Can this function exist (see definition below)?

Question

Can there exist a continuous function f:R->R that sends rationals into irrationals and irrationals into rationals?

I'm confused

Answer 1

yes and here it is :

f : x --> 0

It has al the desired properties you want.


update.
I see now i didnt read well the question....


dulot i dont understand ...

" I make a demonstration by absurd. Let f be such a function. The image f(R) of f is not countable because f is not constant and its image is an interval."

<=== that I dont get

Why can the image of f(R) be not countable ?
sending all rationals to the same irrational ( or to a countable number of rationals) makes the image of f(R) countable in my opinion.

last update ...

digesting ...

http://planetmath.org/encyclopedia/ThereIsNoContinuousFunctionThatSwitchesTheRationalNumbersWithTheIrrationalNumbers.html

Answer 2

Of course, sending rationals to irrationals is easy, just multiply by an irrational number.

But one that brings irrationals into the rationals at the same time?

I don't know.

Answer 3

The answer is no.

I make a demonstration by absurd. Let f be such a function. The image f(R) of f is not countable because f is not constant and its image is an interval. The image of rational numbers is countable because the rational numbers form a countable set. The image of irrationals is countable too because it is included in rational numbers's set. We have :f(R)=f(Q) union f(R \ Q). The union of two countable set is countable. Contradiction.

Answer 4

best I can think of is something like x/(pi) but then thats only for multiples of pi and nothing like the squrt(3).