Continuous function/R/Intermediate value theorem/Constructive/Section

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We want to know what happens to an interval under a continuous function . The values and belong to the image. The Intermediate value theorem tells us that all numbers between and do also belong to the image of the interval.

Let be real numbers, and let be a continuous function. Let be a real number between and . Then there exists an such that .

We consider the situation , and show the existence of such an with the bisection method. For that, we put and , we consider the arithmetic mean , and we compute

If , then we put

and if , then we put

In each case, the new interval is lying inside the initial interval and has half of its length. It fulfills again the condition , therefore we can apply the same defining method again and get recursively a family of nested intervals. Let denote the real number which is determined by these nested intervals. For the lower bounds of the intervals, we have , and this carries over to the limit , due to the criterion for continuity in terms of sequences. Hence, . For the upper bounds, we have , and this again carries over to , so .  Therefore, .

The method described in this proof is constructive and can be used to give an explicit numerical method.

Let be real numbers, and let be a continuous function with and . Then there exists an such that .

This follows immediately from fact.

Let be real numbers, and let denote a continuous function such that and . Then the function has a zero within the interval, due to the Intermediate value theorem. Such a zero can be found by the bisection method, as described in the proof of the Intermediate value theorem. We put and , and the other interval bounds are inductively defined in such a way that and hold. Define and compute . If , then we set

and if , then we set

In both cases, the new interval has half the length of the preceding interval and so we have bisected intervals. The real number defined by these nested intervals is a zero of the function.

We want to determine approximately a zero for the polynomial

with the help of the method given by the Intermediate value theorem. We have and , hence by fact, there must be a zero inside the interval . We compute the value of the function at the arithmetic mean of the interval, which is , and get

Hence, we have to continue with the right half of the interval . The arithmetic mean thereof is . The value of the function at this point is

So now we have to continue with the left part of the interval, namely . Its arithmetic mean is . The value of the function at this point is

Therefore, we know that there is a zero between and .

The existence of arbitrary roots of nonnegative real numbers follows also from the Intermediate value theorem, since, for , the continuous function attains negative and positive values and has therefore also a zero. The proof of fact rests on the method of the Intermediate value theorem, even though continuity is not explicitly used.

Suppose that a regular quadratic table with four table legs is standing on an uneven but stepfree underground. At the moment, it stands on the legs , and the leg does not touch the ground (if we leave in their positions and put down to the ground, then would sink into the ground). We claim that we can bring the table, by turning it around its middle axis, into a position such that it stands on all four legs (we do not claim that the table is then horizontal). To see this, we consider the function which assigns to the angle of rotation, the height of above the ground, when the three other legs are (or would be) on the ground. This height might be negative (if we are on sand, this can be realized, else think of this "ideally“). In degree , the height is positive. In degree , we get a situation which is symmetric to the initial position, but still the legs are supposed to be on the ground. Hence, the height of is now negative. The function has, on the interval , positive and also negative values. Since the function is continuous, by the Intermediate value theorem, it also has a zero.