I looked up the proof for Bézout's identity and it makes sense to me so I am satisifed with this proof - great video by the way, the fact that you were so quick to reply to my comment and work on a video to answer the question tells me you're going far with this channel :)
@wishizuk Жыл бұрын
I am glad to hear that you are satisfied with the proof. I will make a video for the proof of Bézout's identity soon, so stay tuned. Again, thanks for your kind words. It was fun making the video!
@wishizuk Жыл бұрын
As promised, here is the link to the video for the proof of Bezout's Identity: kzbin.info/www/bejne/eYXJiJqJh6qMlcU
@darcash17386 ай бұрын
p|n^k --> p|n*n*n*... (k times). By the extension of euclid's lemma, then, p divides some ai, in there. because they are all just n's, then it divides n. --> p|n. And we can apply this simply for the case of p|n^2 also with p|ab meaning p|a or p|b by the standard case of euclids lemma.
@wishizuk6 ай бұрын
Thank you for your comment! I believe that this video was made right around when we covered Bézout's identity, so this approach felt most natural.
@darcash17386 ай бұрын
@@wishizukyour approach reminds me the most of proving Euclids lemma. Thinking about it, you’re doing division, so you could also use the division algorithm. n = pq + r, p E N, q E Z, 0 0, then, it would not divide evenly -> r = 0, since we know p must divide n^2 evenly - n = p(q), q E Z -> p|n This way would probably be the longest explanation of it. If you do it up to some kth power on n for p|n^k, you’d need binomial theorem and say everything up to the last element has a factor of p, and the last element is r^k, where we can apply the same logic as before of r not having p as a prime factor by the division algorithms restriction
@wishizuk6 ай бұрын
@@darcash1738 Yes, since I used the division algorithm to establish Bezout's Identity, it makes sense.
@BS-bd4xo Жыл бұрын
Great proof. It feels trivial, but the proof has to be more precise ofcourse. But where did you use that p is prime? I feel like it doesn't have to be. Tho then there are some cases where the theorem doesn't hold. Interesting
@wishizuk Жыл бұрын
Thanks for your kind words! I thought that the proof was sufficiently precise, so if you could indicate which parts need more explanation, then I would be able to offer more detailed explanation. The fact that p is prime was used when I went from "p does not divide n" to "gcd(p,n) = 1," which is not true if p is not prime; for example, 6 does not divide 9, but gcd(6,9) = 3.
@BS-bd4xo Жыл бұрын
@@wishizuk Ohh alright. Thanks, I didn't think of that. I didn't mean the proof has to be more precise, I just meant that you can't just say "oh that is just logical". I was also thinking that there are probably even more non-prime numbers for p where this condition would also hold, tho idk how to discribe them. Prolly boils down to the condition p|n, which is circle reasoning lol.
@wishizuk Жыл бұрын
@@BS-bd4xo You are welcome! Yes, the statement holds when the divisor is the product of distinct prime numbers. For example, if 30 (= 2 x 3 x 5) divides n^2, then 30 divides n; however, 4 (= 2 x 2) divides 2^2, but 4 does not divide 2.