# proof of intermediate value theorem

We first prove the following lemma.

If $f:[a,b]\to\mathbb{R}$ is a continuous function   with $f(a)\leq 0\leq f(b)$ then there exists a $c\in[a,b]$ such that $f(c)=0$.

Define the sequences $(a_{n})$ and $(b_{n})$ inductively, as follows.

 $a_{0}=a\quad b_{0}=b$
 $c_{n}=\frac{a_{n}+b_{n}}{2}$
 $(a_{n},b_{n})=\begin{cases}(a_{n-1},c_{n-1})&f(c_{n-1})\geq 0\\ (c_{n-1},b_{n-1})&f(c_{n-1})<0\end{cases}$

We note that

 $a_{0}\leq a_{1}\leq\cdots\leq a_{n}\leq b_{n}\leq\cdots\leq b_{1}\leq b_{0}$
 $(b_{n}-a_{n})=2^{-n}(b_{0}-a_{0})$ (1)
 $f(a_{n})\leq 0\leq f(b_{n})$ (2)

By the fundamental axiom of analysis  $(a_{n})\to\alpha$ and $(b_{n})\to\beta$. But $(b_{n}-a_{n})\to 0$ so $\alpha=\beta$. By continuity of $f$

 $(f(a_{n}))\to f(\alpha)\quad(f(b_{n}))\to f(\alpha)$

But we have $f(\alpha)\leq 0$ and $f(\alpha)\geq 0$ so that $f(\alpha)=0$. Furthermore we have $a\leq\alpha\leq b$, proving the assertion.

Set $g(x)=f(x)-k$ where $f(a)\leq k\leq f(b)$. $g$ satisfies the same conditions as before, so there exists a $c$ such that $f(c)=k$. Thus proving the more general result.

Title proof of intermediate value theorem ProofOfIntermediateValueTheorem 2013-03-22 12:33:56 2013-03-22 12:33:56 yark (2760) yark (2760) 9 yark (2760) Proof msc 26A06