No continuous function satisfies f(f(x)) = minus x

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Is there a function f from the real numbers to the real numbers such that f(f(x)) = minus x for every x? 1. Show that if such a function is continuous, a contradiction follows. 2. Show that if continuity is dropped, such a function does exist.

Accepted Answer

There is no continuous solution. Without continuity, solutions exist. Part 1: continuity forbids it. - f is one-to-one. If f(x) = f(y), then applying f gives f(f(x)) = f(f(y)), so minus x = minus y, so x = y. - A continuous one-to-one function on the reals is strictly monotone. If it went up and then down it would have to take some value twice, by the intermediate value theorem. So f is either strictly increasing or strictly decreasing. - Either way, f applied twice is increasing. Increasing after increasing is increasing; decreasing after decreasing is increasing (two reversals cancel). - Bu…

No continuous function satisfies f(f(x)) = minus x

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