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# The Proof is Trivial! watch

1. 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.
2. (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:

• For all m, n, .
• For all , we have .

Show that the value F given by satisfies and is the smallest member of to do so.
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.
3. (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.
Yes all you said is correct.

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.
4. (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.
But how can be infinite if
5. 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?
6. (Original post by Lord of the Flies)
But how can be infinite if
For example let be all the even numbers.

@ukdragon37 I am still unable to comprehend your definition of . Can we write ? And if this is the case, then is not finite?
I'm not sure what you are trying to do. All am saying is is the same as the set of natural numbers except with the addition of an "infinity" element that is defined to be geq all the naturals (and trivially geq itself). is definitely infinite.
7. (Original post by joostan)
x
Solution 5
I can't even begin to comprehend ukD's question so here's joostan's

part i

and

Equating real and imaginary parts

and as required. Really joostan?

S_1
Consider the expansion of

and

The sum in question is

and

as required

S_2
Similarly,

8. (Original post by Felix Felicis)
Solution 5
I can't even begin to comprehend ukD's question so here's joostan's

part i

and

Equating real and imaginary parts

and as required. Really joostan?

S_1
Consider the expansion of

and

The sum in question is

and

as required

S_2
Similarly,

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
9. (Original post by Felix Felicis)
I can't even begin to comprehend ukD's question so here's joostan's
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 A-level 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.
10. 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.
11. (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
Ha, fair enough

(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 A-level 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.
Sets does look quite interesting, definitely something to look at over the summer
12. (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 A-level 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.
IMHO it looks like a horrendous mess of notation
But still, I agree things like groups, sets and rings seem fascinating and applicable.
13. 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.
14. (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.
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:
Show
is the supremum or least upper bound of . The conditions that satisfies are exactly that of a Scott-continuous function. is of course the lowest transfinite ordinal.
15. (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:
Show
is the supremum or least upper bound of . The conditions that satisfies are exactly that of a Scott-continuous function. is of course the lowest transfinite ordinal.
I can see you being a totally nasty supervisor to the undergrads at Cornell
16. (Original post by Felix Felicis)
I can see you being a totally nasty supervisor to the undergrads at Cornell
They don't have supervisions there, only recitations to medium-sized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads.
17. (Original post by ukdragon37)
They don't have supervisions there, only recitations to medium-sized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads.
Aww You'll have your dictatiorial rise to power some day
18. (Original post by Felix Felicis)
Aww You'll have your dictatiorial rise to power some day
I look forward to it

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.
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.
19. (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.
Let me try again.

Solution 6

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 .
20. Solution 7:

Spoiler:
Show

since is prime.

Now considering

If we have the solution

It is easily verified that there is no solution if

If ,

which clearly has no solutions.

Hence the only solution is

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