The Location of Roots Theorem

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

We will now look at a theorem known as The Location of Roots Theorem which says that given a continuous function $f : I \to \mathbb{R}$ where $I = [a, b]$ is a closed and bounded interval and $f(a) < 0 < f(b)$ or $f(a) > 0 > f(b)$, then $f$ has at least one root on $I$.
 Theorem 1 (Location of Roots): Let $f : I \to \mathbb{R}$ be a continuous function where $I = [a, b]$ is a closed and bounded interval. Then if $f(a) < 0 < f(b)$ or $f(a) > 0 > f(b)$, then $f$ has at least one root on the interval $I$. • Proof: To show that $f$ has at least one root on $I = [a, b]$ we will show that $\exists c \in I$ such that $f(c) = 0$.
• Without loss of generality, assume that $f(a) < 0 < f(b)$. Let $I_1 = [a_1, b_1]$ where $a_1 = a$ and $b_1 = b$. Then let $p_1 = \frac{1}{2}(a_1 + b_1)$. Now if $f(p_1) = 0$, then let $c = p_1$ and so $f(p_1) = f(c) = 0$, and so we have found a root. If $f(p_1) \neq 0$, then it follows from The Order Properties of Real Numbers order axiom 3 that either $f(p_1) > 0$ or $f(p_1) < 0$. If $f(p_1) > 0$ then let $a_2 = a_1$ and let $b_2 = p_1$, and if $f(p_1) < 0$ then let $a_2 = p_1$ and $b_2 = b_1$. Then let $I_2 = [a_2, b_2]$. Therefore $I_2 \subseteq I_1$ and $f(a_2) < 0$ while $f(b_2) > 0$.
• Continue this process. For the $k^{\mathrm{th}}$ interval, if $f(p_k) = 0$ then let $c = p_k$ and we're done, otherwise if $f(p_k) > 0$, then let $a_{k+1} = a_k$ and let $b_{k+1} = p_k$, and if $f(p_k) < 0$ then let $a_{k+1} = p_k$ and let $b_{k+1} = b_k$. Then let $I_{k+1} = [a_{k+1}, b_{k+1}]$ so that $I_{k+1} \subseteq I_k$. If this process terminates at some $n \in \mathbb{N}$, that is $f(p_n) = 0$, then let $p_n = c$. If this process never terminates, then we obtain a sequence of nested, closed, and bounded intervals $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ such that $\forall n \in \mathbb{N}$ we have that $f(a_n) < 0$ and $f(b_n) > 0$.
• So by The Nested Intervals Theorem, for all $n \in \mathbb{N}$ there exists a $c \in I_n$. Therefore $a_n ≤ c ≤ b_n$ for all $n \in \mathbb{N}$. Now the length of each interval $I_n = b_n - a_n = \frac{b-a}{2^{n-1}}$, and so $\lim_{n \to \infty} (b_n - a_n) = \lim_{n \to \infty} \frac{b-a}{2^{n-1}} = 0$ which implies that $\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n$. Therefore by The Squeeze Theorem, since $a_n ≤ c ≤ b_n$ we have that $\lim_{n \to \infty} a_n = c = \lim_{n \to \infty} b_n$.
• Since $f(a_n) ≤ 0$ for all $n \in \mathbb{N}$ then we have that $\lim_{n \to \infty} f(a_n) = f(c) ≤ 0$. Similarly, since $f(b_n) ≥ 0$ for all $n \in \mathbb{N}$ then we have that $\lim_{n \to \infty} f(b_n) = f(c) ≥ 0$. Combining these inequalities we get that $f(c) = 0$, and so $c \in I$ is a root of $f$. $\blacksquare$