1. The problem statement, all variables and given/known data Let f(x) be continuous in the closed interval [0,1]. Suppose further that f(x) assumes rational values only and that f(x) = 1/2 when x = 1/2. Prove that f(x) = 1/2 everywhere. 2. Relevant equations N/A 3. The attempt at a solution If f is continuous at x = 1/2, then for every positive e (epsilon), we can find a positive number d (delta) such that |f(x) - f(1/2)| < e for all x in the domain of f such that |x-1/2| < d. Now, we have f(1/2) - e < f(x) < f(1/2) + e. F assumes rational values only, but we know that between any two rationals numbers, there are an infinite amount of irrational numbers, so there is an irrational value, say c in the open interval f(1/2) - d < x < f(1/2) + d. Since the function is continuous, we will take a smaller epsilon, say e1, with e1 < e. This is done as to "get rid" of the irrational. However, again by the density of the irrationals in the reals, there is another irrational value in that even smaller epsilon-neighborhood. In an interval length of 2en, where en > 0, we can always find an irrational in this interval. So we msut take e = 0 and the inequality becomes f(1/2) = 1/2 < f(x) < f(1/2) = 1/2, meaning that f(x) = 1/2. QED For starters, I'm not sure if this proof is correct. Also, would the inequality 1/2 < f(x) < 1/2 need to be changed to 1/2 <= f(x) <= 1/2 in order for it to imply that f(x) = 1/2?