@BruceLCM Not all functions are bijective, and not all functions are injective, nor surjective. It is your choosing. But if the sets have the same cardinality, a bijection is possible by definition. I think we are on the same page :).
The video said it is always "possible" to make a one-to-one function when the domain and range have the same number of elements but not "necessarily" all functions are one-to-one.
I thought the same before. What u said is so true! Not all the elements in the range of a function have a corresponding value from the domain. In the example given, A and B have the same number of elements. Yes, if we assign them to a function randomly, f: A->B may not be one-to-one, but we can deliberately make it! Say if A={1,2,3,4};B={5,6,7,8}. I can make f(1) = 5, f(2) = 8, f(3) = 7, f(4) = 6. Now, B is entirely covered under f: A->B.
You explain surjective and injective functions from A to B as having, respectively, these properties:
1. B is "covered" by the mapping, in that any element you can find in B is mapped by at least one element in A, possibly more.
2. If we find that two variables representing elements in A, say a & b, are mapped such that f(a) = f(b), we can only conclude that a and b actually represent one element. That is, a = b.
Hi verity seeker, in your illustration of the onto function between {a,b,c,d,e} and {s,t,u,v,w}, the w is missing. Perhaps it sneaked off. Some of the meaning of the onto function might be lost due to w's untimely absense.
Keep up the good work. I am enjoying watching the videos. Cool music. Is there something subliminal going on? I feel myself getting smarter....
Hi, a comment. You state that if two groups are the same then the same binary operator should behave the same way in both groups. You gave a formula expressing this as f (a*b) = f(a)*(b). I can't see why if your intent is to see if operator behaves the same you would do a different thing between the groups. Wouldn't you want to test if f (a*b) in group F = f (A*B) in group G? Now you will see if the same operation gives the same result in two groups?
Hi, this is a good question. I have to ask, though, what are the elements A and B in your question? I think I know what you mean, and I will make an attempt to explain. If I misunderstood you, just ask again.
Ok, here it goes: I have two groups G and G', and a homomorphism f from G to G'. The function f is a correspondence between the groups, so if a is in G, then f(a) is an element in G'. I want to show that f(a)*f(b), which are an element in G' corresponds to a*b in G VIA the function f.
That would mean that f(a*b) = f(a)*f(b), where the first * is the operation in G and the second * is the operation in G'. In your question, however, the operation * is in the group G, and it doesn't involve the operation in G' at all. So what I really want to test is:
Take two elements in G, say a and b and form the element a*b. Now I want to check that the corresponding element to a, namely f(a) multiplied with the corresponding element of b, namely f(b) is the element corresponding to a*b, namely f(a*b). In other words: f(a*b) = f(a)*f(b).
f sends an element from G to G'. But in your example, the elements f(a*b) and F=f(A*B) are bot elements in G'. a,b,A and B are all elements in G for it to make any sense.
Amazing video ! we learned Izomorphizem in class today , and this video made it easier to understand and fun ( with the music and all ). Gee ya think my Proffessor would allow music in class ? :)
Sorry for not answering until now. I would think most basic abstract algebra books have a natural connection to linear algebra. I can recommend "Basic Abstract Algebra" by Bhattacharya, Jain and Nagpaul. There is no direct connection, since group theory is a whole new subject, but linear algebra provides many nice examples, so is required as basic knowledge.
Thank you for taking the time out to make these videos. They tighten up my basic understanding of the concepts. I'm studying finite albelian groups right now and it''s easy to get stuck under all the language. This brings me back on top. Thanks again
Some people are marking comments as spam, and I am sorry about that. I did not notice your comment until now. Thank you. Feel free to ask if you have questions about the topics.
Up till this video, I considered myself more of a calculus/analytical/geometry kind of person, but ... this is just beautiful. Isomorphisms-- wow. It's a whole new way of thinking that opens up so many beautiful new possibilities!
I love abstract algebra, it's like looking for (& finding) crystals in everyday life. (The crystalline reference is because of the algebraic structures--which form crystalline bases in my mind) Thanks for doing this.
I want to create a crazy group. I always made an analogy of creating a group of people, where the BO is "a punch in the arm" or "kick in the leg." I just toy with the idea, but I think it demonstrates the awesomeness of REAL algebra. 8D
No, it is not an equal sign. The groups are not equal, but they are isomorphic - i.e. algebraically equivalent. You can have two groups that are isomorphic but having different elements. An equal sign would mean that their elements are and are called the same, and that the binary operator is the same. You can have two groups with different elements and different binary operators, but still being isomorphic.
Thanks for the clarification. Apparently, if two groups are isomorphic, then the symbols that represent them behave the same; however, the elements and binary operations of the two groups can be very different.
Part 4 was blocked because of some copyright EMI content violation. Please remove it and reload asap.
Kurtlane 2 weeks ago in playlist Basic abstract algebra
I love this series. I'm a music student studying mathematics and music theory, and I want to beef up my math background.
To clarify:
a "binary operation" is the operation that governs a group G, correct? That is, the operation * such that a*b = c, where a, b and c are elements of G?
Also, I would really like to see Part 4, but it's been blocked by EMI "due to copyright issues." Could you please repost it sans song?
Yaldatchayil 2 months ago in playlist VeritySeeker
what happened to video 4?
Tarkvinius 2 months ago
oh, EMI has blocked it. looks like the song you used was copyrighted. could you repost it without the song, please?
Tarkvinius 2 months ago
Why can't we view video #4
screamformealan 5 months ago in playlist More videos from VeritySeeker
sorry, i mean bijective instead of one-to-one
BruceLCM 6 months ago
@BruceLCM Not all functions are bijective, and not all functions are injective, nor surjective. It is your choosing. But if the sets have the same cardinality, a bijection is possible by definition. I think we are on the same page :).
VeritySeeker 5 months ago
The video said it is always "possible" to make a one-to-one function when the domain and range have the same number of elements but not "necessarily" all functions are one-to-one.
BruceLCM 6 months ago
...but to me, this doesn't preclude the possibility of extra elements hanging around in A that don't get mapped to B at all.
My guess here is that the terminology f: A -> B implicitly maps all of A, but doesn't necessarily cover B. So if we have some function:
f: Integers between 4 and 1089 -> Users on youtube
then we expect f to have a value for any such integer, but we needn't expect the image of f to contain every user on youtube. Am I right?
SirBastian200 7 months ago
I thought the same before. What u said is so true! Not all the elements in the range of a function have a corresponding value from the domain. In the example given, A and B have the same number of elements. Yes, if we assign them to a function randomly, f: A->B may not be one-to-one, but we can deliberately make it! Say if A={1,2,3,4};B={5,6,7,8}. I can make f(1) = 5, f(2) = 8, f(3) = 7, f(4) = 6. Now, B is entirely covered under f: A->B.
BruceLCM 6 months ago
These videos are great. I have a question though:
You explain surjective and injective functions from A to B as having, respectively, these properties:
1. B is "covered" by the mapping, in that any element you can find in B is mapped by at least one element in A, possibly more.
2. If we find that two variables representing elements in A, say a & b, are mapped such that f(a) = f(b), we can only conclude that a and b actually represent one element. That is, a = b.
...
SirBastian200 7 months ago
great videos thanks
pbajaj5 7 months ago
whats the name of this song????????
rryking 1 year ago
Thanks and keep it up. This will help me get through a Modern Algebra class.
cazro 1 year ago
Hi verity seeker, in your illustration of the onto function between {a,b,c,d,e} and {s,t,u,v,w}, the w is missing. Perhaps it sneaked off. Some of the meaning of the onto function might be lost due to w's untimely absense.
Keep up the good work. I am enjoying watching the videos. Cool music. Is there something subliminal going on? I feel myself getting smarter....
fergalegallagher 1 year ago
precise and clear
jdra999 1 year ago
The background song is EXTREMELY weird...love it! What's its name? Great videos btw.
giovanni9107 2 years ago
Hi, a comment. You state that if two groups are the same then the same binary operator should behave the same way in both groups. You gave a formula expressing this as f (a*b) = f(a)*(b). I can't see why if your intent is to see if operator behaves the same you would do a different thing between the groups. Wouldn't you want to test if f (a*b) in group F = f (A*B) in group G? Now you will see if the same operation gives the same result in two groups?
voluptuate 2 years ago
Hi, this is a good question. I have to ask, though, what are the elements A and B in your question? I think I know what you mean, and I will make an attempt to explain. If I misunderstood you, just ask again.
Ok, here it goes: I have two groups G and G', and a homomorphism f from G to G'. The function f is a correspondence between the groups, so if a is in G, then f(a) is an element in G'. I want to show that f(a)*f(b), which are an element in G' corresponds to a*b in G VIA the function f.
VeritySeeker 2 years ago
That would mean that f(a*b) = f(a)*f(b), where the first * is the operation in G and the second * is the operation in G'. In your question, however, the operation * is in the group G, and it doesn't involve the operation in G' at all. So what I really want to test is:
VeritySeeker 2 years ago
Take two elements in G, say a and b and form the element a*b. Now I want to check that the corresponding element to a, namely f(a) multiplied with the corresponding element of b, namely f(b) is the element corresponding to a*b, namely f(a*b). In other words: f(a*b) = f(a)*f(b).
f sends an element from G to G'. But in your example, the elements f(a*b) and F=f(A*B) are bot elements in G'. a,b,A and B are all elements in G for it to make any sense.
I hope this was clear. If not, just ask.
VeritySeeker 2 years ago
nice explanation and easy to follow. why is our prof making it more harder for us to understand?
sionmejia 2 years ago
These videos are extremely helpful! Thank you so much!
magestaff567 2 years ago
Amazing video ! we learned Izomorphizem in class today , and this video made it easier to understand and fun ( with the music and all ). Gee ya think my Proffessor would allow music in class ? :)
Lorri279 2 years ago
thanks a lot dude! well done~
xinliw 2 years ago
Really brilliant and useful. I only wish you could include the name of the bands and songs you use in the description.
ballantrae101 2 years ago
Btw thanks for these video's.
oryxfreeride 2 years ago
do u recommend a self study textbook that ties abstract algebra with linear algebra?
jgirken 2 years ago
Sorry for not answering until now. I would think most basic abstract algebra books have a natural connection to linear algebra. I can recommend "Basic Abstract Algebra" by Bhattacharya, Jain and Nagpaul. There is no direct connection, since group theory is a whole new subject, but linear algebra provides many nice examples, so is required as basic knowledge.
VeritySeeker 2 years ago
so what's math all about? learning other people's notation it seems
there must be a better way
Nanumir 2 years ago
Could be.
VeritySeeker 2 years ago
Thank you for taking the time out to make these videos. They tighten up my basic understanding of the concepts. I'm studying finite albelian groups right now and it''s easy to get stuck under all the language. This brings me back on top. Thanks again
zxtonizx 2 years ago 5
Some people are marking comments as spam, and I am sorry about that. I did not notice your comment until now. Thank you. Feel free to ask if you have questions about the topics.
VeritySeeker 2 years ago
Up till this video, I considered myself more of a calculus/analytical/geometry kind of person, but ... this is just beautiful. Isomorphisms-- wow. It's a whole new way of thinking that opens up so many beautiful new possibilities!
pyrofreakingmaniac 2 years ago 4
very helpful.. thank you!
vee3vee3 3 years ago 3
for 2:44, you define a function with two properties, I thought if you said. for part 2.
f(a)=f(c) then, a=c.,it could be better. Just tiny bit confusing with part a. but, i guess that's okay.
claslow 3 years ago
You mean I should use a different letter? Yeah, you may be right about that.
VeritySeeker 3 years ago
I love abstract algebra, it's like looking for (& finding) crystals in everyday life. (The crystalline reference is because of the algebraic structures--which form crystalline bases in my mind) Thanks for doing this.
I want to create a crazy group. I always made an analogy of creating a group of people, where the BO is "a punch in the arm" or "kick in the leg." I just toy with the idea, but I think it demonstrates the awesomeness of REAL algebra. 8D
Chaos7703 3 years ago
So an isomorphism between two groups is essentially an equal sign between those groups.
nemo1620 3 years ago
No, it is not an equal sign. The groups are not equal, but they are isomorphic - i.e. algebraically equivalent. You can have two groups that are isomorphic but having different elements. An equal sign would mean that their elements are and are called the same, and that the binary operator is the same. You can have two groups with different elements and different binary operators, but still being isomorphic.
VeritySeeker 3 years ago
Thanks for the clarification. Apparently, if two groups are isomorphic, then the symbols that represent them behave the same; however, the elements and binary operations of the two groups can be very different.
nemo1620 3 years ago
Really well explained.
Another soon I hope.
lolathedog69 3 years ago
Next video is out now. I've been having computer problems, but I was able to borrow a computer. Thanks for watching.
VeritySeeker 3 years ago
your videos have helped me learn abstract algebra so well thank you for making these videos
MathMikie 3 years ago
I am glad to hear the videos can help you. It is an interesting subject with lots of application. Good luck with your studies.
VeritySeeker 3 years ago
Good video, please continue I enjoy reading them.
checkmatebtch 3 years ago
Thank for for commenting and watching. Next one is out.
VeritySeeker 3 years ago