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A Summer of Maths

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Original post by TwoTwoOne
Slight plug here, I've recently uploaded my notes to (almost) all courses in Part IA, hope they are useful :smile:

http://lishen.wordpress.com/2012/07/16/some-of-my-notes-for-the-part-ia-cambridge-maths-tripos/

The IA Groups is lectured by Saxl again this coming year I think, so these notes should be relevant enough.


That is excellent, thanks! I will have a proper look at them later tonight. :smile:

Edit: Made the notes noticeable in the OP.
(edited 11 years ago)
Original post by TwoTwoOne
Yeah, just finished. It could've gone better (i.e. if I can do numbers) but I'm reasonably happy with it. You starting this year?


Out of interest, how did you find the Metric and Topological Spaces course? :smile:
Do you have any book suggestions, or perhaps any tips on what is useful to know beforehand?
Reply 442
Terribly confusing, and hugely dry, but otherwise an awesome course! :smile:

The only book worth getting is probably Sutherland's Introduction to Met and Top. I don't think you need to know much beforehand, and just stay awake during lectures would be a good tip (you'd surprised how easy it is to let your mind wonder when you start to not understand stuff).
Original post by jack.hadamard
Question(1):

Spoiler


Btw, I'd put Oxford undergraduate problems in spoilers, so that Oxford people can save them for their actual preparation!


Question(2):

The set SS consists of elements a1,a2,...,ana_1, a_2, ..., a_n, with S>2|S| > 2.

Determine whether SS forms a group under \star, defined by

i)  aiaj=am,    m=max(i,j)\ a_i \star a_j = a_m, \ \ \ \ m = \max(i, j)

ii)  aiaj=am,    m=ij+1\ a_i \star a_j = a_m, \ \ \ \ m = |i - j| + 1

where 1i,jn1 \leq i, j \leq n.


Finally got around to writing solutions for your problems. So don't think we wouldn't care, I do appreciate that you post them.

Quick LaTeX-trick: You can use "\quad" and "\qquad" instead of multiple "\"s.:wink:

Question 1:

Spoiler



Question 2:

Spoiler

Reply 444
Original post by DamoclesAustria
I don't think there's a name for that...
A set with a binary operation is a magma....but I'm guessing that you knew that.
Original post by DamoclesAustria
Finally got around to writing solutions for your problems. So don't think we wouldn't care, I do appreciate that you post them.

Quick LaTeX-trick: You can use "\quad" and "\qquad" instead of multiple "\"s.:wink:


I'm glad you like the questions. Well, I thought people's interest dropped a bit, but it is still more than what I expected. :tongue:
The solutions look very good. About LaTeX, I am a bit primitive -- what does the job, that is what I use. I will try to improve on that. :biggrin:

Original post by TwoTwoOne
Terribly confusing, and hugely dry, but otherwise an awesome course! :smile:
The only book worth getting is probably Sutherland's Introduction to Met and Top.


Thanks for the reply, it is helpful! I don't find it dry -- I like the ideas and how concepts generalise, but I don't feel comfortable with any of them either.
I have that book (it seems very good), but after going through the first few chapters, I decided to do more on Analysis; didn't comprehend it well enough.
Reply 446
Original post by TwoTwoOne
Terribly confusing, and hugely dry, but otherwise an awesome course! :smile:

The only book worth getting is probably Sutherland's Introduction to Met and Top. I don't think you need to know much beforehand, and just stay awake during lectures would be a good tip (you'd surprised how easy it is to let your mind wonder when you start to not understand stuff).


Stick with it. I bliltzed through Analysis and found M&T a shock - once you get used to it and something 'clicks', you'll find M&T a rewarding course. I tend to find that I can either see how to do a problem straight away or its impossible for me - don't know how to plug away at a course like M&T...
Original post by shamika
Stick with it. I bliltzed through Analysis and found M&T a shock - once you get used to it and something 'clicks', you'll find M&T a rewarding course. I tend to find that I can either see how to do a problem straight away or its impossible for me - don't know how to plug away at a course like M&T...


That's interesting because, in some respects, I found topological spaces to be more intuitive than say, the real line. I mean continuity in this setting (i.e. the preimages of open sets being open) is much easier to grasp than epsilon delta stuff from real analysis. Though, tbf, I don't really know much about either :tongue:
Original post by BabyMaths
A set with a binary operation is a magma....but I'm guessing that you knew that.

Yeah, but we have commutativity, an identity and inverses as additional properties. So the question should be: Is there a name for a commutative magma with inverses and an identity?

I checked wikipedia: we "almost" have a "loop". A loop is a quasigroup with an identity. In a quasigroup the equations xa=b,ay=bx*a=b, a*y=b have unique solutions. (Note that we can have this property without an identity and hence without inverses and,conversely, we can have inverses but not this uniqueness-property)

In our example, the equation aiax=aja_i*a_x=a_j has exactly one solution iff jij\ge i and exactly two otherwise.
Perhaps, we can do some Analysis alongside Groups.
(It seems people have done quite a bit of it)


How do you find results like the Intermediate Value Theorem?

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _


Question:

i) Suppose g:[0,1][0,1]g : [0, 1] \to [0, 1] is a continuous function.

By considering f(x)=g(x)xf(x) = g(x) - x, show that there exists c[0,1]c \in [0, 1] such that g(c)=cg(c) = c.


ii) What about if g:[a,b][a,b]g : [a, b] \to [a, b] is a continuous functions?



Footnote: Decipher "Cofu Clobi Ibaatib".
Question:

Consider a set SS and a binary operation * on SS.

By assuming that (ab)a=b(a * b) * a = b for all a,bSa, b \in S, show that

a(ba)=ba * (b * a) = b

for all a,bSa, b \in S.
Reply 451
Original post by shamika
Stick with it. I bliltzed through Analysis and found M&T a shock - once you get used to it and something 'clicks', you'll find M&T a rewarding course. I tend to find that I can either see how to do a problem straight away or its impossible for me - don't know how to plug away at a course like M&T...


Yeah, I'm also beginning to see how it could click. It's definitely not a methods course where you can use recipes in anger. More definition grubbing I guess..

Original post by ben-smith
That's interesting because, in some respects, I found topological spaces to be more intuitive than say, the real line. I mean continuity in this setting (i.e. the preimages of open sets being open) is much easier to grasp than epsilon delta stuff from real analysis. Though, tbf, I don't really know much about either :tongue:


Hmmm, that's interesting, I always found the eps-delta to be more intuitive (not initially though), it also generalises to open balls quite nicely. But I do admit the topological definition is neater.

Original post by jack.hadamard


Footnote: Decipher "Cofu Clobi Ibaatib".


Haha, we had fun deciphering this this year (btw the clue is literally staring at you in the face) :smile:
Original post by TwoTwoOne

Hmmm, that's interesting, I always found the eps-delta to be more intuitive (not initially though), it also generalises to open balls quite nicely. But I do admit the topological definition is neater.


The topological definition also makes certain proofs trivial/really tasty. For example, if memory serves me correctly, If you want to prove that the composition of two continuous functions is continuous then the epsilon delta proof is pretty messy but in the topological setting it becomes obvious.
Again, I have very limited experience so it's highly likely I'll see things differently in the light of actually having to do university maths.

Are you looking forward to 2nd year?
Reply 453
Original post by ben-smith
The topological definition also makes certain proofs trivial/really tasty. For example, if memory serves me correctly, If you want to prove that the composition of two continuous functions is continuous then the epsilon delta proof is pretty messy but in the topological setting it becomes obvious.
Again, I have very limited experience so it's highly likely I'll see things differently in the light of actually having to do university maths.

Are you looking forward to 2nd year?


For the composition you could use the sequential definition of continuity, it comes out in a line or 2 I think. I actually don't remember we need to use the eps-delta that often, the sequential definition usually does the trick.

Yes you definitely will, you'll start to like things you had no idea you were interested in (Groups and Geometry for me, so quite looking forward to them).
Original post by jack.hadamard
Question:

Consider a set SS and a binary operation * on SS.

By assuming that (ab)a=b(a * b) * a = b for all a,bSa, b \in S, show that

a(ba)=ba * (b * a) = b

for all a,bSa, b \in S.


Spoiler

Original post by DamoclesAustria

Spoiler



Yup. :tongue: So, the assumption is a sufficient condition for an operation to be associative.
Original post by jack.hadamard
Yup. :tongue: So, the assumption is a sufficient condition for an operation to be associative.


Are you saying the assumption implies associativity? This is not true, as it only implies associativity for special kinds of products (those of the form abaa*b*a).

As a counterexample, take S=a,b,cS={a,b,c} with the operation
aa=a, bb=b,cc=ca*a=a,\ b*b=b, c*c=c, and ab=c=baa*b=c=b*a and so on.
Then we have (ij)i=j=i(ji)(i*j)*i=j=i*(j*i) for every choise of i and j but a(bc)=(ab)ca*(b*c)=(a*b)*c does not hold.
Original post by DamoclesAustria
Are you saying the assumption implies associativity? This is not true.


Very good, indeed! I like the counter-example. :tongue:

Anyone wants to post some questions? How many binary operations are there on a set?
Original post by jack.hadamard
Very good, indeed! I like the counter-example. :tongue:


I didn't come up with it specifically for that. :wink:
It's always good to have a "non-trivial" example for something in the back of your head.

Prove that any subgroup of (Z,+)(\mathbb Z,+) is generated by one integer or, equivalently, is of the form nZn\mathbb Z for an integer n
Original post by DamoclesAustria

Prove that any subgroup of (Z,+)(\mathbb Z,+) is generated by one integer or, equivalently, is of the form nZn\mathbb Z for an integer n.


Very nice question. :smile:

Spoiler




How about making the result stronger?


Question:

Prove that any non-trivial subgroup of (R,+)(\mathbb{R}, +) is cyclic if and only if it contains a smallest positive element.

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