it feels that we are researching on this topic and we are making our own theorem as basic Number Theory is very intuitive. It will improve your channel and teaching skills. Just a suggestion, sir.
@opsschwarz3651
4 жыл бұрын
can't understand last step "=>"
@ScreechOwlx
4 жыл бұрын
it's by definition of X congruent to y mod N. those statements literally mean the same thing
@Icenri
3 жыл бұрын
From kconrad.math.uconn.edu/blurbs/ugradnumthy/crt.pdf If x = c and x = c′ both satisfy x ≡ a mod m, x ≡ b mod n. Then we have c ≡ c′ mod m and c ≡ c′ mod n. Then m | (c−c′) and n | (c−c′). Since gdc(m, n) = 1, the product mn divides c−c′, which means c ≡ c′ mod mn. This shows all solutions to the initial pair of congruences are the same modulo mn. I think since c and c' are interchangeable here we can count them as being the same solution when viewed mod mn.
@m4riel
3 жыл бұрын
people here are overcomplicating things. If: N | (x - y) x - y ≡ 0 (mod N) x ≡ y (mod N) Michael just skipped one redundant step.
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