Abstract Algebra: Group or Not Group? (Integer edition)

[@Socratica]

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[@S2DEliptiK]

Hahaha this is beyond weird in the best possible way. Thanks for making these.

[@Socratica]

Heehehehee thank you !! We're so glad you had fun with it. :)

[@lifecore04]

she speaks english very well

[@solomonirailoa]

Yes she does 😄

[@shantingsim8842]

For multiples of 7 under +? I dont think its a group? I thought the identity supposed to be 0? but 0 is not an element?

[@natedgg]

0 is a multiple of any integer

[@shyamsundersaini6100]

then according to you (N,+) should be a group

[@Kalernor]

Assuming by N you mean Natural numbers? If so then no, it violates the condition of every element having an inverse.

[@tommygunhunter]

Shan, watch again. She explained this clearly. Seven times zero is a multiple of seven

[@nicco2650]

it's so funny to see how she's upset with the no answers.

[@Sexy_sachin]

sorry....if we gave all answers wrong...its you....

[@chummyigbo8844]

Why am i smiling while learning math?

[@debdattachatterjee6353]

chummy Igbo mee too

[@steliostoulis1875]

Why not? Mathematics is beautiful

[@thefuckdidyousaytomelittle7580]

I call ya happy smile

[@missingno9]

The contestants win a better understanding of groups!

[@mathwithrolandosoto8282]

socratica you are a fun professor. I love how you teach math. I am learning algebraic structures now at Florida International University.

[@Socratica]

Rolando Soto thankj you for watching! We're so glad to hear our videos are helpful. We'd love to hear what other kinds of videos you would be interested in.

[@subramaniannk3364]

Hi, Can you do some videos for topology and hilbert spaces?

[@ashishkr.4075]

@@thegrey53 So early! He would have double graduated by now lol

[@seyyedhosseini1998]

this video is more about Abstract Hotgebra.

[@ticTHEhero]

Yeee

[@SahibKhurana]

You're the best maths professor I ever had! :D

[@Socratica]

You are so kind. Thank you for watching! :)

[@Orlando3OH895]

From a UNC Chapel Hill math major, thanks for these videos! I subscribed. I love the enthusiasm from the actor. I wish emotion and humor was used every now and then in my my math courses. Sometimes I feel professors are either bored by the subject themselves or just don't want to be spending their time teaching the class. I think this type of approach is something that should be applied in STEM courses, I'm sure many more students would be interested.

[@v3g499]

God i can’t stop laughing and smiling 😆 I like this

[@sevarchy]

I get why "Odd integers under +" isn't a group. But why have you checked the inverses box? If there's no identity element, can we even talk about inverses? How can 3 have an inverse if there's no 0 in the set? Isn't it true to assume if there is no identity, then there are no inverses?

[@account-ll8ou]

Ecactly!

[@tophbeifong1046]

The property of inverses just talks about whether all elements have inverses or not. Adding them is the property of identity's concern only.

[@Socratica]

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[@Tuuuuusssjjjjjjnrnfnnfnfn]

oh my god..succa a great jolly professor..i want more such videos mam...plz make more such test videos on algebra...great fun!!!!

[@twilightsparkle6756]

+Kevin Small When you ask questions, make sure we can reply to you. (There's no "Reply" button for your comments.) By definition, an even number is a number which you can divide by two leaving no remainder. That is, distribute it evenly into two equal groups. Imagine you have some number of marbles. Take two cups and put one marble in one cup and the other marble in the other cup. Repeat that as long as you can. If you end up with no marbles, you divided them evenly into two equal groups in these cups, which means the number of marbles you started with was *even*. But if you end up with one marble left (and you cannot put it into any cup without putting another one in the other cup), your original number was *odd*. You ended up with a *remainder* of 1 marble. Zero is even, because you can divide it by 2 leaving no remainder. Try it with marbles again: If you started with no marbles to begin with, you're already done, because you have nothing to distribute, and there are equal amounts of marbles in both cups: 0. But after you're done, you also have no marbles left, so there's no remainder (there couldn't be any remainder if you started with nothing), which means you distributed them *evenly*. Therefore 0 is even. Actually, 0, is the most even number, because it can be divided evenly into any other number of groups leaving no remainder, that is, it is divisible by any number. There are some controversies about dividing something into 0 groups, though, so I'll leave it this way. Humans are not ready for the knowledge about division by zero yet. They're still in denial.

[@HilbertXVI]

'still in denial'? Lol

[@mksarav75]

Thank you. I have fallen in love with this series.

[@natevanderw]

I am so playing this in my class this semester when we get to groups

[@technodruid]

I'd like to see more of these group or not group videos!

[@a.blackwater3076]

Lmao, love this. Informative and far from dull, thank youuu

[@schmitty918]

Love how corny and awesome this is. Binge watching now, started Abstract Algebra this week. Thank you!

[@sharavanakumar2737]

Your are my Abstract Algebra teacher!! You are explaining very well!! You have made maths easy for me

[@PunmasterSTP]

I don't think I've seen anything quite like Socratica before, and I mean that in the absolute best possible way! I'm not sure if anyone else has ever seen the "Standard Deviants", which is a educational TV series that began on PBS, but I almost get some of the same (great) vibes!

[@mohmd252]

استفدت منك كثييييير مشكورة ماقصرتي انت تنشرين العلم وهذا اكبر شرف لك اشكرك من اعماق قلبي

[@musamoloi2149]

Iv just started abstract algebra,and I couldn't hear what my professor was saying yesterday about groups..thank you so much, I guess il be needing these videos from now on.as I walk in the math lecture hall😩😩

[@newbornaman3859]

Mee too

[@mathwithrolandosoto8282]

I would be interested in more videos of abstract algebra with some exercises from the book algebra chapter zero which is by a professor from FSU Paolo Aluffi.

[@augustopinochet6841]

Thank you for all your videos they are really very helpful. However I have a question. If the gropu his under "x" or "*" it has as you told 1 as identity element.. I can agree with the second point about inverse.

[@sourashismondal3110]

The jokes are so funny 😂😂 😂😂😆😂😆😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂

[@tantzer6113]

Ha ha... you make me laugh.

[@lazyplayer1]

Please come back

[@shinkwrloggie7579]

Amazing playlist :) The exam preparation is so much more enjoyable with a teacher like you!

[@Love_Hope_from_Above]

Professor Socratica: As a teacher of math, I really enjoy the recent videos on abstract algebra -- especially the "integer edition of the group/not group game." It was so much fun and educational! Could your team produce more videos on: Cyclic groups, order of a group and order of an element (of a group), cosets, permutation/symmetry groups, isomorphism, etc.? Abstract algebra has been a very difficult subject, and your video approach makes the subject come alive! Thx. > Benny L.

[@NathanSmutz]

Kevin Small If you define even as divisible by 2 and 0 = 2*0 I think odd are usually {2n + 1 : n in the integers} so 0 can't be odd.

[@pradityoadi]

Can't stop learning Abstract Algebra...

[@ComputerScienceExplorati-ik6lr]

Liliana's emotional response to things not being a group is priceless. Such disappointment. Muita desapontada.

[@Karthik-ys7mi]

Thank you this video was really helpful and fun!!

[@lifeforever1665]

On multiples of 7 you said yes you are right..... you were horribly wrong

[@MacmodKnx]

This is just amazing :)

[@abiharaj4227]

Very splendid explanation madm I like this too much

[@oborooizamisi1894]

hey, I tot groups under multiplications do not include 0 only 1?

[@zippydipity42]

I can't think of anything that isn't associative; What's an example, and how do I look out for it? Thank you!

[@omarm.7068]

Subtraction and division are non-associative, there are examples here. www.mathwarehouse.com/dictionary/A-words/definition-of-associative-property.php

[@cameronspalding9792]

@1:39 it satisfies 3 out of 4 properties so I wouldn’t say ‘not even close

[@aqleemaansari8732]

Mam make vedio on group Action ....pls mam

[@arijitmishra2435]

Awesome....

[@akd73]

0.51 isnt zero neither even nor odd? that makes even integers under + "not group". Correct me if I'm wrong!

[@pettPette]

+Akshay Ramesh Babu 0 is even because it is divisible by 2. An integer is EVEN if and on if it is divisible by 2.

[@alfredoespinozapelayo]

@1:55 ... why? 1 is multiple of 7 and 1 + 7 = 8, and 8 is not a multiple of 7, i would suppose it is not closed under addition. Please, a little help here

[@MuffinsAPlenty]

1 is _not_ a multiple of 7. Do not confuse "multiple" with "divisor" or "factor". The number m is a multiple of 7 if there exists an integer n so that 7n = m. The number d is a divisor (or factor) of 7 if there exists an integer n so that dn = 7. 1 is a divisor/factor of 7, but not a multiple of 7.

[@palashkhanra7816]

I don't know why I am laughing 😃😃😃.... But really wonderful ma'am ....😊

[@nate4511]

This may be a dumb question but can we have a group that contains a specific subset. consider the following.... Let G be congruent to R x R. Let A be a group of real numbers. Define f: A -> R s.t. f(a) = 2(a)+3 then define T as that subset of R s.t. T = { (a,f(a)): a belongs to A } . Now let Q = G union A union T. G and A are groups but T is a set So my question is....Is Q a group?

[@moularaoul643]

Thank you so much!!!

[@jjhassy]

I really need you in my life

[@filipve73]

(Abstract) Integers under multiplaction? Not a group ? Inverses = false ? If division of zero (non-standard analyse) the group splits in two piece ?

[@kemsekov6331]

This is so strange that I am actually smiling while thinking about groups))) You are so cool)))

[@shyamsundersaini6100]

Mam how can multiple of 7 under adition can be a group.bcz it doesn't have identity.in case of addition identity should be zero .but here in this case zero is not in the set.please reply me and tell me also that (N,+) is a group or not if not then why?? thank you i am waiting for answer.

[@ninosawbrzostowiecki1892]

It would be so cool if you could teach my representation theory course!

[@ozzyfromspace]

wow, surprised I got em all right. So, what I win, Johnny boy? Hehe fun video, and I'm stoked for the rest of the series :)

[@june8430]

Im learning lots from these videos. Why cant university professors hold online lectures like this. All they do is “uhhhhh group duh duh radda dud”

[@sankarbhattacharya1770]

I can't resist writing to say how much I was carried away by it. Really it's magnificent--it gives one a new conception of glories of human mind.

[@asadullahfarooqi254]

hey dear i saw you also add programming videos to your channel if you want courses or tutorials then most welcome i'll make for you i am a developer.. ; ) : )

[@apunkaadda3898]

Mam you are way ahead time....nice lessons ...really loved the way you teach

[@joego5359]

where are you getting the inverses from in multiples of 7? that isn't logical there aren't any negatives

[@_robert__]

Thank you ❤ for this great lesson, will there ever be lessons on other areas of mathematics ?

[@atharvas4399]

Please make videos on Discrete Math, Linear Algebra

[@loldnb5435]

The most weird thing is how i come up with the subject - by just being curious and trying to invent my own math )) and here it is

[@gk-qf9hv]

Is zero an even number?????

[@veselin-penev]

1:15, if the identity element isn't a part of the group (basicly set, becuase it's not a group), isn't it wrong to claim that every element has an inverse?

[@SK-bu1yb]

it could be possible that all the inverses exist, but the identity doesn't. Even then it is not a group because that violates the "closure" property. (Eg: group of odd integers. All inverses exist, but no identity element)

[@MuffinsAPlenty]

I would argue that the concept of inverse doesn't make sense without the concept of identity. But there are algebraists who recast the inverse axiom of a group as a "cancellation" axiom of a group. In other words, instead of having an axiom that inverses exist, you could say, For every g, x, and y in G, if gx = gy, then x = y, and if xg = yg, then x = y. This is a way to get at the spirit of "existence of inverses" (what is the point of inverses anyway? to cancel from both sides of an equation!) without having an identity element and without violating closure.

[@maartenvs01]

Can someone explain to me why Integers under x is a closed operation? Take for example 3 x (1/4) = 0,75. This isn't an integer, right? Where is my reasoning mistake?

[@andypetsch]

Well, Maarten, you multiplied by 1/4. But 1/4 is not an Integer. So, the result does not habe to be an integer as well.

[@maartenvs01]

Thank you!

[@andypetsch]

My pleasure

[@timmy18135]

What about groups that reference themselves? Are those a group?

[@hoangtudaden1304]

in my opinion, if you have an identity it implies that you have it inverse.

[@neloka4313]

No it doesn't. In Z you have a multiplicative identity 1, but 2 doesn't have a multiplicative inverse in Z. In fact you could complete Z to form a field by quotienting Z² by the equivalence relationship (a,b) ~ (a',b') ab' = a'b which encapsulates the fact that a/b = a'/b' ab' = a'b, but if you started with, for example, Z/4Z, you couldn't find any field containing Z/4Z because this ring has divisors of zero, 2 * 2 = 0 and 2 != 0.

[@mugheesghayas2380]

singleton set zero under multiplication forms group or not ?

[@admink8662]

I want to play again.

[@lixu8141]

you scared me :)

[@amritathakur2036]

🎉🎉🎉 super 👍😊

[@saurabhsingh-ow7ue]

thank you madam........

[@nirupadik4068]

Mam if some sets satisfies the axioms of group what it's meaning

[@julianoch]

This was intense.

[@docu73]

Tbh I think we can define 0 to be odd... still it wouldn't make "odd integers with +" a group though...

[@neloka4313]

No we cannot.

[@viktorashistan7207]

chill bro

[@ybc8495]

hahaha

[@morgengabe1]

You guys should do this more often!

[@mortkebab2849]

Multiples of any integer under addition are a group. Interestingly, that includes {0}.

[@Zaurthur]

no one cares if zero doesn't have an inverse, the definition of multiplication already excludes that.

[@tcpjh]

but the inverse is not included in the set of integer numbers, so there.

[@theokeeping1213]

Hello helpful people! I'm confused as to why {Z^+,+} is a group. Firstly I was taught that whilst Z contains 0, Z^+ only contained 1,2,3,4,5... meaning there is no identity. Also, for an element of the positive numbers 'a', a+a^-1=0 let a=3, therefore a^-1=-3 which isn't in the set of positive integers! Please can you explain this to me, I am but a humble high school student!

[@theokeeping1213]

Theo Keeping Now I'm questioning all my beliefs! Odd integers under + must be closed, surely?!

[@MichaelGoldenberg]

How can the odd integers be closed under +? If you add any two odd integers, the result is even. You can prove this by algebra: let x, y be odd integers. Then x = 2j + 1 and y = 2k + 1 for some integers j & k. Adding x + y = (2j + 1) + (2k + 1) and removing parentheses and rearranging terms by the commutative property of addition, you can get (2j + 2k) + 1 + 1. Simplifying yields (2j + 2k) + 2. Now, by the distributive property of multiplication over addition, you can obtain [2(j + k) + 2]. Again, by the distributive property, this is equivalent to 2(j + k + 1). And hence, by definition, this is an even number as was required. Since the result didn't depend on the selection of x, y, j, or k, we have proved that the sum of two arbitrary odd integers is an even integer.

[@1900maniac]

What do you mean by "closed" when you say it is closed under something?

[@skulldyvan]

Closure means that, when you apply an operation to any elements of a set, the result is also within the set. Addition is closed over the positive integers (adding any two of them gives you another positive integer), but subtraction isn't (1 - 2 does not give you a positive integer).

[@azaanahmad6265]

This is the best video I have ever seen

[@putin_navsegda6487]

you are the best prof ever!🥰

[@flayy5750]

this is super amazing!! uwu

[@Vedvart1]

"0 is an even integer" *triggered*

[@shekioio]

just thank you!

[@johnmorales4328]

Hello. Could someone explain why the "multiplicative inverse of an integer is not an integer" by giving an example? 1:39 Thanks!

[@Socratica]

The multiplicative inverse of 2 is 1/2 because 2*(1/2) = 1, and 1 is the identity element for multiplication. The additive inverse of 2 is -2, because 2 + (-2) = 0, and 0 is the identity element for addition. So other than 1 and -1, the multiplicative inverse of any integer is a fraction, not an integer.

[@johnmorales4328]

Huh. That's very interesting. I thought integers included fractions since every integer _can_ be expressed as one by putting it over 1. But I guess the requirements for a number to be considered an integer are very strict. Thank you for your very prompt reply - I wasn't expecting it!

[@johnmorales4328]

Thank you for further clarifying my confusion. It all makes sense to me now. :)

[@예밤-f3k]

wait so 0 is even!?!??!!

[@MuffinsAPlenty]

Yes, 0 is even! We consider an integer n to be even if n = 2m for some integer m. In our case, n = m = 0 works, so 0 is even :)

[@slaozeren8742]

It was so much fun!

[@ihumanity2010]

I think the Odd Integers under + has not inverses since the result of addition of an odd integer with another must produce the identity which we have not one here. Am I right technically?

[@solomonirailoa]

At 1:10, she says that the group of odd integers under + has inverses, as every positive odd integer has a negative counterpart, however there is no identity because 0 is an even number.

[@gmangalam3180]

How multiples 7 is a group

[@kunslipper]

Thank you so much.

[@math13]

wow nice presentation

[@yoskkdkdk]

I living for this

[@3072kelvin]

Vai Braziil

[@ManojKumar-or4gi]

very nice video

[@indiraocampo7086]

Amazing hahaha

[@Tadesan]

Hmmmm.