5/6 = 0 rem. 5 yes, but does -1/6 simply have remainder 5? (I am taking nothing for granted here!) I think when dealing with sets generated by modulus arithmetic there is an issue of |n| to be addressed, or have I missed something?
These videos are one of youtube's best kept secrets. Every abstract algebra student should watch your videos! Keep up the great work! The world needs more people like you! Thank you so much!
Love these vid clips. Please, keep them coming. :thumbup:
I know that these vid clips are overviews of fundamental concepts in Basic Algebra. Have you thought about vid clips that actually walk through specific proofs? One vid clip per proof?
Will you be covering Rings, Fields, Galois Theory? I HOPE SO! :thumbup:
I am not sure how far I will push this basic abstract algebra series, but maybe I will do a series with more advanced topics. If I have an audience, I guess I will continue to define rings and fields too.
I have thought about making videos with specific proofs, yes. I have done a few, but I guess as the topic gets more advanced, I have to dedicate a whole video to certain proofs.
If you know what an isomorphism is (see earlier videos), then an automorphism is nothing more than an isomorphism from a group to itself. If G is a group, then f:G -> G is an automorphism on G if f is an isomorphism.
For your last question, the answer is yes. Eulers formula gives: n^i = e^(i ln(n)) = cos(ln(n)) + i sin(ln(n)). The first equality is by definition, and the second is by Eulers formula.
I hope I wouldn't be too much of a bother if I asked another question.
Is a ring just a field with a multiplicative identity or is there more than that? If so, I would like to know, but I can wait if we are covering it soon.
Just ask. That is why I'm here. It's my pleasure to see people interested in the topic. :)
I think you got the definitions of rings and fields mixed up a little. A ring is not necessarily a field, but a field is always a ring.
A ring has addition and multiplication, and it satisfies certain axioms.
A field is a ring where every element (except 0) has a multiplicative inverse. Multiplication is also commutative (a*b = b*a) in a field, which do not need to be true for rings in general.
Thank you so very much!!! I really do appreciate all of your help. I also would like to apologize for not knowing too much about the subject when I am asking so many questions about it. I probably will ask more questions in the future, but I'll try not to overly disturb you until the next video.
I thought you talked about analysis, but you mean further algebra? I guess you'll be working with rings and fields and possibly modules. Maybe some Galois theory.
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Normally I've got the attention span of a goldfish, but I find myself on Part 12.** yes that's how i feel love the BR music
MrJoshualizardi 3 weeks ago in playlist More videos from VeritySeeker
Normally I've got the attention span of a goldfish, but I find myself on Part 12.
Emperorlawson 2 months ago
Cool
But you are still writing 7! and 1! (seven factorial and one factorial)
Why conflate the issue with ambiguous notation?
TraleeFair 8 months ago
@TraleeFair Nah, I am just being way too excited!!!! :)
VeritySeeker 3 months ago
5/6 = 0 rem. 5 yes, but does -1/6 simply have remainder 5? (I am taking nothing for granted here!) I think when dealing with sets generated by modulus arithmetic there is an issue of |n| to be addressed, or have I missed something?
farfromtheland 1 year ago
@farfromtheland -1 is congruent to 5 modulo 6, yes. Congruence is also defined for negative numbers.
VeritySeeker 1 year ago
Thnx! Your videos are great (and music too ;)
mailoisback 1 year ago
These videos are one of youtube's best kept secrets. Every abstract algebra student should watch your videos! Keep up the great work! The world needs more people like you! Thank you so much!
magestaff567 2 years ago 2
Please do include more advanced stuff like rings and fields, I really love these clips! TKS~
xinliw 2 years ago
I would be so happy if you made more of these videos! :)
UndeadTheta 2 years ago 3
Don't leave all the other truth seekers hanging, we need more!
humby123 2 years ago 3
Sorry to bother you again, but I was wondering if you would consider making a similar series on Non-Euclidean Geometry.
Sorry if I am over-commenting on 1 video.
"I want to know God's thoughts... the rest are details." -Albert Einstein.
kirbylassie 2 years ago
Love these vid clips. Please, keep them coming. :thumbup:
I know that these vid clips are overviews of fundamental concepts in Basic Algebra. Have you thought about vid clips that actually walk through specific proofs? One vid clip per proof?
Will you be covering Rings, Fields, Galois Theory? I HOPE SO! :thumbup:
openuniverse2003 2 years ago
Thank you.
I am not sure how far I will push this basic abstract algebra series, but maybe I will do a series with more advanced topics. If I have an audience, I guess I will continue to define rings and fields too.
I have thought about making videos with specific proofs, yes. I have done a few, but I guess as the topic gets more advanced, I have to dedicate a whole video to certain proofs.
VeritySeeker 2 years ago
At 4:13, I think you mean {-11, -5, 1, 7, 13,...}. Sorry to bother you, but I was just thinking about this.
Thanks again!
kirbylassie 2 years ago
You are absolutely right. Thanks for pointing it out.
VeritySeeker 2 years ago
I'm a Sophomore in high school, but I'm sort of good at math. Your videos are wonderful. By the way, would the answer be 0?
Also, would you be willing to tell me what an automorphism is?
Sorry to bother you, but I do have one more question. Would n (a number) to the i power be cos(ln(n))+isin(ln(n))?
I value your knowledge.
kirbylassie 2 years ago
Hi, yes the answer is 0.
If you know what an isomorphism is (see earlier videos), then an automorphism is nothing more than an isomorphism from a group to itself. If G is a group, then f:G -> G is an automorphism on G if f is an isomorphism.
For your last question, the answer is yes. Eulers formula gives: n^i = e^(i ln(n)) = cos(ln(n)) + i sin(ln(n)). The first equality is by definition, and the second is by Eulers formula.
Hope that helps. Feel free to ask. :)
VeritySeeker 2 years ago
Thank you very much.
I hope I wouldn't be too much of a bother if I asked another question.
Is a ring just a field with a multiplicative identity or is there more than that? If so, I would like to know, but I can wait if we are covering it soon.
kirbylassie 2 years ago
Just ask. That is why I'm here. It's my pleasure to see people interested in the topic. :)
I think you got the definitions of rings and fields mixed up a little. A ring is not necessarily a field, but a field is always a ring.
A ring has addition and multiplication, and it satisfies certain axioms.
A field is a ring where every element (except 0) has a multiplicative inverse. Multiplication is also commutative (a*b = b*a) in a field, which do not need to be true for rings in general.
VeritySeeker 2 years ago
Thank you so very much!!! I really do appreciate all of your help. I also would like to apologize for not knowing too much about the subject when I am asking so many questions about it. I probably will ask more questions in the future, but I'll try not to overly disturb you until the next video.
Thanks again :)
kirbylassie 2 years ago
I thought you talked about analysis, but you mean further algebra? I guess you'll be working with rings and fields and possibly modules. Maybe some Galois theory.
VeritySeeker 2 years ago
That is difficult to answer. What topics have been covered earlier? Did you learn about multivariable analysis?
VeritySeeker 2 years ago
When are we getting a next one!
BrianDiehr 2 years ago
Good vid. look forward to the next.
britishguy444 2 years ago
(First :-P) Nom-nom-nom. Delicious algebra.
eedahl 2 years ago