Maths Uni Chat

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Reply 4500

assmaster

My knowledge of rings is now possibly worse than before.
I appreciate the effort though, Simps.

Reply 4501

Simplicity

harr

Do your notes say something along the lines of o:RxR->R (o representing multiplication)?

No

http://www.maths.manchester.ac.uk/~rs/AlgStrNotes13.pdf

Reply 4502

harr

Simplicity

No

http://www.maths.manchester.ac.uk/~rs/AlgStrNotes13.pdf

I suppose it technically must be closed by the definition of a binary operation (a binary operation must map into the set that it acts on), but I wouldn't have thought that many people would know the exact definition, or bother to think about the exact definition. They're taking a big risk by neither explicitly mentioning closure nor writing o:RxR->R. Did they mention closure when you defined a group?

Reply 4503

Simplicity

harr

Yeah, looking back he did define it that way.

And no he didn't actually mention closure when defining a group. Which, is strange as I'm used to thinking of a group having four axioms and not three.

P.S. To be fair, that sounds like the correct way. Whats the point of saying something closed if by the definition of a binary operation it is closed?

Reply 4504

harr

Simplicity

The point is that you didn't realise that it had to be closed, while you would have known if he'd pointed it out. I'm not saying that he should have given it as an axiom, but I think it should have been mentioned. It wasn't explicitly mentioned when we defined a ring, but the lecturer did write o:RxR->R.

Reply 4505

Hedgeman49

Definitions of group, ring and field as I have been taught:

A set G is a group equipped with binary operation # if:
- Closed (for a,b in G a#b is in G)
- Associative
- Contains identity element
- For any element g in G, g^-1 is in G. (inverse elements). If the operation is + then this element is -g. If * then 1/g.
This group is Abelian if a#b = b#a for all a,b in G.

A ring is a set R equipped with TWO binary operations (usually + and *) and elements 0 and 1 (which can be equal) that satisfy:
- R equipped with + and 0 is an Abelian group.
- Multiplication is closed, associative, has an identity element (1) and must be commutative but does NOT require inverses.
- Distributive (ie. for a,b,c in R, a*(b+c) = ab + ac, etc. etc.)

Finally, R is a field if 0 and 1 are distinct, and each non-zero element has a unique two-sided multiplicative inverse.

P.S. I realise that these definitions differ from country to country (and mathematician to mathematician) and this is just what I've been taught.

P.P.S. I agree that the positive integers equipped with addition is not a group (no inverse).

P.P.P.S. Just wanted to see what it feels like to post like Simps

Reply 4506

Simba

Ah, the wonderful joy of listening to techno/dance music at 1:15am whilst studying binary quadratic forms... Part II number theory is fun (perhaps this should read "loud techno/dance music can make stuff a million times more exciting"!)

Graph theory next year, woo! So neat and interesting, wanted a book on it from the library but someone had removed it... :frown:

Dizmo

The last two years I've gotten firsts and both times my director of studies/tutor has gone "This is a respectable result but you can still do better/this is below your standards." God I hate Cambridge.

Does anyone else have a tutor like this?

I swear he's said something akin to all of the following in the past year:

"LOL DO LOTS OF COURSES"
"NO WAIT FOCUS ON YOUR STRENGTHS"
"NO WAIT DO ALL THE COURSES"
"DON'T WORRY EXAMS ARE JUST THERE TO ENCOURAGE YOU TO STUDY"
"BUT JUST GETTING A FIRST ISN'T GOOD ENOUGH"

The atmosphere here makes me want to just get my degree then get out sometimes.

Congratulations on your first :smile:

.

Hmm, I suppose it depends who your DoS is. Both my DoSs so far have seen me at the start of each term, and that has been pretty it. Perhaps your DoS just likes to try to get involved with students. I suppose at Trinity Hall there are a lot less of you so it's easier for the DoS to take an active interest. My DoS doesn't really seem to make a big deal about what I get in my exams (though I suspect he'd care a lot more if I got a 3rd/worse, eeep, bye bye Trinity...)

Come to think of it, I did email my DoS in the middle of Michaelmas term:

"Analysis II is dreadful, I'm dropping it."
"But analysis II is easy, everyone should do it!"
"Raaaaaaaaaaaaaaar, do not want!" (or something to that effect)

I ended up dropping it after 10 lectures or so, ugh, hated it. Very poorly lectured, and just a subject matter that was kind of bleh :\ . Though I do enjoy metric spaces, and bagged some marks on that stuff in the exam. Kind of a shame since I thought analysis I was fine.

At the start of the second term, I indicated I wanted to take six of the courses, and my DoS asked "are you sure you want to take that many?" I was a little taken aback considering when I looked on his sheet there were people taking all eight, but whatever... Only ended up doing one of them for exams anyway :p:

the applied courses in Lent were disgusting. Vector calculus is a nightmare, makes as little sense as analysis.

Plan for next year is to specialise in quantum mechanics, and hopefully the exam questions won't be yuck like this year. Isn't it ironic how three years ago on MSN we were saying to each other how applied maths is not worth studying and being so very purist :p:

... Ah, the joys of university maths being so different and unexpected...

Reply 4507

Simplicity

harr

To be fair, I was thinking like if <R,+> is a group, then that implies that <R,x> is closed under multiplication. Or something like that. A bit confusing at first though.

Hedgeman49

A ring is a set R equipped with TWO binary operations (usually + and *) and elements 0 and 1 (which can be equal) that satisfy:
- R equipped with + and 0 is an Abelian group.
- Multiplication is closed, associative, has an identity element (1) and must be commutative but does NOT require inverses.
- Distributive (ie. for a,b,c in R, a*(b+c) = ab + ac, etc. etc.)

Finally, R is a field if 0 and 1 are distinct, and each non-zero element has a unique two-sided multiplicative inverse.

P.S. I realise that these definitions differ from country to country (and mathematician to mathematician) and this is just what I've been taught.

I can understand closure for multiplication, but having an identitiy and being commutative. Thats pretty strong. Certainly, are matrices rings from your definition? as matrices aren't normally communtative. Surely, communtativity is a mistake?

Hmm, looking at that I wonder if my definition of field is the same as yours. Note, doesn't your definition fail when R={0}? i.e. the gay case.

As for field.

Definition: A communtative ring R is a ring in which multiplication is communtative.

Definition: A field is a communtative ring with one in which every non- zero element is invertible.

They seem the same. Yeah, they are actually the same. Except, 0 and 1 being distinct can be deduced from definition and 0=1 in R={0}.

P.S. The person who wrote the notes is German so he probably been taught the German way. England is going to lose so badly tomorrow, I'm going to watch but its going to be painful.

P.P.S. Wow, this saved me a lot of thinking time and been really helpful. People on this thread are awesome.

Reply 4508

harr

Simplicity

To be fair, I was thinking like if <R,+> is a group, then that implies that <R,x> is closed under multiplication. Or something like that. A bit confusing at first though.

I gave an example of an abelian group <R,+> where <R,x> isn't closed earlier.

Reply 4509

Simplicity

England is getting killed.

Beaten by Germans again.

P.S. Tbf, England are crap. England will never win the world cup.

Reply 4510

Hedgeman49

Simplicity

Some people do not insist on commutativity in rings... but the definition I have does. By my definition n x n matrices on a ring would not form a ring, but without commutativity they would.

Reply 4511

around

in other brighter news ENGLAND ARE WINNING THE CRICKET which is the only sport which matters

Reply 4512

.matt

Never gotten round to plotting functions of complex variables before - Gamma gets pretty funky when restricted to the line x+it with t small! Here it is with t varying from -0.2 to 0.2:

Wish we had 4 spacial dimensions so that we could plot complex functions properly :frown:

Gamma.gif152.3KB

Reply 4513

Simplicity

.matt

Never gotten round to plotting functions of complex variables before - Gamma gets pretty funky when restricted to the line x+it with t small! Here it is with t varying from -0.2 to 0.2:

Wish we had 4 spacial dimensions so that we could plot complex functions properly :frown:

Can you really think of a complex axis as another dimension?

I don't see how its like R^4. Like be R^3+ a complex axis.

P.S. Higher dimensions are probably overrated.

P.P.S. Also, 4 spacial dimensions, don't we already have 4 spatial dimensions.

Reply 4514

.matt

Simplicity

I've always thought of C as pretty much the same as R^2 (f(x+iy) = (x,y) is an isometry with the standard metrics on each space).

So we'd basically need a way of plotting functions R^2 -> R^2. To fully show the graph of a function f:R^n -> R^m, (x,f(x)), we need n+m dimensions. For parametric plots you can get away with having fewer, but information is lost about which point goes to where. e.g. for a function R -> R^3, you could take the 'graph' just to be the image of the domain: r_n(t) = (cos(nt), sin(nt),1) then has the same image for all n>1 when t runs over some interval of length 2 pi.

I meant 4 dimensions that I can see :frown:

Reply 4515

Simplicity

I was wondering does anyone on here don't think 0.9 recurring=1?

Simplicity

I was wondering does anyone on here don't think 0.9 recurring=1?

Of course it doesn't!

Reply 4518

harr

Simplicity

I was wondering does anyone on here don't think 0.9 recurring=1?

Is this because of the Newcomb's Paradox thread? Would you pick both boxes or only the closed one?

(Regarding what you said in that thread, I personally find it more obvious that 0.999...=1 than 0.333...=1/3.)