The intermediate value theorem

Theorem

The intermediate value theorem

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Statement

Let \( a \lt b \) be real numbers and let \( f : [a,b] \to \mathbb{R} \) be continuous on the closed interval \([a,b]\) (continuous at every interior point, and one-sidedly continuous at the endpoints). Let \( y \) be any real number strictly between \( f(a) \) and \( f(b) \); that is, either \( f(a) \lt y \lt f(b) \) or \( f(b) \lt y \lt f(a) \). Then there exists a point \( c \in (a,b) \) such that \( f(c) = y \). Equivalently: a continuous real-valued function on an interval attains every value between any two of its values.

Why it matters

The intermediate value theorem (IVT) is the first theorem in analysis that converts qualitative information (a sign change, continuity) into an existence statement (a solution of an equation). It underwrites the existence of \( n \)-th roots, real roots of odd-degree polynomials, fixed points of maps of an interval into itself, and the bisection algorithm of numerical analysis. It is the one-dimensional shadow of a deep topological fact: continuous maps preserve connectedness.

It is also the theorem that reveals what the real numbers are for. Over \( \mathbb{Q} \) the statement is false, so any proof must invoke completeness. IVT is, in a precise sense, an avatar of the least-upper-bound axiom on the space thread: it is where the geometry of the line (no gaps) first meets the calculus of change.

Hypotheses

Hypothesis.\( f \) is continuous on all of \([a,b]\). Drop it at even one point and the conclusion dies: the sign function \( f(x) = -1 \) for \( x \lt 0 \), \( f(x) = 1 \) for \( x \geq 0 \) on \([-1,1]\) satisfies \( f(-1) = -1 \lt 0 \lt 1 = f(1) \) but never takes the value \( \tfrac{1}{2} \) (or \(0\) is taken, but \( \tfrac12 \) is not); the single jump at \(0\) destroys everything.

Hypothesis.The domain is an interval. On the disconnected set \( [-2,-1] \cup [1,2] \), the continuous function \( f(x) = x \) satisfies \( f(-2) \lt 0 \lt f(2) \) but never vanishes: the "path" from \(-2\) to \(2\) is broken and the value \(0\) falls through the gap.

Hypothesis.\( y \) lies strictly between \( f(a) \) and \( f(b) \). Without strict betweenness the interior conclusion \( c \in (a,b) \) can fail: for \( f(x) = x^2 \) on \([0,1]\) and \( y = 0 \), the only preimage is the endpoint \( c = 0 \). (The weaker claim "some \( c \in [a,b] \)" survives for \( y = f(a) \) or \( y = f(b) \), trivially.)

Hypothesis.The codomain scale is the complete field \( \mathbb{R} \). This is invisible in the statement but load-bearing in the proof. Work over \( \mathbb{Q} \): the function \( f(x) = x^2 - 2 \) is continuous on \( [1,2] \cap \mathbb{Q} \) (with the metric definition of continuity, verbatim), and \( f(1) = -1 \lt 0 \lt 2 = f(2) \), yet \( f \) has no rational zero since \( \sqrt{2} \notin \mathbb{Q} \). IVT is not a theorem of "continuity alone"; it is a theorem about continuity on a line with no gaps. Indeed IVT (for all continuous functions) is equivalent, over the ordered-field axioms, to Dedekind completeness.

Proof

We prove: if \( f : [a,b] \to \mathbb{R} \) is continuous and \( f(a) \lt y \lt f(b) \), then \( \exists\, c \in (a,b) \) with \( f(c) = y \). The mirrored case \( f(b) \lt y \lt f(a) \) follows by applying this case to \( -f \) and \( -y \) (Step 8). The engine of the proof is the least-upper-bound (completeness) axiom for \( \mathbb{R} \); the tactic is to walk right as far as \( f \) stays below \( y \), and show the supremum of that walk is forced to land exactly on \( y \).

Assume \( f(a) \lt y \lt f(b) \). Define \[ S = \{\, x \in [a,b] : f(x) \lt y \,\}. \]

Definition of an auxiliary set; no axiom used yet. \( S \) collects every point where \( f \) is still below the target value. A

\( S \neq \emptyset \) and \( S \) is bounded above by \( b \).

\( a \in S \) because \( f(a) \lt y \) by assumption; every element of \( S \) lies in \([a,b]\), hence is \( \leq b \). A

\[ c := \sup S \ \text{exists, and} \ c \in [a,b]. \]

Least-upper-bound axiom (Dedekind completeness of \( \mathbb{R} \)): every nonempty set bounded above has a supremum. Since \( a \in S \) we get \( c \geq a \); since \( b \) is an upper bound and \( c \) is the least one, \( c \leq b \). This is the only place completeness enters — and it is exactly the step that fails over \( \mathbb{Q} \). B

Local sign-preservation lemma. If \( f \) is continuous at \( p \in [a,b] \) and \( f(p) \neq y \), then there exists \( \delta \gt 0 \) such that \( f(x) - y \) has the same sign as \( f(p) - y \) for all \( x \in [a,b] \) with \( |x - p| \lt \delta \).

Proof of lemma. Put \( \varepsilon = |f(p) - y| \gt 0 \). Continuity at \( p \) (the \( \varepsilon\)–\(\delta \) definition) gives \( \delta \gt 0 \) with \( |f(x) - f(p)| \lt \varepsilon \) whenever \( x \in [a,b] \), \( |x - p| \lt \delta \). If \( f(p) \gt y \): \( f(x) \gt f(p) - \varepsilon = y \). If \( f(p) \lt y \): \( f(x) \lt f(p) + \varepsilon = y \). At an endpoint the same argument runs one-sidedly. \(\square\) C

Suppose \( f(c) \lt y \). Then \( c \lt b \), and the lemma gives \( \delta \gt 0 \) with \( f(x) \lt y \) for all \( x \in [a,b] \cap (c - \delta, c + \delta) \). Pick \( x_0 = \min\left( c + \tfrac{\delta}{2},\, \tfrac{c+b}{2} \right) \); then \( x_0 \in [a,b] \), \( x_0 \gt c \), and \( f(x_0) \lt y \), so \( x_0 \in S \) — contradicting that \( c \) is an upper bound of \( S \).

First, \( c \neq b \) because \( f(b) \gt y \) while we supposed \( f(c) \lt y \). Then Step 4 with \( p = c \). The contradiction is with "upper bound" in the definition of supremum: we found a member of \( S \) strictly to the right of \( \sup S \). Hence \( f(c) \lt y \) is impossible. B

Suppose \( f(c) \gt y \). Then \( c \gt a \), and the lemma gives \( \delta \gt 0 \) (shrunk so that \( c - \delta \gt a \)) with \( f(x) \gt y \) for all \( x \in [a,b] \cap (c - \delta, c + \delta) \). Hence no point of \( S \) lies in \( (c - \delta, c] \), so \( c - \delta \) is an upper bound of \( S \) — contradicting that \( c \) is the least upper bound.

First, \( c \neq a \) because \( f(a) \lt y \) while we supposed \( f(c) \gt y \). Then Step 4 with \( p = c \). Carefully: \( x \in S \) requires \( f(x) \lt y \), which the lemma forbids on \( (c-\delta, c] \); and \( S \) has no points \( \gt c \) anyway (Step 3), so every element of \( S \) is \( \leq c - \delta \). The contradiction is with "least" in the definition of supremum. Hence \( f(c) \gt y \) is impossible. C

\[ f(c) = y, \quad \text{and} \quad c \in (a,b). \]

Trichotomy of the order on \( \mathbb{R} \): Steps 5 and 6 eliminate \( f(c) \lt y \) and \( f(c) \gt y \), leaving \( f(c) = y \). Interiority: \( c \neq a \) since \( f(a) \lt y = f(c) \), and \( c \neq b \) since \( f(b) \gt y = f(c) \). A

Mirrored case: if \( f(b) \lt y \lt f(a) \), apply Steps 1–7 to \( g = -f \) and target \( -y \).

\( g \) is continuous (algebra of continuous functions), \( g(a) = -f(a) \lt -y \lt -f(b) = g(b) \), so some \( c \in (a,b) \) has \( g(c) = -y \), i.e. \( f(c) = y \). \(\blacksquare\) A

Result

\[ f : [a,b] \to \mathbb{R} \ \text{continuous}, \quad y \ \text{strictly between} \ f(a) \ \text{and} \ f(b) \ \implies \ \exists\, c \in (a,b): \ f(c) = y. \]

Reading. A continuous function cannot jump over a value. If its graph starts below a horizontal line and ends above it, the graph must cross the line somewhere strictly inside the interval — though the theorem tells you nothing about where, or how many times.

Scope. Applies to any continuous real-valued function on any interval of \( \mathbb{R} \) (apply the statement to a closed subinterval joining the two points in question). It requires the completeness of \( \mathbb{R} \): it fails over \( \mathbb{Q} \) and on disconnected domains. It is purely an existence statement — no uniqueness, no formula, no constructive locator (though bisection extracts an approximation from the proof idea).

Corollaries & converses

Bolzano's theorem. If \( f \) is continuous on \([a,b]\) and \( f(a) f(b) \lt 0 \), then \( f \) has a zero in \( (a,b) \). (Take \( y = 0 \).)
Existence of \( n \)-th roots. For \( \alpha \gt 0 \) and \( n \in \mathbb{N} \), the map \( x \mapsto x^n \) is continuous with \( 0^n \lt \alpha \lt (1+\alpha)^n \), so \( \alpha \) has a positive \( n \)-th root.
Odd-degree real polynomials have a real root. \( p(x) \to \pm\infty \) as \( x \to \pm\infty \) forces a sign change on some \([-M, M]\); apply Bolzano.
Fixed-point theorem on an interval. Every continuous \( f : [a,b] \to [a,b] \) has a fixed point: apply Bolzano to \( g(x) = f(x) - x \) (the one-dimensional Brouwer theorem).
Continuous images of intervals are intervals. Equivalently, continuous maps preserve connectedness in \( \mathbb{R} \).
Continuous injections on an interval are strictly monotone (Problem 5), hence have continuous inverses — the foundation of the inverse-function theory in one variable.
Converse is FALSE. Having the intermediate-value property (a "Darboux function") does not imply continuity: \( f(x) = \sin(1/x) \) for \( x \neq 0 \), \( f(0) = 0 \), has the IVT property on every interval but is discontinuous at \(0\). Extreme case: Conway's base-13 function attains every real value on every interval yet is continuous nowhere. Darboux's theorem says every derivative has the intermediate-value property, yet derivatives can be discontinuous — so "Darboux" is genuinely weaker than "continuous".

Fails without

Without continuity (one jump suffices): \( f : [-1,1] \to \mathbb{R} \), \( f(x) = -1 \) for \( x \lt 0 \) and \( f(x) = 1 \) for \( x \geq 0 \). Here \( f(-1) = -1 \lt \tfrac12 \lt 1 = f(1) \), but \( f(x) = \tfrac12 \) has no solution: the range is \( \{-1, 1\} \).
Without completeness of the underlying field: \( f : [1,2] \cap \mathbb{Q} \to \mathbb{Q} \), \( f(x) = x^2 - 2 \). Continuous by the same \( \varepsilon\)–\(\delta \) definition, \( f(1) = -1 \lt 0 \lt 2 = f(2) \), but \( f(c) = 0 \) forces \( c = \sqrt{2} \notin \mathbb{Q} \). In the proof, \( \sup S \) simply does not exist in \( \mathbb{Q} \).
Without a connected domain: \( f(x) = 1/x \) on \( [-1, 0) \cup (0, 1] \) is continuous at every point of its domain and satisfies \( f(-1) = -1 \lt 0 \lt 1 = f(1) \), yet \( 1/x = 0 \) has no solution. (This is also the classic trap: "\( f \) changes sign, so it has a root" is false when the domain has a hole.)

Common errors

Asserting the converse. "\( f \) takes all intermediate values, so \( f \) is continuous" — false; see the Darboux counterexamples above. IVT is a one-way implication.
Claiming uniqueness. IVT gives at least one \( c \), never exactly one. \( \sin \) takes the value \( \tfrac12 \) infinitely often on \( [0, 100\pi] \). Uniqueness needs an extra input, usually strict monotonicity.
Applying it across a singularity. Concluding \( 1/x \) or \( \tan x \) has a root from a sign change across \( 0 \) or \( \pi/2 \): the function is not continuous (indeed not defined) on the whole closed interval, so the hypothesis fails.
Treating it as constructive. Writing "by IVT, \( c = \dots \)" with a formula. IVT locates nothing; if you need digits, you must run bisection or another algorithm — the theorem only guarantees the search will not be in vain.
Confusing IVT with the mean value theorem. IVT is about values of \( f \) (continuity only); MVT is about values of \( f' \) (differentiability needed). "Intermediate" refers to the \( y \)-axis, not to slopes.
Checking endpoints only and forgetting the hypothesis is about the whole interval. Continuity must hold at every point of \([a,b]\); one bad interior point (see the sign function) voids the theorem even if \( f \) is beautifully continuous elsewhere.

Discussion

Historically, IVT marks the moment analysis stopped trusting pictures. Through the eighteenth century the statement was regarded as geometrically self-evident — a curve crossing a line obviously meets it. Bernard Bolzano's 1817 paper, whose title announces a "purely analytic proof", is the founding document of rigorous analysis: he saw that the theorem needs a proof, that the proof needs a precise definition of continuity, and that it must rest on a completeness principle for the reals (his version: a form of the least-upper-bound property, proved via what we now call Bolzano–Weierstrass reasoning). Cauchy gave a bisection-style argument in his 1821 Cours d'analyse. The modern statement "continuous images of connected sets are connected" is the twentieth-century topological distillation.

The theorem sits at the junction of the space and change threads. As a statement about change, it says a continuously varying quantity cannot skip a value — the physical intuition behind "at some instant the temperature was exactly \( 20^\circ \)". As a statement about space, it is the fact that \( [a,b] \) is connected and continuity preserves connectedness; the same skeleton, run in higher dimensions, becomes the Brouwer fixed-point theorem and the Borsuk–Ulam theorem, of which the antipodal example below is the one-dimensional case.

Algorithmically, the proof's dichotomy becomes the bisection method: evaluate \( f \) at the midpoint, keep the half with the sign change, repeat. After \( n \) steps the root is trapped in an interval of length \( (b-a)/2^n \) — guaranteed linear convergence with no smoothness assumptions, which is why bisection remains the fail-safe of numerical root-finding even though Newton's method is faster when it works.

Foundationally, IVT is a sensitive instrument. Over an arbitrary ordered field \( F \), "every continuous \( f : [a,b] \to F \) with a sign change has a zero" is equivalent to Dedekind completeness of \( F \), so IVT can serve as an axiomatization of \( \mathbb{R} \). In constructive mathematics (Bishop-style, without the law of excluded middle) the exact IVT is not provable: the decision "\( f(c) \lt y \) or \( f(c) \gt y \) or \( f(c) = y \)" in Steps 5–7 is non-algorithmic, and one can build weak counterexamples from functions that hover near the target value. What survives constructively is the approximate IVT (\( \forall \varepsilon \gt 0\ \exists x : |f(x) - y| \lt \varepsilon \)) and the exact version for locally non-constant functions. In reverse mathematics, IVT for continuous functions with modulus is provable in the base system \( \mathsf{RCA}_0 \), which is strictly weaker than what Bolzano–Weierstrass needs — a reminder that "uses completeness" admits fine gradations. Common misconceptions. IVT does not say the function passes through the value once; it does not say where; it does not require differentiability; and its converse fails badly — the intermediate-value property is so far from characterising continuity that a nowhere-continuous function (Conway base 13) can have it in the strongest possible form.

Worked examples

Example 1. The equation \( x^5 + x - 1 = 0 \) has exactly one real solution, and it lies in \( (0,1) \).

Let \( f(x) = x^5 + x - 1 \). Then \( f \) is continuous on \( [0,1] \).

Polynomials are continuous on \( \mathbb{R} \) (algebra of continuous functions: sums and products of the continuous functions \( x \mapsto x \) and constants). A

\[ f(0) = -1 \lt 0, \qquad f(1) = 1 \gt 0. \]

Direct evaluation; this exhibits a sign change over \([0,1]\). A

By the intermediate value theorem (Bolzano form, \( y = 0 \)), there exists \( c \in (0,1) \) with \( f(c) = 0 \).

All hypotheses verified in Steps 1–2: continuity on the closed interval, and \( 0 \) strictly between \( f(0) \) and \( f(1) \). B

Uniqueness: for \( x_1 \lt x_2 \), \[ f(x_2) - f(x_1) = \left( x_2^5 - x_1^5 \right) + \left( x_2 - x_1 \right) \gt 0, \] since \( t \mapsto t^5 \) is strictly increasing on \( \mathbb{R} \). So \( f \) is strictly increasing and has at most one zero; combined with Step 3, exactly one, and no zero outside \( (0,1) \) since \( f \lt 0 \) on \( (-\infty, 0] \) and \( f \gt 0 \) on \( [1, \infty) \) by monotonicity.

Strict monotonicity of odd powers (or note \( f'(x) = 5x^4 + 1 \geq 1 \gt 0 \) if derivatives are available — but the algebraic argument keeps the example derivative-free). Note the division of labour: IVT gives existence, monotonicity gives uniqueness; IVT alone never gives uniqueness. B

\[ \exists!\, c \in (0,1) : \ c^5 + c - 1 = 0. \]

Reading. A sign change plus continuity manufactures a root out of nothing; strict monotonicity then pins it down as the only one.

Scope. The template — evaluate at two points, cite continuity, invoke IVT, add monotonicity for uniqueness — solves every "show this equation has a (unique) solution" problem at this level.

Example 2 (antipodal temperatures). Let \( T : \mathbb{R} \to \mathbb{R} \) be continuous and \( 2\pi \)-periodic (temperature at angle \( \theta \) on a circular wire). Then there exist two diametrically opposite points with equal temperature: \( \exists\, \theta_0 : T(\theta_0) = T(\theta_0 + \pi) \).

Define \( g(\theta) = T(\theta) - T(\theta + \pi) \) on \( [0, \pi] \).

\( g \) is continuous as a difference of continuous functions (the second being a composition of \( T \) with the continuous shift \( \theta \mapsto \theta + \pi \)). A

\[ g(\pi) = T(\pi) - T(2\pi) = T(\pi) - T(0) = -\bigl( T(0) - T(\pi) \bigr) = -g(0). \]

Periodicity gives \( T(2\pi) = T(0) \); the rest is algebra. This antisymmetry is the key structural observation: the quantity we want to kill takes opposite signs at the two ends. B

If \( g(0) = 0 \), take \( \theta_0 = 0 \) and we are done. Otherwise \( g(0) \) and \( g(\pi) = -g(0) \) are nonzero with opposite signs, so \( 0 \) lies strictly between them.

Case split on trichotomy; in the second case the strict-betweenness hypothesis of IVT is verified exactly. A

By the intermediate value theorem applied to \( g \) on \( [0, \pi] \), there exists \( \theta_0 \in (0, \pi) \) with \( g(\theta_0) = 0 \), i.e. \( T(\theta_0) = T(\theta_0 + \pi) \).

Hypotheses checked in Steps 1 and 3. Note the standard manoeuvre: an equation between two quantities is converted into a zero of their difference so that Bolzano's form applies. B

\[ \exists\, \theta_0 \in [0, \pi) : \ T(\theta_0) = T(\theta_0 + \pi). \]

Reading. On any circular wire, however wildly the temperature varies, some pair of antipodal points agrees exactly. Continuity plus the built-in antisymmetry of "difference across a diameter" forces a crossing.

Scope. This is the \( n = 1 \) case of the Borsuk–Ulam theorem; the same difference trick proves the fixed-point corollary and Problem 4 below. It needs genuine continuity of \( T \) all the way round the circle.

Problems

Show that the equation \( \cos x = x \) has a solution in \( (0, \tfrac{\pi}{2}) \). Is it unique in \( \mathbb{R} \)?
Solution

Let \( f(x) = \cos x - x \), continuous on \( [0, \tfrac{\pi}{2}] \) (difference of continuous functions). Then \( f(0) = 1 \gt 0 \) and \( f(\tfrac{\pi}{2}) = 0 - \tfrac{\pi}{2} \lt 0 \). Since \( 0 \) lies strictly between \( f(\tfrac{\pi}{2}) \) and \( f(0) \), IVT gives \( c \in (0, \tfrac{\pi}{2}) \) with \( \cos c = c \).

Uniqueness on \( \mathbb{R} \): for \( x \lt 0 \) we have \( \cos x \geq -1 \gt x \) once \( x \lt -1 \), and on \( [-1, 0) \), \( \cos x \gt 0 \gt x \); so no solution with \( x \lt 0 \). For \( x \gt \tfrac{\pi}{2} \), \( \cos x \leq 1 \lt \tfrac{\pi}{2} \lt x \); no solution there. On \( [0, \tfrac{\pi}{2}] \), \( f \) is strictly decreasing: if \( x_1 \lt x_2 \) then \( \cos x_1 \geq \cos x_2 \) (cosine is non-increasing on \( [0, \pi] \)) and \( -x_1 \gt -x_2 \), so \( f(x_1) \gt f(x_2) \). A strictly decreasing function has at most one zero. Hence the solution (the Dottie number, \( c \approx 0.739 \)) is unique in \( \mathbb{R} \).
Prove that every polynomial of odd degree with real coefficients has at least one real root.
Solution

Let \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 \) with \( n \) odd and \( a_n \neq 0 \); dividing by \( a_n \) we may assume \( a_n = 1 \). For \( x \neq 0 \) write \[ p(x) = x^n \left( 1 + \frac{a_{n-1}}{x} + \dots + \frac{a_0}{x^n} \right). \] Set \( R = \max\bigl(1,\ 2(|a_{n-1}| + \dots + |a_0|)\bigr) \). For \( |x| \geq R \) we have \( |x|^{n-k} \geq |x| \) for each \( k \leq n-1 \), so \( |a_k / x^{n-k}| \leq |a_k| / |x| \), and therefore \[ \left| \frac{a_{n-1}}{x} + \dots + \frac{a_0}{x^n} \right| \leq \frac{|a_{n-1}| + \dots + |a_0|}{|x|} \leq \frac{1}{2}, \] hence the bracket lies in \( [\tfrac12, \tfrac32] \) and \( p(x) \) has the sign of \( x^n \). Since \( n \) is odd, \( p(R) \gt 0 \) and \( p(-R) \lt 0 \).

\( p \) is continuous on \( [-R, R] \) (polynomials are continuous), and \( 0 \) lies strictly between \( p(-R) \) and \( p(R) \), so by IVT there is \( c \in (-R, R) \) with \( p(c) = 0 \). (No analogue for even degree: \( x^2 + 1 \) has no real root.)
Let \( f : [a,b] \to [a,b] \) be continuous. Prove \( f \) has a fixed point. Show by example that it need not be unique, and that the conclusion fails for the open interval \( (0,1) \).
Solution

Let \( g(x) = f(x) - x \), continuous on \( [a,b] \). Since \( f(a) \in [a,b] \), \( g(a) = f(a) - a \geq 0 \); since \( f(b) \in [a,b] \), \( g(b) = f(b) - b \leq 0 \). If \( g(a) = 0 \) or \( g(b) = 0 \) we have a fixed point at an endpoint. Otherwise \( g(a) \gt 0 \gt g(b) \), and by IVT there is \( c \in (a,b) \) with \( g(c) = 0 \), i.e. \( f(c) = c \).

Non-uniqueness: \( f(x) = x \) on \( [0,1] \) fixes every point. Failure on \( (0,1) \): \( f(x) = x/2 \) maps \( (0,1) \) into \( (0,1) \) continuously, but \( f(x) = x \) forces \( x = 0 \notin (0,1) \) — the endpoint sign-trap needs the closed interval. (This is the one-dimensional Brouwer fixed-point theorem; compactness and convexity of the domain are both doing work.)
(Universal chord, half-length case.) Let \( f : [0, 2] \to \mathbb{R} \) be continuous with \( f(0) = f(2) \). Prove there exists \( x \in [0, 1] \) with \( f(x) = f(x+1) \).
Solution

Define \( g(x) = f(x+1) - f(x) \) on \( [0,1] \); \( g \) is continuous (difference and composition-with-shift of continuous functions). Then \[ g(0) + g(1) = \bigl( f(1) - f(0) \bigr) + \bigl( f(2) - f(1) \bigr) = f(2) - f(0) = 0, \] so \( g(1) = -g(0) \). If \( g(0) = 0 \), take \( x = 0 \). Otherwise \( g(0) \) and \( g(1) \) have strictly opposite signs, so \( 0 \) is strictly between them, and IVT yields \( x \in (0,1) \) with \( g(x) = 0 \), i.e. \( f(x) = f(x+1) \).

(Remark: the same telescoping argument works for chords of length \( \tfrac{1}{n} \) on \( [0,1] \) for every \( n \in \mathbb{N} \); for non-reciprocal lengths \( \lambda \in (0,1) \), \( \lambda \neq \tfrac1n \), there are counterexamples — the universal chord theorem of Lévy.)
(Harder.) Let \( f : [a,b] \to \mathbb{R} \) be continuous and injective. Prove that \( f \) is strictly monotone on \( [a,b] \).
Solution

Since \( f \) is injective, \( f(a) \neq f(b) \); assume \( f(a) \lt f(b) \) (otherwise replace \( f \) by \( -f \), which is continuous and injective, and strict monotonicity transfers). We claim \( f \) is strictly increasing.

Claim 1: for all \( x \in (a,b) \), \( f(a) \lt f(x) \lt f(b) \). Suppose instead \( f(x) \lt f(a) \) for some \( x \in (a,b) \) (the case \( f(x) \gt f(b) \) is symmetric; equality is barred by injectivity). Choose any \( y \) with \( f(x) \lt y \lt f(a) \); then also \( y \lt f(a) \lt f(b) \). Apply IVT on \( [a, x] \): \( y \) is strictly between \( f(a) \) and \( f(x) \), so \( \exists\, c_1 \in (a, x) \) with \( f(c_1) = y \). Apply IVT on \( [x, b] \): \( y \) is strictly between \( f(x) \) and \( f(b) \), so \( \exists\, c_2 \in (x, b) \) with \( f(c_2) = y \). Then \( c_1 \lt x \lt c_2 \) but \( f(c_1) = f(c_2) \), contradicting injectivity. This proves Claim 1.

Claim 2: \( f \) is strictly increasing. Let \( a \leq x_1 \lt x_2 \leq b \). By Claim 1 applied on the interval \( [x_1, b] \) — whose endpoint values satisfy \( f(x_1) \lt f(b) \) by Claim 1 (or trivially if \( x_1 = a \)) — every interior point \( x_2 \in (x_1, b) \) satisfies \( f(x_1) \lt f(x_2) \lt f(b) \); and if \( x_2 = b \) then \( f(x_1) \lt f(b) = f(x_2) \) directly. To justify the endpoint hypothesis of this re-application: \( f(x_1) \lt f(b) \) holds by Claim 1 when \( x_1 \in (a,b) \), and by assumption when \( x_1 = a \). Hence \( f(x_1) \lt f(x_2) \) in all cases, which is strict increase. \( \blacksquare \)

(Consequence: the inverse \( f^{-1} : f([a,b]) \to [a,b] \) exists and is continuous — monotone bijections between intervals have monotone, hence continuous, inverses. This is the standard route to continuity of \( \sqrt[n]{x} \), \( \arcsin \), \( \log \), all ultimately resting on IVT.)