Can you give some example of an irrational number that can be proved to be irrational with this theorem?

Theorem. Given $a\in \mathbb{R}$, if there exists a sequence of integers $u_n,v_n \rightarrow \infty$ such that:

  • $a$ is not equal to any $u_n/v_n$.
  • The quotien sequences aproximates $a$ in this sense: $$\lim_{n\rightarrow \infty}v_na-u_n=0 $$

then the number $a$ is irrational.

The proof is in:

  • see Belgi's answer in that link. It can be very useful – Chinnapparaj R Oct 7 at 14:45
  • 1
    Think of a number between 0 and 1 whose decimal expansion only has 0s and 1s, with the distance between consecutive 1s growing very very fast. – Andrés E. Caicedo Oct 7 at 14:59
  • There is no proof of that theorem at the link. There are counter examples by Darth Geek and Belgi. – Steve B Oct 7 at 15:39
  • @SteveB: yes there is it: "the proof is very simple: if $a=p/q$ and $a\neq u_n/v_n$ then $v_na-u_n$ is a non-zero rational with denominator at most $q$. Therefore it is at least $1/q$ in absolute value, and can't go to 0" – Wilem2 Oct 7 at 15:43
  • Related questions:… – Wilem2 Oct 7 at 15:45
up vote 2 down vote accepted

First, there are some nice examples like $$ e=\sum_{n\ge0}\frac1{n!} $$ or Liouville-like numbers, mentioned in the answers by Wilem2, that can be easily proved to be irrational using the theorem, but for which typically there are simpler irrationality proofs: For $e$, we quickly get that $$0<n!\bigl(e-\sum_{k\le n}\frac1{k!}\bigr)<1$$ for all large enough $n$, so $e$ cannot be rational, because otherwise for $n$ large enough the number in between 0 and 1 would be an integer. For Liouville-like numbers, the many 0s in between consecutive 1s readily imply the number does not have a periodic decimal expansion, so it should be irrational.

Now, one can prove that given any irrational $r\in\mathbb R$ there is a sequence as in the theorem converging to $r$. This is not quite obvious. Instead, I'll refer you to the theory of continued fractions. The two points to make are that if $|r-p/q|<1/q^2$ then $|qr-p|<1/q$ (so, if $q\to\infty$, then $|qr-p|\to0$) and that for any two consecutive convergents to $r$, in fact one satisfies $0<r-p/q<1/q^2$. The convergents to $r$ are the rational numbers that are obtained by iterating the following recursive procedure:

Start with $r$ irrational, let $r_0=\lfloor r\rfloor$. Note that $0<r-r_0<1$, so it is $1/x$ for some $x>1$, and we can repeat the same procedure with $x$ instead of $r$ and to obtain $r_1\in\mathbb N$ and $y>1$ such that $x=r_1+1/y$. For instance, if $r=\sqrt2$, you get $r=1+(\sqrt2-1)=1+\frac1x$, where $x=\frac1{\sqrt2-1}=\sqrt2+1=2+\frac1x$, so $$ \sqrt2=1+\frac1{2+\frac1{2+\frac1\ddots}}, $$ and the convergents to $\sqrt2$ are the rationals that appear along the way, by stopping the procedure after finitely many times, that is, the sequence $$ 1, 1+\frac12=\frac32,1+\frac1{2+\frac12}=\frac75,1+\frac1{2+\frac1{2+\frac12}}=\frac{17}{12},\dots$$

In general, if $p_0/q_0,p_1/q_1,\dots$ are the convergents to $r$, we have that $(p_n,q_n)=1$ for all $n$ and $$ p_0/q_0<p_2/q_2<\dots<r<\dots<p_3/q_3<p_1/q_1, $$ so $p_0/q_0,p_2/q_2,\dots$ provides a sequence as desired (one can easily check just from the inequalities that $q_n\to\infty$).

That said, it is perhaps a bit unsatisfactory to use continued fractions to illustrate the theorem, because of course if $r$ has an infinite sequence of convergents, then it is irrational (by the Euclidean algorithm!), so we already know $r$ is irrational before we even exhibit the relevant sequence. That said, in certain cases, there are nice recursive relations that allow us to find $p_{n+2},q_{n+2}$ in terms of $p_n,p_{n+1},q_n,q_{n+1}$, so we can easily exhibit sequences as in the theorem, thus witnessing the irrationality of many $r$. (Of course, for some $r$, there is no nice recursive way of obtaining such sequences unless we have access to $r$ to begin with.)

Indeed, for instance for $\sqrt2$, we have $p_0=1=q_0$, $p_1=3,q_1=2$ and, in general $p_{n+2}=2p_{n+1}+p_n$ and $q_{n+2}=2q_{n+1}+q_n$.

In general, for a given $r$, rather than always using $2$, we use $r_{n+2}$, where the integers $r_0,r_1,r_2,\dots$ are the integer parts obtained through the procedure as described above. For many irrationals $r$ there does not seem to be a nice formula for these numbers $r_n$. But there are nice formulas for $e$ and for any quadratic irrational.
(It is actually an interesting open problem whether there are nice patterns for, say $\root3\of2$ or $\pi$.)

  • Thank you very much! It only reminds to check that $|r-p/q|<1/q^2$ for the sequence given by the continued fraction – Wilem2 Oct 8 at 21:59

Let's proof that $e=\sum_{n=0}^{\infty}\frac{1}{n!}$ is irrational with this method:

Define $v_n=n!$ and $u_n=n!(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!})$

It's direct to see that $u_n$ is an integer, because expanding the product, each term is an integer:


And also, $u_n$ tends to infinity cause each term is positive and the leading term is $n!$

Let's proof the satements of the theorem:

Clearly, $e\neq\frac{u_n}{v_n}$ because $(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+...)\neq(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!})$

And for the second statement:

$x_n=v_ne-u_n=n!(\sum_{i=0}^\infty \frac{1}{i!})-n!(\sum_{i=0}^n \frac{1}{i!})=n!(\sum_{i=n+1}^\infty \frac{1}{i!})=\sum_{i=n+1}^\infty \frac{n!}{i!}>0$

Now, each term of the last serie is limited by


and so $x_n=\sum_{i=n+1}^\infty \frac{n!}{i!}\leq\sum_{i=n+1}^\infty\frac{1}{(n+1)^{i-n}}=\sum_{k=1}^{\infty}\frac{1}{(n+1)^{k}}=\frac{1}{n+1}(\frac{1}{1-\frac{1}{n+1}})=\frac{1}{n}\rightarrow 0$

So we can apply the theorem to conclude that $e$ is irrational.

  • The theorem is an overkill in this case, though. – Andrés E. Caicedo Oct 7 at 16:28
  • @AndrésE.Caicedo: You are totally right, it's a recycled proof. This is why I won't put this as an answer of the question... – Wilem2 Oct 7 at 16:38
  • 1
    $v_n$ = 1, 2, 6, 24, ... $u_n$ = 1, 3, 10, 41, ... That makes $x_n$ = 1.7183, 2.4366, 6.3097, 24.239. $x_n$ does not converge to 0. – Steve B Oct 7 at 18:24
  • @SteveB : Corrected, thanks – Wilem2 Oct 7 at 18:35

With the idea of @AndrésE.Caicedo, I will try to proof the irrationality of the Liouville Constant:


With $v_n=10^{n!}$ and $u_n=10^{n!}\sum_{i=1}^{n}10^{-i!}$.

Clearly, $v_n$ is an integer that tends to infinity.

For $u_n$, note that $u_n=\sum_{i=1}^{n}10^{n!-i!}$ and the power $n!-i!$ is an integers satisfying $n!-i!\geq 0$ for $i\leq n$, so each term of the serie is an integer and also, $u_n$ tends to infinity caused by the leading term $10^{n!-1}\rightarrow \infty$.

To check the hypothesis:

  • $L \neq\frac{u_n}{v_n}$ because $L$ have more terms in the serie: $L=\sum_{i=1}^{\infty}10^{-i!}\neq \sum_{i=1}^{n}10^{-i!}=\frac{u_n}{v_n}$
  • $x_n=v_nL-u_n=...=\sum_{i=n+1}^{\infty}10^{n!-i!}$. It can't be difficult to proof that this last serie is $$\sum_{i=n+1}^{\infty}10^{n!-i!} <2\cdot 10^{n!-(n+1)!}=2\cdot 10^{-n\cdot n!}\rightarrow 0$$ But because I don't have the argument, here is another:

For $i>n$ one have $n!-i!\leq (i-1)!-i!\,=\, -(i-1)\cdot(i-1)! \leq -(i-1)$

So $n!-i! \leq -(i-1)$ and hence: $$x_n=\sum_{i=n+1}^{\infty}10^{n!-i!}\leq \sum_{i=n+1}^{\infty}10^{-(i-1)}=\frac{1}{10^n}\sum_{i=0}^{\infty}10^{-i}=\frac{1}{10^n}\frac{1}{1-\frac{1}{10}}=\frac{1}{10^n}\frac{10}{9}\longrightarrow_{n \rightarrow\infty} 0$$

By the theorem, L is irrational.

  • The idea inside is that, as @AndrésE.Caicedo pointed, the term $x_n=10^nL-u_n$ have its first non-zero digit in its decimal expansion too far (more than $n$-far) from the unit digit, so as $n$ grows, $x_n$ tends to zero. – Wilem2 Oct 7 at 21:54

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