Intermediate value theorem/Bisection method/3/Method

Let ${}a\leq b$ be real numbers, and let ${}f\colon [a,b]\rightarrow \mathbb {R}$ denote a continuous function such that ${}f(a)\leq 0$ and ${}f(b)\geq 0$. Then the function has a zero within the interval, due to the Intermediate value theorem. Such a zero can be found by the bisection method, as described in the proof of the Intermediate value theorem. We put ${}a_{0}=a$ and ${}b_{0}=b$, and the other interval bounds are inductively defined in such a way that ${}f(a_{n})\leq 0$ and ${}f(b_{n})\geq 0$ hold. Define ${}x_{n}={\frac {a_{n}+b_{n}}{2}}$ and compute ${}f(x_{n})$. If ${}f{\left(x_{n}\right)}\leq 0$, then we set

$a_{n+1}:=x_{n}\,\,{\text{ and }}\,\,b_{n+1}:=b_{n},$

and if ${}f{\left(x_{n}\right)}>0$, then we set

$a_{n+1}:=a_{n}\,\,{\text{ and }}\,\,b_{n+1}:=x_{n}.$

In both cases, the new interval ${}[a_{n+1},b_{n+1}]$ has half the length of the preceding interval and so we have bisected intervals. The real number ${}x$ defined by these nested intervals is a zero of the function.