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?

  • 6
    $\begingroup$ What is "naturally occurring"? Fundamental examples of transcendental numbers like $\sum 10^{-n!}$ (Liouville's constant) are most obviously irrational by the non-repeating criterion. $\endgroup$ – Matthew Towers Jul 19 '16 at 19:59
  • $\begingroup$ I can easily get $\sqrt2$ to be non-repeating. Working on non-terminating. $\endgroup$ – Emily Jul 19 '16 at 20:11
  • 1
    $\begingroup$ @Emily: non-terminating is easy! To wit: Suppose $\sqrt 2$ terminates, and let $d$ be its least significant non-zero decimal digit. Then the least significant non-zero decimal digit of $2$ must be $d^2 \bmod 10$, which is never equal to $2$. Contradiction. $\endgroup$ – TonyK Jul 19 '16 at 20:32
  • 1
    $\begingroup$ I think for purposes of this question, any number that is deliberately constructed by created a non-terminating decimal expansion can't be counted as "naturally occurring". So that would include Liouville's number. $\qquad$ $\endgroup$ – Michael Hardy Jul 19 '16 at 23:02
  • 1
    $\begingroup$ I'd put the question the other way around, Michael: are there any naturally occurring irrationals such that we can say anything useful at all about their decimal representations? I'd say the answer, at our present state of knowledge, is no – no one has anything useful to say about the decimal representation of $\pi$ or $e$ or $\sqrt2$ or $\log_23$ or any such number. $\endgroup$ – Gerry Myerson Apr 17 '18 at 2:48

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.


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.

  • 1
    $\begingroup$ From the examples Michael Hardy gave for "naturally occurring number", I suspect that any number defined pretty much only in terms of its decimal expansion doesn't qualify. $\endgroup$ – Dave L. Renfro Jul 19 '16 at 20:12
  • $\begingroup$ This was already stated in the first paragraph of the question. $\qquad$ $\endgroup$ – Michael Hardy Jul 19 '16 at 23:07
  • $\begingroup$ Given that Liouville's constant is possibly the easiest number to prove to be transcendental I would like to argue that it is "naturally occurring". But, ok, this is Michael's game :-) $\endgroup$ – Jyrki Lahtonen Jul 20 '16 at 19:52

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.

  • 3
    $\begingroup$ The trouble I'd have (but I'm not Michael Hardy) is that your proof seems to be just the standard proof that $\sqrt{2}\neq\frac{m}{n}$ except that you are writing $10^k-1$ instead of $n$ - so, just a typographical variant. But perhaps I have misunderstood a subtlety here. $\endgroup$ – Martin Kochanski Jul 19 '16 at 20:28
  • $\begingroup$ Ah, yes. The standard approach I refer to is when you have a repeating decimal, you let the "repeating" part be equal to some $m$, and multiply by a big enough power of $10$ to get a fraction that is equal to some number over $10^k-1$. Example, take $x = 0.142857\ldots$. Let $x = 0.142857$ exactly. Then, $10^6 x -x = 142857 \implies x = \frac{142857}{10^6-1}$. $\endgroup$ – Emily Jul 19 '16 at 20:32
  • $\begingroup$ This is the non-reduced rational representation of the decimal expansion for a repeating number, so I'm just showing it can't be of this form. $\endgroup$ – Emily Jul 19 '16 at 20:36
  • $\begingroup$ The problem here is this amounts to a conjunction of two things: $(1)$ the standard proof that repeating decimals represent numbers of the form $a/b$ for some $a,b\in\mathbb Z$, and $(2)$ the standard proof that $\sqrt 2$ is not of that form. Gluing those two things together doesn't amount to showing that the decimal expansion is non-repeating INSTEAD OF showing that the number is not of the form $a/b$ for some $a,b\in\mathbb Z$. $\qquad$ $\endgroup$ – Michael Hardy Jul 19 '16 at 23:14

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.

  • 1
    $\begingroup$ you know that proving Apery's constant $\sum_{n=1}^\infty \frac{1}{n^3}$ is irrational has its own theorem proved in 1978 ? and it is not known if $\sum_{n=1}^\infty \frac{1}{n^{2k+1}}, k \ge 2$ is rational. so your argument doesn't work for $\sum_{n=1}^\infty \frac{1}{n^2}$. And see my answer above for how your argument works for $e-1 = \sum_{n=1}^\infty \frac{1}{n!}$ $\endgroup$ – reuns Jul 19 '16 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.