A Second Proof of the Location of Roots Theorem

We looked at one way to prove the very important Location of Roots theorem on The Location of Roots Theorem page. We will now present another proof of the result

Theorem 1 (The 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 $I$.

• Proof: Without loss of generality assume that $f(a) > 0$. Then $f(b) < 0$. Since $f$ is continuous on $[a, b]$ there exists a $\delta > 0$ such that $f(x) > 0$ on the subinterval $[a, a + \delta]$ and $f(x) < 0$ on the subinterval $[b - \delta, b]$. Define the set $A$ as follows:

(1)

\begin{align} \quad A = \{ x \in [a, b] : f(x) > 0 \} \end{align}

• Now since $A$ is a bounded subset of $\mathbb{R}$ and is nonempty since $a \in A$ we have by the completeness property of the real numbers that $\sup A$ exists. Let $c = \sup A$. Note that $a < c < b$ since $a < a + \delta \leq c$ and $c \leq b - \delta < b$. We claim that $f(c) = 0$.

• Suppose not. Then there are two cases to consider.

• Case 1: Suppose that $f(c) > 0$. Then by the continuity if $f$ there exists an $\delta^* > 0$ such that $f(x) > 0$ on $[c - \delta^*, c + \delta^*]$. But then $[c, c + \delta^*] \subset A$ and $c \leq c + \epsilon \in A$, so $c$ is not an upper bound of $A$ which is a contradiction.

• Case 2: Suppose that $f(c) < 0$. Then by the continuity of $f$ there exists a $\delta^* > 0$ such that $f(x) < 0$ on $[c - \delta^*, c + \delta^*]$. But then $[c - \delta^*, c + \delta^*] \not \subset A$ so $c - \epsilon$ is an upper bound for $A$ that is less than $c$ which is a contradiction.

• Therefore $f(c) = 0$, i.e., $f$ has at least one root on $I$. $\blacksquare$