Theory of numbers: Euclid's theorem
This lecture is part of an online undergraduate course on the theory of numbers.
We discuss Euclid's proof that there are infinitely many primes, and give a few variations of it showing that there are infinitely many primes in certain arithmetic progressions.
A couple of typos pointed out in the comments:
Sheet 1 at the bottom: 13x39 should be 13x139
Sheet 4 (14:00): 2^2^6+1 should be 2^2^5 +1
For the other lectures in the course see • Theory of numbers
Пікірлер: 19
I agree so much about his lexicography discussion
Thanks Prof. Borcherds. Please keep on doing lectures like that. They calm me down during these strange times.
I notice "Euler" is mentioned instead of "Euclid", several times. It's a mistake I make and nice to see one of the top mathematicians in the world has the same issue :)
Another mistake at 14:00 : 2^(2^6) + 1 is not divisible by 641. It should be 2^(2^5) + 1 (which equals 641×6700417). Instead, 2^(2^6) + 1 can be prime factorized into 274177×67280421310721.
@richarde.borcherds7998
3 жыл бұрын
Thanks! This is indeed a mistake by me.
@wangjason8400
3 жыл бұрын
@@richarde.borcherds7998 Thank you professor for your video series! Really appreciate it.
So much jargon. Is it not easier to say that the list of primes is endless because the lowest factor greater than 1 of p!+1 must be a prime number and must be greater than p?
Thankyou
I hate to correct Prof. Borcherds, but 1807 = 13*139 (39 is not prime, or coprime to 1806).
@brettaspivey
3 жыл бұрын
Oh no! Euclid's algorithm doesn't work!
@dneary
3 жыл бұрын
@@brettaspivey 🤣 Clearly not what I meant, but thank you for the laugh.
@SaveSoilSaveSoil
3 жыл бұрын
@Dave I hate to correct Dave, but you clearly meant 1807 when you wrote 1806.
@dneary
3 жыл бұрын
@@SaveSoilSaveSoil You are correct, thank you.
@richarde.borcherds7998
3 жыл бұрын
Thanks! I've added a note about it.
for 3(p1 to pn)-1, if it's a prime, then it's also 3n+2, that's good. if it's not a prime, it must contain a prime factor greater than (p1 to pn), if the factor is in 3n+2 that's also good. if it's in 3n+1, then we have (3m+1)k = 3(p1 to pn) - 1 3mk+k=3(p1 to pn) - 1 so k is -1 in Z3. Since 3m+1 is greater (p1 to pn) so k is no more than 3 and k is positive integer. k can only be 2. then 3 *m*2 = 3(p1 to pn) - 3. m*2 = (p1 to pn) - 1 . left is even right is odd since p1 =2 the 1st prime. so contradiction. is my idea right? i hope anyone can help to check
8:21 "A lot of examples in politics that I'm not gonna mentioned... is Pluto a planet?" I think I almost hear "Is a fetus a human being?" lol
Use a dot for multiplication!!!!
Why are you using numerals and glyphs, thats not euclid you silly goose