I would even go so far as to say "This is standard multiplication, only we have shifted all the names for all the numbers one unit to the right", and BAM! there is no need for further proof. However, I am SO happy that he _didn't_ go this route. Because while it is an elegant solution, it doesn't help introductory algebra students a single bit in understanding the nitty-gritty of how to show something is a group. It just solves this one exercise with one neat trick that we were able to pull out of a hat precisely because we *aren't* introductory students any more, without shedding any light on how to solve the next one.
Justified as fuck. Taking physics where we have one object that the lecturer uses as functions from two different spaces, a vector, a constant and an operator. All using one object. Nobody has any idea what fucking set this object belongs to, it belongs to the intersection of all of them as far as we're concerned
As a CS student I was actually surprised that I could follow till the end. This was just straighforward algebra with minimal knowledge of group theory.
never done any abstract algebra but i really liked this video, when you strip away the fancy words it's just a simple riddle a middle schooler could solve with some thought
We are given: a∘b := a+b+ab, for a,b∈ℝ\{-1} The whole exercise becomes a lot easier if you notice that a∘b = (a+1)(b+1)-1 This expression also makes it easy to see that ∘ is commutative, although it is not needed to show that it is a group. First we need to show closure, that is, for a,b∈ℝ\{-1}, a∘b∈ℝ\{-1}. In symbols: (a≠-1 ∧ b≠-1) ⇒ a∘b≠-1 Which is equivalent (by contrapositive) to proving: a∘b=-1 ⇒ (a=-1 ∨ b=-1) ...which is easy to show: a∘b = -1 ⇒ (a+1)(b+1)-1 = -1 ⇒ (a+1)(b+1) = 0 ⇒ a+1=0 ∨ b+1=0 ⇒ a=-1 ∨ b=-1 To prove that it is a group we need to show: ① Associativity ② Neutral term ③ Inverses ① Associativity a∘(b∘c) = (a∘b)∘c a∘(b∘c) = a ∘ [(b+1)(c+1)-1] = (a+1) [(b+1)(c+1)-1+1] - 1 = [(a+1)(b+1)] (c+1) - 1 = [(a+1)(b+1)-1+1] (c+1) - 1 = [a∘b+1] (c+1) - 1 = (a∘b)∘c ② Neutral term a∘e = e∘a = a a∘e = e∘a = a ⇒ (a+1)(e+1)-1 = a ⇒ (a+1)(e+1) = a+1 [since a≠-1 we can safely divide by (a+1)] ⇒ e+1 = 1 ⇒ e = 0 ③ Inverses a∘c = c∘a = e [=0] a∘c = 0 ⇒ (a+1)(c+1)-1 = 0 ⇒ (a+1)(c+1) = 1 ⇒ c+1 = 1/(a+1) ⇒ c = 1/(a+1)-1 ⇒ c = -a/(a+1)
Randomly popped up in my recommended. I was an avid watcher 4-5 years and you were part of the reason I got into math and just wanted to say thank you! In the mean time started and am nearing the end of my bachelors, although it seems I was bitten by the algebra brainworms instead of the analysis ones.
the bonus problem is easy if you first notice that circ is commutative and write it as: -3 circ 2 circ x = 17, then compute -3 circ 2 and then apply the inverse, getting x.
suppose a+b+ab=-1, we can rewrite the equation as a(1+b)=-(1+b), which implies at least one of a or b must be -1. This is a contradiction, so a+b+ab cannot be -1 and thus the operation is closed.
The first proof is solved far more easily by simply noticing that a+b+ab=-1 implies that (a+1)(b+1)=0, which is clearly a contradiction since there are no zero divisors in R and neither a nor b are equal to -1 by assumption.
This problem becomes trivial if you notice that if you define f(x):=x+1 (and thus f^{-1}(x)=x-1) then aºb=ab+a+b=ab+a+b+1-1=(a+1)·(b+1)-1=f(a)·f(b)-1=f^{-1}(f(a)·f(b)) where · is the usual real multiplication. As (R\{0},·) is a group and f(-1)=0 you get automatically that (R\{-1},º) is a group because you can translate º in terms of ·.
I took abstract algebra as an elective because it has the word 'algebra' in it, not knowing what I was getting into. The class started with 8 of us, within three weeks it was down to 5. I couldn't navigate the proofs so I would memorize about 50 of them knowing two or three might show up on an exam, it was brutal and not uncommon to spend about 70hrs on preparation. Next to advanced calc this is the most difficult undergrad coursework.
The first problem is not a contradiction. However, what the proof shows is that if -1 is an element, then either a or b must be -1 in the operation. As -1 must be used to reach itself through the operation, by removing -1 it cannot be formed by any other two reals.
f it indeed. OF indeed. this brings back fond memories of a course in modern algebra i took as an undergraduate with Prof Staknis at Northeastern University back in the late 60s. ty Pappa Flamme.
the last exercise is much easier to solve by using what we called in our class the cancelative property of group operations; you can just multiply both sides of the equation on the left by the inverse of -3 and then multiply both sides on the right by the inverse of 2.
If you can find it, "Introduction to MODERN ALGEBRA" by John L. Kelley (Official Textbook for the "Continental Classroom" NBC TV early-morning TV program broadcast on weekdays in 1960 in the U.S.) It is my favorite introduction to "Modern Algebra" (Abstract Algebra).
by the way, I misread your 2 as a z for the last question and was confused how you found a single number for x. I got x = -(10-z)/(1+z) which gives -4 if you plug in x = 2.
You should make an in depth video on how to write proofs, and not just a "p -> q" video, but like what the typical grammar is, where we get axioms from, etc.
I think there's a slightly easier way to show closure. Since an and b are not -1, a=a'-1 and b=b'-1 for some nonzero a' and b'. You then have a\circ b=a+b+ab=(a'-1)+(b'-1)+(a'-1)(b'-1)=a'-1+b'-1+a'b'-a'-b'+1=a'b'-1, and since both a' and b' are nonzero we must have a'b' is nonzero, thus a'b'-1≠-1 so a\circ b=a'b'-1≠-1 and we are done. Edit: I edited my comment due to a mistype I noticed in my math.
Everyone says real analysis is more difficult than abstract algebra. To some degree that is true since the proofs in real analysis are all over the place, but the content is very intuitive and has tons of geometric intuition. I just started learning abstract algebra and its seriously the weirdest subject Ive ever seen. The proofs are pretty straightforward but developing deep intuition is difficult and the motivation given by most textbooks is so hard to hold onto. The subject was supposedly developed to determine solvability of polynomials by radicals which is awesome, but that was galois theory and the standard course starts with the broader structures groups with galois theory at the 30th chapter or something. So it works inwards in a sense and that just bugs me because trying to relate these ideas of groups, cyclic groups, permutation groups, etc to polynomials seems like such a reach. Also, it has so much number theory which is okay but that was something that caught me off guard entirely, but the examples Zn and U(n) actually help to get a feel for groups. If anyone has advice for motivating the study of these insanely abstract things please enlighten me because I tend to be stupid about that kind of insight.
actually cyclic groups are used in organic chemistry or something and actions of a group describs many thing in real life such as possible permutation of a cube and such things. you could look it up if you are interested. altho imho it is called "abstract" for a reason. you put any kind of axioms you want to work with and just go with the flow. i'd give the exemple of boolean groups. you want find any intuition behind it unless you are studying logic yet they are fun to work with.
For the final problem, you can use the definition of the inverse. Find it for the numbers on LHS and then apply to RHS and cancel on LHS to get the answer without algebraic rearranging.
At 7:00 you used cancellation, which does not hold in general (however it does over R). For instance, in Z/6Z multiplication is not cancellative, so you can have a * b = c * b with a ≠ b (take a = 2, b = 3, c = 4, for instance).
Seems easy enough. I'm gonna try to put it under the read more in case anyone else wants to take a crack at it. This should be enough room for the read more thingy to appear. Suppose b=-1. -1=-a/(1+a). Multiply both sides by -(1+a). 1+a=a. Subtract a from both sides. 1=0. Contradiction, therefore b≠-1.
He did tho? He showed that any a circ b cannot be -1 at the start. The inverse element b is just a element of the set R\{-1}. By definiton it cannot be -1.
@@theinappropriateworddope4210 no, sorry, that's not enough. You can't define b to be in the set when it is already chosen (even if not yet found). And b was not defined by operations that we have proven guarantee an element in the set, so that is not enough either. The guy above you made a nice quick proof though 😊
@theinappropriateworddope4210 consider a function f(i) that maps a positive index i to a corresponding even number (0→0, 1→2, 2→4, etc.) and a negative index i to a corresponding odd number (-1→1, -2→3, -3→5, etc.). In this case, f(i) = 2*|i|+2*sign(i)-1. Consider applying this to the set of positive integers. ∀i∈ℕ, f(i) = 2*i ∈ℕ. However, it is not necessarily the case that ∀i∈ℕ f⁻¹(i) ∈ℕ. For example, f⁻¹(5) = -3∉ℕ. Thus it is not always the case that a function mapping a set to itself will always have an inverse that maps the set to itself.
The abstract algebra situation is crazy
aint no party like abstract algebra party
As someone who identifies as an audience I can confirm that my retention is no longer than 69 seconds
are you suffering from nocturnal edumissions? don't skip this ad, we have the solution!
The proofs become much easier if you notice that a circ b = a + b + ab = (a + 1)(b + 1) - 1. You also get commutativity for free.
I would even go so far as to say "This is standard multiplication, only we have shifted all the names for all the numbers one unit to the right", and BAM! there is no need for further proof.
However, I am SO happy that he _didn't_ go this route. Because while it is an elegant solution, it doesn't help introductory algebra students a single bit in understanding the nitty-gritty of how to show something is a group. It just solves this one exercise with one neat trick that we were able to pull out of a hat precisely because we *aren't* introductory students any more, without shedding any light on how to solve the next one.
commutative simply follows by the commutativity of addition and multiplication and makes the entire proof+exercise so much easier
Well, commutativity is free anyway.
You released this literally the day after I finally started teaching myself group theory lol. Impeccable timing. Great video!
Embrace the algorithm. It has embraced you.
Ahhh yes, the classic "you get no points for not telling me what 'a' and 'b' are elements of." Abstract algebra can be a bit brutal lol
Justified as fuck. Taking physics where we have one object that the lecturer uses as functions from two different spaces, a vector, a constant and an operator. All using one object. Nobody has any idea what fucking set this object belongs to, it belongs to the intersection of all of them as far as we're concerned
People that turn away from abstract algebra videos are missing out. Best field of mathematics.
This truly is an algrebruh moment
This comment has no response? Let me fix that!
As a CS student I was actually surprised that I could follow till the end. This was just straighforward algebra with minimal knowledge of group theory.
1:30 and I already don't understand anything but I'll watch it anyway.
15:10 actually now I feel like I can follow along.
Good thing you did everything step by step.
i need more abstract algebra from papa flammy
never done any abstract algebra but i really liked this video, when you strip away the fancy words it's just a simple riddle a middle schooler could solve with some thought
I FUCKING LOVE ABSTRACT ALGEBRA, FINITARY MONADS OVER SET RAHHHHH
We are given:
a∘b := a+b+ab, for a,b∈ℝ\{-1}
The whole exercise becomes a lot easier if you notice that
a∘b = (a+1)(b+1)-1
This expression also makes it easy to see that ∘ is commutative, although it is not
needed to show that it is a group.
First we need to show closure, that is, for a,b∈ℝ\{-1}, a∘b∈ℝ\{-1}. In symbols:
(a≠-1 ∧ b≠-1) ⇒ a∘b≠-1
Which is equivalent (by contrapositive) to proving:
a∘b=-1 ⇒ (a=-1 ∨ b=-1)
...which is easy to show:
a∘b = -1
⇒ (a+1)(b+1)-1 = -1
⇒ (a+1)(b+1) = 0
⇒ a+1=0 ∨ b+1=0
⇒ a=-1 ∨ b=-1
To prove that it is a group we need to show:
① Associativity
② Neutral term
③ Inverses
① Associativity
a∘(b∘c) = (a∘b)∘c
a∘(b∘c)
= a ∘ [(b+1)(c+1)-1]
= (a+1) [(b+1)(c+1)-1+1] - 1
= [(a+1)(b+1)] (c+1) - 1
= [(a+1)(b+1)-1+1] (c+1) - 1
= [a∘b+1] (c+1) - 1
= (a∘b)∘c
② Neutral term
a∘e = e∘a = a
a∘e = e∘a = a
⇒ (a+1)(e+1)-1 = a
⇒ (a+1)(e+1) = a+1
[since a≠-1 we can safely divide by (a+1)]
⇒ e+1 = 1
⇒ e = 0
③ Inverses
a∘c = c∘a = e [=0]
a∘c = 0
⇒ (a+1)(c+1)-1 = 0
⇒ (a+1)(c+1) = 1
⇒ c+1 = 1/(a+1)
⇒ c = 1/(a+1)-1
⇒ c = -a/(a+1)
BTW, i just want to point out that the proof of closure is a proof by contrapositive, NOT by
contradiction; people often confuse the two.
Randomly popped up in my recommended. I was an avid watcher 4-5 years and you were part of the reason I got into math and just wanted to say thank you! In the mean time started and am nearing the end of my bachelors, although it seems I was bitten by the algebra brainworms instead of the analysis ones.
Very nice, I remember your username! :) Glad you've made it this far, proud of you :))
the bonus problem is easy if you first notice that circ is commutative and write it as:
-3 circ 2 circ x = 17, then compute -3 circ 2 and then apply the inverse, getting x.
or you could do inverse of -3 circ 17 circ inverse of 2 = x
I opened the video and closed it again at 1:09.
I watched it in full right after.
clean 69 like he predicted
I think that is what they intended to purposefully confirm. :D
suppose a+b+ab=-1, we can rewrite the equation as a(1+b)=-(1+b), which implies at least one of a or b must be -1. This is a contradiction, so a+b+ab cannot be -1 and thus the operation is closed.
As someone who has a AP Calculus AB test in 2 days, I can confirm that I definitely understand all of this
The first proof is solved far more easily by simply noticing that a+b+ab=-1 implies that (a+1)(b+1)=0, which is clearly a contradiction since there are no zero divisors in R and neither a nor b are equal to -1 by assumption.
This problem becomes trivial if you notice that if you define f(x):=x+1 (and thus f^{-1}(x)=x-1) then aºb=ab+a+b=ab+a+b+1-1=(a+1)·(b+1)-1=f(a)·f(b)-1=f^{-1}(f(a)·f(b)) where · is the usual real multiplication. As (R\{0},·) is a group and f(-1)=0 you get automatically that (R\{-1},º) is a group because you can translate º in terms of ·.
I took abstract algebra as an elective because it has the word 'algebra' in it, not knowing what I was getting into. The class started with 8 of us, within three weeks it was down to 5. I couldn't navigate the proofs so I would memorize about 50 of them knowing two or three might show up on an exam, it was brutal and not uncommon to spend about 70hrs on preparation. Next to advanced calc this is the most difficult undergrad coursework.
The first problem is not a contradiction. However, what the proof shows is that if -1 is an element, then either a or b must be -1 in the operation. As -1 must be used to reach itself through the operation, by removing -1 it cannot be formed by any other two reals.
f it indeed. OF indeed. this brings back fond memories of a course in modern algebra i took as an undergraduate with Prof Staknis at Northeastern University back in the late 60s. ty Pappa Flamme.
the last exercise is much easier to solve by using what we called in our class the cancelative property of group operations; you can just multiply both sides of the equation on the left by the inverse of -3 and then multiply both sides on the right by the inverse of 2.
If you can find it, "Introduction to MODERN ALGEBRA" by John L. Kelley (Official Textbook for the "Continental Classroom" NBC TV early-morning TV program broadcast on weekdays in 1960 in the U.S.) It is my favorite introduction to "Modern Algebra" (Abstract Algebra).
this was fun. you should upload more of this kind of video
by the way, I misread your 2 as a z for the last question and was confused how you found a single number for x. I got x = -(10-z)/(1+z) which gives -4 if you plug in x = 2.
You should make an in depth video on how to write proofs, and not just a "p -> q" video, but like what the typical grammar is, where we get axioms from, etc.
I think there's a slightly easier way to show closure. Since an and b are not -1, a=a'-1 and b=b'-1 for some nonzero a' and b'. You then have a\circ b=a+b+ab=(a'-1)+(b'-1)+(a'-1)(b'-1)=a'-1+b'-1+a'b'-a'-b'+1=a'b'-1, and since both a' and b' are nonzero we must have a'b' is nonzero, thus a'b'-1≠-1 so a\circ b=a'b'-1≠-1 and we are done.
Edit: I edited my comment due to a mistype I noticed in my math.
That is a beautyfull argument, because with it you can't only prove clausure, but also all the other group propieties.
@ I agree, this substitution can kind of help give a feel as to what’s going on
Everyone says real analysis is more difficult than abstract algebra. To some degree that is true since the proofs in real analysis are all over the place, but the content is very intuitive and has tons of geometric intuition. I just started learning abstract algebra and its seriously the weirdest subject Ive ever seen. The proofs are pretty straightforward but developing deep intuition is difficult and the motivation given by most textbooks is so hard to hold onto. The subject was supposedly developed to determine solvability of polynomials by radicals which is awesome, but that was galois theory and the standard course starts with the broader structures groups with galois theory at the 30th chapter or something. So it works inwards in a sense and that just bugs me because trying to relate these ideas of groups, cyclic groups, permutation groups, etc to polynomials seems like such a reach. Also, it has so much number theory which is okay but that was something that caught me off guard entirely, but the examples Zn and U(n) actually help to get a feel for groups. If anyone has advice for motivating the study of these insanely abstract things please enlighten me because I tend to be stupid about that kind of insight.
Honestly, as you learn more math and work with these structures more, you just naturally develop a sense of understanding/intuition for them.
actually cyclic groups are used in organic chemistry or something and actions of a group describs many thing in real life such as possible permutation of a cube and such things. you could look it up if you are interested. altho imho it is called "abstract" for a reason. you put any kind of axioms you want to work with and just go with the flow. i'd give the exemple of boolean groups. you want find any intuition behind it unless you are studying logic yet they are fun to work with.
I neeed more abstract algebraa nowwww
I'm an engineer and I've always been curious about abstract algebra. Thanks for sharing
The audience retention for channels covering Spinors and Lie algebras doesn't seem bad.
LET'S GO 2 years of studying group theory all for this moment
Abstract Algebra: ❌️
Abstract Brainrot: ✅️
excited for this class
For the final problem, you can use the definition of the inverse. Find it for the numbers on LHS and then apply to RHS and cancel on LHS to get the answer without algebraic rearranging.
I imagine that's the "group theory hack" that the person writing the question wanted you to use
probably
Abstract algebra is my favourite algebra
You reminded me of some galois theory I took at second year of university
Oh shit Flammy doing abstract algebra?! No way
Finally a video interesting enough for me
papa flammy
So who are these ppl that have an attention span longer than 69 seconds?
.
.
.
Squirrel!!!
Where’s the guy that was looking for abstract algebra help and he clicked on this?
When I look back at uni exams I see them easy nowadays, but at the time they seem very hard. How experience helps.
At 7:00 you used cancellation, which does not hold in general (however it does over R). For instance, in Z/6Z multiplication is not cancellative, so you can have a * b = c * b with a ≠ b (take a = 2, b = 3, c = 4, for instance).
the operation was on R \ {-1} not on any other set
my favorite part was the proof of the eidentity property of the group :)
please can you make a video about reimannian curvature and stuff like that and tensors
That's not a video, that's a video *series* . Eigenchris and Math the Beautiful both have pretty good ones, if you want that.
*this is ASMR for me*
_good night_ flammy boi
great video!!!
thx =)
Loving the shirt.
I don't know if this is too pedantic, but you never showed that the inverse couldn't be -1.
Seems easy enough. I'm gonna try to put it under the read more in case anyone else wants to take a crack at it.
This should be enough room for the read more thingy to appear.
Suppose b=-1. -1=-a/(1+a). Multiply both sides by -(1+a). 1+a=a. Subtract a from both sides. 1=0. Contradiction, therefore b≠-1.
He did tho? He showed that any a circ b cannot be -1 at the start. The inverse element b is just a element of the set R\{-1}. By definiton it cannot be -1.
@@theinappropriateworddope4210 no, sorry, that's not enough. You can't define b to be in the set when it is already chosen (even if not yet found). And b was not defined by operations that we have proven guarantee an element in the set, so that is not enough either. The guy above you made a nice quick proof though 😊
@theinappropriateworddope4210 consider a function f(i) that maps a positive index i to a corresponding even number (0→0, 1→2, 2→4, etc.) and a negative index i to a corresponding odd number (-1→1, -2→3, -3→5, etc.). In this case, f(i) = 2*|i|+2*sign(i)-1. Consider applying this to the set of positive integers. ∀i∈ℕ, f(i) = 2*i ∈ℕ. However, it is not necessarily the case that ∀i∈ℕ f⁻¹(i) ∈ℕ. For example, f⁻¹(5) = -3∉ℕ. Thus it is not always the case that a function mapping a set to itself will always have an inverse that maps the set to itself.
f(i) is actually 2|x|+(sign(2x+1)-1)/2 I made a mistake but it's fine cuz we never used that definition anyways.
Computer Science guy here: I’m so cooked 😅
Proof by taking a sharp look hahahaha
I remember studying it at university
Geil, nach 8 Stunden eine Aufgabe von 5 geschafft 🎉🎉🥳
Honestly I expected the equation at the end to be solved by circing with the circular inverse of ((-3) circ 2)...
You could have used the inverse property to solve the last one
Where Papa Flammy OF?
How can you write WLOG for without loss of generality but not write Sps for suppose ?
NEW. MATH.
Proof by multivariable interpolation
Great techers!
My abstracr algebra final is next week. Wish me luck 🤞
shouldn't you also prove that the inverse for each element still lies in the group?
Lovely introduction to abstract algebra! Now I gotta go learn more
Indians be like: we've been studying this since 10th grade
@@adrian0667 no this is in 11th get corrected noob
I agree. They are annoying.
Yeah. Their bios are like sci fi novels.
You missed to prove that 0•n = 0 for all n !!!
🎉🎉❤
epic
私もおでこが広いです
Did you really need all the work to find an identity? Didn't you already show 0 was the identity?
calculus = shit