A familiar student reaction to the Intermediate Value Theorem. After all, if a function is “continuous,” it can’t jump from place to place without stopping in between. Or,
“Real functions are like x2or sin(x). Those step thingies don’t really matter.”
This cherished theorem usually falls flat in beginning calculus because, I would argue, students see it as so obvious as not to merit discussion. And it’s not their fault; the theorem that justifies the word “continuous” strikes most students as unimportant because they rarely see it do anything other than confirm their long-held intuition about what “continuous” ought to mean. If a continuous function is positive somewhere and negative somewhere else then, sure, it has a root in between. But this familiar “application” is of course just a restatement—or a special case—of the original result.
So what kinds of applications are there? For starters, how do you cut a cake in half? What you don’t do is find the center and cut through it. Rather you mentally move a knife across the top until the area on the left of the knife looks to be about the same as the area on the right—a simple application of our old friend. Does the cake have to be round you ask? Nope. If S is any closed figure in the plane, then there is a line in any given direction that bisects the area of S. (A “closed figure” is a set bounded by a simple closed curve.)
To prove this claim, we can construct a standard coordinate system with the y-axis parallel to the chosen direction. For each x, let l(x) be the line through x and perpendicular to the x-axis. L(x), the area of S to the left of l(x), is a continuous function of x as is R(x), the area to the right of l(x). Hence D(x) = R(x) – L(x) is continuous. For a line to the left of S, D(x) = Area of S, and for a line to the right of S, D(x) = – (Area of S). By the Intermediate Value Theorem there is an intermediate line for which D(x) = 0 and L(x) = R(x).
If that were the whole story there would be no story. After all, we’ve really just beaten a simple theorem to a pulp and not learned much except that the Intermediate Value Theorem may be part of our DNA. Let’s look a bit further.
If S is a closed figure in the plane, then in fact there are two perpendicular lines that divide the figure into four “quadrants” of equal area. To see why this is so, let l(α) be a line that makes an angle α with the x-axis and, appealing to the previous argument, assume it bisects the area of S. Clearly l(α) and l(α + π/2) cut S into four quadrants. We’ll label them in the usual counterclockwise fashion and designate their areas A1(α), A2(α), A3(α) and A4(α). Since A1(α) + A2(α) = A3(α) + A4(α) and A1(α) + A4(α) = A2(α) + A3(α), we have at once that A1(α) = A3(α) and A2(α) = A4(α).
The difference D(α) = A2(α) – A1(α) is continuous, because each component is, and A1(α + π/2) = A2(α) and A2(α + π/2)= A3(α) = A1(α). Therefore D(α) changes sign between α and α + π/2, and there is an angle for which A2 = A1 (=A3 = A4).
A better known example is the “Ham Sandwich” Theorem: Given a piece of ham and a piece of bread (in the plane), it is always possible to cut both in half with one slice of a knife. Intuitive? Obvious? The proof combines the ideas explored in the previous arguments—give it a try.
Here is a less familiar example: There is a square (not just a rectangle) that circumscribes any figure S in the plane in the sense that S lies inside the square and each side of the square contains a boundary point of S (possibly a vertex). To prove this one, let l(α) be a line tangent to S in direction α and with S on the left of l(α).The lines l(α), l(α + π/2), l(α + π) and l(α + 3π/2) define a rectangle R(α) circumscribing S. Let L(α) be the “length” of R(α), the dimension parallel to l(α), and W(α) the “width” of R(α), the dimension perpendicular to l(α). Since W(α) = L(α + π/2), applying the Intermediate Value Theorem to L(α) – W(α) proves the theorem. As you visualize the rectangle R(α) changing dimension, you can “see” the sought-after square.
The Intermediate Value Theorem won’t matter unless the instructor makes it matter, so here’s a final problem to ponder: Consider a planar set where the maximum distance between any two points is 1. Find the side length of the smallest regular hexagon that is guaranteed to contain any such set. (And be sure to check out the Zip-line section of The Playground in this issue.)
About the author: Bruce Peterson is Charles A Dana Professor of Mathematics and College Professor Emeritus at Middlebury College. His fondness for the Intermediate Value Theorem stems from a lifelong advocacy of geometry in general. He also has an avid interest in ornithology.
Aftermath essays are intended to be editorials and do not necessarily reflect the views of the MAA. Contact information is available here.