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A continuous function on an interval attains every value between its end values; proved by bisection, and false over the rationals.
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By the end of this lesson you will be able to state and prove the intermediate value theorem by bisection, use it by turning an equation into a root and finding a sign change, and build the auxiliary function an application needs — including the fixed point argument. You will also be able to say which hypothesis each standard counterexample removes, and why the converse of the theorem is false.
Continuity at a point, the sign-preservation lemma from the last lesson, and the nested interval theorem from unit 1. This lesson is the first theorem that needs continuity at every point of an interval, and it is proved from exactly those three things.
A root of a function is a point where it takes the value zero. A sign change across an interval means the values at the two ends have opposite signs. An auxiliary function is one built so that the question being asked becomes a question about its roots.
The intermediate value theorem. Let $f$ be continuous on $[a,b]$ and let $y$ lie between $f(a)$ and $f(b)$. Then $f(p) = y$ for some $p \in [a,b]$.
It is enough to prove the case $y = 0$ with $f(a) < 0 < f(b)$; the general case follows by applying that to $f - y$.
Proof by bisection. The interval $[a,b]$ has a sign change across it. Bisect: at the midpoint $m$, either $f(m) = 0$ and the proof ends, or the sign changes across one of the halves — keep that one. Repeating gives nested closed intervals $[a_n, b_n]$, each with a sign change across it and with lengths $(b-a)/2^n \to 0$. They meet in a single point $p$. Both $a_n \to p$ and $b_n \to p$, so by continuity $f(a_n) \to f(p)$ and $f(b_n) \to f(p)$; the first sequence is never positive and the second never negative, so $f(p) \le 0 \le f(p)$.
It is an existence theorem. It produces a point and says nothing about how many, or where, or how to compute one — except that the proof happens to be an algorithm, which is why bisection is a real numerical method.
The converse is false. A function can take every intermediate value and be discontinuous. Every hypothesis in this subject is doing work somewhere in the proof. The way to know a theorem is to know which example breaks when a hypothesis is dropped.
Another way: picture
A curve starting below a horizontal line and finishing above it must cross the line. The picture is exactly right here, and the proof's only job is to say why 'must cross' is a theorem about $\mathbb{R}$ rather than about pictures — over the rationals the same curve can step across a gap.
Another way: steps
To use the theorem:
Step 1 is the one that takes invention; everything after it is checking.
Over $\mathbb{Q}$. Let $f(x) = x^2 - 2$ on the rationals between $1$ and $2$. It is continuous, $f(1) = -1 < 0 < 2 = f(2)$, and it has no rational root. So the theorem is not a fact about continuity; it is a fact about the completeness of $\mathbb{R}$, entering through the nested interval theorem.
Without continuity. The step function that is $-1$ below $0$ and $1$ from $0$ onwards changes sign on $[-1, 1]$ and is never zero. It skips every value strictly between $-1$ and $1$.
On a set that is not an interval. On the domain $[-2,-1] \cup [1,2]$, the continuous function $f(x) = x$ changes sign and has no root. Connectedness of the domain is doing work that the word 'interval' quietly carries.
Each failure removes one hypothesis and produces a counterexample, which is the only reliable way to know what a theorem is saying.
| Goal | Auxiliary function | Sign check |
|---|---|---|
| a root of $x^3 - x - 1$ | itself | $f(1) = -1$, $f(2) = 5$ |
| a solution of $\cos x = x$ | $\cos x - x$ | positive at $0$, negative at $1$ |
| a fixed point of $f : [0,1] \to [0,1]$ | $f(x) - x$ | at least $0$ at $0$, at most $0$ at $1$ |
| an $n$-th root of $c > 0$ | $x^n - c$ | negative at $0$, positive at $1 + c$ |
The last row is worth noticing: the existence of $n$-th roots, which unit 1 obtained from the completeness axiom directly, falls out of the theorem in one line. The same axiom is behind both, and the theorem is the more convenient packaging.
Reading it backwards. Taking every intermediate value does not imply continuity. Darboux's theorem says every derivative has the intermediate value property, and derivatives can be discontinuous.
Expecting uniqueness. The theorem says a root exists, not that there is one. A continuous function can cross a level any number of times, and a sign change only guarantees an odd number of crossings counted properly.
Applying it on an open or disconnected domain. The hypothesis is continuity on a closed interval, and a domain in two pieces defeats it with the simplest function there is.
Does $x^3 - x - 1$ have a real root? It is a polynomial, so it is continuous everywhere.
Continuity, checked rather than assumed.
At $x = 1$ the value is $-1$; at $x = 2$ it is $5$.
A sign change across a closed interval.
So a root lies between $1$ and $2$. Bisecting narrows it to any accuracy asked for, and the proof is that procedure.
Existence, with a method attached.
Solve $\cos x = x$. As written it is not a root problem, so set $g(x) = \cos x - x$.
Turn the equation into a root.
$g(0) = 1 > 0$ and $g(1) = \cos 1 - 1 < 0$, and $g$ is continuous.
Signs at two points.
So there is a solution between $0$ and $1$. Almost every use of this theorem starts by inventing the right $g$.
The invention is the step that matters.
Write $p(x) = a_n x^n + \cdots$ with $n$ odd and $a_n > 0$, without loss of generality.
Normalise the leading coefficient's sign.
For large positive $x$ the leading term dominates and $p(x) > 0$; for large negative $x$, since $n$ is odd, $p(x) < 0$.
A sign change, found at large arguments.
$p$ is continuous, so it has a root between those two points. Evenness would break it: $x^2 + 1$ has no real root, and the odd degree is exactly what supplies the sign change.
A continuous $f$ has $f(0) < 0 < f(6)$. Put the steps of the bisection proof that $f$ has a root between them into order.
Number the steps in order (write the number in the box):
A root is bracketed in an interval of length $8$. After $4$ bisections, how long is the bracket?
Answer:
Match each situation to what may be concluded from it.
| a root strictly between the two points | every value between the two end values is attained | a point where the function equals its input | nothing: this is the false converse | |
|---|---|---|---|---|
| $f$ continuous with $f(0) < 0 < f(8)$ | ||||
| $f$ continuous on the whole interval from $0$ to $8$ | ||||
| $f$ continuous, mapping the interval from $0$ to $8$ into itself | ||||
| $f$ takes every value between $f(0)$ and $f(8)$ |
$f$ is continuous on the interval from $0$ to $1$, with $f(0) = -2$ and $f(1) = 2$. Give the set of values the theorem guarantees $f$ attains.
This task has no paper form; do it on a device.
Here is an argument about a function $g$ on the interval from $0$ to $4$. Mark the one step that is not a theorem.
This task has no paper form; do it on a device.
Build the proof that a continuous function from the interval $[0,1]$ into itself has a point where it equals its input.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A continuous $f$ has $f(0) < 0 < f(4)$. Put the steps of the bisection proof that $f$ has a root between them into order.
Number the steps in order (write the number in the box):
You can prove the theorem by bisection, use it to bracket a root, and build the auxiliary function an application needs. Say in your own words why the theorem fails over the rational numbers. Next: the other thing continuity on a closed bounded interval buys.
10. Your turn: show that every odd-degree polynomial has a real root, step 3