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# The Proof is Trivial! watch

1. (Original post by KongShou)
ISL 2012 A1

Knew it looked familiar
2013 actually it was the first ISL problem I solved.
2. (Original post by Renzhi10122)
2013 actually it was the first ISL problem I solved.
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
3. (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
Yeah. The reason I remember this being the first ISL question I did (or at least, knowingly doing cos they use them in FST) is because I only did my first one a month ago -_-
4. *Prove there are irrational numbers such that is rational.
Spoiler:
Show
Not the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.
5. (Original post by poorform)
*Prove there are irrational numbers such that is rational.
Spoiler:
Show
Not 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.
6. (Original post by metaltron)
Spoiler:
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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.
I may be a fool, but isn't undefined, for instance?
7. (Original post by poorform)
*Prove there are irrational numbers such that is rational.
Spoiler:
Show
Not the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.
It's a very nice problem I'll give the pretty solution I know.

Spoiler:
Show
Let . If is rational, we're done. Otherwise, let . We're assuming wlog that x is irrational; then x^y is 2.
8. (Original post by Smaug123)
It's a very nice problem I'll give the pretty solution I know.
Spoiler:
Show
Let . If is rational, we're done. Otherwise, let . We're assuming wlog that x is irrational; then x^y is 2.
That is also the solution I know. Very nice.
9. (Original post by Smaug123)
It's a very nice problem I'll give the pretty solution I know.

Spoiler:
Show
Let . If is rational, we're done. Otherwise, let . We're assuming wlog that x is irrational; then x^y is 2.
(Original post by poorform)
*Prove there are irrational numbers such that is rational.
Spoiler:
Show
Not the hardest problem but a really interesting one in my opinion. Hopefully it hasn't been posted before.
An easy direct example is to take , for n a positive integer which is not a square, or a power of 2.

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)
10. (Original post by Smaug123)
I may be a fool, but isn't undefined, for instance?
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
11. (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
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 requires only that this function is increasing and continuous on (1, 2) which is very easy to prove
12. (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
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 non-direct ways to solve this problem.
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 non-direct ways to solve this problem.
It's not completely over the top at all.
14. (Original post by TheMagicMan)
It's not completely over the top at all.
Mathematicians hey, if you can't find one of them (transcendental numbers), find lots of them!
15. Haven't had a proper look through this thread so this might be too easy:

Problem 497 *:

Find all positive integers such that
16. (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
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

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,
17. (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,
Nice, although I did it considering prime factors but the same kind of argument.

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?
18. Came across this problem online the other day and thought it deserved to be here. Hopefully this hasn't been posted before.

Find .*
19. (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 .*
This is a pretty cool question.

Solution:

Spoiler:
Show

Define . Note that . This is crucial. Then:

The pattern is relatively easy to spot if you try a couple of small values (N=1, 2) first.

20. (Original post by DJMayes)
This is a pretty cool question.

Solution:
Spoiler:
Show

Define . Note that . This is crucial. Then:

The pattern is relatively easy to spot if you try a couple of small values (N=1, 2) first.

Nice! Are you studying maths at university?

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