The intermediate value theorem states that if a continuous function attains two values, it must also attain all values in between these two values. Intuitively, a continuous function is a function whose graph can be drawn "without lifting pencil from paper." For instance, if is a continuous function that connects the points and , then there must be some between and where also some where , another where , etc. For this function there is an for any value between and .
Intermediate Value Theorem
For two real numbers and with , let be a continuous function on the closed interval Then for every between and there exists a number with .
The idea of the proof is as follows: take the two points and on the graph of the continuous function . These points demarcate a segment of the graph which connects the two points, as in the picture above. Since this segment has no holes or breaks in it (by continuity of the function), it must pass through each "level curve" where .
Note that the guaranteed by the theorem may not be unique; the theorem implies that there is at least one such that but there may be more than one, depending on and
The statement of the theorem has multiple requirements, all of which are necessary for the conclusion to hold. Here is an illustrative example:
For the function , over which of the following intervals does the intermediate value theorem guarantee a root:
In order for the intermediate value theorem to guarantee a root on a specified interval , not only must the function be continuous on the interval, but 0 must be contained between and Let's check the values of and
For the first interval the values returned by are both positive which do not sandwich 0, meaning the intermediate value theorem does not guarantee a root.
For the second interval the values returned by are on either side of 0, which seems to suggest that has a root on the interval However, it's important to note that has a discontinuity at meaning the intermediate value theorem does not hold. Indeed, does not have an -intercept on the interval
For the last interval the values returned by are on either side of 0, which implies that has a root on the interval This is confirmed by the intermediate value theorem because is continuous on
As in the above example, one simple and important use of the intermediate value theorem (hereafter referred to as IVT) is to prove that certain equations have solutions. Consider the following example:
Does the equation have solution(s) ? If so, how many solutions does it have?
We study the function . Note that and . Thus, by the IVT, there must be some such that , i.e. .
One root of the equation has been identified. Is this the only root? Note that is everywhere non-negative, so is increasing monotonically. Hence, can only have one root.
Is there a solution to where
At , we have
At , we have
So the IVT implies that there is a solution to in the interval .
Suppose that is continuous on and . Let be any positive integer, then prove that there is some number such that
Consider the set of numbers .
Let be such that is the largest number in . Suppose that and .
Then , and .
By the Intermediate value theorem, there is with , so that , or as desired.
Finally, if the largest number in is , then the same argument works with chosen such that is the minimum number in .
Note that if is both the largest and smallest number in , then they are all the same and .
Suppose that is continuous on and . Let be the hyperreal unit, then prove that there is some number such that
First, assume is not constant on . The result holds trivially if it is.
Since is not constant, there exists a such that is a maximum (or a minimum, but assume for now that it is a max; a min is handled similarly).
Now, by the Intermediate Value Theorem, if , there exists a such that .
If , instead.
Thus, , as wanted.
If , .
A real-valued function is said to have the intermediate value property if for every in the domain of , and for every
there exists some such that .
The intermediate value theorem states that if is continuous, then has the intermediate value property. Is the converse of this theorem true? That is, if a function has the intermediate value property, must it be continuous on its domain?
Since it can detect zeroes of functions, the IVT is an important tool for the analysis of continuous functions. However, through some clever contortions, IVT can give even more impressive results. For instance, one can prove the Borsuk-Ulam theorem in dimension 1. This theorem states that for any continuous real-valued function on a circle, there is some point on the circle such that takes the same value at and at the point on the circle directly opposite to the antipode of
This implies that on any great circle of the globe, any continuously varying information will take on the same value at some two antipodal points. For instance, there must exist two antipodal points on the equator at which the air temperature is the same.
Let denote a circle, and suppose that is a continuous function. Then, there exists such that .
Note: Here, by the circle , we mean the set of vectors in of length precisely 1. In other words, is the set of points such that .
There is a function given by . Composing this with gives . Define a function by
where we take if .
In particular, and have opposite signs. Thus, by the IVT, there is some such that . This means .
Let . Since , we conclude
Any die is modeled by some polyhedron. If the polyhedron is completely symmetric in the sense that any face can be taken to any other face via a rigid motion, then the die will be fair; when the die is rolled, the probability of landing on any face will equal the probability of landing on any other face.
Do there exist fair dice that are not completely symmetric?
Hint: Start with a prism whose cross-sections are regular -gons. Now consider the dual polyhedron , the polyhedron whose vertices are the centers of the faces of the original prism. This looks like two pyramids with regular -gon cross-sections that have been glued together at their bases. Now, can you modify to obtain a fair die that isn't completely symmetric?
One standard proof of the intermediate value theorem uses the least upper bound property of the real numbers that every nonempty subset of with an upper bound has a least upper bound. This is an important property that helps characterize the real numbers—note that the rational numbers don't have the least upper bound property consider e.g. the set of all rational numbers less than For a discussion of this property, see the Infimum/Supremum wiki. Here is a proof of the intermediate value theorem using the least upper bound property.
Let be a continuous function. Let be a number between and Suppose without loss of generality that and consider
Then is nonempty since and it has an upper bound namely so there is a least upper bound. Call that least upper bound
Suppose Then let By continuity, there is a such that implies But implies for all in that range, so no 's in that range lie in So is an upper bound for as well, which is a contradiction of the "leastness" of
Suppose . Then let By continuity, there is a such that implies But implies for all in that range, so every in that range lies in So for instance is in , which is a contradiction since is an upper bound.
The conclusion is that as desired.
In fact, the intermediate value theorem is equivalent to the least upper bound property. Suppose the intermediate value theorem holds, and for a nonempty set with an upper bound, consider the function that takes the value on all upper bounds of and on the rest of Then is not continuous by the intermediate value theorem it takes on the values and but never and it is straightforward to show that a point where is discontinuous must be a least upper bound for
The intermediate value theorem can also be reframed and generalized in terms of connected sets in Recall that a connected set is a set which is not a disjoint union of two open subsets. The continuous image of a connected set is connected, so the image of a continuous function on a closed interval is connected, and thus must contain every point between any two points in the image. Here is a formal translation of this idea, adapted from the wiki on connected sets:
Let be a continuous function, and any real number between and We want to show that there is a such that Because is connected, the image of is connected. If then the sets and are disjoint nonempty sets which are open in the subspace topology on And their union is which is a contradiction.
Note that the least upper bound property is used in a relatively subtle way: it is needed in order to prove that the domain, the closed interval is connected. Any proof of the intermediate value theorem must appeal to a property equivalent to the least upper bound property, which uses the completeness of the real numbers.
This framing in terms of connected subsets explains why the intermediate value theorem does not generalize easily to continuous functions whose image lies in for the connected subsets of are just the intervals, but the connected subsets of are potentially much more complicated.