# Example of a concrete irrationality test

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: https://math.stackexchange.com/q/898420

• see Belgi's answer in that link. It can be very useful – Chinnapparaj R Oct 7 '18 at 14:45
• 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 '18 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 '18 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 '18 at 15:43
• Related questions: math.stackexchange.com/questions/787382/… – Wilem2 Oct 7 '18 at 15:45

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 $$0 for all large enough $$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 $$0. The convergents to $$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 $$0, so it is $$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 $$p_0/q_0 so $$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 '18 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:

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

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

$$\frac{n!}{i!}=\frac{1}{(n+1)(n+2)\cdot...\cdot(n+(i-n))}<\frac{1}{(n+1)^{i-n}}$$

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 '18 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 '18 at 16:38
• $v_n$ = 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 '18 at 18:24
• @SteveB : Corrected, thanks – Wilem2 Oct 7 '18 at 18:35

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

$$L=\sum_{n=1}^{\infty}10^{-n!}=0.11000100000000000000000100000000000000000...$$

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 '18 at 21:54