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A Summer of Maths

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Reply 520
Original post by jack.hadamard
Here is an old chestnut.


What is the expression

Unparseable latex formula:

\displaystyle 1 + \frac{1}{\displaystyle 1 + \frac{1}{1 + \frac{\displaystyle 1}{\displaystyle 1 + \frac{1}{1 + \cdots}}}}}}



equal to as a number?


Call the expersion x, re-arange it a bit to get rid of the ... bit and I end up with a quadratic. The answer must be greater than 1 so (1+sqrt(5))/2 is the answer.
Original post by jack.hadamard
Here is an old chestnut.


What is the expression

Unparseable latex formula:

\displaystyle 1 + \frac{1}{\displaystyle 1 + \frac{1}{1 + \frac{\displaystyle 1}{\displaystyle 1 + \frac{1}{1 + \cdots}}}}}}



equal to as a number?


purely off the top of my head:

Spoiler

Reply 522
Original post by nuodai
I like this one; it leads to one of those "when you see it you'll **** bricks" moments. Even more so when you discover that it's equal to 1+1+1+\sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}.


No suprise really. This number has a habbit of appearing in all sorts of weird places.
Original post by jack.hadamard
Here is an old chestnut.


What is the expression

Unparseable latex formula:

\displaystyle 1 + \frac{1}{\displaystyle 1 + \frac{1}{1 + \frac{\displaystyle 1}{\displaystyle 1 + \frac{1}{1 + \cdots}}}}}}



equal to as a number?


Damn - seeing this felt like a nice glass of scotch.
Lets make it interesting. :tongue:


How about

1+2+3+4+54321\displaystyle \frac{\displaystyle 1 + \frac{\displaystyle 2 + \frac{\displaystyle 3 + \frac{\displaystyle 4 + \frac{\cdots}{5}}{4}}{3}}{2}}{1}

as a number?
Original post by jack.hadamard
Lets make it interesting. :tongue:


How about

1+2+3+4+54321\displaystyle \frac{\displaystyle 1 + \frac{\displaystyle 2 + \frac{\displaystyle 3 + \frac{\displaystyle 4 + \frac{\cdots}{5}}{4}}{3}}{2}}{1}

as a number?


Spoiler

Here are a couple of problems from Martin Liebeck's book A Concise Introduction to Pure Mathematics.

Question(1):

Spoiler



Question(2):

Spoiler



Question(3):

Spoiler



Question(4):

Work out the union n=1An\displaystyle \bigcup_{n=1}^{\infty} A_n, and intersection n=1An\displaystyle \bigcap_{n=1}^{\infty} A_n, for each of the following AnA_n for nNn \in \mathbb{N}.

i) An={xR  x>n}A_n = \{ x \in \mathbb{R}\ |\ x > n\}

ii) An={xR  1n<x<2+1n}A_n = \{ x \in \mathbb{R}\ |\ \frac{1}{n} < x < \sqrt{2} + \frac{1}{n}\}

iii) An={xR  n<x<1n}A_n = \{ x \in \mathbb{R}\ |\ -n < x < \frac{1}{n}\}

iv) An={xQ  21nx2+1n}A_n = \{ x \in \mathbb{Q}\ |\ \sqrt{2} - \frac{1}{n} \leq x \leq \sqrt{2} + \frac{1}{n}\}


Question(5):

Notice that the number of subsets of {1,2,3,...,n}\{1, 2, 3, ..., n\} of size rr is given by (nr)\displaystyle \binom{n}{r}.

By considering subsets, prove by induction that r=0n(nr)=2n\displaystyle \sum_{r=0}^{n} \binom{n}{r} = 2^n.
Reply 527
Original post by jack.hadamard
Answer to the bit in bold: NO!

As it was mentioned earlier, on several occasions, these problems are (already) a lot more advanced than they should be.


Did you have a look at Cambridge's Mathematics Workbook? Oxfords problems are in the OP (first post in this thread).
In case you find these reasonably straightforward, how about STEP papers (in particular STEP III)?

(Feel free to ask any questions about any of the above problems)


I could not find a reading list for Mathematics on Nottingham's website, but probably you could.

Alternatively, either of the two books under category Introductory/First-year books may be helpful to you.
Today (at some point), I will post couple of problems from these two books, so that one can try a piece of them before buying. :tongue:

Also, I think Li Shen's notes on Numbers and Sets (an introductory course) can be quite useful.



Am I? Perhaps, your's is better. :tongue:


Thanks this has restored my confidence a lot!
I think I did see a reading list somewhere I will have to have look again :smile: and thanks ill try a few questions but I will be undoubtedly rusty aha!

Do you think its worth going through a few of the step questions out of that booklet everyone talks about? (silkos I think)
Original post by ben-smith

Spoiler



Yup. Nice, eh? :tongue:

EDIT:

The interested can have a look at Continued fractions, and Engel expansions. :smile:
(edited 11 years ago)
Original post by skibur

Do you think its worth going through a few of the step questions out of that booklet everyone talks about? (silkos I think)


Oh, forgot about it. Yes, definitely. I think you can find the two booklets on the STEP website.
I think the one with title "Advanced Problems in Mathematics" contains a bit more interesting problems, but you can start with the other one.
Reply 530
Original post by jack.hadamard
Here are a couple of problems from Martin Liebeck's book A Concise Introduction to Pure Mathematics.

Question(1):

Spoiler



Question(2):

Spoiler



Question(3):

Spoiler



Question(4):

Work out the union n=1An\displaystyle \bigcup_{n=1}^{\infty} A_n, and intersection n=1An\displaystyle \bigcap_{n=1}^{\infty} A_n, for each of the following AnA_n for nNn \in \mathbb{N}.

i) An={xR  x>n}A_n = \{ x \in \mathbb{R}\ |\ x > n\}

ii) An={xR  1n<x<2+1n}A_n = \{ x \in \mathbb{R}\ |\ \frac{1}{n} < x < \sqrt{2} + \frac{1}{n}\}

iii) An={xR  n<x<1n}A_n = \{ x \in \mathbb{R}\ |\ -n < x < \frac{1}{n}\}

iv) An={xQ  21nx2+1n}A_n = \{ x \in \mathbb{Q}\ |\ \sqrt{2} - \frac{1}{n} \leq x \leq \sqrt{2} + \frac{1}{n}\}


Question(5):

Notice that the number of subsets of {1,2,3,...,n}\{1, 2, 3, ..., n\} of size rr is given by (nr)\displaystyle \binom{n}{r}.

By considering subsets, prove by induction that r=0n(nr)=2n\displaystyle \sum_{r=0}^{n} \binom{n}{r} = 2^n.


what chapter you on :eek: or have you finished it already :colone:
Original post by Rahul.S
what chapter you on :eek: or have you finished it already :colone:


Well, I picked those from two, or three, of the chapters. Otherwise, I finished it last summer.
Reply 532
Original post by jack.hadamard
Well, I picked those from two, or three, of the chapters. Otherwise, I finished it last summer.


moist :biggrin:
Original post by Rahul.S
moist :biggrin:


I think I advertised it enough, but it is a very good and well structured book; i.e. don't jump to the above questions. :smile:
Tomorrow (again at some point, perhaps the next day :biggrin:), I will add some Groups questions from Geoff Smith's book. :tongue:
Reply 534
Original post by jack.hadamard
I think I advertised it enough, but it is a very good and well structured book; i.e. don't jump to the above questions. :smile:
Tomorrow (again at some point, perhaps the next day :biggrin:), I will add some Groups questions from Geoff Smith's book. :tongue:


just started the book....its on the reading list. most of the ideas ive visited when doing STEP :tongue:
Original post by nuodai
...


Does every finite subset of N\mathbb{N} form a cyclic group? (exclude \varnothing) :smile: It would be great if you know a result, or a theorem, that would imply the above, or show it is false.
Reply 536
Original post by jack.hadamard
Does every finite subset of N\mathbb{N} form a cyclic group? (exclude \varnothing) :smile: It would be great if you know a result, or a theorem, that would imply the above, or show it is false.


Under what operation? N\mathbb{N} has no subgroups under addition and only one under multiplication. (Indeed it doesn't make sense to talk about 'subgroups of N\mathbb{N}' since N\mathbb{N} isn't a group under the expected operations.)
(edited 11 years ago)
Original post by nuodai
Under what operation?


That's the punchline, it does not matter. In fact, I don't even need to know what the operation is to use it, all I need is existence. :tongue:
Reply 538
Original post by jack.hadamard
That's the punchline, it does not matter. In fact, I don't even need to know what the operation is to use it, all I need is existence. :tongue:


Well in that case, it doesn't matter that your underlying set is N\mathbb{N}. There are groups of all cardinalities, and so you can stick a group structure on any set you like (be it a subset of N\mathbb{N} or otherwise). For instance I can declare {2,3,613} to be a group by, for instance, saying that 3 is the identity, that 2·2=613, 613·613=2, 2·613=3=613·2, 3·2=2=2·3, 613·2=613=2·613. (Then the group is isomorphic to Z/3Z\mathbb{Z}/3\mathbb{Z}.) More specifically, if ANA \subset \mathbb{N} is any subset and (G,)(G,*) is a group with A=G\left| A \right| = \left| G \right| then any bijection f:AGf : A \to G induces a group structure (A,)(A, \cdot) on AA by defining ab=f1(f(a)f(b))a \cdot b = f^{-1}(f(a) * f(b)). This works when AA is any set, not just a subset of N\mathbb{N}!

Edit: More relevantly to a question, any countable set can form a cyclic group, but it doesn't have to (unless its cardinality is a prime integer).
(edited 11 years ago)
Original post by nuodai
This works when AA is any set, not just a subset of N\mathbb{N}!


Well, I need it in case of N\mathbb{N}, since I'm looking at some sequences of naturals.


What I thought is, if I labelled subset as {a1,a2,...,ap}\{ a_1, a_2, ..., a_p\}, and defined

aiaj=a(i×pj)a_i * a_j = a_{(i \times_{p} j)}

then I think, I get a cyclic group. I assume the operation is associative, but it follows from the associativity of multiplication modulo.

Then, if I define f(ai)=if(a_i) = i, I get that

f(aiaj)=f(a(i×pj))=i×pj=f(ai)×pf(aj)f(a_i * a_j) = f(a_{(i \times_{p} j)}) = i \times_{p} j = f(a_i) \times_{p} f(a_j)

and this is clearly bijective, so it is an isomorphism to Zp\mathbb{Z}_p.


In this case I get a cyclic group, for all sets with a prime number of elements.
However, I need this to be true for every number of elements, and if it helps, they will be naturals.

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