Corollary of bezout's lemma Proof

[@MatheusNasi]

I really love when you make Number Theory videos. In my opinion, it's one of the most beautiful topics in Maths

[@gooball2005]

oh, so another way to put it would be to say that in order for N to divide just B, all the common factors between N and A*B must lie in B, so A has to be coprime with N.

[@darkmage35]

A counterexample that makes the NO answer even more obvious is n=6, a=2, b=3.

[@MathForLife]

Yay! We need more number theory!:D

[@znarwhal4530]

Double "YAY!" 0_0

[@blackpenredpen]

Yes!! yay and YAY!!!

[@axelreispereiravaz1699]

Awsome video again ! Number theory is also one of my favorites topics in maths ! I studied this lemma/theorem in the past. Isn't that the Gauss' lemma/theorem ? 🤔 Anyway, your work is amazing ! I hope you'll do more videos about number theory ! ^^

[@mohammediqbal6779]

axel reis pereira vaz I think it is

[@rissalatahmed6719]

Beautiful explanation

[@mohammediqbal6779]

I think it has more to do with the Gauss' theorm, which is the base of the Bezout's theorem?

[@RyanLucroy]

Juat wondering, whether you could have proved it this way: Look at the prime factorization of a and b. As gcd(n, a) = 1, but n|ab, you can conclude that all prime factors of n must be contained within the prime factorization of b, as n and a don't have any in common and therefore n|b.

[@samuel0100]

Fredo Lucroy But, we have not proved the prime factorization yet. en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

[@samuel0100]

Fredo Lucroy I am thinking if the prime factorization is not unique, whether or not your logic is still correct.

[@samuel0100]

Fredo Lucroy n|ab ab = nk If ab and nk have different prime factorizations, then there may exist some prime factors of n, that are not contained within the prime factorization of a or b.

[@zhongyuanchen8424]

Allow me to share Some interesting arguments that I used to convince myself about trusting some math theorems. What I call proof below is simply dialog though. All rules of differentiation can be viewed as special cases of the multivariable chain rule. All examples are trivial, but it is a good point to view about differentiation when things gets complicated. The alternating series test asserts that a sequence converge if the sequence is monotone decreasing. Proof: For a bounded function that is monotone decreasing in the interval [N,∞] for some N large enough, the alternating sum is bounded by the sum of it’s discrete derivative, which is a difference in its antiderivatives at N, and ∞. The antiderivatives is the original function. Although this could go wrong as long as the function alternate slightly. Differentiation with respect to different variable commute for k times as long as the function is k times differentiable for all variables at a point z. Proof: since a k times differentiable function at a point z has a k degree taylor series, which is a power series, it is easy to see that differentiation commute for simple power series. Clearly, a taylor series contains enough information to determine it’s derivatives about any point. The generalization of taylor series into higher dimension is easily proved by parametrizing a straight line in space and taylor expand it as usual. If f(x) and g(x) are two N degree polynomials with N same distinct roots with the coefficients of the Nth power being the same, then f(x) and g(x) are the same. Proof: f(x)-g(x) is then a N-1 degree polynomial and has at most N-1 roots. The reason for why is trivial even in complex domain. Thus if f - g is forced to be zero at N points by the definition, f-g=0 everywhere.

[@oscarobioha595]

Hey...Can u do sum of all positive integers

[@blackpenredpen]

-1/12

[@h4ck3r211]

Congratulations, I like your videos :) Can you some day do the integral of sqrt(1-x²/(a-x²))dx where a is just a constant? It would be so nice!

[@kalyanguhoroy514]

What is the n th derivative of x^n ??

[@larzperson9601]

Good video :)

[@xaxuser5033]

Of course no and a big nooooo If n|ab and gcd(a,n)=1 then n|b must be true

[@okoyoso]

What area of math is your specialty? What have you done research in?

[@Traw-ve7qf]

6:12 if b/ky proof wrong

[@anilkumars6438]

How to solve this question?? (0.05)^ log base sqrt (20) (0.1+0.01 +0.001 +...)

[@anilkumars6438]

Yes, I got it. 👍

[@blackloop1861]

Can you do a video about 1+2+3+4...=-1/12

[@alanturingtesla]

He did.

[@blackloop1861]

Woow great

[@MrBastibro]

Any link please, can’t find it

[@blackloop1861]

Me to 😭😭😭😭😭

[@simonbohnen1384]

Crazy Drummer Would be really nice if you could post the link here!

[@felixzwick5055]

Yaaay

[@gammafirion9162]

You’re doing so much number theory lately

[@willful759]

did you just unlisted a video where you proved that if ka = kb (mod n) and gcd(k,n) = 1 then a = b (mod n) ? and you did it because you used this lemma?

[@blackpenredpen]

ivan pineda no. I made it private since I made a mistake

[@willful759]

blackpenredpen ok, great video btw, you really helped me with my calculus class, and I love your videos, keep the good work, I'm waiting for a corrected version of that video ;)

[@akhilkumarvarma3257]

Unless n is a prime number,u missed it

[@InZiDes]

Prop. If n | ab then n | gcd(a,n) gcd(b,n)

[@soumyachandrakar9100]

Number Theory.....YaY!!!!!

[@konstanty8094]

4 2 2

[@blackpenredpen]

Yay! That's the easiest!

[@konstanty8094]

I chose it for that reason.

[@kmac5912]

Solve for n if n^n=(n+1)!

[@blackpenredpen]

Can we Use gamma function?

[@kmac5912]

As long as you can get the answer

[@kmac5912]

There is

[@kmac5912]

a solution just not a integer

[@kmac5912]

It’s OK