Solution 3
The condition is equivalent to . After the substitutions and , we have to prove that . Now let . Hence , . Finally, integrating by parts, for example, , we see that the equality holds true.

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 08042013 12:09

und
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 08042013 12:32
(Original post by ukdragon37)
As expected, the problems I'm interested in are more quaint than the usual Here's one that's strange but rather easy once you get it.
Problem 6**/***
Let where we define for all , . For any set we write to denote the smallest member of such that for all . (Hence .)
Let be a function such that:
Show that the value F given by satisfies and is the smallest member of to do so.
I see why that is the case but the problem for me here will be clarity and rigour in how I present the proof.Last edited by und; 08042013 at 12:38. 
ukdragon37
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 08042013 12:43
(Original post by und)
Just to clarify... is the set of natural numbers up to and including , and is the greatest element of any subset of or in case of the empty set it is 1?
I see why that is the case but the problem for me here will be clarity and rigour in how I present the proof.
EDIT: Actually just a slight clarification/correction: suppose is infinite but does not include , in this case is still clearly so your definition that it is the "the greatest element of any subset of " is not technically right.Last edited by ukdragon37; 08042013 at 12:46. 
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 08042013 13:41
(Original post by ukdragon37)
Actually just a slight clarification/correction: suppose is infinite but does not include , in this case is still clearly so your definition that it is the "the greatest element of any subset of " is not technically right. 
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 08042013 13:42
One simple number theory problem.
Problem 7*
Let be a prime number. Find all triples of positive integers such that .
@ukdragon37 I am still unable to comprehend your definition of . Can we write ? And if this is the case, then is not finite? 
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 08042013 14:07
(Original post by Mladenov)
@ukdragon37 I am still unable to comprehend your definition of . Can we write ? And if this is the case, then is not finite?Last edited by ukdragon37; 08042013 at 14:11. 
Felix Felicis
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 08042013 14:38
(Original post by joostan)
x
I can't even begin to comprehend ukD's question so here's joostan's
Last edited by Felix Felicis; 08042013 at 18:50. 
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 08042013 14:42
(Original post by Felix Felicis)
Solution 5
I can't even begin to comprehend ukD's question so here's joostan's

ukdragon37
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 08042013 14:45
(Original post by Felix Felicis)
I can't even begin to comprehend ukD's question so here's joostan's 
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 08042013 14:48
Partial Solution 6
The first assertion is true since , and
I shall edit, if I manage to prove that is the smallest element such that .
By induction, we obtain . Now suppose that there is such that:
;
.
Then there is with . Therefore, for every , we have  contradiction.Last edited by Mladenov; 08042013 at 15:30. 
Felix Felicis
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 08042013 14:49
(Original post by joostan)
The first part was a warm up, to get you thinking about De Moivre's Theorem, some people need a hint Felix  Lol same about ukD's
(Original post by ukdragon37)
Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on Alevel knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists. 
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 08042013 14:54
(Original post by ukdragon37)
Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on Alevel knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
But still, I agree things like groups, sets and rings seem fascinating and applicable. 
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 08042013 15:04
Problem 8*
A harder version of problem 0. Prove that for any n, it is possible to fine n consecutive integers such that none of them are prime powers. 
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 08042013 15:04
(Original post by joostan)
IMHO it looks like a horrendous mess of notation
But still, I agree things like groups, sets and rings seem fascinating and applicable.
Spoiler:Showis the supremum or least upper bound of . The conditions that satisfies are exactly that of a Scottcontinuous function. is of course the lowest transfinite ordinal. 
Felix Felicis
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 08042013 15:16
(Original post by ukdragon37)
To be fair, I am trying to sneak in a few interesting concepts, which is why I had to expend notation defining them. But this kind of question is very much like what a nasty university examiner (or even interviewer ) would give you  define some new concepts and see what you can make of them.
Spoiler:Showis the supremum or least upper bound of . The conditions that satisfies are exactly that of a Scottcontinuous function. is of course the lowest transfinite ordinal. 
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 08042013 15:19
(Original post by Felix Felicis)
I can see you being a totally nasty supervisor to the undergrads at Cornell 
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 08042013 15:21
(Original post by ukdragon37)
They don't have supervisions there, only recitations to mediumsized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads. 
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 08042013 15:39
(Original post by Felix Felicis)
Aww You'll have your dictatiorial rise to power some day
(Original post by Mladenov)
Partial Solution 6
The first assertion is true since , and
I shall edit, if I manage to prove that is the smallest element such that .
By induction, we obtain . Now suppose that there is such that:
;
.
Then there is with . Therefore, for every , we have  contradiction. 
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 08042013 17:12
(Original post by ukdragon37)
Nearly there, you want to make the last bit clearer. For example it is not a contradiction if f is a constant function. But I'm glad someone is able to look beyond the notation and make quite a bit of headway.
Solution 6
Firstly, , and lead to
Secondly, by induction, we obtain . Consequently, is a directed complete partial order, and its supremum is obviously .
If is a constant, let for some . Then obviously the smallest such that is . Furthermore, .
Now suppose that is not a constant function. Suppose also that there is such that:
;
.
Then there is with . Therefore, for every , we have , which implies that is not the supremum of . 
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 08042013 17:32
Solution 7:
Last edited by j.alexanderh; 09042013 at 11:18.
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