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[ZetaX]

By already knowing.

[MagmaMcFry]

From where? The IMO?

[ZetaX]

In the end, all but very few persons know it originally from the IMO, maybe with some steps in between 
For a couple of reasons, I watch the IMO regularily and look at the problems. But I wasn't there this year, even if it was less than 1000 miles away (I had too much to do to take an offer to be corrector/coordinator there). You can find everything online, though, the problems are normally posted on a certain forum (and sometime even on the official IMO page) within the hour the exams end.

[ed boy]

What is the IMO?

[MagmaMcFry]

International Mathematics Olympiad.

[ed boy]

Here's one that I've been struggling with for a week or two, but I can't find a solution.

"Is it possible to define a real continuous function f such that f(x) is rational if and only if f(x+1) is irrational?"

I suspect the answer is no, but I cannot prove it. I think the proof has something to do with countability of rationals and uncountability of irrationals, but the density of both of them is scuppering my proof attempts.

[MagmaMcFry]

I suspect the answer is yes, although I have no idea why.

[Another]

A Cantor function is a non constant continuous function that is rational outside of Cantor set. Its construction gave me the following idea:

Let f(x)=π*x on [0; 1]. Let f(2)=4. Take a random number x₁ in (0; 1) and if it is rational (irrational) - define f(x₁+1) as any random rational (irrational) number in the middle third between π and 4. Take a second random number x₂ and define f(x₂+1) to be a random number in the middle third between f(x₁+1) and the value of f(x) at the other side of the interval to which x₂+1 belongs with appropriate rationality.
In general at ith step x_i+1 will land between some x_k+1 and x_l+1 (or 1, or 2). There will always be an interval of finite length between f(x_k+1) and f(x_l+1) (or π, or 4) in which we will be able to choose either rational or irrational number in its middle third part for f(x_i+1).

All random choices above can be substituted with some well defined rules but that would leave uncountable many points never to be visited.

I am not sure if uncountable number of steps is allowed in construction of functions and suspect that the axiom of choice is required to make a function based on the above idea defined. If allowed - the function f(x) will be defined on [0; 2] and be not only continuous but also monotonous just like Cantor function but (like the question mark function) never constant.

Since it is still unknown (unproved) whether π-e is rational or irrational and there is no general irrationality testing algorithm - I suspect that an explicit construction of required function is not possible. As in the function in question probably exists but is not strictly definable.

[MagmaMcFry]

Every function f that is generated this way actually has a well-defined set of rules that does not require the axiom of choice, and every function value can definitely be calculated in countably infinite time. So the values of f(x) are in the countable set S := {f(x₁+1), f(x₂+1), ...} for almost every x in (1, 2). But the function is continuous, so the uncountably many values in [π, 4] that don't crop up in S also have to be equal to f(x+1) for some particular x in (0, 1). But what can you say about f(x) in these cases?
You just reduced the problem from (1, 2) to a Cantor-like set within (1, 2), which is also uncountable, so you haven't actually solved anything.
But I had this idea too at first...

[Another]

...But the function is continuous, so the uncountably many values in [π, 4] that don't crop up in S also have to be equal to f(x+1) for some particular x in (0, 1). But what can you say about f(x) in these cases?
...

I know that a countable set can not be continuous and that kills direct approach. I still hoped that it would be possible to place additional conditions while choosing the values for f(x_i) to force the resulting continuous function to converge to points of right rationality everywhere outside of S.

I was not thinking clearly the first time and now I see that any strictly monotonous function on some interval must take countably many rational values and a non-constant (it can't be constant because it has to take both rational and irrational values) continuous function on the corresponding x+1 interval must take uncountably many irrational values. So the required function can not be strictly monotonous on any interval.

[MagmaMcFry]

Now I see that any strictly monotonous function on some interval must take countably many rational values and a non-constant (it can't be constant because it has to take both rational and irrational values) continuous function on the corresponding x+1 interval must take uncountably many irrational values. So the required function can not be strictly monotonous on any interval.

Actually, this only says that f is either non-differentiable or constant in one of the two intervals [a, b] and [a+1, b+1] for all a and b.
But yes, that's kinda the problem here.

[Another]

Either non-differentiable or constant is exactly the same as not strictly monotonous everywhere. And since we can go f(x+1)->f(x+2) that is for both/any intervals.

It can still be constant almost everywhere (in mathematical sense) but that doesn't help at all in determining if it can exist.

[Trapezohedron]

Uhh, how does one simplify complex & rational fractions with variables properly? Oh, and mind giving me a refresher in factoring?

Thanks.

[MagmaMcFry]

Uhh, how does one simplify complex & rational fractions with variables properly? Oh, and mind giving me a refresher in factoring?

Are you talking about polynomials and polynomial fractions?

[Trapezohedron]

Uh yes. I forgot to mention that.