2013 actually it was the first ISL problem I solved.
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Renzhi10122
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 09062015 17:14

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 09062015 18:00
(Original post by Renzhi10122)
2013 actually it was the first ISL problem I solved. 
Renzhi10122
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 09062015 18:07
(Original post by KongShou)
Whoa nice! I didnt really bother with that questions because iirc it was really messy with a case bash at the end right? I dont even remember the first ISL question ive done haha 
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 09062015 22:25
*Prove there are irrational numbers such that is rational.
Spoiler:ShowNot the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.Last edited by poorform; 09062015 at 22:27. 
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 09062015 22:54
(Original post by poorform)
*Prove there are irrational numbers such that is rational.Spoiler:ShowNot the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.Spoiler:Show
Suppose not, let X be the set of positive irrational numbers. Then define f: X > Q by f(x) = the rational number y such that x^y = 2. If z =/= x then f(x) =/= f(z) since for f(x) = x^y is increasing or decreasing for fixed y =/= 0. Then f is an injection, so X is countable, contradiction.Last edited by metaltron; 09062015 at 22:56. 
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 10062015 10:04
(Original post by metaltron)
Spoiler:Show
Suppose not, let X be the set of positive irrational numbers. Then define f: X > Q by f(x) = the rational number y such that x^y = 2. If z =/= x then f(x) =/= f(z) since for f(x) = x^y is increasing or decreasing for fixed y =/= 0. Then f is an injection, so X is countable, contradiction. 
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 10062015 10:07
(Original post by poorform)
*Prove there are irrational numbers such that is rational.Spoiler:ShowNot the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.

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 10062015 13:11
(Original post by Smaug123)
It's a very nice problem I'll give the pretty solution I know. 
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 10062015 13:38
(Original post by poorform)
*Prove there are irrational numbers such that is rational.Spoiler:ShowNot the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.
Then , and x and y are irrational
And finally the nicest proof I know follows if you establish the following: 'If we have rational then x is an integer or irrational' (which is kind of obvious if you think about it)Last edited by TheMagicMan; 10062015 at 13:50. 
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 10062015 20:14
Last edited by metaltron; 10062015 at 20:18. 
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 10062015 20:48
(Original post by metaltron)
I don't think so, as long as pi^y = 2 for some y. f must be defined since we're doing a proof by contradiction, so we've assumed that x^y is never rational if x and y are both irrational, so if x is irrational and x^y = 2 then y is rational. The direct proofs are also nice of course
In particular, the proof that argues on requires only that this function is increasing and continuous on (1, 2) which is very easy to proveLast edited by TheMagicMan; 10062015 at 21:09. 
metaltron
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 10062015 21:05
(Original post by TheMagicMan)
You argument is certainly valid, but it uses a much stronger result than what we are trying to prove so it sits a bit uncomfortably. It's sort of like using a contour integral and the Residue Theorem to work out : it's certainly possible but is it desirable?
In particular, the proof that argues on [tex]x^x/tex] requires only that this function is increasing and continuous on (1, 2) which is very easy to prove
Consider also that when we prove that transcendental numbers exist it is easier to use a countability argument than to construct a transcendental number explicitly, so I think its nice to see direct and nondirect ways to solve this problem. 
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 10062015 21:13
(Original post by metaltron)
I wouldn't say my proof is completely over the top. After all some intuition for the original problem is there are so many more irrationals than rationals, so you cannot find a rational power for every irrational number to make it x^y = 2, which might suggest countability as a route to a solution.
Consider also that when we prove that transcendental numbers exist it is easier to use a countability argument than to construct a transcendental number explicitly, so I think its nice to see direct and nondirect ways to solve this problem. 
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 10062015 21:17
(Original post by TheMagicMan)
It's not completely over the top at all.Last edited by metaltron; 11062015 at 00:35. 
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 11062015 23:10
Haven't had a proper look through this thread so this might be too easy:
Problem 497 *:
Find all positive integers such that 
Renzhi10122
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 12062015 18:55
(Original post by Gawain)
Haven't had a proper look through this thread so this might be too easy:
Problem 497 *:
Find all positive integers such that
Let where
We rewrite the equation as
Suppose , then where both sides must be integer. Thus but and are coprime so . But , contradiction, so
Now we have where both sides are integer so . Thus,
Thus, where
If .
If ,there are no solutions
If
If
For , by considering for , and , so we have found all solutions being,
Last edited by Renzhi10122; 12062015 at 18:57. 
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 12062015 19:54
(Original post by Renzhi10122)
Suppose . We have so clearly, . Now consider when . From the original equation, since must be an integer more than 1.
Let where
We rewrite the equation as
Suppose , then where both sides must be integer. Thus but and are coprime so . But , contradiction, so , for , and , so we have found all solutions being,
Here are some more shorter questions I found interesting:
1) Show that any palindromic number with an even number of digits in base n is divisible by n+1.
2) Let be a permutation of the sequence 1,2,3,4,5,6 such that for all is not a permutation of
How many such permutations are there?
3) Not as quick
a)
By considering an equilateral triangle side length equal in area to 3 equilateral triangles side length show that is irrational.
Hint: Using the four triangles described above how can you construct another equilateral triangle side length
b) Prove that is irrational by considering pentagons.
c) Can we extend the proofs above to show that either is an integer or irrational?Last edited by Gawain; 12062015 at 20:02. 
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 01072015 14:39
Came across this problem online the other day and thought it deserved to be here. Hopefully this hasn't been posted before.
Find .* 
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 01072015 20:28
(Original post by poorform)
Came across this problem online the other day and thought it deserved to be here. Hopefully this hasn't been posted before.
Find .*
Solution:

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 01072015 21:10
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