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?

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    $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – Larry Wang Aug 9 '10 at 6:38
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    $\begingroup$ Let me try to fix your intuition: when you restrict yourself to denominators which are powers of 10, you make the problem of rational approximation unnecessarily difficult because you added a constraint that doesn't need to be there. You can reach this conclusion without knowing anything about continued fractions. $\endgroup$ – Qiaochu Yuan Aug 9 '10 at 7:42
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    $\begingroup$ 3/1 is closer to X than 3.001 = 3001/1000 which has a higher denominator , where X = 3.000000000000014. $\endgroup$ – T.. Aug 9 '10 at 8:22
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    $\begingroup$ @Qiaochu: thanks for that, I guess 22/7 is going to be closer to a whole range of real numbers than both 314/100 and 315/100. One of them happens to be pi. $\endgroup$ – Douglas S. Stones Aug 9 '10 at 12:17
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    $\begingroup$ That is not an important implication at all. $\endgroup$ – Kevin H. Lin Aug 21 '10 at 11:59

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.

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    $\begingroup$ And of course, as Kaestur's comment points out, Wikipedia has all this and much, much more. $\endgroup$ – Vaughn Climenhaga Aug 9 '10 at 6:43
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    $\begingroup$ @Vaughn: It's still worthwhile to have it here! I posted the link just as a quick reference for anyone who wants to write a real answer. $\endgroup$ – Larry Wang Aug 9 '10 at 15:58
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    $\begingroup$ Wait a minute! This is inaccurate: "22/7 is better approximation than 314/100 because 22/7 appears in the CF while 314/100 does not". The more accurate statement about continued fractions is "if m/n is a convergent, then it is a closer approximation than a/b, where (a,b) = 1 and b <= n". The restriction on denominator is important. For instance 31415/10000 is a better approximation than 22/7 to pi. I don't think it is a convergent of the CF though. $\endgroup$ – Aryabhata Aug 20 '10 at 22:49
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    $\begingroup$ @Vaughn: Maybe, but I don't think that really answers the question that was asked. $\endgroup$ – Aryabhata Aug 21 '10 at 1:06
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    $\begingroup$ @Vaughn. Only a mind reader would infer that that is the definition of 'better approximation' from what you have written in your answer. Besides, it is a totally unnatural definition of 'better approximation', IMO. I believe OP (and others who answered this) were only looking to compare the error terms in determining which is better approximation. Yes OP mentioned denominators, but this is just convoluted, IMO. Anyway, I suggest you edit your answer and make it clear. $\endgroup$ – Aryabhata Aug 21 '10 at 3:45

Well, just measure $|\pi - 22/7|$ and $|\pi-3.14|$ ...

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    $\begingroup$ -1 for smartass answer $\endgroup$ – I. J. Kennedy Dec 10 '10 at 15:48
  • $\begingroup$ measure or calculate? $\endgroup$ – David Nov 6 '14 at 14:35

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

$$\frac{22}{7} - \pi < \frac{1}{700}$$


$$2\pi > 2(\frac{22}{7} - \frac{1}{700})$$

$$2\pi > \frac{22}{7} + \frac{22}{7} - \frac{2}{700}$$

$$2\pi > \frac{22}{7} + \frac{2200}{700} - \frac{2}{700}$$

$$2\pi > \frac{22}{7} + \frac{2198}{700}$$

$$2\pi > \frac{22}{7} + \frac{314}{100}$$

Thus we have that

$$0 < \frac{22}{7} - \pi < \pi - \frac{314}{100}$$


It only seems odd to you because you are used to representing numbers in base 10. What if you used base 7?

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    $\begingroup$ You'd be constantly confused in supermarkets? $\endgroup$ – walkytalky Aug 19 '10 at 18:18
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    $\begingroup$ 7 is still smaller than 100 in base 7... $\endgroup$ – Douglas S. Stones Aug 20 '10 at 1:55
  • $\begingroup$ Yes, but 3.1 would look nicer than 313/101. $\endgroup$ – Dan Aug 21 '10 at 4:05

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