Introduction to number theory lecture 1.
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- Опубліковано 20 лип 2024
- This lecture is the first lecture of my Berkeley math 115 course "Introduction to number theory"
For the other lectures in the course see • Introduction to number...
This lecture gives a survey of some of the topics covered later in the course, mainly about primes and Diophantine equations.
The textbook is "An introduction to the theory of numbers" by Niven, Zuckerman, and Montgomery (5th edition).
Richard won the fields medal in 1998 and he’s here on UA-cam giving us free lectures in number theory. I earned a masters in math at Illinois state some 10 plus years ago and I literally forgot a lot coz my current work has nothing to do with the things I studied. This channel brings back a lot of memories and I have started going through my notes again. Thanks 😊
@@username8644 it was a struggle finding work after a pure math degree. You have to move towards data science or more applied “new sciences” like Machine learning and Artificial intelligence to get work. Do a lot of internships when offered. Also probability and statistics present better opportunities
Does not seem to be doing the best of jobs...
Im so lucky 😳
What is your current work?
@@galenspikesmusic Software QA engineer.
Mediocre math professor will make a 45-min number theory intro insufferable, but masters like Prof Borcherds can make it accessible to a high schooler. Much respect.
yah! Right you are!
I must say I really like this camera setup with handwriting. It makes it so much easier for the lecturer to do math.
Very Intresting lecture. Presented with a rare pedagogic ability and a touch of humour. Thanks Prof!
There's a slight error at 8:36. 2047 = 23 x 89.
He is probably only field medalist whose course lectures are available here on UA-cam
How do you find primes?
Eratosthenes algorithm
How many primes are there.
Euclid Theorem: For any finite set of primes, there is a new prime which is not in the set.
Proof: In fact, for the set of primes I = {p_1,...,p_n}, any prime factor of the number
p_1*...*p_n + 1 gives such a new prime p_{n+1}, i.e., p_1*...*p_n + 1 = p_{n+1}*Q for some Q.
In fact, suppose contrary that p_[n+1} is in I, then 1 is decomposed.
(E.g., if p_[n+1} = p_1, then p_1*...*p_n + 1 = p_{n+1}*Q = p_1*Q, so 1 = p_1*(p_2*...*p_n + Q).)
So it contradicts.
Example, For a set of primes {3,5}, we can find a new prime as a factor of 3*5 +1 (= 2^3), so we get a prime 2 as a new prime.
Mersenne prime: If 2^n -1 is prime, then n is prime.
Proof. If n is not prime, then 2^n -1 is also not prime.
In fact, 2^{ab} -1 = (2^a-1)(2~{ab-a} + ...+1), I am not sure.
Fermat number: 2^n + 1, 2^2^m + 1
Open question: Are there infinite prime numbers of the form 2^n + 1, 2^2^m + 1??
Gauss discovered that if the number of sides of polygon is Fermat prime, then the polygon can be draw by ruler and compass.
Proposition. There are no polynomials f(n) so that its value is always prime for natural number.
In fact, let f(n) be a polynomial and suppose that f(0) = k >1, then f(k) is divided by k.
In case of f(0) = 1, by considering f(n+k), we can reduce it to f(0) > 1.
So it does not seem to be a simple way to generate primes.
How many primes are there less than x?
Gauss Pi(x) =~ x/ log(x)
Informally,
Chance of n being prime ~ 1/ log(n)
Chance of 2^n-1 begin prime ~ 1/ log(2^n-1) = k/n
:
For 2^n-2, probability argument is also available but 2^n-2 is not prime for all n, so probability argument is sometimes ...
23:44 Riemann
Li(x) = logarithm integral of x = integral of 1/log(x) which diverge at x = 1...
Pi'(x) is new for me.
Pi'(x) # prime number being less than x in which p^n counts as 1/n
Pi'(10) = # prime number being less than 10 (p^n counts as 1/n)
= #{ 2,3,5,7 } + #{ 4,9 }
= 4 + 2*(1/2) = 5 ??
Pi'(x) can be represented by Li(x) and zeros of Riemann zeta function.
31:50 In product representation of zeta function, he showed how to find (1/90)^s in the product and at the calculation he forgot "to the s power".
Diophantine equation
Solutions of integers for equations
such as x^2+y^2 = z^2, 27x+11y = 1, x^4+y^4+z^4 = t^4,
x^2=61y^2+1,....
Hilbert's 10th problem: Is there an algorithm to solve all diophantine equations?
Answer: No Possible in general by Robinson, et al.
This introductory lecture was absolutely fascinating for me as I'm only in my 2nd year of a maths degree right now. Thank you for this.
Is it a good idea to spend time watching this series? I'm currently on 1st year math
@@piotrskalski1477honestly watch you want to, time spent practicing math is time spent practicing math
Currently taking this course, I'm so excited!
Thank you for your excellent lectures. These are some of the best math lectures I have heard/seen.
is that your real name?
My truest gratitude and deepest bow to Mr Borcherds (no disrespect for the title instead of Prof., Dr., etc., as all of those are still understating the greatness from yours) for the selfish-less sharing to all laymen as I am here.
n^2 + n + 41 is prime up to n = 39, but 40^2 + 40 + 41 is divisible by 41.
Can be quickly shown by:
40^2 + 40 + 41 = 40 (40 + 1) + 41 = 40 * 41 + 41 = 41 ( 40 +1) = 41^2
I read somewhere that the equations such as the one above that produce a lot of primes are tied to another field. But I can not remember the actual relation
@@ingiford175 FWIW, it has to do with so-called Heegner numbers and the values of the j-function (not my area however). Wikipedia has a little bit: en.wikipedia.org/wiki/Heegner_number#Consecutive_primes
@@toddtrimble2555 That was the exact page I was at like 3-4 years ago. Thanks.
"Solving the Riemann Hypothesis is the easy bit - getting someone to read your proof is the hard part!" 😂😂😂😂 Sounds like a Douglas Adams joke. 😂😂
Fact bro 😂
A correction. At 16:08 while it's correct that f(n) is now divisible by k it does not follow that it is not a prime: it can be equal to k and k might be a prime. A more precise argument would be to choose sufficiently large x such that f(x*k) is greater than k (we can always do that since highest power term will dominate the others and we presume its coefficient is positive). It's still divisible by k and now we know it is not a prime since it is larger than k.
I had a little chuckle at "0 and 1 are not prime because 0 is zero and 1 is a unit [a divisor of one]"
Thanks for sharing your knlowledgement with us. Hugs from Brazil
I remember my class of elementary theory of numbers was the first part of the modern abstract algebra course; it began with the definition of divilibility and then developed propositions related to linear combinations of product of numbers; there was an order in the discussion. Here, we go from classical theory of numbers to quickly use analisis for theory of numbers... no good for a first class...
Just a minor correction. At 15:23 the polynomial is not prime at n = 40.
Was gonna say, n=41 is definitely a prime. thx for the clarification.
Nice. Gotta hop on this while it’s FRESH
That looks like it generalises to:
Positive integers are "1 over their factor count" of a prime, so 11 is half a prime, and 12 is 1/6 of a prime :)
Thank you very much professor!
Both, "a" and "b" need be odd for the numerator to be divisible by 2^a+1
Seems it's sufficient to let b be odd, Since 2^(ab) + 1 = (2^a + 1 - 1)^b = K(2^a + 1) - (-1)^b + 1. It's equivalent to (-1)^b + 1 mod (2^a + 1).
@@jehovah0121 yep absolutely, its sufficient to just let b be odd!
Great lecture! I wanted to get a good intro to number theory since I want to study cryptography in more detail. This was a very engaging presentation!
Thanks!
Be sure to catch the mistakes.
Fun topic! Thanks!
Minor correction at 10:34: the condition should be "b odd" instead of "a odd"
Thank you, I was also looking at it.
Or instead, professor Borcherds could correct it like if a is odd then 2^(ab)+1 is divisible by (2(^b) + 1).
@@tahminatabassum9454 yeah same here I am also looking for this
still do not see it. If b is odd then we have 2^(ab)+1 = 2^(a(2k+1)) + 1 = (2^2ka )* (2^a) + 1 is not divisible by 2^a + 1. Where am i wrong?
@@yannicko.5936So, I suppose you might already know that x-y divides x^n-y^n whenever n is a positive integer. If n happens to be odd, then it also holds that x+y divides x^n+y^n, because you can write x+y as x-(-y) and x^n+y^n as x^n-(-y)^n. And that's almost it: let n:=b, x:=2^a and y:=1 in the above divisibility rule, and you get that 2^a+1 divides (2^a)^b+1^b=2^(ab)+1 whenever b is odd.
thanks a lot, it 's fantastic !
Does this course get as far as the riemann zeta fn and stuff like that?
Hi professor, unsure how to reach you aside from your videos. I have a burning question not necessarily related to this video but one I'd really appreciate your thoughts on. It seems that any analysis humans have ever made always relies on a framework that buids on our notion of linear order (x < y). We can't seem to escape linear order. Even complex numbers which have no natural order can be thought of as an infinite union of linearly ordered sets. In any ideas that can be represented in terms of graphs, linear order is implicit in the depth of nodes and notions of distances
I'm a little anxious that there might be structures that can't be examined within the framework of linear order. Structures that "counting" can't shed light on. Are we capable of perceiving those structures at all? Do they even exist? What tools do we have to discover and study them?
I was following his course on Abstract Algebra and my reaction was 'wooh old fella can teach, at last a good algebra course'. Old fella on YT happens to have a Fields Medal, I feel very ashamed, but thankful nonetheless
thank you so much, i admire you
I read in class 9 , and I am preparing for IOQM . 🦖
The part at 10:40 is really not clear to me, even if I "think about it a bit". Can someone please explain why 2^(ab)+1 is composite for odd a?
Well, you see, a+b divides a^n + b^n for all odd n natural numbers... thus you can set b =1, and a = 2^b then you will get 2^b + 1 divides 2^ab + 1 so it becomes composite...
My university had number theory (same textbook) as a 300 level course... ouch.
Number theory is 300!
Thank you!
Sir, 23 times 199 equals 4577, not 2047. 2047 divided by 23 equals 89.
has anyone tried the Lagrange interpolation to find the "prime polynomial" ??
Fermat saw that shit got so mad and said nah it only generates primes
Thank you
minor correction, the professor's name is actually Roberto González
Re(ρ)
They are equivalent by the functional equation of the zeta function.
He's also including the trivial zeroes (negative even integers).
ah, well, he is including the trivial zeros!
Why are some videos inaccessible? It says there are 49 videos on the playlist, but only 35 are accessible. Or is it just me?
P.S. Absolutely amazing lectures, can't stand the fact that it's not possible to watch all of them:)
I know people can produce several videos and then release them on a schedule. Looks as if this channel is releasing a new video every two days.
@@gummansgubbe6225 yes, i think you're right. Thank you!
thanks
Minor correction at 9:00 is that 2^11 -1 = 2047 = 23*89, doesn't change the sentiment though!
Glad to know that is the start of at least 20-hours course
hi can i watch is course
Many universities should've used these videos for undergraduate courses. But, they have to let their own professors give lectures so that these professors can make a livelihood, but students have to bear the poor education they received.
Wouldn't be better to merge both number theory courses? They seem pretty similar. Or could you explain why are you making two separate lists of lectures? Thanks
thanks!!!!!!!!!!
I konow the new path of findings primenumbers....but I didn't study mathematics properly. What do I do?
Use nonassociative structures. Universals, please. Archimedes used infinitesimals rigorously. 'Zeroes of the Zeta function, which I will describe in a minute ...!' Please use Gougu and not Pythagoras! ((i)to(i))to(i) is contravariant. This modifies the Zeta function!
Excuse me sir what books we use to study number theory
Description says use Niven
thanks!!!!!!!!!!!!!!!!!
4:51 I don’t get this. What is Pn+1 was a number like 45, that’s not a prime
Ok
Yess! I got the book
Thank you 😇 you're the best 💓
Oh hey, it's Sylvester's sequence!
There are two types of mathematicians. Those who can count and those who cannot.
what books you use ????
Niven et al. It's already off copyright, easy to download.
"there is actually no point in taking logs to base 10"
american engineers are fuming rn
He said in pure math. Engineering is not pure math, it is an application of math.
yeeeeeeee
2 to the power of 11 minus 1 is not 2047; it's 4577, and 23x199 is not 2047 as well, but 4577.
You’re nearly correct. 2^11 - 1 = 2047 but 2047 = 23 x 89 and hence not prime.
@@markelkins no doubt. But why 23x199? I stopped for a moment thinking I must have miscalculated
Zero and 1 are a unit? 1:30
What he meant was that 0 is just nothing, (or the absence of any units), and 1 is just the unit that were counting and making everything out of with numbers.
You can think of the number 5 for instance as just being 1 1 1 1 1, like tallies as units.
He didn't say that
Zero is the identity for addition, one for multiplication. The natural numbers can be generated starting with the empty set.
How different is this course from the one u did before ??
I think that this will contain more explicit information and include some graduate course content. The previous is mostly undergraduate.
Besides, Borcherds' videos always have new math when giving a lecture on an old topic.
@@autumnsthree8609 yeah thats true . I Just finished some lectures from that series so was confused which series to take. If this one has some graduate topics that would be better as i already have done an introductory course.
This is an undergrad course, not many prerequisites
It is wrongly claimed that 2047 is 23 times 199. It is in fact 23 times 89. It is 4567 that is 23 times 199.
errata at 08:38, 2047=23*89
I don't mean to be a bother, but could you use another color than the orange marker? It's a bit hard to see. I feel terribly for complaining about such a trivial thing, because I very much appreciate you giving these lectures on UA-cam for all of us who wish to learn! 🙏
Anyway, thank you so much for posting these! It's really very kind of you!
Oh, I was also giving a look at Hilbert's 11th problem. Actually, I should probably start with an easier problem (I'm still getting calibrated, it's been a decade since I was in school), do you have any suggestions?
Number theory - a definition please, what is number fundamental properties.... I'll known the thoughts of God. The rest is details.
why is this importaint ?
Why is anything important?
im sorry, you lost me at nothing + nothing = 1 lol
This series is similar to the first on number theory, but will include examples of applied number theory in nonlinear transgender bathroom power dynamics in a cis-normative post-colonial patriarchal paradigm. Should be exciting!
thanks
thanks!!!!!!!!!!
thanks!!!!!!!!!!!!!!!!!!!!!!!!!
thanks
thanks
thanks
thanks
thanks!!!!!!!!!!!!!!!!!!!!!!