Proof: By Induction

Let $[a,b]$ be a closed real interval.
Let $f:[a,b]\to\mathbb R$ be a continuous real function with $f(a) < 0$ and $f(b) > 0$ (or $f(a) > 0$ and $f(b) < 0$). Without loss of generality, assume $f(a) < 0 < f(b)$.
We will prove by induction that there exists a sequence $(I_n)_{n\in\mathbb N}$ of closed real intervals $I_n:=[a_n,b_n]$ with the following properties:
- Case $(i)$ $[a_n,a_n]\subset [a_{n-1},b_{n-1}]$ for $n\ge 1$.
- Case $(ii)$ $b_n - a_n = 2^{-n}(b-a)$.
- Case $(iii)$ $f(a_n) \le 0$ and $f(b_n)\ge 0$ for all $n\in\mathbb N$.
Base case $n=0$
- For $[a_0,b_0]:=[a,b]$ the properties $(i)$ to $(iii)$ are obviously fulfilled.
Induction step $n\rightarrow n + 1$.
- Let $[a_n,b_n]$ be an interval, for which the properties are fulfilled. Set $m:=(a_n + b_n)/2$. $m$ is the "middle" of the interval $[a_n,b_n]$.
- We set now \[[a_{n+1},b_{n+1}]:=\begin{cases} [a_n,m]&\text{if }f(m)\ge 0,\\ [m,b_n]&\text{else.} \end{cases}\]
- Obviously, the properties $(i)$ to $(iii)$ are fulfilled.
- It follows that the sequence $(a_{n})_{n\in\mathbb N}$ is monotonically increasing and bounded and the sequence $(b_{n})_{n\in\mathbb N}$ is monotonically decreasing and bounded.
- Therefore, because of the monotone convergence theorem, both sequences are convergent and we have \[\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=c\] for some $c\in [a,b]$.
- Because $f$ is continuous on the interval $[a,b]$, we have \[\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}f(b_n)=f( c).\]
- Because convergence preserves upper and lower bounds for sequence members and because of $(iii)$, we have \[f( c) \le 0\text{ and } f( c)\ge 0,\] which means that $f( c)=0$.
- Thus we have found a root value of the function $f$.
  ∎

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References

Bibliography

Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983

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Proof: By Induction

References

Bibliography