An Irrationality Criterion

Recall Corollary 4 from this post:

Corollary 4. If S is a subgroup of (\mathbb R,+), then the following are equivalent:

(i) S is well-ordered;

(ii) S is not dense;

(iii) S is cyclic.

Let \alpha be a non-zero real number and take S=\langle 1,\alpha\rangle to be the subgroup of (\mathbb R,+) generated by 1 and \alpha. This is a cyclic group if and only if \exists \beta, \langle 1,\alpha\rangle=\langle \beta\rangle
\Leftrightarrow 1=m\beta', \alpha=n\beta' for some m,n\in\mathbb Z and \beta'
\Leftrightarrow \alpha/1=n/m, i.e. \alpha is rational.

Now by the above corollary, S is dense iff it is not well-ordered, i.e. iff

(*)\qquad\forall\varepsilon>0,\exists m,n\in\mathbb Z, 0<m\alpha-n<\varepsilon.

So we have the following criterion for irrationality:

Criterion 1. \alpha is irrational if and only if (*) holds.

Note that S is dense in \mathbb R iff S/\mathbb Z=\{\{n\alpha\}:=n\alpha-\lfloor n\alpha\rfloor : n\in\mathbb Z\} is dense in (0,1). So we deduce:

Criterion 2. \alpha is irrational \Leftrightarrow\forall\varepsilon>0,\exists n\in\mathbb Z, 0<\{n\alpha\}<\varepsilon.

We demonstrate how this may be useful by proving that certain types of numbers are irrational.

Proposition 1. Let a_1,a_2,\dots be a sequence of non-zero integers such that

(\dagger)\qquad\displaystyle S=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\dots exists, and

(\ddagger)\qquad\displaystyle\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots\to 0 as n\to\infty.

Then S is irrational.

Proof. We have

\displaystyle \left\{a_1\cdots a_nS\right\}=\left\{\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots\right\}.

Multiplying by -1 if necessary, we can take the expression in brackets on the right to be positive and it tends to 0 as n\to \infty. Moreover, if




which cannot happen infinitely often as the left hand side tends to zero. So the conclusion follows by Criterion 2. \square

Proposition 2. If a_1,a_2,\dots is a sequence of non-zero integers such that |a_1|\le |a_2|\le\dots and \displaystyle\lim_{n\to\infty}|a_n|=\infty, then

\displaystyle S=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1a_2a_3}+\dots

exists and is irrational.

Proof. It suffices to show that (\dagger) and (\ddagger) above hold.

Convergence follows easily from the ratio test, so (\dagger) holds. Now




\displaystyle =\frac{1}{|a_{n+1}|-1}\to 0 as n\to \infty,

i.e. (\ddagger) holds. \square

Some special cases of Proposition 2 are particularly interesting:

Corollary 1. \displaystyle \sum_{n=0}^\infty\frac{1}{(n!)^k} is irrational for all positive integers k.

Proof. Take a_n=n^k. \square.

Corollary 2. e is irrational.

Proof. Take k=1 in Corollary 1. \square

Corollary 3. \sin 1 and \cos 1 are irrational.

Proof. Take a_1=1 and for n>1, a_n=-(2n-2)(2n-1) for sine, a_n=-(2n-3)(2n-2) for cosine. \square

Corollary 4. I_0(2) and I_1(2) are irrational, where I_\alpha(x) is the modified Bessel function of the first kind.

Proof. Taking k=2 in Corollary 1 shows that I_0(2) is irrational. Taking a_n=n(n+1) shows that I_1(2) is irrational. \square

Corollary 5. e^{\sqrt{2}} is irrational.

Proof. If it is rational, then so is e^{-\sqrt{2}}, and so is

\displaystyle \cosh(\sqrt 2)=\frac{e^{\sqrt{2}}+e^{-\sqrt 2}}{2}=\sum_{n=0}^\infty\frac{2^n}{(2n)!}.

Taking a_n=n(2n-1) in the above shows that this is false. \square

Corollary 6. Let k>1 be an integer and F_0,F_1,F_2,\dots the Fibonacci sequence. Then

(i) \displaystyle \sum_{n=0}^\infty\frac{1}{k^{F_n}} is irrational.

(ii) \displaystyle \sum_{n=0}^\infty\frac{1}{F_{k^n}} is irrational.

Proof. (i) Take a_n=k^{F_n} and use F_0+F_1+\dots+F_n=F_{n+2}-1.

(ii) Take a_n=F_{k^n}/F_{k^{n-1}}. \square

1 Comment

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One response to “An Irrationality Criterion

  1. Pingback: More on Irrationality | Samin Riasat's Blog

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