Consider the compact interval $[a,b], a<b$ and a continuous real valued function $f$ . If $f(a).f(b)<0$ then there exists $c\in(a,b)$ such that $f(c)=0$

WLOG consider $f(a)<0$ and $f(b)>0$ . The other case can be proved using $-f(x)$ which will also verify the theorem's conditions.

consider $a_1 = \frac{a+b}{2}$ , three cases can occur:

• $f(a_1)=0$ , in this case the theorem is proved $c=a_1$
• $f(a_1)>0$ , in this case consider the interval $I_1 = (a,a_1)$
• $f(a_1)<0$ , in this case consider the interval $I_1 = (a_1,b)$

so starting with an open interval $I_0 = (a,b)$ we get another open interval $I_1 \subset I_0$ with length half of the original $|I_1| = \frac{|I_0|}{2}$ .

Repeat the procedure to the interval $I_n$ and get another interval $I_{n+1}$ .

We can thus define a succession of open intervals $I_n$ such that $I_{n+1} \subset I_n$ , $|I_n|=2^{-n}|I_0|$ , such that $I_n = (a_n,b_n)$ and $f(a_n)<0<f(b_n)$ .

The succession $c_{2n} = a_n, c_{2n+1}=b_n$ is Cauchy by construction since $m>n \implies |c_m-c_n|<2^{-[n/2]}|I_0|$ .

$c_n$ is therefore convergent $c_n\to c\in [a,b]$ , and since $a_n$ and $b_n$ are sub-successions, they converge to the same limit.

$f$ is continuous in $[a,b]$ so $x_n \to x \implies f(x_n) \to f(x)$

By construction

$f(a_n)<0$ and $f(b_n)>0$ so in the limit $\lim_{n \to \infty} f(a_n) = f(\lim_{n \to \infty} a_n)= f(c)\le 0$ and $\lim_{n \to \infty} f(b_n) = f(c) \ge 0$ .

So there exists $c \in [a,b]$ such that $0 \le f(c) \le 0 \implies f(c)=0$ .

But since $f(a).f(b)<0$ , neither $f(a)=0$ nor $f(b)=0$ and since $f(c)=0$ , $c \in (a,b)$