The way you explain these concepts is perfect. After watching your videos I understand efortlessly where the theorems come from. When listening to other lectures I'm usually able to follow through as well but I need to concentrate hard all the time and It's tiring. Watching you feels like watching Richard Feynman. Thanks a lot!!!
This demonstration is really cool. One could also arrive at the same point by thinking: if my number n has a prime p among its factors, then no multiple of p can be coprime to n. But, during counting, a multiple of any number k occurs once every k numbers. So you can expect n/k multiples of k from 1 to n. Naturally, this also applies to prime numbers. This means that n - n/p numbers won't share a factor p with n. If you extend this line of reasoning to all prime factors of n, you get exactly the formula for the totient function. For example: 24 has 2 and 3 among its prime factors. So no multiples of 2 or 3 can be coprime to 24. But 1/2 of natural numbers are multiples of 2, since even numbers occur every two consecutive numbers. And 1/3 of natural numbers are multiples of 3, since a multiple of 3 occurs every three consecutive numbers (yeah, I know it isn't rigorous of me to speak of halves and thirds of infinite sets, but, please, bear with me). So one half of the numbers between 1 and 24 will share a factor of 2 with it, and one third of the numbers will share a factor of 3. So one half and two thirds of numbers, respectively, *won't* share a factor of 2 or 3 with 24. So we find that 24 * (1/2) * (2/3) = 8 numbers will be coprime to it. But this is precisely the definition of the totient function!
You do have to pay attention when talking about that kind of stuff! The process that you're describing is similar to an alternate method to prove that φ(ab)=φ(a)φ(b). The important thing to remember is that it's possible for a number to be BOTH a multiple of 2 and a multiple of 3, so we can't immediately assume that we remove 1/2 in the first step and 1/3 in the second step. Ultimately the proof does work out, but you have to prove that after removing all the multiples of 2, a third of the REMAINING numbers are multiples of 3, and so on. It's an interesting exercise to think about how to prove that part!
@@MuPrimeMath Check out these guys video for #SoME. Astoundingly, they found a formula for the Totient function without needing to know a number's prime factors. ruclips.net/video/isf6k_RVV6M/видео.html
The way you explain these concepts is perfect. After watching your videos I understand efortlessly where the theorems come from. When listening to other lectures I'm usually able to follow through as well but I need to concentrate hard all the time and It's tiring.
Watching you feels like watching Richard Feynman. Thanks a lot!!!
This channel is great. Just found it during the megafav number event. Keep up the great work!
This demonstration is really cool. One could also arrive at the same point by thinking: if my number n has a prime p among its factors, then no multiple of p can be coprime to n. But, during counting, a multiple of any number k occurs once every k numbers. So you can expect n/k multiples of k from 1 to n. Naturally, this also applies to prime numbers. This means that n - n/p numbers won't share a factor p with n. If you extend this line of reasoning to all prime factors of n, you get exactly the formula for the totient function.
For example: 24 has 2 and 3 among its prime factors. So no multiples of 2 or 3 can be coprime to 24. But 1/2 of natural numbers are multiples of 2, since even numbers occur every two consecutive numbers. And 1/3 of natural numbers are multiples of 3, since a multiple of 3 occurs every three consecutive numbers (yeah, I know it isn't rigorous of me to speak of halves and thirds of infinite sets, but, please, bear with me). So one half of the numbers between 1 and 24 will share a factor of 2 with it, and one third of the numbers will share a factor of 3. So one half and two thirds of numbers, respectively, *won't* share a factor of 2 or 3 with 24. So we find that 24 * (1/2) * (2/3) = 8 numbers will be coprime to it. But this is precisely the definition of the totient function!
You do have to pay attention when talking about that kind of stuff! The process that you're describing is similar to an alternate method to prove that φ(ab)=φ(a)φ(b). The important thing to remember is that it's possible for a number to be BOTH a multiple of 2 and a multiple of 3, so we can't immediately assume that we remove 1/2 in the first step and 1/3 in the second step.
Ultimately the proof does work out, but you have to prove that after removing all the multiples of 2, a third of the REMAINING numbers are multiples of 3, and so on. It's an interesting exercise to think about how to prove that part!
clearest video i've seen so far 👍
Left-handed crew
oh wait..
_(accidentally exposes himself)_
Very nice. Gauss theorem next! :)
Factorization is weird, still very neat, also there were only two splices that I saw, also very nice
I would like a vídeo about Gauss theorem.
Nice vídeo.
Watching it from Mexico.
Best of the best ❤️♥️
How do you calculate the value of the euler's toilent function without having to find the factorization of n?
This is probably not possible, since the value of the totient function depends on its prime factors.
@@MuPrimeMath Check out these guys video for #SoME. Astoundingly, they found a formula for the Totient function without needing to know a number's prime factors. ruclips.net/video/isf6k_RVV6M/видео.html
Love from india
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