Reminder to self: A magma is a set M with an operation, M x M -> M. A semigroup is a magma that is associative. A monoid is a semigroup with identity. A group is a monoid with inverses.
I've got spare time so I've started teaching myself from Linear Algebra Done Right over the weekend (this is my second pass, and a friend recommended it), these videos could not come at a better time in all honesty.
Warmup Question 2: By induction n=1 then observable its true given for n=k, (a^-1ba)^(k+1) = (a^-1ba)^(k)*(a^-1ba) = (a^-1b^(k)a)*(a^-1ba) = (a^-1b^(k+1)a)
Warmup Question 1: e=aa=aea=abba=(ab)(ba) so ab and ba are inverses. But we know inverse of any element is itself due to aa=e and inverses are unique (proven in video) so it means ab must equal ba
Note that for the final proof, 0*0=0 does NOT serve as a counter example...because 0 is NOT a member of the classic structure associated with multiplication, which is R/{0}...and, implicitly, these propositions assume that 'a' is a member of said group.
In Brazilian Portuguese we do not have an equivalent name for magma. We simply say that the set is closed under the operation (o conjunto é fechado para a operação)
24:17 I think you assumed what you were trying to prove in this part of the proposition. The correct proof is that _a_ ^{-1} being the inverse of _a_ means by definition that _a a_ ^{-1} = _e_ = _a_ ^{-1} _a,_ but we can flip the equations around to give _a_ ^{-1} _a_ = _e_ = _a a_ ^{-1}, which by definition means that _a_ is an inverse of _a_ ^{-1}, so since inverses are unique we thus have ( _a_ ^{-1} )^{-1} = _a._
10 months later: Yes, it's the most probable option; set of all permutations with the operation of composition is in fact a group and prof. Penn referenced the inverses as probably the inverse functions, because he said that they are implied by the bijectivity of permutations. Of course inverse functions are inverse element for operation of composition. Hope it helps
i got some kind of confusion... the neutral element is unique.. but as a part of the set it has to match an inverse even so? .. and if the inverse is itself could be rigth? i found this a little bit contradictory
The neutral element is the identity and from that follows, that it is its own inverse. But this relationship is not equivalent, so you can have an element that is its own inverse but does not equal the identity. Ex: G* := {1, -1, i, -i} 1 is the identity, thus its own inverse -1 is its own inverse but not the identity
It took me like half an hour to figure that one out, it is actually 132, remember to write the starting number and the first step of your permutation and you should get it
took me at least a half hour too. Another vote for (1 3 2) . Tricky stuff. Check out th-cam.com/video/UI-SWiyaArE/w-d-xo.html very carefully. Another good video: th-cam.com/video/MpKG6FmcIHk/w-d-xo.html
Also note that (3 1 2) is in fact the same as (1 2 3). Since in a group we have a^2 = a if and only if a is the identity (last result in the video), and since (1 2 3) is not the identity of the group S_3 (or any other symmetric group on more than 3 letters), we could not have (1 2 3)^2 = (1 2 3). (Just another way to check the correctness of the calculation.)
No... just multiplication. The would-be-identity-element (namely 0) is not in the set. -1 + 1 = 0, but 0 is not in {1,-1,-i,i}. It isn't even a magma under addition, (1) + (i) = 1+i, which isn't even in the set. So the set isn't closed under +.
9:49 In your definition of a group you essentially have 4 conditions: closed, associativity, identity, and inverse all with respect to a binary operation. But the standard definition of a group only has 3 conditions: Associativity, Identity, Inverse all with respect to a binary operation. Does that mean those 3 conditions imply closedness?
well, the "closedness" is often omitted, because the definition of a group operation implicitly states the "closedness" condition (* : G x G ----> G), so explicitly adding it to the axioms is superfluous.
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)
How come you don’t include 0 in the natural numbers? I think it’s much easier, notation wise, if we reserve Z+ for the positive integers and we let N be the set of nonnegative integers.
In this world nothing can be said to be certain, except death, taxes and math people arguing about 0 in ℕ. Personally I’m team « 0 in ℕ » but that’s the kind of debate people have strong opinions and that can go on forever. Michael thinks 0 is not in it, and he won’t change his mind because of TH-cam comments lol
Not sure about the inclusion of magmas, semigroups and monoids in this course (at least introducing the terminology for all of them); seems like abstraction for abstraction's sake. and if you insist on breaking down groups in this way then it sort of obliges you to do the same thing for vector spaces later on, which would involve defining vector spaces as modules over a field. You _could_ justify the inclusion of semigroups/monoids by going super abstract and considering semimodules over semirings (a semiring being a ring without additive inverses, and a semimodule being an abelian semigroup equipped with a scalar multiplication from a semiring, satisfying the usual properties (a+b)v = ab + bv, a(v+w) = av + aw, and a(bv) = (ab)v). But "pulling the legs off the spider" in this way isn't really appropriate in a linear algebra course; semimodules will still form a category (that is, composition of morphisms will be associative and there will be identity morphisms) which is enriched over itself (that is, the set of morphisms between two semimodules will itself be a semimodule, although one will have to be careful will left, right, and two-sided semimodules when the semiring is non-commutative), but apart from these basic properties, the theory of (semi)modules is very different to the theory of vector spaces.
Reminder to self:
A magma is a set M with an operation, M x M -> M.
A semigroup is a magma that is associative.
A monoid is a semigroup with identity.
A group is a monoid with inverses.
Nice
In 2020, with all the lockdowns, Norway had to introduce the Non-Abel Prize, because so many mathematicians weren't commuting.
🤣
I've got spare time so I've started teaching myself from Linear Algebra Done Right over the weekend (this is my second pass, and a friend recommended it), these videos could not come at a better time in all honesty.
I’m taking a linear algebra course that’s heavily relying on Linear Algebra Done Right, so this series is also amazing for me
Warmup Question 2:
By induction
n=1 then observable its true
given for n=k, (a^-1ba)^(k+1) = (a^-1ba)^(k)*(a^-1ba) = (a^-1b^(k)a)*(a^-1ba) = (a^-1b^(k+1)a)
Warmup Question 1:
e=aa=aea=abba=(ab)(ba) so ab and ba are inverses. But we know inverse of any element is itself due to aa=e and inverses are unique (proven in video) so it means ab must equal ba
Warm up Q1:
ab = (bb)ab(aa) = b(ba)(ba)a = bea = ba
10:54 ... "my operation will be addition" ;-)
7:18 and a monad is just a monoid in the category of endofunctors
11:00 your operation is multiplication not addition (again being a pendant because you did write down multiplication).
Note that for the final proof, 0*0=0 does NOT serve as a counter example...because 0 is NOT a member of the classic structure associated with multiplication, which is R/{0}...and, implicitly, these propositions assume that 'a' is a member of said group.
3:16 You say -5, but write -2.(-2 is correct just being pedantic).
Just a note that the link for Linear Algebra Done Right is incorrect. It points to the first book in the list.
@7:30 Do we get left- and right-monoids from left- and right-identity elements? Does this begin to hint at commutation?
Will all the other course videos from main channel be included on this channel eventually?
31:02
In Brazilian Portuguese we do not have an equivalent name for magma. We simply say that the set is closed under the operation (o conjunto é fechado para a operação)
good, I think the term magma is rubbish and unnecessary xD (there is the alternative term "binar" though which I do like)
@@schweinmachtbree1013 in Portuguese we can say operação binária
24:17 I think you assumed what you were trying to prove in this part of the proposition. The correct proof is that _a_ ^{-1} being the inverse of _a_ means by definition that _a a_ ^{-1} = _e_ = _a_ ^{-1} _a,_ but we can flip the equations around to give _a_ ^{-1} _a_ = _e_ = _a a_ ^{-1}, which by definition means that _a_ is an inverse of _a_ ^{-1}, so since inverses are unique we thus have ( _a_ ^{-1} )^{-1} = _a._
oh wait no your proof is totally fine. just consider my comment as an alternate proof then, oops
@@schweinmachtbree1013 how is his proof correct? can you please elaborate
@@ahzong3544 the second equals sign follows by using e = x^{-1}x with x being a^{-1}
@@schweinmachtbree1013 I see now, thank you so much
9:34 Isn't it enough to say that there is only the reverse left (or right)?
presumably function composition among invertible functions can also be said to be a group?
I suppose it would have to be invertible functions f: X -> X
Wait, is that just the symmetric groups?
What is the group operation for the symmetric group? Is it composition of permutations?
10 months later:
Yes, it's the most probable option; set of all permutations with the operation of composition is in fact a group and prof. Penn referenced the inverses as probably the inverse functions, because he said that they are implied by the bijectivity of permutations.
Of course inverse functions are inverse element for operation of composition.
Hope it helps
i got some kind of confusion... the neutral element is unique.. but as a part of the set it has to match an inverse even so? .. and if the inverse is itself could be rigth? i found this a little bit contradictory
The neutral element is the identity and from that follows, that it is its own inverse. But this relationship is not equivalent, so you can have an element that is its own inverse but does not equal the identity.
Ex: G* := {1, -1, i, -i}
1 is the identity, thus its own inverse
-1 is its own inverse but not the identity
@28:05 I think the permutation (1 2 3)^2 = (3 1 2) not (1 3 2)?
It took me like half an hour to figure that one out, it is actually 132, remember to write the starting number and the first step of your permutation and you should get it
took me at least a half hour too. Another vote for (1 3 2) . Tricky stuff.
Check out
th-cam.com/video/UI-SWiyaArE/w-d-xo.html
very carefully.
Another good video:
th-cam.com/video/MpKG6FmcIHk/w-d-xo.html
Also note that (3 1 2) is in fact the same as (1 2 3). Since in a group we have a^2 = a if and only if a is the identity (last result in the video), and since (1 2 3) is not the identity of the group S_3 (or any other symmetric group on more than 3 letters), we could not have (1 2 3)^2 = (1 2 3). (Just another way to check the correctness of the calculation.)
@@paulshin4649 good point! Yes I agree with all of you, coming back and looking at this I was evaluating cycles incorrectly.
12:17 so the set is a group with respect to multiplication AND addition.
Edit: Not Addition.
No... just multiplication.
The would-be-identity-element (namely 0) is not in the set. -1 + 1 = 0, but 0 is not in {1,-1,-i,i}.
It isn't even a magma under addition, (1) + (i) = 1+i, which isn't even in the set. So the set isn't closed under +.
Are your examples from a text book. I swear I have seen a few of these before
9:49 In your definition of a group you essentially have 4 conditions: closed, associativity, identity, and inverse all with respect to a binary operation. But the standard definition of a group only has 3 conditions: Associativity, Identity, Inverse all with respect to a binary operation.
Does that mean those 3 conditions imply closedness?
No, the standard definition includes all 4, according to wikipedia
well, the "closedness" is often omitted, because the definition of a group operation implicitly states the "closedness" condition (* : G x G ----> G), so explicitly adding it to the axioms is superfluous.
Do you really need to use associativity to prove that inverses are unique? I am thinking it is not necessary...
(Shoes and socks)(socks and shoes) very very funny ekekek. Also, we need LaTeX support in the comment section.
I might be wrong, but this could work: th-cam.com/video/PvUrbpsXZLU/w-d-xo.html
P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)
How come you don’t include 0 in the natural numbers? I think it’s much easier, notation wise, if we reserve Z+ for the positive integers and we let N be the set of nonnegative integers.
In this world nothing can be said to be certain, except death, taxes and math people arguing about 0 in ℕ.
Personally I’m team « 0 in ℕ » but that’s the kind of debate people have strong opinions and that can go on forever. Michael thinks 0 is not in it, and he won’t change his mind because of TH-cam comments lol
Not sure about the inclusion of magmas, semigroups and monoids in this course (at least introducing the terminology for all of them); seems like abstraction for abstraction's sake. and if you insist on breaking down groups in this way then it sort of obliges you to do the same thing for vector spaces later on, which would involve defining vector spaces as modules over a field.
You _could_ justify the inclusion of semigroups/monoids by going super abstract and considering semimodules over semirings (a semiring being a ring without additive inverses, and a semimodule being an abelian semigroup equipped with a scalar multiplication from a semiring, satisfying the usual properties (a+b)v = ab + bv, a(v+w) = av + aw, and a(bv) = (ab)v). But "pulling the legs off the spider" in this way isn't really appropriate in a linear algebra course; semimodules will still form a category (that is, composition of morphisms will be associative and there will be identity morphisms) which is enriched over itself (that is, the set of morphisms between two semimodules will itself be a semimodule, although one will have to be careful will left, right, and two-sided semimodules when the semiring is non-commutative), but apart from these basic properties, the theory of (semi)modules is very different to the theory of vector spaces.
Ok buddy
24:22 r u using what u want to prove ? mmmm