# How do you prove that $\pi < \sqrt{10}$?

In the course of solving a puzzle, I had cause to assert that $\pi <\sqrt{10}$. I mean, that’s just true: I know that $\sqrt{10} \approx 3.16$ and $\pi \approx 3.14$; I also know that $\pi < \frac{22}{7}$ and that $\left(\frac{22}{7}\right)^2 < 10$. But these beg the question. How would I go about proving the inequality without resorting to “I just know”?

### Polygons

My first thought is, can I construct a polygon with a perimeter of $\sqrt{10}$ that lies outside of the unit circle?

An $n$-gon is made up of isosceles triangles, each of which has an angle at the apex of $\frac{2\pi}{n}$ and – if I’m putting it just outside of a unit circle – the triangles have a height of 1. The base of the triangle is… *scribbles*… $2\tan\left(\frac{\pi}{n}\right)$, so the polygon’s perimeter is $2n \tan\left(\frac{\pi}{n}\right)$. Are there any “nice” tangents I can draw on?

I’m drawn to the 24-gon, because $\tan\left(\frac{\pi}{24}\right) = \sqrt{6}+\sqrt{2}-\sqrt{3}-2$. I don’t know that this will lead anywhere, but the numbers look ok.

I know that $24\tan\left(\frac{\pi}{24}\right)>\pi$, by construction; if it’s smaller than $\sqrt{10}$, then the proposition is proved.

I spot a couple of factors: $\tan\left(\frac{\pi}{24}\right) = (\sqrt{2}-1)(\sqrt{3}-\sqrt{2})$. So, if I can show that $24(\sqrt{2}-1)(\sqrt{3}-\sqrt{2}) < \sqrt{10}$, we’re good.

I wonder if I can use a conjugate trick? I can multiply both sides by $\sqrt{3}+\sqrt{2}$, for example, and wonder, does $24(\sqrt{2}-1) < \sqrt{30} + \sqrt{20}$?

Rearranging, does $24\sqrt{2} < \sqrt{30} + \sqrt{20} + 24$?

(I recognise that I’m doing the thing we always tell students not to do, and starting from the answer. It’s ok. I’m a professional. More to the point, I’m trying to convince *myself*, knowing that I could write up the proof properly later).

Now: I want an upper bound on the left and a lower bound on the right. What’s $24\sqrt{2}$? It’s the same as $\sqrt{1152}$. That’s less than (sensei assures me) $34 - \frac{1}{17}$.

What’s an upper bound on $\sqrt{30}$? I know that $54^2 = 2916$, so $\sqrt{3000}< 54 + \frac{7}{9}$ and $\sqrt{30} < 5.4 + \frac{7}{90}$. I can probably improve that with a second decimal place if I need to.

Similarly, $\sqrt{20} < 4.4 + \frac{4}{55}$.

So far, I know that $24\sqrt{2} < 34 - \frac{1}{17}$ and that $\sqrt{30}+\sqrt{20}+24 < 33.8 + \frac{4}{55} + \frac{7}{90}$.

So now all I need to do is find out whether $0.2 < \frac{1}{17}+\frac{4}{55} + \frac{7}{90}$! We could, of course, work it out exaclty. But we’re as well to estimate: $\frac{1}{17}> \frac{1}{20}$; $\frac{4}{55} > \frac{1}{13}$ and $\frac{7}{90}>\frac{1}{13}$. Lastly, $\frac{2}{13}+\frac{1}{20} = \frac{53}{260}$, which is definitely larger than 0.2.

Whew! So, assuming all of my logic and arithmetic is correct, we can work backwards and prove that $\sqrt{10}> \pi$. I won’t do that, because writing up is tedious.

There are, without doubt, other ways of doing it. I’ve not explored complex numbers, calculus or continued fractions, which all look like reasonable approaches. If you’ve got another way, let me know!