# Continuous function/R/Intermediate value theorem/Fact/Proof

We consider the situation , and show the existence of such an with the bisection method. For that, we put and , we consider the arithmetic mean , and we compute

If , then we put

and if , then we put

In each case, the new interval is lying inside the initial interval and has half of its length. It fulfills again the condition , therefore we can apply the same defining method again and get recursively a family of nested intervals. Let denote the real number which is determined by these nested intervals. For the lower bounds of the intervals, we have , and this carries over to the limit , due to the criterion for continuity in terms of sequences. Hence, . For the upper bounds, we have , and this again carries over to , so . Therefore, .