Basic abstract algebra, pt.3

[slherman9205]

Thanks for the videos! these are really going to help on my test!

[VeritySeeker]

@johnwpurcellx Thanks, mate. If someone can watch this on a Saturday evening and have something to think about on Monday, well - then I am happy ;).

[cmonington]

In the last part of the video, how does 2*x=3 where (a*b)=a*a*b*b and x = -1/2. I don't understand how x=-1/2. Can anyone explain this please.

[tqoftu1]

@cmonington Since we have 2*x = 3 and (a*b) = a*a*b*b, we can rewrite the first 2*x as 2+2+x+x=3 which can be rewritten as 4+x+x=3

4+x+x+-4=3+-4

x+x=-1

So clearly the only solution is x=-1/2

I hope this helps.

[pbajaj5]

Thanks so much, you rock!!!

[RurouniDizeru]

@gorgolyt so i didn't know that conditions for properties were sometimes called axioms. but i accept that now.i realise something.if i really didn't now what i was talking about then why did VeritySeeker agree with me when i explained what i meant?think about that

[RurouniDizeru]

the conditions for (G,*) to be a group are NOT called axioms. axioms are self evident facts used to develop theorems and algorithms. for example the Axiom of Extensiont which states that two sets are equal if they have the same elements. then this theorem follows:

Two sets A and B are equal iff A is a subset of B and B is a subset of A.

PLEASE emend that error

[VeritySeeker]

@RurouniDizeru Conditions in defintions are sometimes called axioms.

[RurouniDizeru]

@VeritySeeker but they're not axioms in the true sense. its a misuse of the word axiom.check out Euclid's Elements for his axioms on plane geometry for example

[VeritySeeker]

@RurouniDizeru No, that is true. They are not axioms in that sense. I agree.

[VeritySeeker]

@RurouniDizeru True. They are not axioms in that sense. They are nothing like the axiom of choice, for example.

[RurouniDizeru]

@VeritySeeker i don't know that axiom. what is it?

[VeritySeeker]

@RurouniDizeru Hi again. It is a fundamental axiom in set theory. It says that given any set S of disjoint non-empty sets, there exist a set with at least one element in common with all the sets in S. In other words, there exist a function that chooses an element in every set in S. Hence axiom of choice.

[gorgolyt]

@RurouniDizeru for use in later life: if you don't know very well what you're talking about, don't try talk about it, especially if you're going to pontificate.

[RurouniDizeru]

@gorgolyt if it seemed like i was pontificating i apologise .i'm okay with the misuse of the word axiom cause i can't do anything about it. i do know what i'm talkin about .my mathematics professor never called any condition to satisfy a property an axiom so that's why is why i didnt.

[gorgolyt]

@RurouniDizeru an axiom is just any unproven statement or definition. the axioms of arithmetic include things like a + (b + c) = (a + b) + c and these define a particular structure. the axioms of group theory include things like a*e = e*a and these define another structure.

[otreborh]

It is not clear for me the analogy between the non-existance of the solution of the equation in the set of integers, and the set Z (with the product operation) not being a group. What is the relation between solution for equations and groups?

[VeritySeeker]

@otreborh Hi, the equation has to have a solution in the group. The equation a*2=1 has to have a solution inside Z, if the set Z together with multiplication can be a group. But inside Z there does not exist a multiplicative inverse to the element 2. It should be 1/2, but 1/2 is not an element in Z.

[shivbhaktajoshi]

good 

[aliasbirt]

So by group theory do you mean ring theory? as we have only talked about ring theory in my class

[VeritySeeker]

@aliasbirt No, a ring is something else. A ring has two operators, while a group has one. A ring is an additive group together with another operator and some axioms.

[malanis1]

name of song lol?

[knighttango]

this is awesome i say!!!

[munited369]

u save me dude!! thanks a mil 4 the videos!!! love thm!!

[interted]

i dont get something. you say that the 2*x=3 can't be solved because (Z,multiplication) is not a group. why?

[VeritySeeker]

Because there exist no integer a such that 2*a=3. The element 2 has no inverse in the set of integers with multiplication.

[Tikeslar]

Z under multiplication doesn't have an inverse for every element.

Remember the identity is 1, so you should have an element a in Z which satisfies 2*a=1... and you don't, since 1/2 isn't an element in Z.

[apurvamisra]

@interted : It is so because "(Z,multiplication) is not a group" implies that not all the conditions for it being a group are not satisfied.

here, there is no x which multiplied to 2 will be mapped into 3

@VeritySeeker: Thanks a lot for such a humane treatment of the subject

[HotTomKat]

i have a degree in math, and barely squeeked through Abstract Algebra wondering...did I learn anything? I've learned more from your videos than I did from my torturous semester! Thanks for taking the time to make these videos!

[mnhtapu]

Great job!

Name of proof method : Reductio ad absurdum

Mechanics: Assume opposite, get a ridiculous/ contradictory result.

[chrisdesilva66]

clear example proving Z is not a group under multiplication - thanks

[stylefighterdks]

great!

[skinnylee]

thanks for this series of videos! Just wished that I tried looking for this 3 months ago haha

[ILoveYouToThe9999999]

Condition 1 is not necessary to be checked for a set to be group. It is enough to check the last three conditions. If the second condition is satisfied, then this means that the first condition must be satisfied.

[VeritySeeker]

Yes, there have been a discussion about this earlier in the comment section, and condition 1 is redundant because we assume * is a binary operator. However I decided to let this condition be an axiom to check, as it is common that people forget to check this. Condition 2, however, does not imply condition 1.

[Aishasiddiqa100]

can u give me  a (general) proof of ur claim

[Aishasiddiqa100]

can u give me a general proof for ur claim

[VeritySeeker]

@Aishasiddiqa100 What claim?

[tsliyan]

Thanks a lot, watch all three videos. Getting a clear understanding of the basics are very important prior to studying a subject. This helps a lot. Keep up the good work.

[pioneerrr]

Thanks for your videos. They explain things quite clear. Thanks again!

[eedahl]

If multiplication doesn't have an inverse on Z, do I remember correctly if that makes Integers a ring?

[VeritySeeker]

Yes, that is true. The integers is a ring with ordinary addition and multiplication. The elements do not need multiplicative inverses to be a ring. If all elements have multiplicative inverses, then it is called a division ring.

[achrispens]

Oh my! You were worried that no one would watch these videos.

These are great. Thanks for taking the time to put these together.

[Death2AII]

Question... in integer multiplication you stated the identity element is integer 1, and the condition that a*1=1*a=a must hold true for any a of the set of integers. If a=0, then any integer, including 1, will produce an identity. Hmmm?

[VeritySeeker]

This is a very good question and a common one too. The number 0 does create problems for us, doesn't it ;).

But the thing is that the integers together with multiplication is NOT a group, and so hence we can find an example where the axioms are violated. My point was that the element 1 is our ONLY candidate for an identity. But it is not an identity, since 0*1 = 0, and not 1. It is not a group.

[agentredlum]

Should a*1 = 1 ? I'm not sure what you are implying here with the statement 0*1 = 0, and not 1 1 is not disqualified by this statement because it leaves 0 unchanged which is the purpose of the identity element. However, I think you have very nicely disqualified 0 as a candidate for the identity element under the stated conditions ;- )

[VeritySeeker]

Yes, 0 is disqualified. I guess it should have read "But 0 is not an identity, since..." The sentence before says 1 is our only candidate, though, what is what I meant. ;)

[pish4fish]

would 0 create a null (sub)set, or a (sub)set with one element being 0?

[VeritySeeker]

I am not sure what you mean. 0 is certainly a subset of any set that contains 0. But I don't think I understand your question?

[gregghutchence]

0 is an element in the set. it is not the empty set. i think make a distinction between a set with 0 elements and the set with the element 0 in it?

[chirho23]

Methods of proof which is very important in abstract algebra should have been included in the introduction.

[VeritySeeker]

At this level you can maybe talk about methods of proofs, which would be confirming axioms. But there is no general method of doing a proof.

[zzzzzzd]

you are a legend, when i get home i will finish watching the rest of them, thank you so much you are awesome

[nonsentence]

this video was very helpful. i remember proving vector spaces, not long ago, and this follows in a very clear and accessible way.

[justwaving]

Excellent stuff!

[rehcaethtam]

Dear VeritySeeker.

Thanks for your efforts to make those useful videos about abstract Algebra.

For the time being, I've read part 1 to 3.

I will continue studying the whole video series.

To understand more abstract algebra is one of my dreams, and so thank you very much again.

Math learning and teaching mania

rehcaethtam

[chordsequencer001]

I believe I have correctly realized that not any one subset of the power set of a universe set S can serve as an identity element, maybe the NULL SET can...and the P(S) of S definitely does not consitute a group

[VeritySeeker]

A power set with union is not a group, and neither with intersection. They are however both binary operators. You can make a power set of a set into a group by letting the binary operator be the symmetric difference. The symmetric difference between sets X and Y is all elements in the union that is not in the intersection. The empty set will then be the identity. Not sure if I am being clear...

[britcruise]

your math vid is a dope show!

[zugimon21]

this is amazing, i love you, please keep up the good work

[trystanlea]

Thanks again, really good pace, very clear.

[VeritySeeker]

Thanks trystanlea, for watching and commenting.

[checkmatebtch]

I am interested in pure mathematics and am self studying linear algebra as of now.. Please continue this work. Your presentation is very CLEAR.

[VeritySeeker]

Thanks for your comments. I am glad it is clear.

It is very nice that you started to study linear algebra on your own. It is also a very good place to start. For many it is the first meeting with somewhat abstract concepts, like vector spaces.

[notoriousvcm]

Yes, unfortunately, for others, like me, it is nothing but hell. IFF you have a BAD Instructor, that can't even prove stuff her self.

Yep, she could not prove that a nonzero element "a" in Zm is a zero divisor IFF "a" and "m" are not relatively prime.

I didn't know what that was trying to say, so I asked her to help me... well... she couldn't. So, I still dont know.

Thankfully, there is VeritySeeker on YouTube!

That way, I hope to some day prove it to my self.

Thanks again.

[JustinDude5]

These are good. I like them :)

KEEP THEM COMING!!

[VeritySeeker]

Thank you for watching, JustinDude5. I will make more.