Ï€ does go on forever (unpredictably)

Posted by: Gary Ernest Davis on: January 26, 2011

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Reuters new agency reported on January 20, here 2011 that a Japanese systems engineer, Shigeru Kondo, had used a home-built computer to calculate $pi$ toÂ trillions of digits and therebyÂ broken the existing Guinness Book of Records entry for calculating $pi$ .

A sentence toward the end of the report caught my eye:

“Calculating a more accurate pi, which is believed to go on forever, has been a challenge for scholars for thousands of years, ever since the parameter was used in ancient Egypt.” (My emphasis)

$pi$ is not a rational number

Of course the decimal expansion of $pi$ goes on forever, because $pi$ is not a rational number. That is, there are no integers $m,n$ for which $pi=frac{m}{n}$ .

This was first proved by Johann Heinrich Lambert in 1761.

The irrationality of $pi$ means not only does the decimal expansion of $pi$ go on forever, but that it never repeats from some point on.

Of course we are familiar with rational numbers whose decimal expansions repeat, such as $frac{1}{3}=0.333333ldots$ , $frac{1}{7}=0.142857142857142857ldots$ and $frac{611}{4950}=0.12343434ldots$ which repeats after the first two decimal places.

Equally, any decimal that eventually repeats is a rational number.

The reason for this is fairly straightforward.

First, if $x$ is a number that has a repeating decimal after, say, the $n^{th}$ decimal place then $10^n imes x$ is a repeating decimal.

Second, repeating decimals $y$ are rational numbers – just multiply $y$ by $10^p$ where $p$ is the length of the repeating block, to get $10^p imes y= extrm{ [repeating block] } + y$ .

For example, if $x = 0.21356565656ldots$ then $1000 imes x = 213+0.565656ldots$ and if $y= 0.565656ldots$ then $100y=56.565656ldots = 56+y$ so $99 imes y = 56$ . So $y$ is a rational number, and so therefore is $x$ .

All this means that not only does the decimal expansion of $pi$ go on forever, it never eventually repeats.

$pi$ is transcendental

The same is true, of course, for $sqrt{2}$ which is an irrational number.

What distinguishes $pi$ from a number such as $sqrt{2}$ is that the latter is algebraic.

A number $heta$ is “algebraic” if there is a polynomial

$p(x)=a_0+a_1x+a_2x^2+ldots+a_nx^n$

all of whose coefficients $a_i$ are integers, for which

$p( heta )=0$

Of course, in the case of $sqrt{2}$ such a polynomial is $p(x)=x^2-2$ .

For $pi$ there is no such polynomial. This is what is means for $pi$ to be a transcendental number: $pi$ is not algebraic.

This was first proved by Ferdinand von Lindemann in 1882.

Algebraic numbers are in many respects easier to understand than non-algebraic (= transcendental ) numbers. In fact, Leopold Kronecker denied the existence of transcendental numbers (including $pi$ ).

Are the digits of $pi$ random?

This is a very difficult question to answer unless we have a workable definition of “random”.

Gregory Chaitin and Andrey Nikolaevich Kolmogorov came up with a definition of “algorithmic randomness” which has to do with the non-compressibility of the description of a number. While this is an interesting and useful idea it does not seem to correspond exactly to what most people think of as randomness.

In practice, a sequence of numbers – such as the digits of $pi$ – is NOT random if it fails one of a batch of tests for randomness.

What this means in practice is that we cannot prove that a sequence of numbers is random, only that it is not random.

If a sequence of numbers passes all known tests for randomness we tentatively conclude that the sequence is behaving like a random sequence.

A 2001 research report by Paul Preuss at Berkley, indicates that the sequence of digits of $pi$ does behave as if it were random.

So not only, do the digits of $pi$ go on forever, they do not eventually repeat, and appear to all intents and purposes, to be random.