Lec 4 | MIT 6.042J Mathematics for Computer Science, Fall 2010
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- Опубликовано: 4 янв 2025
- Lecture 4: Number Theory I
Instructor: Marten van Dijk
View the complete course: ocw.mit.edu/6-0...
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
MIT, you rule. You understand that knowledge should remain free and open. So many universities have their lessons under paywalls or lock and key. Thanks for making the world a better place.
Lies again? MRT MIT
Knowledge is free and open already
tell that to Aaron Swartz
This is actually a very clever lecture. I didn't think it was possible to introduce state machines, divisibility, gcd's, and Euclid's Algorithm in one problem which looks so trivial on the surface.
After reading the lecture note first, I found it's much easier to follow the professor when watching the video than the first time when I didn't read.
where can i get the notes ??
@@sababaig5220 MIT has them on the website for the 2010 version of this course
@sababaig5220 ح٩
Do u mean readings and recitations ? Please reply as i only found them n not notes@@mikeriordan3428
this is the solution
I see a lot of negative comments here! And initially I too supported all the comments, but it took me a while to understand that the problem is not with the lecturer, but its with the view. We do not have a view of all the blackboards at a time, and hence its hard for us to refer back and forth.
And the way out to this is, start taking notes of all the things he writes on the blackboard. It will appear tedious in the beginning and will take a little long to complete this lecture, but I am sure you will come back to thank me later! Worked for me, and it will work for you as well. Have fun :)
I love you man!
You can also take screenshots
Professor Dijk, thank your for another great lecture in discreet mathematics and the use of EUCID Algorithm from number theory.
Number theory is one of the oldest mathematics on this earth and for the last few years we know that it's application is growing in this world.
@Abhay Pai
Without being formal: if x is a bunch of m's and y is a bunch of m's and a is a bunch of m's, then any linear combination of those 3 is a bunch of m's.
So a bit more formal now: There exist i,j,h so that m.i = x, m.j = y, m.h = b.
Now examine:
x + y - b = m.i + m.j - m.h = (i + j - h).m => m | x + y - b
Man this lecturer seems like a smart dude and also a really nice guy but teaching wise the other lecturer is so much better.
I appreciate the open courseware.
Here, what he didn't mention is m has to be the highest common factor between a and b. Otherwise, you can get anything using a and b. Like in the example around 31:00, 1 divides both 33 and 55 but we take m to be 11 and not 1 because 11 is the highest common factor.
thank you! I came to the comments section to find just this. You can notice this if you think about the problem a bit but it may be confusing at first
that's GCD itself right? he did mention GCD?
m doesn't have to be the GCD, but every divisor a and b have in common devides the result. At this time in the lecture you can only use it for condiction but one example (like 11) is enough. The GCD would always work for such contradictions
It's obviously true that any m must be divisible by 1
This guy is very sincere in his job, but coming here after watching the first 3 lectures, the difference with Prof. Leighton is obvious. Prof. Leighton tries to engage the whole class throughout and proofs are derived collaboratively between him and the class
I think when the content starts to get involved you will be dead inside anyway😂
@fermisurface2616 pardon me, I did not write any of that stuff which you are claiming. All I said was Prof. Leighton has a different style and that style appeals to many including me.
@fermisurface2616thanks for your opinion. It's an interesting one.
(1) Fill the 5-gal tank
(2) Transfer 3 gallons from the 5-gal tank into the 3-gal tank
(3) Empty out the 3-gal tank
(4) Transfer the remaining 2 gallons from the 5-gal tank to the 3-gal tank
(5) Fill the 5-gal tank
(6) Use the 5-gal tank to provide the remaining 1 gallon into the 3-gal tank.
Your 5-gal tank now has 4 gallons
I think John didn't get it. Andrew Tang that's actually the process he also described at the start of the theorem.
@@joek398 More efficient than filling the 3-gallon jug 3 times
For those needing the course materials: ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/
Is there an answer sheet for the problem sets? The survival of humanity 25 years into the future may depend on this.
+Sean Farley Sorry, no answers were provided for this version of the course. But there is an older version that does have problem sets with answers: ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2005/. Hopefully this will be of some help to you and to the survival of humanity. :D
+MIT OpenCourseWare
Thanks! Now our future robot overlords will be error free!
+Sean Farley Made my day :D (And perhaps my future :D)
@@mitocw i made a request to you guys about a question in the problem set 2, it's in the comment section of the previous video. Could you please give it a look over?
@@mitocwCould you pls state the topic of the lecture instead of "Mathematics for Computer Science" in the title of each lecture? That will be really helpful for revising a specific topic. Thanks! Great lecture!
I think most people are having an issue with the way he arrived at the Theorem/Invariant for the first problem(water jugs). He somewhat seems to have arrived at it intuitively, but you could arrive at the Theorem/Invariant by looking at 2 or more examples of the problem with 'a' and 'b' being changed.
This professor's lecture is speaking the truth. If you are a little bit lazy at a late night to grasp the complete lecture it doesn't work unless you go back and back twice and thrice. It's asking me for my full consciousness (and attention). What can I do more than that? watching in a desktop where we are facing 12 hours unavailability of electric power all over the day and it's only possible at a late night.
After reading the mateials, it's more clear and more organized.
Do you have any link for materials, cause there's nothing given in the description box?
personal reference, restart at 00:37:00 after going through the material till. 4.3
This video shows the connection between Linear Algebra and Number Theory
@31:46 Doesn't m=1 divide both 55 and 33 (why did we chose m=11). So we should be able to reach anything divisible by m=1 according to the proof. (I know it actually can not. Please point out anything I missed in the proof and not give a "proof by example")
EDIT:
I think I understand why. The theorem proves that the invariant condition has to be true FOR ALL m, s.t. m|a and m|b. So in the example, we need to check for both m = 1, 11.
And then we can conclude if gcd(a,b) divides it then we can reach the state.
gcd is the "GREATEST" common divisor. We're looking at the largest integer that can divide both a & b.
the theorem only proves that each (x, y) will have to be divisible by m; it doesn't say all multiples of m are solutions or reachable by the state equation. in other words, the theorem only eliminates what x and y can not be, it doesn't actually tell you what they can become.
53:20 thats where i got to know m not alone ,,,,the class is also clueless,,,,,,and even so nobody asked a single question
I could keep track of what he's tryng to say ,,,,,,i lost it after 35 mins
thanx anyways MIT
the silence comforts me
The lecture is disorganized. Like, he jumps from one thing to another, without ever completely explaining what the thing was about, and how does it link to the next one (He never proved the corollary at 35:27). I have to do more into understanding, and putting things in place. But still, once I did get all things right, I think he was pretty great. Thanks MIT :D
+Paritosh Mehta Here's my idea: Hand out gatorade or water or juice at the beginning of the lecture. The rule is every time you don't understand wtf he's talking about, take a sip. After 35 or 40 minutes of this, the whole room should be sipping water.
Also, I really liked this guy's style. He immediately struck me as a talented professor and extremely fluent in mathematics. The only problem is our understanding. Really difficult to follow after a certain point when he stops explaining and focuses on getting all the writing on the board. We're not automatically seeing the explanation in your math the way you can see it written.
the corollary is assumed because the invariant that is being proven is that any number m which divides both a and b will divide all of the possible states of jugs a and b after n moves. the greatest common denominator of a and b necessarily divides both a and b, so it also divides all possible states of a and b. it doesnt need to be proven, it's trivial.
m(which divides a and b) divides any results is already proven. so there is no need to proof that corollary. because gcd(a,b) divides a and b.
Better read lecture note before watching the lecture. I guess that is what your supposed to do.
42:43 but in previous class professor said avoid using a word "obvious".
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exactly my thought
Thanks for free content MIT. I really appreciate it.
Just an opinion, the instructor from the previous lectures is more competent instructor.
I have an impression that this instructor passes really fast over important points - like what theorem we are going to prove and passes really slow over content that seems to be simple.
Now I'm not an export, I don't judge anyone, just an impression.
Man, Tom Leighton, the other lecturer for this class is so much clearer at explaining; this is unfortunate.
3:45 That Windows XP screen... nostalgia kicks in.
(Celebration) praise MIT professors, authorities and others for excellent public education resources, Open Course Ware.
I'll read MIT course notes this summer to help me understand proofs, strong induction, number theory, mathematic notations, logic, applications of math, and other concepts, such as prepositions.
Thanks people.
At 1:48 , he said 1, 2 , 3 and so on , thus , disregarding the negative integers , they are also integers .
And really OCW is a very good program .
If you did not understand the point at 31:00 about a = 33 and b = 55 not being able to give 4, here is what was meant
If m|a and m|b then m|(any linear combination of a and b), this statement is always true, but can you write 4 as a linear combination of 33 and 55, the answer is no.
Then what was the Student trying to say by both numbers being divisible by 11, well if 11|33 and 11|55 then there linear combination should be divisible by 11 but 4 is not divisible by 11 => so it cannot be a linear combination of 33 and 55.
Damn my university doesn't eve want to publish the reading list before one pays the corse.
No wonder the US is miles ahead in front of everyone else.
haha, i started to take notes as if i was in class ( i forgot that i didnt need to copy this down) .
This is amazing, it is very clear, i can go back and rewind and check if something wasnt clear. I LOVE THIS !!!!
Minute 54 didn't understand why prof says t'+u diff 0 so r>b.. and why t'+u is equal to 0
I learned a lot; thank you for uploading the video and thank you, professor.
An amazing lecture! Thanks
somewhere I see Dr. strang here but these lectures has some harsh comments which I feel is not true and Thanks for making the class interesting with the example
I think the problem can be formulated as drawing a star-shaped graph, with “b" ordered nodes, and vertices being drawn from a starting node by moving by “a” to next node… ”a” and “b" must be relative primes in order for all nodes to have connection... Each graph-drawing move is equivalent to 4 bucket moves… Would that simplify the proof?...
Great lecture, now I understand what a state machine is.
Amazing lecture. Thank you so much
A similar solution is: (3,0)->(0,3)->(3,3)->(1,5)->(1,0)->(0,1)->(3,1)->(0,4) done :)
IK this comment is old but: I always see your comments on all the videos I watch, and you've commented on mine too. It seems that we both self-study extremely similar stuff. I hope all is well man!
Okay so from what I gather, the really only very misleading thing about the lecture is the assumption that m is automatically the gcd, where in reality m can be any number. From the start of the lecture m|a is being defined to exist iff there exists and integer k such that a = mk. By this definition if we have m|a and m|b, m can technically be 1, because there exists an integer k in both statements such that a = mk' and b = mk'' where k' and k'' are different values of k corresponding to each statement respectively
Throughout the lecture, what I did was assume for m to be the the gcd for all statements, and then the lecture made sense. There should be a proof where m should be bounded to the gcd otherwise certain scenarios doesn't really work out.
Edit: at 51:27 you can hear that everyone is lost when the lecturer ask "is there any questions about this?"
Yeah I came to conclude the same thing when he said that 3 and 5 are only divisible by the common factor 1, but he only considered an m of 3 when he was talking about jugs of sizes 3 and 6 gallon
Why dont they rub the board before writing?
oh gosh, his accent is so subtle!
If you'd sat in on some classes I've been in you'd understand just how dense lecturers' accents can get!
about theorem 2, he only proved that the remainder after executing the algorithm is a linear combination of a and b, but he didnt prove that any value between 0 and b can be reached.
i think he also missed one assumption in theorem 2, gcd(a,b)=1. if we take a=3 b=6, then 0< L=5 < b=6 cannot be reached because five is not divisible by 2.
While you sit for calculating by using a State Machine. The bomb will go off.
Write faster!
I have a question about the last conclusion that the GCD is the smallest linear combination of a and b, how do you know every linear combination is included in the Euclid's calculations?
gcd(a,b) divides any reachable results and any combinations of a,b can be those results. Thus, gcd(a,b) divides any comb of a,b. While gcd(a,b) is a combination of a,b, so gcd is the smallest combination.
Number theory has the word average that means sums of values/total # of values
Wait, so this opening video from DH was OK to watch, but the one in Lecture 1 isn't?
I like the other teacher, One in first 3 lectures more
I like women
Ahem we are not talking about that dude
@@nikolaykolev5125
Watch lecture 5 first if you need more motivations to study number theory.
I think his explanation was good. I don't get how people find it so easy to crib. It's obvious this prof's first language isn't English and if you refer to the accompanied readings as the lecture progresses, understanding shouldn't be so hard. You can't expect to be spoon-fed all the time. Put in some effort for yourself.
were did he proved that m\a and m\b 28:45
A little bit difficult lecture with the new professor . How many hours does it take to watch this lecture and fully understand it ?
1/2
Happy New Years!
Happy New Year but 2021 🤗
At 19:00.Shouldn't the conditions of the states be:
x + y b (instead of >=)
x + y a (instead of >=)
?
I guess x+y = b is a common case between the two transitions so it is more appropriate to put the equal sign in both.
if you put x+y = b, both transitions will give (0, b).
Can someone explain me what was 'm' in the original example (a=3, b=5) and if it's m=1 then why didn't the next example (a=33, b=55) allowed m=1 by this logic?
+Érchegyi Dávid Or does m have to be the greatest common divisor of a and b?
+Érchegyi Dávid correct, the gcd would be 11 for that example.
The theorem is true for any m that divides both a and b => every linear combination of 33 and 55 always has to be divisble by all the common divisors of a and b and one of those happens to be 11.
This course just got me hooked to MIT OCW!
This is amazing!
Why is it assumed that m|a or m|b in the invariant property during our induction step hypothesis?
The theorem he's proving is
"If [ m|a and m|b ], then [ m | any result ]"
To prove such implication, he assumes [ m|a and m|b ] and shows that [m|any result] logically follows by the Principle of Induction.
This is method # 1 for proving an implication.
Did I answer your question?
at 34:40 he says that if gcd(a,b) = 1 then a,b are primes
but this actually not right as gcd(9,8) = 1
did i misunderstood his words?
He say a,b are relatively prime not prime
if gcd(a,b) = 1 then a,b are relatively primes
Any factor of 9 cannot divide 8 or vice versa hence it is the relatively prime .
I could not get the proof by induction at the very beginning.
Generally, the steps to be taken for proof by induction are:
1. Inductive basis
2. Induction Hypothesis
3. Inductive Proof/Conclusion
I got the point that m | 0, m | a, m | b, m | x, m | y, but what about m | x+y , m | x+y -a, m | x+y-b ???? :\
+Ben Jones The professor explained it 10s later that they can be included because they are the linear combinations of the others.
Someone have some other material that is more clear that cover this lecture?
Where is materiak
At 56:00 Why couldn’t he set the restriction of s being greater than 0 before defining the linear relationship? Why did he have to redefine s’ and then set the restriction?
I had a question from lecture 4 by prof Marten Van Dijk
Sir 33 and 55 both are divisible by 1 so as in earlier case where we have 3 gallon jug and 4 gallon jug we can take 4 gallon out of it but why not in case of 33 gallon and 55 when 1 is divisible by both?????
First, for the in-class problem, we had a 5-gallon jug and a 3-gallon jug (you said we had a 4-gallon jug and a 3-gallon jug, but I guess it's a typing mistake).
The problem he proved is
"If m is a common divisor of a and b, then m is a common divisor of any result of the game"
For the 5-jug/3-jug problem, we have that the ONLY common divisor for 5 and 3 is 1. So, any multiple of 1 (smaller or equal than 5, obviously) would be a reachable result for the game.
Now, for the 33-jug/55-jug problem, we have many common divisors and ALL OF THEM should satisfy the theorem.
You're right, 1 is a common divisor of 55 and 33, and also divides 4. However, 11 divides 55 and 33 but does not divide 4, therefore, 4 is not a possible result for the 55/33 game. QED.
Did it help?
@@felipe1055 "Now, for the 33-jug/55-jug problem, we have many common divisors and ALL OF THEM should satisfy the theorem." This is a good explanation.
How is it apparent though? I guess it's like "for any m, where m|a and m|b". And then moving to 33/55 problem, the first theorem doesn't hold and so it's refined?
@@felipe1055 Your remark """ The problem he proved is "If m is a common divisor of a and b, then m is a common divisor of any result of the game"
For the 5-jug/3-jug problem, we have that the ONLY common divisor for 5 and 3 is 1. So, any multiple of 1 (smaller or equal than 5, obviously) would be a reachable result for the game.""" is incorrect. This statement "If m is a common divisor of a and b, then m is a common divisor of any result of the game" is equivalent with (m|a ∩ m|b) -> [ (x,y) is a state/result -> ( m|x ∩ m|y) ]. This statement "So, any multiple of 1 (smaller or equal than 5, obviously) would be a reachable result for the game." is equivalent with (1|a ∩ 1|b) -> [ (1|x ∩ 1|y) -> (x,y) is a state/result ]. Please note the reserve on the direction inside the "[ ]". This is actually a flaw in the entire prove. We can not say any linear combination of a and b is a result. In fact Lemma 4.1.3 here (ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_chap04.pdf ) only proves that any result is a linear combination but not vice versa. Unless we can prove (x,y) is a state/result ( m|x ∩ m|y) (note the double direction here), we can not use the prove shown in the class to state that "any linear combination of a and b is a result". We can however show that if something is not a linear combination of a and b, then it will not be a result/state. For instance, since 4 is not a liner combination of 33 and 55, it can not be a result. This is using the fact that (m|a ∩ m|b) -> [ (x,y) is a state/result -> ( m|x ∩ m|y) ] is equivalent with (m|a ∩ m|b) -> [ NOT ( m|x ∩ m|y) -> (x,y) is not a state/result ] (because any conditional statement is equivalent with its contrapositive, which can be verified through Truth Table). EDIT: the
Does anyone know what handout they are talking about ? Do you know where can I find it ?
Course materials can be found on MIT OpenCourseWare at: ocw.mit.edu/6-042JF10. Best wishes on your studies!
@@mitocw Thanks so much !
I think there is an error in the water gallon prove?
are they assuming there is an other empty( with no measurement ) jug to keep the 1 gallon of water?
This is exactly the trick,, that you get away with only 2 jars
Is that invariant in the first proof actually just the predicate, and he accidentally wrote it wrong, or am I an idiot?
can anyone tell me what the name of movie that the professor show his students is?
The clip is from the movie "Die Hard 3: With a Vengeance".
4:59 which series or movie ???? anyone???
he literally said it's from Die Hard
Oh, So he's the guy from 1st Recitation
dat windows media player on windows xp...
Algorithm at 42:20
Nobody teaches better than zlatan ibrahimovic 😎 He is too good understood everything
Please help. How do I prove m is the "greatest" common divisor? At 1:11:30, it seems m can be any "lesser" common divisor in those 3 argument.
what will happen if there are 3 or more jugs??
6:00 Problem Solution:
Fill 5 gallon jug. Empty it into 3 gallon jug. 2 left in 5 gallon jug.
Throw away content of 3 gallon jug.
Pour 2 gallon in 3 gallon jug.
Fill 5 gallon jug again and pour out water from it into 3 gallon jug till its filled.
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I am having really a hard time understanding what 'm' actually is. Can anyone help on this?
m is just a variable. the invariant that is being proven for this game is that for any number m that divides both a and b, m can divide any possible state of the jugs. for example m can equal 1 in any possible case because all integers are divisible by 1; this holds true to our original statement, because one divides any a, any b, and any combination of a and b (it's one, duh). because that's a bit general though and doesnt really tell us anything interesting, we take that deduction and we extrapolate that the greatest common denominator also will divide any possible state because it satisfies m in the equation by dividing both a and b. m itself can stand for 1, it can stand for the greatest common denominator, and it can stand for other numbers too.
say you had a 12 gallon jug for a and a 24 gallon jug for b. m in that case can equal 1, 2, 3, 4, 6, or 12, because they all divide both 12 and 24. but, the most useful of all of those and the one that tells us the most about the game is the greatest common denominator 12, which is the corollary at 35:00 or so. if we look at this same example of a 12 gallon jug and a 24 gallon jug, using 1 for m doesnt tell us anything. we just think, well then the combination can be any number from 0-36, which is wrong. using 12 for m though, the greatest common denominator, tells us, well actually, we can only have multiples of 12, because if 12 divides both a and b then it MUST divide any combination due to our invariant. so now we've gone from 37 possible states, to only 4: 0, 12, 24, and 36.
+thecombinearecoming Thanks for the explanation :)
'm' is the gcd(a,b), where a and b are the jug capacities.
Can someone explain exactly why a 6 and 3 gallon jug wouldn't let you get 4 gallons. I understand it intuitively: you need a remainder to get you to 4. But can someone provide a mini proof?
In the lecture, he proved the theorem
"If [ m|a AND m|b ], then [ m | any result ]".
By "any result", he means any state ( x , y ) of the game.
(So, if [ m | any result ], then [ m|x AND m|y ] ).
In your question, let's say, a = 6 and b = 3. Also, we want to get 4 gallons in the 6-gallon jug, therefore, we want to reach the state x = 4.
Following the theorem, any number m, which divides a and b, must also divide x.
In this case, 3 divides a = 6 and b = 3 but does not divide x = 4. Therefore, x = 4 is not a feasible result. Namely, it is not possible to reach x = 4 in any state of the game. QED.
Did it help?
I find it interesting that the answer that came to my mind for an expression for r was s'a = r (mod b) (with = actually having 3 lines for modulo congruence) and that if you work through you get s'a = r + kb => r = s'a -kb which would mean that k is the same as u in this case. I dont I think it's interesting finding a value for this abstract integer coefficient.
The only diff b/w this and college is that we don't get a degree on paper.
MIT- ♥♥♥ :D
SDixit95 Depends on how much you apply yourself. Watching the lectures is only a fraction of the experience; you have to do the corresponding assignments on the website to really learn something.
Where can I find 18.781 course videos?
If not on youtube, try edx
Hopefully someone sees this post, I need help understanding. If someone could explain where he got the equation sa + tb I'd appreciate it. Maybe its me. My furthest level of college mathematics is Math1B (integral calculus at Berkeley) is that enough that I should completely understand his mathematics?
you may want to think the theorem as " any L gallons that can be written as sa + tb (s, t are integers) can be reached in the game"
he was just proving that any linear combination will be divisible by some number m if m|a, m|b, so lets say you have 2a + 3b it ll still be divisible by m. but he generalizes linear combination and uses sa + tb , to help himself prove correctness of this stmnt.
Thank you.
From where to study state machine?
As you can see, professor van Dijk doesn't exactly have the highest ratings, unlike professor Leighton:
www.ratemyprofessors.com/ShowRatings.jsp?tid=1902774
I object to your statement in this matter. However, I do agree MIT videos are of immense help.
The reason why I object to your statement is that I have two issues regarding his lecture. First is his accent. This is not a mockery but his pronunciation of words are misleading. Second, I am oppose to the clearness. I find his demonstration to be an excellent example but when he discuss the theorem it started to become vague.
I have always wondered who created the reference angle and why they created it. I've asked many people about this and no one seems to know.
Where I'm gonna get these handouts from?
The course materials are available on the MIT OpenCourseWare site at: ocw.mit.edu/6-042JF10
Actually, the divisibility handout is not currently included with the course materials.
It is not included. There are certain materials on class are not available.
i am clueless.
He should have excluded the case m=0 at 0:09:46
What........... Number theory is a study of integers
he isnt that guy of Silent Hill 3?
What happened to the other professor?
I gotta say, that whole scene in the movie was just dumb. love how he said it had to be "precisely 4 gallons". uhhh, yeah.
What's the book?
The text is Mathematics for Computer Science, which is available in the Readings section of the OCW course site: ocw.mit.edu/6-042JF10.
at 8:59 the transcript says "define" but the professor said "divide". I die =))))
Corrected. Thanks for your feedback!
@@mitocw there's another occurrence about 30 seconds later. Also "mainly" should be "namely" :)
This teacher isn't as awesome as the other one. Didn't like this lecture. Took to much time to understand. A lot of things are not fully explained. And the logic of proof is confusing. In the poof of Euclidean algorithm he uses the idea that m divides linear combination of a and b, and then proves that m divides linear combination of a and b with Euclid algorithms. Also statement of invariants feels a bit out of the blue. And jug example feels unconcluded. Should have started the lecture with divisibility if the rest relies so heavily on it. Thanks for uploading the lectures, obviously, but I hope the rest of course is as great as the beginning. Coz I kind of had my hopes up.
I agree, there is lot of dependent stuff pushed into 1 lecture
I feel so sorry for his students, pay a huge amount of money and get like this teacher, he is teaching himself not students, at least I'm watching this for free which is good, luckily there is a textbook, 100 times better than this "lecture". But again MIT OpenCourseWare thank you so much for uploading the course you are the best.
Guys help me out here. This lecture doesn't dazzle me unlike the two previous lectures because:
a) Something wrong with the lecturer
b) Something wrong with the lecture
c) I already know where this is going (groups, rings, fields, cryptography, RSA).
d) The combination of above?
I don't wanna sound like an ass, I just don't understand why does it take me 3 days to get through this lecture?
It happens to all of us brah we are not Euler, Erdos or Tao. So be patient and as they say "keep calm and carry on".
Adam M. Yeah but it shouldn't happen when you're the smartest man in the universe Rick Sanchez
int the multiverse lol
Am I the only one who read his comment with Rick's voice ?
Pulverizer sounds like villain from a super hero movie
❤️💜❤️
I am so confused