Intermediate Value Theorem

If a temperature sensor reads 10°C at dawn and 24°C at noon, it must have read exactly 17°C at some moment in between — assuming the temperature changed continuously. The Intermediate Value Theorem formalises this everyday observation and turns it into a tool for proving existence of roots and fixed points without ever computing them explicitly.

Statement

Theorem (Intermediate Value Theorem, IVT). Let $f : [a, b] \to \mathbb{R}$ be continuous. If $y$ is any value strictly between $f(a)$ and $f(b)$, then there exists $c \in (a, b)$ such that $f(c) = y$.

An equivalent formulation useful in practice: if $f(a)$ and $f(b)$ have opposite signs (i.e.\ $f(a) \cdot f(b) < 0$), then $f$ has a root in $(a, b)$.

Proof via bisection

Without loss of generality, assume $f(a) < y < f(b)$.

Define a sequence of nested intervals $[a_n, b_n]$ by bisection, starting with $[a_0, b_0] = [a, b]$:

• Let $m_n \coloneqq \dfrac{a_n + b_n}{2}$ be the midpoint.
• If $f(m_n) < y$, set $[a_{n+1}, b_{n+1}] \coloneqq [m_n, b_n]$.
• If $f(m_n) \geq y$, set $[a_{n+1}, b_{n+1}] \coloneqq [a_n, m_n]$.

By construction, $f(a_n) < y \leq f(b_n)$ at every step, and the interval length satisfies

$$b_n - a_n = \frac{b - a}{2^n} \;\to\; 0.$$

The completeness of $\mathbb{R}$ (nested interval property) guarantees a unique point $c \in \bigcap_{n=0}^{\infty} [a_n, b_n]$.

Since $a_n \leq c \leq b_n$ and $b_n - a_n \to 0$, both $a_n \to c$ and $b_n \to c$. Continuity of $f$ then gives $f(a_n) \to f(c)$ and $f(b_n) \to f(c)$. Because $f(a_n) < y$ for all $n$, taking the limit gives $f(c) \leq y$. Because $f(b_n) \geq y$ for all $n$, taking the limit gives $f(c) \geq y$. Therefore $f(c) = y$. $\square$

Applications

Root-finding for polynomials

Every odd-degree real polynomial has at least one real root. A degree-$(2k+1)$ polynomial $p$ satisfies $p(x) \to +\infty$ as $x \to +\infty$ and $p(x) \to -\infty$ as $x \to -\infty$, so for large enough $R > 0$, $p(-R) < 0 < p(R)$. The IVT gives a root in $(-R, R)$.

Example. The polynomial $p(x) = x^3 - 2$ is continuous, satisfies $p(0) = -2 < 0$ and $p(2) = 6 > 0$. By the IVT there exists $c \in (0, 2)$ with $c^3 = 2$ — this is the existence proof for $\sqrt[3]{2}$.

Fixed-point existence

Corollary (Brouwer in one dimension). Every continuous $f : [0, 1] \to [0, 1]$ has a fixed point: some $c \in [0, 1]$ with $f(c) = c$.

Proof. Define $g(x) \coloneqq f(x) - x$. Since $f$ maps $[0,1]$ into $[0,1]$, we have $g(0) = f(0) \geq 0$ and $g(1) = f(1) - 1 \leq 0$. If either is zero, then $0$ or $1$ is a fixed point. Otherwise $g(0) > 0 > g(1)$, and the IVT gives $c \in (0, 1)$ with $g(c) = 0$, i.e.\ $f(c) = c$. $\square$

The bisection algorithm

The proof is constructive: each step halves the interval containing a crossing point. After $n$ steps the root is located to within $(b - a)/2^n$. This is the bisection method — one of the most reliable root-finding algorithms in numerical computing, with guaranteed linear convergence.

def bisect(f, a, b, tol=1e-9):
    assert f(a) * f(b) < 0, "f must have opposite signs at a and b"
    while (b - a) / 2 > tol:
        m = (a + b) / 2
        if f(m) == 0:
            return m
        elif f(a) * f(m) < 0:
            b = m
        else:
            a = m
    return (a + b) / 2

Summary

• Intermediate Value Theorem: a continuous $f$ on $[a, b]$ takes every value between $f(a)$ and $f(b)$.
• Proof: bisect the interval at each step, maintaining opposite signs at the endpoints; completeness of $\mathbb{R}$ pins down a crossing point, and continuity forces it to equal $y$.
• Root-finding: every odd-degree polynomial has a real root; specific roots like $\sqrt[3]{2}$ exist by applying the IVT to $x^3 - 2$.
• Fixed points: every continuous self-map of $[0, 1]$ has a fixed point — proved by applying the IVT to $f(x) - x$.
• Bisection method: the proof is algorithmic — repeated halving locates any root to precision $\varepsilon$ in $O(\log(1/\varepsilon))$ steps.