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.

The Location of Roots Theorem

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 $\lim_{n \to \infty} a_n = c = \lim_{n \to \infty} b_n$.

• Now since $f$ is continuous at $c$ and since $(a_n)$ and $(b_n)$ are sequences from the domain $I$ we have by the Sequential Criterion for the Continuity of a Function that $\lim_{n \to \infty} f(a_n) = f(c) = \lim_{n \to \infty} f(b_n)$.

• 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 $f$. $\blacksquare$