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Message: [QUOTE="Jacob Fugal, post: 4525727"] WARNING: Just more math geeking ahead. If you don't care for this section of the thread, just skim on past... :) A nitpick, it should be: u (1 - u) + v * (b - v) + w * (b^2 - w) >= u (1 - u) + b^2 - 1 rather than a strict less than. Doesn't affect the outcome of the proof, since the following inequality is still correct. In the case where v = 0 and w = 1 u (1 - u) + v * (b - v) + w * (b^2 - w) = u (1 - u) + 0 * (b - 0) + 1 * (b^2 - 1) = u (1 - u) + b^2 - 1 For those following along at home, it might be hard to see why it must be less in all other cases -- I had a hard time with it for a while. Imagine v > 0. v * (b - v) > 0 since b > v. So that term can only make the expression larger. Also imagine w > 1. w * (b^2 - w) = w * b^2 - w^2 = b^2 + (w - 2) * b^2 + b^2 - w^2 = b^2 + (w - 2) * b^2 + (b - w) (b + w) Since b > w, b - w > 0. Also, b + w > b >= 2. So (b - w) (b + w) >= 2 > 1: w * (b^2 - w) > b^2 + 1 + (w - 2) * b^2 Now since w >= 2, (w - 2) * b^2 >= 0, and we have: w * (b^2 - w) > b^2 + 1 And that term can only increase as well. So as Michael showed, g(x) < x for all x with three or more digits in the respective base. Jacob Fugal [/QUOTE]