PlanetMath: proof of intermediate value theorem

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proof of intermediate value theorem

(Proof)

We first prove the following lemma.

If $f:[a,b] \to \mathbb{R}$ is a continuous function with $f(a) \le 0 \le f(b)$ then there exists a $c \in [a,b]$ such that .

Define the sequences and inductively, as follows.

$\displaystyle a_0 = a \quad b_0 = b$

$\displaystyle c_n = \frac{a_n + b_n}{2}$

$\displaystyle (a_n, b_n) = \begin{cases}(a_{n-1}, c_{n-1}) & f(c_{n-1}) \ge 0 \\ (c_{n-1}, b_{n-1}) & f(c_{n-1}) < 0 \end{cases}$

We note that

$\displaystyle a_0 \le a_1 \le \cdots \le a_n \le b_n \le \cdots \le b_1 \le b_0$

$\displaystyle (b_n - a_n) = 2^{-n}(b_0 - a_0)$

(1)

$\displaystyle f(a_n) \le 0 \le f(b_n)$

(2)

By the fundamental axiom of analysis $(a_n) \to \alpha$ and $(b_n) \to \beta$ . But $(b_n - a_n) \to 0$ so $\alpha = \beta$ . By continuity of

$\displaystyle (f(a_n)) \to f(\alpha) \quad (f(b_n)) \to f(\alpha)$

But we have $f(\alpha) \le 0$ and $f(\alpha) \ge 0$ so that $f(\alpha) = 0$ . Furthermore we have $a \le \alpha \le b$ , proving the assertion.

Set where $f(a) \le k \le f(b)$ . satisfies the same conditions as before, so there exists a such that . Thus proving the more general result.

"proof of intermediate value theorem" is owned by yark. [ full author list (2) | owner history (1) ]

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Cross-references: axiom, sequences, continuous function

This is version 6 of proof of intermediate value theorem, born on 2002-04-04, modified 2006-11-03.
Object id is 2813, canonical name is ProofOfIntermediateValueTheorem.
Accessed 9320 times total.

Classification:

26A06 (Real functions :: Functions of one variable :: One-variable calculus)

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