Is there a non-zero rational number q such that sin(q) is rational (radian measure)?

How about a transcendental number with transcendental sine?

First of all, since the set of algebraic numbers is countable, and since the sine function is injective in [0,pi/2], the image of the trancedentals must be non-countable, and thus must contains at least one number which is not algebraic, hence, there is a trancedental the image of which is also trancedental.

Abount the rational numbers: I don't know tha answer yet, but my guess is there is no such number, the intuition this: define an equialence relation on real numbers such that two numbers are equivalent if they differ by a rational number (a ~ b if and only if a-b=r, r rational). then the set of equivalency classes can be shown to be non-countable. one of these classes is the class of rational numbers. now, for one rational to have a rational sine, the image of the rationals under sine (which is a countable set) must have at least one number is the equivalency class of rationals.

Posted by ronen on 2005-05-02 21:44:54