# 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 $$(i)$$ $$[a_n,a_n]\subset [a_{n-1},b_{n-1}]$$ 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 $$f$$.

Github: ### References

#### Bibliography

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