## Niven’s proof that the trigonometric and inverse trigonometric functions are irrational for rational non-zero arguments

In the late 1940s, Niven discovered a way to apply Hermite’s method to the cos function.  The following proof that cos x is irrational for non-zero rational x is adapted from that given in Niven’s book on irrational numbers [1].  I have tried to simplify the proof and make the motivation a little clearer.

Once again, we begin with the fact that if x is a real number

x is irrational if there exist integers h and k such that 0<|kx – h|< ε for any given real number 1>ε>0

Now consider the integral ∫sin(ht)Fn(t)dt with h a positive integer and F(t) a polynomial in t with integer coefficients, and whose degree is 4n.  If we integrate repeatedly by parts we obtain

$h^{4n+1}\int \sin ht F_n(t)dt=\cos ht\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(t)+\\\indent\sin ht\sum_{(k+1)/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(t)$

where in the first and second sums k is permitted to take only even and odd values respectively.  Now suppose we take

$F_n(t)=(1-t)^{2n}(1-(1-t)^2)^n$

Since this is an even polynomial in (1-t), its odd derivatives are zero at t=1.  Hence the second sum vanishes when t=1, and since sin ht=0 when t=0, if we integrate from t=0 to t=1 we obtain

$h^{4n+1}\int_0^1 \sin ht F_n(t)dt=\cos h\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(1)-\\\indent\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(0)=(\cos h)k_n-h_n$

with hn, kn integers.  $F_n(t)$ has the properties that 1≥ $F_n(t)$≥0 if 1≥t≥ 0, and (see lemma below) $F^{(k)}_n(1)$ and $F^{(k)}_n(0)$ are divisible by n! so we have

$h\frac{(h^4)^n}{n!}\int_0^1\sin htF_n(t)dt=(\cos h)k_n^\prime+h_n^\prime$

with k’n(h) = kn(h)/n! and h’n(h) = hn(h)/n! integers.

The absolute value of the integral is less than 1.   (h4)n/n! can be made arbitrarily small by taking n large enough, so for all n greater than some sufficiently large value, nε say, the LHS of the equation can be made less in absolute value than any given positive real number ε.

To finish we need only show that for any given real number r and some n≥nε , rk’n+h’n≠0.  This must be the case if r is irrational so we suppose r = p/q with p, q>0 and g.c.d.(p,q) = 1.  rk’n+h’n cannot be zero if pk’n+qh’n is not zero.

By the Lemma, $F^{(k)}_n(1)/n!$ and hence k’n is divisible by n+1.  We take n even.  By the Lemma, all the terms $F^{(k)}_n(0)/n!$ in the sum

$h^\prime_n=\sum_{k/2=0}^{2n}(-1)^kh^{4n-k}F^{(k)}_n(0)$

are divisible by n + 1 with the exception of $F^{(n)}_n(0)$.  We take n + 1 greater than nε , and such that it contains a prime factor that is not a prime factor of either h or q.  Then qh’n is not divisible by n + 1.  If n + 1 divides pk’n but not qh’n , their sum cannot be zero.

The fact that cos h is irrational for a positive integer h implies that cos x is irrational for all non-zero rational numbers x.  Let x be positive and rational and equal to a/b, with a, b positive integers.  Then as a consequence of the recursive trigonometric identity

$\cos nt = 2\cos t \cos(n-1)t-\cos (n-2)t$

cos nt is a polynomial in cos t, and so cos bx = cos a is a polynomial in cos x.  If cos x is rational, then cos a is rational, so by the counterpositive, since cos a is irrational, cos x is irrational.  Since cos (-x)=cos x, the result is true also for negative rational x.

The following is also due to Niven.

If sin x were rational for rational x ≠ 0, so would be $1- 2\sin^2x=\cos 2x$.  Thus sin x is irrational for rational x ≠ 0.

If tan x were rational for rational x ≠ 0, so would be

$(1-\tan^2x)/(1+\tan^2x)=\cos 2x$

Thus tan x is irrational for rational x≠0.

As a consequence of the following

Let f(x) be a function whose argument is a set of real numbers and let it have an inverse function f-1(y) where y is the set of numbers f(x).  Let f be such that if x is rational then f(x) is irrational.  Then if y is rational f-1(y) is irrational.

the inverse trigonometric functions can be shown to be irrational for non-zero rational x.  Since $\arctan 1 = \pi/4$, $\pi$ is necessarily irrational.

Niven provides a proof along similar lines that $\pi^2$ is irrational.  However Lambert’s continued fraction method can also be used to prove the same result, and I will provide this proof in a later post.

Lemma:  Let

$F_n(t)=(1-t)^{2n}(1-(1-t)^2)n=(1-t)^{2n}t^n(2-t)^n$

Then $F^{(k)}_n(1)$ is divisible by n+1! and $F^{(k)}_n(0)$ is divisible by n!  $F^{(k)}_n(0)$ is divisible by n+1! when k ≠ n, but not when k = n if n+1 is not a power of two.

Proof$F^{(k)}_n(t)=(-1)^k\frac{d^kF_n(t)}{d(1-t)^k}$.  Since $F_n(t)$ is a polynomial in $1-t$ whose term of least degree is $(1-t)^{2n}$, $F^{(k)}_n(1)=0$ when k<2n.  When 2n≤ k≤ 4n, $F^{(k)}_n(t)=b_kk!+f(1-t)$.  n+1! divides k!, so $F^{(k)}_n(1)/(n+1!)$ is an integer.

$F_n(t)$ is a polynomial in t whose term of least degree is tn, so $F^{(k)}_n(0)=0$ when k < n.  $F^n_n(0) = 2^nn!$ so if n+1 is not a power of two, it is not divisible by n+1!  When k > n, $F^{(k)}_n(t)$ = akk! + f(t), where f(t) is either zero or has a factor t.  Hence $F^{(k)}_n(t)$ is divisible by k! and hence by n+1!

[1] I Niven, ‘Irrational Numbers’, Wiley, 1956 pp 16-19