This proof of the math fact of Rolle,

I wrote it down; here was my goal:

Use just words with one part.

(So it won't sound too smart.)

Please tell me if you find a hole.

**The math fact of Rolle:**

Let *f * be a map from a closed length of the reals (the length from *a* to *b*) to the set of all reals. (A length of the reals is called closed if both the least and most large points are in the length.) If the slope of the graph of *f* can be found at all points from *a* to *b*, and *f(a)* is the same as *f(b)*, then the graph of *f* is flat (the slope is naught) at a point more than *a* and less than *b*.

**Proof of the math fact of Rolle:**

First, it could be the case that *f* is the same on all reals from *a* to *b*. Then the graph of *f *is flat on all reals from *a* to *b*. We are done!

If that is not the case, then there must be a place *d* where *f(d)* is more than *f(a)* or *f(d)* is less than *f(a)*. We can look at just the first case, for the last case is much the same.

We will now use a math fact about maps with graphs that do not jump: if there is a map like this on a closed length of the reals, the graph must reach a high and a low in the length.

We can use this math fact if we know that the graph of *f* does not jump. To see that the graph of *f* does not jump, we use this math fact: if the slope of the graph of a map can be found at all points, then the graph of the map does not jump. We know that we can find the slope of the graph of *f* at all points, so we may say that the graph of *f* does not jump.

We now know that *f* must reach a high and a low in the length from *a* to *b*. Since there is a *d* such that *f(d)* is more than *f(a) (*which is the same as *f(b))*, we know that the high must not be at *a* or *b*. Thus the high is at a point *c* more than *a* and less than *b*.

We will show that the graph must be flat at *c*. How do we find the slope of the graph at this point? In this way: for some small *h*, take *f(c)* from *f(c+h)*. Now put this on top of *h*. We keep this up as *h* gets more and more close to naught. Since *c* is the high point of the graph, *f(c+h)* is less than *f(c)* for each small *h*, whether *h* is more or less than naught. So the top part must be less than naught for all *h*.

If the top part of a thing is less than naught and the low part is more than naught, then the thing is less than naught. If the top part of a thing is less than naught and the low part is less than naught, then the thing is more than naught.

We know that we can find the slope of the graph of *f* at all points from *a* to *b*, so we will get the same slope when *h* starts less than naught and gets more and more close to naught that we do when *h* starts more large than naught and gets more and more close to naught. In the first case, we get things that are less than naught. In the last case, we get things that are more than naught. So things that are less than naught and things that are more than naught are both on their way to the same thing. The one way this works is if the slope at *c* is naught.

We are done!

*This is a response to the challenge issued by David Roberts on Google+ and the n-Category Café to write proofs using only one-syllable words. (I wrote about the challenge for the AMS Blog on Math Blogs.) I know Roberts was looking for people like Terry Tao or Ben Green to explain the Green-Tao theorem in monosyllables (good luck!), but I hope he doesn't mind my much more pedestrian contribution. For a multisyllabic explanation of Rolle's theorem, which is a special case of the Mean Value Theorem, see its page on cut-the-knot.*