This seems counterintuitive, but $22/7$ is closer to $\pi$ than $3.14=314/100$ which has a significantly greater denominator.
Why is $22/7$ a better approximation for $\pi$ than $3.14$?
This has important implications: e.g. Should "$\pi$-day" be the $14^{th}$ of March or $22^{nd}$ of July?
On
It only seems odd to you because you are used to representing numbers in base 10. What if you used base 7?
On
Just for fun...
Here is a proof that $\displaystyle \frac{22}{7}$ is a better approximation than $\displaystyle 3.14$.
First we consider the amazing and well known integral formula for $\displaystyle \frac{22}{7} -\pi$ (for instance see this page: Proof that 22/7 exceeds pi).
$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \frac{22}{7} -\pi$$
We will to show that
$$0 < \frac{22}{7} -\pi < \pi - 3.14$$
That $\displaystyle 0 < \frac{22}{7} - \pi$ follows trivially from the above integral.
We will now show that
$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \frac{1}{700}$$
We split this up as
$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx = \int_{0}^{\frac{1}{2}}\frac{x^{4}(1-x)^{4}}{1+x^2} + \int_{\frac{1}{2}}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx$$
The first integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle 0$ and the second integral can be upper-bounded by replacing $\displaystyle x$ in the denominator with $\displaystyle \frac{1}{2}$.
Thus we have that
$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx + \int_{\frac{1}{2}}^{1} \frac{4x^{4}(1-x)^{4}}{5}dx $$
Now $$\int_{0}^{\frac{1}{2}}x^{4}(1-x)^{4}dx = \int_{\frac{1}{2}}^{1}x^{4}(1-x)^{4}dx$$ as $\displaystyle x^4(1-x)^4$ is symmetric about $\displaystyle x = \frac{1}{2}$
It is also known that $$\int_{0}^{1}x^{4}(1-x)^{4}dx = \frac{1}{630}$$ (see the above page again)
Thus we have that
$$\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^2}dx < \frac{1}{2*630} + \frac{4}{5*2*630} = \frac{1}{700}$$
Thus we have that
$$\frac{22}{7} - \pi < \frac{1}{700}$$
i.e
$$2\pi > 2(\frac{22}{7} - \frac{1}{700})$$
$$2\pi > \frac{22}{7} + \frac{22}{7} - \frac{2}{700}$$
$$2\pi > \frac{22}{7} + \frac{2200}{700} - \frac{2}{700}$$
$$2\pi > \frac{22}{7} + \frac{2198}{700}$$
$$2\pi > \frac{22}{7} + \frac{314}{100}$$
Thus we have that
$$0 < \frac{22}{7} - \pi < \pi - \frac{314}{100}$$
It has to do with the continued fraction expansion of $\pi$. Suppose $[a_1, a_2, \ldots]$ is the continued fraction of an irrational number $\alpha$ -- that is, if $a_n$ is the (essentially unique) sequence of natural numbers such that if we define partial convergents by $x_1 = a_1$, $x_2 = a_1 + 1/a_2$, $x_3 = a_1 + 1/(a_2 + 1/a_3)$, $x_4 = a_1 + 1/(a_2 + 1/(a_3 + 1/a_4))$, and so on, then $\alpha = \lim_{n\to\infty} x_n$. Then the partial convergents $x_n$ are rational numbers that approximate $\alpha$ better than anything that is not a partial convergent, in the following sense: a rational number $\frac pq$ satisfies the inequality $|\alpha - \frac pq| < \frac 1{2q^2}$ if and only if $\frac pq$ is one of the convergents $x_n$. (One could, of course, come up with different notions of what constitutes a "good" approximation.)
The continued fraction expansion of $\pi$ is $[3,7,15,1,292,1,1,\dots]$, so the first few convergents are $3$, $\frac{22}{7}$, $\frac{333}{106}$, $\frac{355}{113}$, etc. Thus $\frac{22}{7}$ is a better approximation than $\frac{314}{100}$ (in the above sense) because it appears in the list of partial convergents, while $\frac{314}{100}$ does not.
Incidentally, the approximation $x_n$ is best when the coefficient $a_{n+1}$ is quite large, so the size of $a_5 = 292$ means that $x_4 = \frac{355}{113}$ is a particularly good approximation.
At the risk of self-promotion, I wrote a brief exposition of all this in a bit more detail -- you can find it on my website if you're interested, at http://www.math.psu.edu/climenha/contfrac.html.