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<p/2. (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.


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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}}\right|+\frac{1}{|a_{n+1}|-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.

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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

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