# On normal numbers and e

Posted by Jason Polak on 30. January 2018 · Write a comment · Categories: number-theory · Tags: ,
A real number is simply normal in base $b$ if the frequency of each base $b$ digit in the first $n$ digits tends to a limit as $n$ goes to infinity, and each of these limits is the same.

In other words, a real number is simply normal in base $b$ if each digit appears with equal frequency. It may not be the case of course that the frequency is well-defined: as you keep computing more and more decimal digits of a number, the frequency of any given digit could keep oscillating.

A real number is normal in base $b$ if the frequency of each length $k$ string of numbers exists and is equal to $1/b^k$.

A rational number can be simply normal in some base. For example, 0.12345678901234567890…. is simply normal in base ten. Rational numbers can’t be normal, since their decimals repeat. Irrational numbers can be normal, and in fact almost every number is normal in the sense of the Lebesgue measure. Champernowne proved that the number

0.123456789101121314…

is normal to every base. Mathematicians believe that $e$ and $\pi$ are normal but no one has ever proved it, or found a proof that they are not normal. It’s tempting to look at some numerical evidence. For example, here are the frequencies of the digits in the first 301029 decimal digits of $e$:

 0 30057 1 30283 2 29785 3 30316 4 30206 5 30044 6 30375 7 30065 8 30055 9 29843

Counting these took about 2.4 seconds. Looks like the digit ‘6’ occurs the most often, whereas ‘2’ occurs the least often. Is this about what we’d expect? Here’s the result of a $\chi^2$-test for goodness of fit, the null hypothesis being that the probability of each digit is the same:

X-squared = 11.309, df = 9, p-value = 0.2551

Looks like there’s no reason to be alarmed. Of course, since the normality and simple normality of a number depends only on the “tail” of its expansion, just testing a finite number of digits can be misleading. The rational number defined by the first 301029 digits of $e$ is not normal or simply normal.

The frequencies for all two digit combinations starts off like this:

 00 2957 01 3035 02 3040 03 2978

You probably don’t want to look at the other 96 frequencies. But a $\chi^2$-test reveals the following:

X-squared = 108.06, df = 99, p-value = 0.2506

Although this is rather convincing that $e$ is normal, none of this proves anything. If it is, you’d expect 12345 to occur about three times in the first 301029 digits. In fact it does, the first time starting at the 166412th digit. The sequence 99999 only occurs once, starting at the 290471st digit. Even though the odds are against it, the starting six digit sequence 271828 actually does occur once, at the 252474th digit. Maybe that will help you memorise $e$?