And now for the punchline! Today we’ll show that, for large enough values of ,
completing the proof of the irrationality of .
First, let’s show that is positive when . We know that is positive for . But I claim that is too. Remember that
and are clearly positive when is positive; and is also positive when . From here we simply note that if a function is positive over an entire interval, the integral of the function over that interval will be positive as well.
For the second part, note first that for ,
Why is this? Well, clearly (since ), and also $latex a – bx 0$) and hence , so we conclude that
This doesn’t yet include the , but notice that multiplying by can only make things smaller, since is at most . Now, here’s the slightly sneaky part: I claim that we can make as small as we want by making big enough. Why is this? Notice that we can rewrite it as
Now, —the “denominator” of —might be very large. It might have fourteen million zillion digits. But no matter how big is, there will of course be an integer which is bigger than , so . And then , and , and so on… of course, multiplying by something less than one makes things smaller. And it might take a really really long time to cancel out the enormous product , but if we just wait patiently it will get smaller and smaller… and eventually there will come some for which
Actually, even this isn’t quite small enough: we want the integral from to of to be less than 1. But that’s not a problem; to ensure that we can just pick big enough so that (if the graph of fits inside a by box, then its integral on this interval must be less than the area of the box).
Voila! An integral which is an integer absurdly between 0 and 1, all because we assumed was rational.
The inescapable conclusion, which probably would have driven the ancient Greeks crazy, is that is irrational!