Are there any natural proofs of irrationality using the decimal characterization?

Question

Mathematicians typically define rational number to mean quotient of two integers. It is not hard to show that a number is rational by that definition if and only if its decimal expansion terminates or repeats. Let us call that the “decimal characterization” of rationality. The proof of that characterization of rational numbers is obviously just as applicable to other bases as it is to base $10$.

Question: Are there any proofs of the irrationality of $\pi$ or $e$ or $\sqrt 2$ or $\log_2 3$ or any other naturally occurring number that use the decimal characterization, showing directly that the decimal expansion does not terminate or repeat, and that are at least as simple as any proof that uses the characterization that is conventionally taken to be the definition?

Answer 1

As @m_t_ says, just write down a formula for any number that has a non-repeating decimal expansion and you've shown that number is irrational. For example $0.1010010001000010000010000001\ldots$ Such a number is obviously immediately irrational from inspection of the decimal expansion.

Answer 2

Let's take a shot at this.

Suppose $\sqrt2$ had repeating digits. By the standard approach, we can find integers $m, k$ such that

$$\sqrt2 = \frac{m}{10^k-1}.$$

Squaring both sides and expanding, we have

$$m^2 = 2\color{red}{(10^k-1)^2}.$$

We see that the term in red cannot have a factor of $2$; therefore, the integer $m^2$ appears to have a single factor of $2$, in violation of the unique factorization theorem.

Therefore, $\sqrt2$ cannot have repeating digits.

Suppose then that $\sqrt2$ has terminating digits. We can then find integers $m, k$ such that

$$\sqrt2 = \frac{m}{10^k}.$$

As before, $m^2 = 2\cdot \color{red}{10^{2k}}.$

The term in red has an evidently even number of factors of $2$. This shows that $m^2$ has an odd number of factors of $2$, in violation of the unique factorization theorem.

Answer 3

This is a suggestion rather than a proof, but it's too long for a comment.

What about $$\zeta(2)=\sum_1^\infty\frac{1}{k^2}=\frac{\pi^2}{6}\text{?}$$

If it is rational, it has period $n$. But there exists a number $N$ such that the term $\frac{1}{N^2}$ has a period $m$ which is not a divisor of $n$. If it were the only period-$m$ term in the series, we would have a contradiction, which by reductio ad absurdum means that $n$ cannot exist, and that $\zeta(2)$ can't have a period and so must be irrational.

The hand-waving part of this argument, which I'm sure can be made solid, is that the sum of all period-$m$ terms in the series is itself of period $m$. (It is only necessary, strictly speaking, that some values of $m$ should exist for which this is true: for instance, $m=2p$).

A similar argument with factorials instead of squares would show the irrationality of $e^1=e$. Less hand-waving would then be needed because the terms get smaller so fast that there is no possibility of the period-$m$ terms coalescing into something with a shorter period.

Answer 4

I just learnt this, it is not about the decimal expansion, but I think it is clearly related to your question :

A simple proof that $e$ is irrational is using its factorial expansion $2+\sum_{n=2}^\infty \frac{1}{n!}$.

The theorem is that $$\sum_{n=1}^\infty \frac{a_n}{n!}, \qquad a_1 \in \mathbb{Z},\quad a_n \in \{0, \ldots, n-1\}$$ is irrational if and only if $a_n >0$ infinitely often and $a_n < n-1$ infinitely often.

The proof is as follows :

• any real number can be written as $$\sum_{n=1}^\infty \frac{a_n}{n!}, \qquad a_1 \in \mathbb{Z},\quad a_n \in \{0, \ldots, n-1\}$$ with the same algorithm as for the decimal expansion : $$a_n = \lfloor x_n n!\rfloor, \qquad x_{n+1} = x_n- \frac{a_n}{n!}$$

• with this algorithm, every rational number has a finite expansion, so if $x$ has a finite expansion then the algorithm finds it. More generally, if applied to $x = \sum_{n=1}^\infty \frac{c_n}{n!}, c_1\in \mathbb{Z},c_n \in \{0, \ldots, n-1\}$ where $c_n < n-1$ infinitely many times, then the algorithm finds back $a_n = c_n$.

• when applied to $e-1$, the algorithm finds $a_n = 1$, thus $e-1$ doesn't have any such finite expansion, hence it is irrational.

The conclusion is that the factorial expansion (more generally the Cantor series expansion, if someone knows a better reference ?) is much easier than the base $N$ expansion for proving the irrationality of a real number.