I have spent two days to find the right video to understand!!!! Thank you sooooooooooooo much! Keep posting such informative and comprehensive videos! You are a LEGEND!
everybody teaches math a different way and everyone learns math a different way. Anyone can learn math but everyone has a specific way in which they will fully understand the material as an overarching package. that completely depends on how and when different concepts are introduced during the course of the learning process. This might not work for everyone but it definitely introduces things in a way that is easy and simple to understand. sometime some people teaching a way where things are thrown in in an as-a-matter-of-fact way. I have tried learning through my professor and while i do understand some concepts i dont understand the material as an overarching package. An example is U(x). it was thrown in in my class and given a place in groups but i dint understand where it came from and what it meant for the bigger picture but the way you introduced it so organically was amazing almost like a brain ninja. i found myself seeing the justification of the introduction of this new concept into the world of groups and abstract algebra as a whole and by god it is beautiful to see. it brings me back to joy of learning calculus from different angle and understanding that they are not different forms of calc but the same picture seen through different lenses. if you ever find yourself having trouble with a math subject its not that its not for you its that you havent found someone who unravels it in a way where its a seamless path of understanding instead of facts and definitions thrown at you and expected for you to connect the dots. great series man!
How to show that there exists a* ∈ U(n) which was left out for us to do in 13:55 (here I wrote a* instead of b as the element denoting the inverse of "a") We need to show: 1 ≤ a* < n and gcd(a*,n)=1 First, show that 1 ≤ a* < n : (following from the video's proof about the existence of the integer b ∈ ℤ) Since a, n ∈ ℤ are given, by the Division Algorithm, there exist unique integers q and a* such that b=qn+a* with 1 ≤ a* < n. Here we can easily see that a* ≡ b (mod n) Second, show that gcd(a*,n)=1 Suppose not. Let gcd(a*,n) = d > 1 There exist integers m and f such that n=md and a*=fd. (following from the video's statement that n ❘ ab-1, ) since d ❘ n , we see that d ❘ ab-1 , so it should be the case that ab≡1(mod d). However, ab = a(qn+a*) = a(q(md)+fd) = (aqm+af)d ≡ 0 (mod n) CONTRADICTION
Anyone can help me, to find what are the binary properties in the set. 🥺🥺🥺 I'm confused where to find it. Thank you so much. Eventually, I'm answering mod 7
I have spent two days to find the right video to understand!!!! Thank you sooooooooooooo much! Keep posting such informative and comprehensive videos! You are a LEGEND!
We are blessed that we have youtube as our one tutor❤️
i came with the doubt. clearly this video clears my doubt, thank you sir
everybody teaches math a different way and everyone learns math a different way. Anyone can learn math but everyone has a specific way in which they will fully understand the material as an overarching package. that completely depends on how and when different concepts are introduced during the course of the learning process. This might not work for everyone but it definitely introduces things in a way that is easy and simple to understand. sometime some people teaching a way where things are thrown in in an as-a-matter-of-fact way. I have tried learning through my professor and while i do understand some concepts i dont understand the material as an overarching package. An example is U(x). it was thrown in in my class and given a place in groups but i dint understand where it came from and what it meant for the bigger picture but the way you introduced it so organically was amazing almost like a brain ninja. i found myself seeing the justification of the introduction of this new concept into the world of groups and abstract algebra as a whole and by god it is beautiful to see. it brings me back to joy of learning calculus from different angle and understanding that they are not different forms of calc but the same picture seen through different lenses. if you ever find yourself having trouble with a math subject its not that its not for you its that you havent found someone who unravels it in a way where its a seamless path of understanding instead of facts and definitions thrown at you and expected for you to connect the dots. great series man!
you just made me realize something that i will type the whole page, it's also about the understanding, this series also cleared everything to me too
Thank you so much for your clarity, it is so helpful!
using a simple example of urs i was able to understand in 5 mins what i have been trying hard to study for a week thanks
Units modulo n? More like "Ultimately magnificent lecture; thanks again!" 👍
How to show that there exists a* ∈ U(n) which was left out for us to do in 13:55
(here I wrote a* instead of b as the element denoting the inverse of "a")
We need to show: 1 ≤ a* < n and gcd(a*,n)=1
First, show that 1 ≤ a* < n :
(following from the video's proof about the existence of the integer b ∈ ℤ)
Since a, n ∈ ℤ are given, by the Division Algorithm, there exist unique integers q and a* such that b=qn+a* with 1 ≤ a* < n.
Here we can easily see that a* ≡ b (mod n)
Second, show that gcd(a*,n)=1
Suppose not. Let gcd(a*,n) = d > 1
There exist integers m and f such that n=md and a*=fd.
(following from the video's statement that n ❘ ab-1, ) since d ❘ n , we see that d ❘ ab-1 , so it should be the case that ab≡1(mod d).
However, ab = a(qn+a*) = a(q(md)+fd) = (aqm+af)d ≡ 0 (mod n) CONTRADICTION
so clear! Thanks a lot!
clear and to the point.. very nice
Thank you sir , it is easy to understandable ✨✨
videos are awesome bro😃
you are such a brilliant........
You are brilliant :) !!
Thank you!!
Does this theorem only for multiplication? Not for addition?
LOVE YOU BROOOOO
This was a very helpful video, thank you very much you did help a lot. If you ever need help, let me know. I give some good hand job.
####Excellent superb sir.
Will u pls tell me a Normal group defination in one meaningful line?
A group is a set of objects under a single operation that satisfies the axioms of closure, associativity, identities, and inverses.
Thank you so much Sir!!!!
very good,,very understandable
five bloody stars! I finally understand!!!!
Thank you sir
what are the subgroups of multiplication modulo 11?
The trivial group and Z11 itself. There are no others since 11 is prime.
How do we know b belongs to U?
thank you very much sir :)
Thanks 👍
2 is prime number so hw can we generlise tht prime numbers can included in this group
but 2 and 8 are not relatively prime
Sir, please solve this in video.find the order of each element in the group z5 - {[O]}, . 5
Thank you sir...
Im not clear how to find inverse of 1357..?
look at what multiplies those numbers and gives you back the identity, in this case 1. So 3 * 3 = 9, 9 mod 8 = 1. Therefore the inverse of 3 is 3
Anyone can help me, to find what are the binary properties in the set. 🥺🥺🥺 I'm confused where to find it. Thank you so much.
Eventually, I'm answering mod 7
Binary properties. ..
Is it union intersection , subtraction ?
😮
Is it applicable only for multiplication??
U(8) concept
You can do it for addition but there’s a slightly different approach
But you haven't proved that gcd(b,n)=1 while finding inverse
He left that part to us
Units mod n has already excluded any b that does not satisfy this
Tq sir it so help ful to understand multiplicative madulo
Sir ji hidi se bataiye
Thank you sir