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# Generating the rationals watch

1. http://www.bmoc.maths.org/home/bmo1-2005.pdf

I've attempted question 5, but it seems like one of those questions where you can quickly work out how to do it, but it takes ages to write it down.
And I'm pretty sure I've left out some checks, but can't find them. If anyone could confirm this proof is valid or find part that need clarification, that would be great..

Proof

Define a function s: Qn(0,1) --> N \ {1,2}
r=a/b --> a+b
Here a and b have no common factors (apart from +/- 1) and are both positive.

Then s is well defined and surjective.
Therefore, we are done if we can prove the following lemma:

For all n in {3,4,...}, we have r in S for all r in Qn(0,1) such that
s(r)=n

Proof of Lemma

We prove by (strong) induction on n. The base case is clear from condition i). Assume we have the result for 3,4,...,n-1 (where n>3).
Take an r in Qn(0,1) such that s(r)=n. Note that r is not 1/2 (since n>3). We need to show that r in S.

We can write r=a/b with b>a>0 and a and b have no common factors, and a+b=n.

Case a) 1>a/b>1/2

Then 2a>b, so a>b-a, so 1>(b-a)/a>0.
We observe that a and b-a have no common factor (since a and b have no common factors). So s((b-a)/a) is defined and
s((b-a)/a)=b=n-a<n
By the induction hypothesis, (b-a)/a is in S. But then the first part of condition ii) says that
a/b = 1/((b-a)/a +1) is in S.

Case b) 1/2>a/b>0

Then b>2a, so b-a>a, so 1>a/(b-a)>0.
Since a and b-a have no common factors (as noted above), we have that s(a/(b-a)) is defined and s(a/(b-a))=b<n.
By the induction hypothesis, a/(b-a) is in S. But then the second part of condition ii) says that
a/b = (a/(b-a))/(a/(b-a) + 1) is in S

So we are done, by induction.
2. bump
3. (Original post by Drederick Tatum)
For all n in {3,4,...}, we have r in S for all r in Qn(0,1) such that
s(r)=n
By that, do you mean: "Let n be in {3,4,...}. Then, given any with s(r)=n, we have that "? It's slightly difficult to parse the statement as it stands...
4. Yes
5. What kind of maths level is it? I applied to G100 course since I have enjoyed maths all my life, but the questions asked on this forum about maths are just too hard for me. Kind of scares me.
6. (Original post by AntiFlag)
What kind of maths level is it? I applied to G100 course since I have enjoyed maths all my life, but the questions asked on this forum about maths are just too hard for me. Kind of scares me.
Don't Panic, AntiFlag. BMO1 is an challenge set by the UK Maths Trust (UKMT) to which about 1300 high scorers from the Senior Maths Challenge taken in many schools are invited. Blue question paper last 5th Nov. BMO1 has six questions, which get harder from 1 to 6.

The results have just appeared - about 100 students scored 31 or more out of 60. So Q5 will be accessible to a couple of dozen A level students who are already capable of doing problems at about Cambridge IA Maths Numbers and Sets level.
7. The questions are quite difficult, and quite a lot different from school maths, so I wouldn't worry about not being able to do them.

The people that tend to do well also tend to have done a lot of training in these kinds of problems asked.

To put it in perspective, I took the BMO 1 paper 6 years ago and scored 4/50. I'm now having another look at some of the problems having graduated from Cambridge in maths, and I still find them difficult and cant do plenty of them.

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Updated: January 17, 2010
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