23:17 End of the proof left for the viewer 25:32 Homework 25:35 It should be 111 ≡ 0 (mod 37) 26:26 Good Place To Stop
@terryendicott29393 жыл бұрын
Three times 3 is nine and seven times 3 ends in one. You're just being picky.
@crazy4hitman7553 жыл бұрын
@@terryendicott2939 😂what
@izziek862 жыл бұрын
@@terryendicott2939 accuracy is not the same as pickiness in maths. 91 is not congruent to 0 modulo 37. If someone were to try to show that it is then they would have an awful lot of trouble since it is incongruent. A proof is a proof, not an approximation.
@infopek32212 жыл бұрын
At 23:17, proof of the following statement: If a ≡ b (mod n), then a^m ≡ b^m (mod n), m∈ℕ Suppose a ≡ b (mod n) ⇒ a - b = nk, for some k∈ℤ ⇒ a = nk + b ⇒ a^m = (nk + b)^m Using the binomial expansion, we get a^m = (m 0)*(nk)^m + (m 1)*(nk)^(m-1)*b + … + (m m)*b^m, where (x y) means x choose y. If we reduce the equation mod n, all of the terms on the RHS reduce to (m m)*b^m (mod n), since every term containing an n is ≡ 0 (mod n). Also in the video it is proven that if a ≡ b (mod n) and c ≡ d (mod n), then a + b ≡ c + d (mod n). Using this fact, we get that a^m ≡ (m m)*b^m (mod n), but we know (m m) = 1 ⇒ a^m ≡ b^m (mod n). Q.E.D. Please correct me if my proof is wrong.
@ZEPHYRZHANG-mg8ziАй бұрын
he proved the multiplication though so exponentiation is just doing that a bunch of times if the exponent is a natural number
@ZEPHYRZHANG-mg8ziАй бұрын
Idk
@georgesadler78302 жыл бұрын
Professor Michael Penn, this is a solid and well-organized video/lecture on Modulo Arithmetic in Number Theory Eight. These mathematical topics has been around forever.
@erwinfinder91823 жыл бұрын
you're my new teacher and figurative father figure.
@MultiCarlio3 жыл бұрын
Idem
@anshumanagrawal3463 жыл бұрын
fff
@anshumanagrawal3463 жыл бұрын
Thank you so much for making this, I like number theory problems a lot, I watch a lot of those you post on your channel, and while I was able to understand the concept of modular arithmetic by watching those videos mostly some things I always wondered about. I was hoping you would make a video explaining all this
@elihowitt41073 жыл бұрын
Hi Michael, great series. I was recently studying number theory independently (with the help of your old series among other things) and I came across a beautiful fact I suggest you include in the current series : the equivalence of CRT and Lagrange interpolation. #suggestion
@thatkindcoder75102 жыл бұрын
Does CRT mean the Chinese remainder theorem? If so, how is it equivalent to Lagrange interpolation? That sounds cool
@agrimmittal Жыл бұрын
Bro's a magician when it comes to explaining mathematics to dumb kids like me.
@murali96493 жыл бұрын
Wish you speedy recovery from the ailment and surgery Mike. With the best wishes from 🇮🇳 India. All the best for your dedication to Maths and teaching skills.
@weisanpang71739 ай бұрын
The property that ab (mod n) is congruent to ( a mod n) (b mod n) is widely used, but not mentioned in your series.
@kevinmartin77603 жыл бұрын
I always find it incongruous (pardon the pun) that when using the congruence relation one always states the modulus explicitly pretty much each time, but when using the [x] equivalence class notation, the modulus is rarely stated except as an overall context to a whole bunch of math, rather than, say, having it be a subscript to the [x] syntax e.g. [x]_n.
@anshumanagrawal3463 жыл бұрын
The 1st question of the warm up is wrong, I think.
@thomasdemilio616410 ай бұрын
yes, but I think it's wrong on purpose. There should be said "state wether the followings are true"
@bosorot3 жыл бұрын
last question . I got = 5 mod (7) and 13 mod(25)
@anshumanagrawal3463 жыл бұрын
Me too
@kamilkostya86002 жыл бұрын
+1
@thomasdemilio616410 ай бұрын
great job! That's a really nice property :)
@mahes92633 жыл бұрын
Can you proof that every combination and permutation is an integer?
@nowshad8432 жыл бұрын
I like the art on his t-shirt, we really need someone who will save the earth from anyone who will harm it.
@khaledchatah34253 жыл бұрын
We could go much further and give examples about cyclic groups and such structures.
@karl1310582 жыл бұрын
Wouldn't the "equivalence relation" thing be much shorter by using the "congruent when having the same remainder when dividing" proven just before?! Any relation based on "having the same something" is an equivalence relation because "being the same" is one, trivially!
@user-pw6qe7ur4q3 жыл бұрын
What did you do to your arm?
@oida100003 жыл бұрын
27 equiv 5 (mod n) --> 22 equiv 0 (mod n) --> n | 22 ---> n in {1, 2, 11, 22} But we can ignore 1 and 2 as the have no remainder 5. So n=11 or 22.
@Zoulou6023 жыл бұрын
Can’t ignore 1 and 2 since the presence of remainder is not a requirement for congruence
@oida100003 жыл бұрын
@@Zoulou602 true
@juanmolinas3 жыл бұрын
Greetings proffesor, great class!..
@christophersedlak11472 жыл бұрын
thanks!
@dipankarsharma89892 жыл бұрын
Sir for warm up exercise answer for question 2nd is n=2and 11 right???
@thomasdemilio616410 ай бұрын
well that should be all the divisors of 22, therefore 1,2,11 and 22
@koenth23593 жыл бұрын
Warning: listening to this with headphones on causes motion sickness.
@megauser85123 жыл бұрын
lol
@mohammadsalehi7653 жыл бұрын
What happened to your hand? :(
@goodplacetostop29733 жыл бұрын
Dupuytren's contracture surgery
@samharper58813 жыл бұрын
He was falling and saw the ground rushing towards him and thought that's not a good place to stop so put his hand out and it wasn't great.
@anshumanagrawal3463 жыл бұрын
Some conjecture
@anshumanagrawal3463 жыл бұрын
@@samharper5881 😂😂
@mohammadsalehi7653 жыл бұрын
@@goodplacetostop2973 Feel better soon
@a_llama3 жыл бұрын
oof hope your hand is ok!
@arsenzatikyan3 жыл бұрын
Your lectures and problems always interesting, inspiring, rigorously proven, very useful, informative, beautiful and pleasantly unpredictable.
@noelani9763 жыл бұрын
Watching from Nigeria!
@adrianamor84723 жыл бұрын
For the last exercise, you need to calculate some successive factorials mod n, that's fairly easy to do, considering that k! is congruent to k (k-1)! (mod n), so you just take the previous factorial that you already calculated mod n and multiply by k, and try to reduce large numbers to their negative (mod n) which will make it much easier. For example 3! + 4! = 6 + 4! ≡ -1 - 4 (mod 7) ≡ 2 (mod 7)
@milanrebic96503 жыл бұрын
MMO 1988 GB?
@batner3 жыл бұрын
16:35 it is unclear how the set {-2,-1,0,1} give enough remainders mod 4 to cover the integers. The set of [3] is missing.
@whiskymac31033 жыл бұрын
-1(mod 4) is congruent to 3(mod 4). The set covers the integers if there are 4 terms and the terms are congruent to 0(mod 4), 1(mod 4), 2(mod 4), and 3(mod 4)
@batner3 жыл бұрын
@@whiskymac3103 Thanks. I can see now that by definition its completely correct -1-3=-4 (divisible by 4) this -1 congruent to 3 mod 4. What got me confused is that I don't understand modulo with negatives.
@cinemonini3 жыл бұрын
Hi why it has so less views
@iooooooo13 жыл бұрын
I think it might not have been public. I just got a notification now, two days after you wrote this comment.
@cinemonini3 жыл бұрын
@@iooooooo1 ohhh
@cinemonini3 жыл бұрын
@@iooooooo1 you can get every video in a playlist of number theory
@shreyan13623 жыл бұрын
@@cinemonini those are unlisted but can be seen thru the playlists but people usually don't look up for the playlists so yeah less views until they are listed
@agrimmittal Жыл бұрын
Because this is not what usually kids watch..?
@onceuponfewtime3 жыл бұрын
nice
@i18nGuy3 жыл бұрын
around 20:50 you claim 7=3(mod 4) and 2 divides 4 so 7=3(mod 2). However, 3 can't be mod 2 since it can only be 0 or 1 and 7 = 1(mod 2). Of course we can go a further step and reduce 3(mod 2) to 1 (mod 2). d dividing n partitions the remainders into larger equivalence classes. e.g. 5 = 1(mod 4) but is also 1 (mod 2) so 5 is then equivalent to 7. Anyway, seems odd (no pun intended) that you just left it as 3(mod 2) w/o further comment.
@iabervon3 жыл бұрын
You're confusing the remainder from the division algorithm with congruence (mod n). Computer programming calls the operator the produces the remainder "mod", but the math notation is about two numbers having the same remainder, and is not an operator at all: you put (mod n) at the right end of a line to say that all your triple equals signs on that line mean that the two sides have the same remainder from division by n.
@i18nGuy3 жыл бұрын
@@iabervon ok thanks. Yes it is about congruence. My bad.