# Tag Archives: irrational number

## Chebyshev polynomials and algebraic values of the cosine function

Let $p$ be an odd prime and let $T_n(X)$ be the $n$-th Chebyshev polynomial of the first kind. By Lemma 1 on page 5 of this paper, $T_p(X)/X$ is irreducible over $\mathbb Q$. Now

$\displaystyle T_p(\cos\frac{\pi}{2p})=\cos\frac{\pi}{2}=0$,

so the minimal polynomial of $\cos(\pi/2p)$ over $\mathbb Q$ is $T_p(X)/X$. (Note that we are slightly abusing terminology here. By definition, the minimal polynomial is a monic polynomial. But let us drop precision for the sake of brevity.)

By definition (and induction, if you may), $\deg(T_n)=n$. Hence we deduce the following:

Proposition 1. $\cos(\pi/2p)$ is an algebraic number of degree $p-1$.

Since $T_p(X)/X$ consists of only even powers of $X$, $T_p(\cos(\pi/2p))$ can be expressed as a polynomial in $\cos(\pi/p)=2\cos^2(\pi/2p)-1$. Therefore:

Proposition 2. $\cos(\pi/p)$ is an algebraic number of degree $(p-1)/2$.

A nice corollary of this is Niven’s theorem:

Corollary (Niven’s theorem). If $\theta\in (0,\pi/2)$ is a rational multiple of $\pi$ such that $\cos\theta$ is rational, then $\theta=\pi/3$.

Proof. Note that if $\cos\theta$ is rational, then so is $\cos(n\theta)=T_n(\cos\theta)$ for any positive integer $n$. So, (reduction 1) without loss of generality, $\theta=k\pi/p$ for some odd prime $p$ and positive integer $k. (The case $p=2$ is dealt with trivially.)

Further, multiplying by the inverse of $k\pmod p$, (reduction 2) one may assume $k=1$ without any loss of generality.

Now being rational is the same as being algebraic of degree $1$. Hence, by the above, we must have $p=3$ and the conclusion follows immediately. $\square$

Exercise. Find all rational multiples $\theta\in(0,\pi/2)$ of $\pi$ such that $\cos\theta$ is a quadratic irrational.

Filed under Algebra, Number theory

## More on irrationality

Recall some of the irrationality criteria that we discussed in the last post. We showed that

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.

In particular,

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.

Having proven these, it is natural to ask whether every irrational number can have such a representation. Interestingly, work has already been done on this. The Engel expansion of a positive real number $x$ is a unique expansion of the form

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

where $\{a_n\}$ is a non-decreasing sequence of positive integers. Every positive rational number has a finite Engel expansion, and $x$ is irrational if an only if this expansion is infinite.

In this post we shall slightly improve our previous results.

Proposition 3. Let $a_1,a_2,\dots$ and $b_1,b_2,\dots$ be sequences of non-zero integers such that

$(\dagger')\qquad\displaystyle S=\frac{b_1}{a_1}+\frac{b_2}{a_1a_2}+\frac{b_3}{a_1a_2a_3}+\dots$ exists, and

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

Then $S$ is irrational.

Proof. The same argument in the proof of Proposition 1 applies. $\square$

Proposition 4. If $a_1,a_2,\dots$ and $b_1,b_2,\dots$ are sequences of non-zero integers such that $|a_1|\le |a_2|\le\dots$ and $\displaystyle\lim_{n\to\infty}|b_n/a_n|=0$ then

$\displaystyle S=\frac{b_1}{a_1}+\frac{b_2}{a_1a_2}+\frac{b_3}{a_1a_2a_3}+\dots$

exists and is irrational.

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

Convergence follows easily using the ratio test. So we show that $(\ddagger')$ holds.

We have, for sufficiently large $n$,

$\displaystyle\left|\frac{b_{n+1}}{a_{n+1}}+\frac{b_{n+2}}{a_{n+1}a_{n+2}}+\frac{b_{n+3}}{a_{n+1}a_{n+2}a_{n+3}}+\dots\right|$

$\displaystyle\le\left|\frac{b_{n+1}}{a_{n+1}}\right|+\left|\frac{b_{n+2}}{a_{n+2}}\right|\frac{1}{|a_{n+1}|}+\left|\frac{b_{n+3}}{a_{n+3}}\right|\frac{1}{|a_{n+1}a_{n+2}|}+\dots$

$\displaystyle\le\left|\frac{b_{n+1}}{a_{n+1}}\right|+\frac{1}{|a_{n+1}|}+\frac{1}{|a_{n+1}|^2}+\frac{1}{|a_{n+1}|^3}+\dots$

$\displaystyle =\left|\frac{b_{n+1}}{a_{n+1}}\right|+\frac{1}{|a_{n+1}|-1}$

$\displaystyle\le\left|\frac{b_{n+1}}{a_{n+1}}\right|+\frac{2}{|a_{n+1}|}$

$\displaystyle\le 3\left|\frac{b_{n+1}}{a_{n+1}}\right|\to 0$ as $n\to\infty$.

So $(\ddagger')$ holds, as desired. $\square$

As an immediate corollary we get:

Corollary 1. Let $f:\mathbb R\to\mathbb R$ have an infinite Taylor expansion about $x=0$ that converges for $|x|\le 1$. If $f^{(n)}(0)\in\mathbb Z\,\forall n\ge 0$ and $f^{(n)}(0)=o(n)$ as $n\to\infty$, then $f(1/k)$ is irrational $\forall k\in\mathbb Z\backslash\{0\}$.

Taking, for example, $f(x)=\rho\exp(x)+\mu\sin(x)+\nu\cos(x)$ yields:

Corollary 2. Let $\rho,\mu,\nu$ be integers, not all zero. Then $\rho\exp(1/k)+\mu\cos(1/k)+\nu\sin(1/k)$ is irrational $\forall k\in\mathbb Z\backslash\{0\}$.

In particular, $e\pm\sin 1, e\pm\cos 1,\sin 1\pm\cos 1, e\pm\sin 1\pm\cos 1$  etc are all irrational.

Filed under Analysis, Number theory

## 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.

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

$\displaystyle\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots=0$

then

$\displaystyle\frac{1}{a_{n+2}}+\frac{1}{a_{n+2}a_{n+3}}+\frac{1}{a_{n+2}a_{n+3}a_{n+4}}\dots=-1$

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

$\displaystyle\left|\frac{1}{a_{n+1}}+\frac{1}{a_{n+1}a_{n+2}}+\frac{1}{a_{n+1}a_{n+2}a_{n+3}}+\dots\right|$

$\displaystyle\le\frac{1}{|a_{n+1}|}+\frac{1}{|a_{n+1}a_{n+2}|}+\frac{1}{|a_{n+1}a_{n+2}a_{n+3}|}+\dots$

$\displaystyle\le\frac{1}{|a_{n+1}|}+\frac{1}{|a_{n+1}|^2}+\frac{1}{|a_{n+1}|^3}+\dots$

$\displaystyle =\frac{1}{|a_{n+1}|-1}\to 0$ 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 $a_n=F_{k^n}/F_{k^{n-1}}$. $\square$

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Filed under Analysis, Number theory