Proof: By Induction
(related to Theorem: Intermediate Root Value Theorem)
 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 \((i)\) \([a_n,a_n]\subset [a_{n1},b_{n1}]\) for \(n\ge 1\).
 Case \((ii)\) \(b_n  a_n = 2^{n}(ba)\).
 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 VerĂ¤nderlichen", Vieweg Studium, 1983