## Irrationality of pi: the integral that wasn't

And now for the punchline! Today we’ll show that, for large enough values of $n$, $0 < \int_0^\pi f(x) \sin (x)\, dx < 1,$

completing the proof of the irrationality of $\pi$.

First, let’s show that $f(x) \sin(x)$ is positive when $0 < x < \pi$. We know that $\sin(x)$ is positive for $0 < x < \pi$. But I claim that $f(x)$ is too. Remember that $\displaystyle f(x) = \frac{x^n(a - bx)^n}{n!}.$ $n!$ and $x^n$ are clearly positive when $x$ is positive; and $a - bx$ is also positive when $x < \pi = a/b$. 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 $0 < x < \pi$, $\displaystyle f(x) \sin(x) < \frac{\pi^n a^n}{n!}.$

Why is this? Well, clearly $x^n < \pi^n$ (since $x < \pi$), and also $latex a – bx 0$) and hence $(a - bx)^n < a^n$, so we conclude that $\displaystyle f(x) = \frac{x^n(a - bx)^n}{n!} < \frac{\pi^n a^n}{n!}.$

This doesn’t yet include the $\sin(x)$, but notice that multiplying by $\sin(x)$ can only make things smaller, since $\sin(x)$ is at most $1$. Now, here’s the slightly sneaky part: I claim that we can make $\frac{\pi^n a^n}{n!}$ as small as we want by making $n$ big enough. Why is this? Notice that we can rewrite it as $\displaystyle \frac{\pi^n a^n}{n!} = \frac{\pi a}{1} \cdot \frac{\pi a}{2} \cdot \frac{\pi a}{3} \dots \frac{\pi a}{n}.$

Now, $a$—the “denominator” of $\pi$—might be very large. It might have fourteen million zillion digits. But no matter how big $a$ is, there will of course be an integer $z$ which is bigger than $\pi a$, so $\frac{\pi a}{z} < 1$. And then $\frac{\pi a}{z + 1} < 1$, and $\frac{\pi a}{z + 2} < 1$, 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 $\frac{\pi a}{1} \cdot \frac{\pi a}{2} \dots \frac{\pi a}{z}$, but if we just wait patiently it will get smaller and smaller… and eventually there will come some $n$ for which $\displaystyle \frac{\pi a}{1} \cdot \frac{\pi a}{2} \dots \frac{\pi a}{z} \cdot \frac{\pi a}{z + 1} \dots \frac{\pi a}{n} < 1.$

Actually, even this isn’t quite small enough: we want the integral from $0$ to $\pi$ of $f(x) \sin(x)$ to be less than 1. But that’s not a problem; to ensure that we can just pick $n$ big enough so that $f(x) < 1/\pi$ (if the graph of $f(x)$ fits inside a $\pi$ by $1/\pi$ 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 $\pi$ was rational.

The inescapable conclusion, which probably would have driven the ancient Greeks crazy, is that $\pi$ is irrational! Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 8 Responses to Irrationality of pi: the integral that wasn't

1. Thomas van Putten says:

Thanks for showing this.

The thing I don’t understand – but mind you, my math is at the most very rusty – is where the real Pi comes in. As fas as I can tell it is still an abstract ratio of a/b. So what this seems to prove is that any fraction of two integers is irrational. Which seems to me to be counterintuitive, to say the least.

But as mentioned before, I’m not very good at this, Pi is probably hidden in the sinuses, as usual.

2. Ashwin Sawant says:

You are right – it is hidden in the trig functions. We started by assuming pi=a/b. pi had a special meaning which we plugged in when introducing the trig functions in the proof. We never allowed a/b to stand for anything other than pi.

3. Jeffo says:

TvP: Pi is used essentially in Step 5, when Cos Pi = -1 is evaluated. To sidestep that would require a proof that Cos of the rational number is an integer. It is also used when it is asserted that the integral of f Sin b/w 0 and Pi is positive, but the same is true for many rational numbers.

4. Brent says:

Excellent question! I actually had to think about it for a little bit to be sure I knew the right answer. If we take any ratio a/b whatsoever, we could define f(x) in the same way, and the values of all its derivatives would still be integers at both 0 and a/b; we could define F(x) in the same way and the antiderivative of f(x)sin(x) would still be F'(x) sin(x) – F(x) cos(x). The place where things would go wrong—where it matters that we are actually talking about pi—is in evaluating the integral of f(x)sin(x). When we evaluated the integral from 0 to pi, it simplified a lot since sin(pi)=0 and cos(pi)=-1. If we were using some other random a/b we would be stuck with some terms of sin(a/b) and cos(a/b) and we would not be able to conclude anything about the integral being an integer.

5. Dave says:

Excellent proof, Brent. Thanks again!

6. Michael Scott Cuthbert says:

This was an excellent series! My compliments to the writer — haven’t taken a math class in 16 years, yet it was all completely clear and simple to follow. (glad that you didn’t make me dig up what a Taylor series was — you had me scared at one point).

7. Brent says:

Michael: thanks! Glad you enjoyed it!

8. danas says:

Thanks! Very nice and clear explanation of the proof 🙂