these are really benificient videos for all standard but in part 1 once binary operation is defined there is no need to state the first condition .if u find any mathematical mistake in my text pls. inform me. thanx
@TradingTutor In tis case the binary operator + has a meaning. It is ordinary addition. All other binary operations are called * (unless otherwise defined).
You're better than my math teacher. He makes the subject super hard and I've only watched 2 of your videos and understand what you are talking about. Thank you
thank you ... its very nice you re doing these videos because there is not a lot of material on the web on group theory and abstract algebra and if there is its often too abstract and formal:)
you defined binary operator incorrectly.here's what it actually is:
A binary operator * on a set S is a relation from S X S to S. you should have defined a relation from set A to set B also.if it was a function then it would be closure of S would immediately follow. and hence the definition of closure would be unnecessary. which it is not. think about it.
@RurouniDizeru A binary operator on a set S, is a function from the cartesian product of S to S. For every element in SxS, there is a unique element in S. Of course, all functions are also relations.
@RurouniDizeru I think I see what you mean now. You mean that since I use the function-definition, I do not need to include closure in the definition of a group? If that is the case, I agree. It is not necessary to include closure, since the binary operator is a function. If I defined it as a relation, I would have to include closure. The reason for defining it like this, is because many students forget to check the binary operation. The way I defined it is anyway well-defined.
However... your examples for "binary" operator are at best confusing.
Readers should be aware that such an operator is termed BINARY because it takes exactly TWO parameters as input and returns a single result. This point does not come across quite so clearly after the definition is given.
However... your definition of "binary" operator is at best confusing.
Readers should be aware that such an operator is termed BINARY because it takes exactly TWO parameters as input and returns a single result. This point does not come across quite so clearly.
These are awesome videos, thank you for posting them. I'm working on a paper/presentation on Galois Theory and the insolubility of the quintic right now and I have to say that you've given one of the clearest, most readable definitions of a group that I've come across. Keep up the good work.
This is great! After reading through pages of theorys and proofs and getting no understanding of whats going on, this video has blown my mind. And I like the music it makes it seem intense, if your one of those people who needs a library to study just press the mute button.
Brilliant. I find videos like these are FAR AND AWAY more educational than any class with a forlorn professor and hiked fees. Keep it up. Yours are some of the few videos that deserve a rightful place in youtube.
Isn't the '1)' of the definition of group unnecessary? If you already said that there's a binary operation over G called *, so it's not necessary to mention that 'if a,b are in G then a*b is in G too', since * is a function from G to G.
i think that eventually * being closed is assumed because it does become redundant, however there are some functions that are not closed and the answer to a*b is not in G. as shown with 2*x=3, if the operation is multiplication and you are wroking in a group of intergers, then it is not in the group.
that last comment was meant as a reply to helene. I'd check myself in an abstract book but I'm fairly certain I'm right
these are great videos... Thankk you so much! i have a big test tomorrow and your really helping me remember and get this stuff... my prof likes to go over peoples heads alot -_-
By definition a function is closed in that sense, because once you determine which is your codomain (in this case G) and your rule of correspondence (in this case, for (a,b)in G, *(a,b)=c for some c in G), yor function CANNOT send an element of the domain (GXG) to an element that is not in G.
Absolutely - you can do it, but then, by definition, it is not a binary operator. We want to work within the "universe" (group) we are dealing with. But, yes, you're right, and in your example we could choose a=1 and b=2, and the result would be 1/2, which is not in Z. We could not choose b=0, because we cannot divide by 0 (not defined), so hence division would not even be a function, let alone a binary operator on ZxZ. Division is what we can call a partial function. Thanks for watching.
I will repeat: if you say that ' * ' is a binary operation in G, what you mean is that * is a function that goes from GXG to G. I think that is pretty clear. Then, you HAVE to send every ordered pair of elements in GXG to a unique element in G...to G!. If you cannot do that, then you DON'T have a binary operation. So, in your example, your '*' is NOT a binary operation, because 'infinity' doesn't belong to Z. Did you get it? It's just a matter of definition, for god sake!
Yes, but you already said that it was a binary operation, then, it goes from GXG to G. Then, as 1.5 is not in G, * cannot be a binary operation. Period
thanks for the video but could you consider changing the music? I don't think it is appropriate. I feel like if I were in the disco, jaja. Maybe some classical =)
Thank you so much for these vids. I actually find them a lot more engaging and educational compared to my lectures at uni. You are really helping me and also making the subject more intersting for me, thanks again. please keep making them
why does a+a+b+b make sense? you define + as a binary operator. i think you should write (a+a)+(b+b) or (a+(a+b))+b etc. we know they are the same - so we can say a+a+b+b makes sense - due to associative property of +. but for the sake of simplicity you should have mentioned it.
Sure, it is because of associativity of +, you are right. But I am not sure if mentioning it would make it simpler for the audience (at least not for those that are just being introduced to abstract algebra).
You are right. As '+' (usual sum) is defined as a binary operator it should be applied only to pairs of elements. So, as we haven't proved that '+' (the usual sum) has the associative property (independently if it is or it is not a binary operator), a+a+b+b doesn't make sense.
I think the author of the video should at least mentioned it for those who are a little more formal.
answer to 2*x=3 in the set of rational numbers for the operator a*b=a+a+b+b would be x=-1/2? if so i think i did get it hehe i have never worked with these kind of notation before.
You guys should not go to deep into the redundant definition thing - in fact there is a hole bunch of "reduced" definitions for groups - but you loose clarity.
So the "closed axiom" is a perfect example - most beginners are not that aware that you can follow this from the definition of the operation and forget to show it - just like you tend to forget to check the induction-start ;)
I think this video is neatly and creatively done. The author combine music with math and in psychological point of view, this helps makes boring concept alive by making the brain stay awake while going through the concepts. It is nice though.
Stupid question: Isn't the requirement that G be "closed" a redundant one? The definition of the binary operation ensures that G is closed (since by definition of "*", * : A x A --> A )?
This function takes two elements of the set G and produces another in G. =>(G,*) is intrinsically closed under this operation. An axiom is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. Therefore, its truth is taken for granted, and serves as a starting point for deducing and inferring other truths. Conclusion:The requirement that G be "closed" is redundant.
Yes, but it is contained in the property of being a binary operation. So, for definition purposes it's the same if you add it or not. But for sure, it's less elegant and sophisticated if you add it.
I am not quite sure what you mean, but a group is not a set that that defines a binary operator. A binary operator is defined ON the set. I think I might have gotten you wrong and that my answer doesn't make sense to what you want to say?
Thank you. I hope it is understandable. The subject can get very abstract and difficult, but I do believe that it is possible to present the basics in a way so that people can understand what is really going on.
Awesome man! Simply awesome!! I LOVE YOU!!!
musiclearner9222 3 months ago
@musiclearner9222 Tanks, mate :).
VeritySeeker 3 months ago
these are really benificient videos for all standard but in part 1 once binary operation is defined there is no need to state the first condition .if u find any mathematical mistake in my text pls. inform me. thanx
Aishasiddiqa100 4 months ago
@Aishasiddiqa100 That is true. I have commented on it previously. :)
VeritySeeker 3 months ago
What I don't understand is: at 2:32 you write *(a,b) = a + b, but you didn't define what + is.
TradingTutor 4 months ago
@TradingTutor In tis case the binary operator + has a meaning. It is ordinary addition. All other binary operations are called * (unless otherwise defined).
VeritySeeker 3 months ago
@VeritySeeker But how is 'ordinary addition' defined then?
TradingTutor 3 months ago
Comment removed
MaThWa92 10 months ago
where is pt 3 on examples of groups?
melbarose85 11 months ago
ax=b where a, b in ring R, has a solution for x for all a, b if a is invertible.
aznlalaland 1 year ago
You're better than my math teacher. He makes the subject super hard and I've only watched 2 of your videos and understand what you are talking about. Thank you
KEMIBA1 1 year ago
thank you ... its very nice you re doing these videos because there is not a lot of material on the web on group theory and abstract algebra and if there is its often too abstract and formal:)
MrZizizzi 1 year ago
you defined binary operator incorrectly.here's what it actually is:
A binary operator * on a set S is a relation from S X S to S. you should have defined a relation from set A to set B also.if it was a function then it would be closure of S would immediately follow. and hence the definition of closure would be unnecessary. which it is not. think about it.
RurouniDizeru 1 year ago
@RurouniDizeru A binary operator on a set S, is a function from the cartesian product of S to S. For every element in SxS, there is a unique element in S. Of course, all functions are also relations.
VeritySeeker 1 year ago
@VeritySeeker i see both definitions in use. the same kind of thing happens with the construction of the natural numbers
RurouniDizeru 1 year ago
@RurouniDizeru I think I see what you mean now. You mean that since I use the function-definition, I do not need to include closure in the definition of a group? If that is the case, I agree. It is not necessary to include closure, since the binary operator is a function. If I defined it as a relation, I would have to include closure. The reason for defining it like this, is because many students forget to check the binary operation. The way I defined it is anyway well-defined.
VeritySeeker 1 year ago
PS: Thanks for uploading
Monkeyradar 1 year ago
Completed video #2, now onto the third. I'll comment on the last one when I finish it so that you don't get useless comment updates in your inbox.
Monkeyradar 1 year ago
This has been flagged as spam show
Why do it? Because we CAN!!
signinname41 1 year ago
i find your clip so idiot-friendly... this is a very amazing way to teach abstract in a concrete way...
Herman
hermanlagon1 1 year ago
Very educational videos!
However... your examples for "binary" operator are at best confusing.
Readers should be aware that such an operator is termed BINARY because it takes exactly TWO parameters as input and returns a single result. This point does not come across quite so clearly after the definition is given.
ZeeNwar 1 year ago
Very educational videos!
However... your definition of "binary" operator is at best confusing.
Readers should be aware that such an operator is termed BINARY because it takes exactly TWO parameters as input and returns a single result. This point does not come across quite so clearly.
ZeeNwar 1 year ago 2
im sure just four or five videos of this kind is a little more worthier than a week of college...
thank you so much...
knighttango 1 year ago
Real Numbers - set
Function is sqrt (xy)
himmes314159 2 years ago
These are awesome videos, thank you for posting them. I'm working on a paper/presentation on Galois Theory and the insolubility of the quintic right now and I have to say that you've given one of the clearest, most readable definitions of a group that I've come across. Keep up the good work.
RouletteRog 2 years ago
verity seeker, your very smart in the way you introduced the binary operation and hence the closure property of groups. I like the style
analyitic1 2 years ago
Thank you soooooo much! For sure, I will continue to watch the video.
:)
blackpheele 2 years ago 5
Dude,
I'm gonna watch your video series through.
rather I am d/l-ing them now....
Cos i need to know enuf abstract algebra to pass my stupid exam.... :(
So...Keep 'em coming bro!!
hulk13in 2 years ago
This is great! After reading through pages of theorys and proofs and getting no understanding of whats going on, this video has blown my mind. And I like the music it makes it seem intense, if your one of those people who needs a library to study just press the mute button.
notnameless 2 years ago
I appreciate ur effort very much and keep it up.
@ those who r whimpering about the music-- do u tie up ur hands at the time of watching the video? Can't u simply mute? It's matter of one click.
mnhtapu 2 years ago
dude...good resource but...need to get rid of the music...highly distracting...i get over it by muting it though
neon37 2 years ago
You are a very good internet teacher. Thank you for all the help. =) I also like your choice of music.
arcanthith 2 years ago
This is really great! Thank you for making them!
tachyonfield 2 years ago 2
Brilliant. I find videos like these are FAR AND AWAY more educational than any class with a forlorn professor and hiked fees. Keep it up. Yours are some of the few videos that deserve a rightful place in youtube.
humby123 2 years ago 7
Isn't the '1)' of the definition of group unnecessary? If you already said that there's a binary operation over G called *, so it's not necessary to mention that 'if a,b are in G then a*b is in G too', since * is a function from G to G.
HeleneSchopenhauer 2 years ago
i think that eventually * being closed is assumed because it does become redundant, however there are some functions that are not closed and the answer to a*b is not in G. as shown with 2*x=3, if the operation is multiplication and you are wroking in a group of intergers, then it is not in the group.
pish4fish 2 years ago
that last comment was meant as a reply to helene. I'd check myself in an abstract book but I'm fairly certain I'm right
these are great videos... Thankk you so much! i have a big test tomorrow and your really helping me remember and get this stuff... my prof likes to go over peoples heads alot -_-
pish4fish 2 years ago
By definition a function is closed in that sense, because once you determine which is your codomain (in this case G) and your rule of correspondence (in this case, for (a,b)in G, *(a,b)=c for some c in G), yor function CANNOT send an element of the domain (GXG) to an element that is not in G.
HeleneSchopenhauer 2 years ago
well you can send an element of (GXG) to an element not in G...
for example let * represent division...
*(a,b)= a/b = c
let a,b => Z , we still find element c equals infinity(not in Z) for b=0 !!!!!
knighttango 1 year ago
Absolutely - you can do it, but then, by definition, it is not a binary operator. We want to work within the "universe" (group) we are dealing with. But, yes, you're right, and in your example we could choose a=1 and b=2, and the result would be 1/2, which is not in Z. We could not choose b=0, because we cannot divide by 0 (not defined), so hence division would not even be a function, let alone a binary operator on ZxZ. Division is what we can call a partial function. Thanks for watching.
VeritySeeker 1 year ago
@knighttango
I will repeat: if you say that ' * ' is a binary operation in G, what you mean is that * is a function that goes from GXG to G. I think that is pretty clear. Then, you HAVE to send every ordered pair of elements in GXG to a unique element in G...to G!. If you cannot do that, then you DON'T have a binary operation. So, in your example, your '*' is NOT a binary operation, because 'infinity' doesn't belong to Z. Did you get it? It's just a matter of definition, for god sake!
HeleneSchopenhauer 1 year ago
@pish4fish
Did you check it? Because obviously you wasn't right.
HeleneSchopenhauer 1 year ago
@pish4fish
Yes, but you already said that it was a binary operation, then, it goes from GXG to G. Then, as 1.5 is not in G, * cannot be a binary operation. Period
HeleneSchopenhauer 1 year ago
thanks for the video but could you consider changing the music? I don't think it is appropriate. I feel like if I were in the disco, jaja. Maybe some classical =)
HeleneSchopenhauer 2 years ago
This videos are very helpful! I was looking for some recorded uni lectures but this short ones are much more efficient. Keep it up!
Semidicht 2 years ago
This has been flagged as spam show
Regarding: 2*x = 3
Gripe 1
Assertion: There is no number x when multiplied by 3 yields 2.
Proof. You cannot show me this number. Therefore this number cannot exist.
DoctorNumLuck 2 years ago
Hehe, yes proof by incompetence, right?
VeritySeeker 2 years ago
Thank you so much for these vids. I actually find them a lot more engaging and educational compared to my lectures at uni. You are really helping me and also making the subject more intersting for me, thanks again. please keep making them
zzzzzzd 2 years ago
Ahhh a fellow zzz:er. All praise to zzz.
zzzbbb123456 2 years ago
why does a+a+b+b make sense? you define + as a binary operator. i think you should write (a+a)+(b+b) or (a+(a+b))+b etc. we know they are the same - so we can say a+a+b+b makes sense - due to associative property of +. but for the sake of simplicity you should have mentioned it.
thanks for the vids. theyre motivating
allahcc666 3 years ago
Sure, it is because of associativity of +, you are right. But I am not sure if mentioning it would make it simpler for the audience (at least not for those that are just being introduced to abstract algebra).
Thanks for commenting.
VeritySeeker 2 years ago
You are right. As '+' (usual sum) is defined as a binary operator it should be applied only to pairs of elements. So, as we haven't proved that '+' (the usual sum) has the associative property (independently if it is or it is not a binary operator), a+a+b+b doesn't make sense.
I think the author of the video should at least mentioned it for those who are a little more formal.
HeleneSchopenhauer 2 years ago
answer to 2*x=3 in the set of rational numbers for the operator a*b=a+a+b+b would be x=-1/2? if so i think i did get it hehe i have never worked with these kind of notation before.
kavudiego 3 years ago
That is correct :).
VeritySeeker 3 years ago
You guys should not go to deep into the redundant definition thing - in fact there is a hole bunch of "reduced" definitions for groups - but you loose clarity.
So the "closed axiom" is a perfect example - most beginners are not that aware that you can follow this from the definition of the operation and forget to show it - just like you tend to forget to check the induction-start ;)
DigtBrain2 3 years ago
i like your simple break down of the basics of abstract algebra! great sequence of ideas from the topic =D
beenpimped31 3 years ago
nice vid and nice dog !
linuxva 3 years ago
I think this video is neatly and creatively done. The author combine music with math and in psychological point of view, this helps makes boring concept alive by making the brain stay awake while going through the concepts. It is nice though.
lpredd 3 years ago
Thanks.
uj8r48u584548j 3 years ago
Stupid question: Isn't the requirement that G be "closed" a redundant one? The definition of the binary operation ensures that G is closed (since by definition of "*", * : A x A --> A )?
nemo1620 3 years ago
Yes you are right about that. But I wanted it as an axiom because it is one of the things we need to check.
VeritySeeker 3 years ago
A binary operation on a set G is a function
*:GxG -> G
This function takes two elements of the set G and produces another in G. =>(G,*) is intrinsically closed under this operation. An axiom is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. Therefore, its truth is taken for granted, and serves as a starting point for deducing and inferring other truths. Conclusion:The requirement that G be "closed" is redundant.
jppereyra 3 years ago
Yes, it is redundant.
VeritySeeker 3 years ago
@VeritySeeker
Yes, but it is contained in the property of being a binary operation. So, for definition purposes it's the same if you add it or not. But for sure, it's less elegant and sophisticated if you add it.
HeleneSchopenhauer 1 year ago
can we say., groups are some sets that defined some closed binary operation?
Saying closed under specific binary operation means it could be a just simple set for some other binary operation.
well, i am saying redundantly. it's just part of def. i think
claslow 3 years ago
I am not quite sure what you mean, but a group is not a set that that defines a binary operator. A binary operator is defined ON the set. I think I might have gotten you wrong and that my answer doesn't make sense to what you want to say?
VeritySeeker 3 years ago
Consider the system or structure [P(x), U], if you unite a set with it's subset this operation behaves as multiple identity elements
chordsequencer001 3 years ago
The power set P(X) of a set X together with the binary operator U (union) is not a group, because union does not support the axiom of inverses.
VeritySeeker 3 years ago
Is it possible for a group to have 2 identity elements?
chordsequencer001 3 years ago
nope
oaties87 3 years ago
In video number 3 we show that an identity is unique.
VeritySeeker 3 years ago
This is really really helpful. Thank you :)
mishy88 3 years ago
yeraaaaah chemical brothers!!
elephantman2222 3 years ago
very good, very understanable
eul01er 3 years ago
Thank you. I hope it is understandable. The subject can get very abstract and difficult, but I do believe that it is possible to present the basics in a way so that people can understand what is really going on.
VeritySeeker 3 years ago
Thanks for making a very very nice mathematics video.
I'm looking forward to your next videos.
logikung 3 years ago
Thank you logikung. The next video is out, and there will be more.
VeritySeeker 3 years ago