A Journey Through Niven’s Theorem


To understand this post, you’ll need to know \sin \alpha+\beta = \sin\alpha\cos\beta+\cos\alpha\sin\beta and \cos \alpha+\beta=\cos\alpha\cos\beta-\sin\alpha\sin\beta. A good grasp on Algebra is also nice. This might be helpful, but knowing these are not necessary.


One morning I was thinking of the following configuration:

Consider \triangle OA_1A_2 with OA_1=3, A_1A_2=4, and OA_2=5. Then for all n, construct \triangle OA_nA_{n+1} such that \triangle OA_nA_{n+1}\sim \triangle OA_1A_2.

It doesn’t take long to figure out that this is very easy to express with polar coordinates (or complex, whatever flavor you desire). If O=(0,0) and A_0=(3,0), then in polar coordinates, A_n=(3\cdot (\frac{5}{3})^n,n\theta) where \theta=\arcsin\frac{3}{5}.

Then the realization strikes; none of the triangles “overlap” each other! In other words, there is no pair of integers p,q such that O,A_p,A_q are collinear. What does this mean? Well, there are no integers p,q such that p\theta=q\theta+2\pi k for some integer k. This basically means that there is no rational number k such that \theta = 2\pi k.

The Meat

We claim that \frac{\arcsin \frac{3}{5}}{\pi} is irrational. This is the same as claiming \frac{\arccos \frac{4}{5}}{\pi} is irrational.

Suppose otherwise. Then there must be some positive integer n such that \cos(n\theta)=0. But notice that \cos (n\theta) is a polynomial about \cos \theta. Similarly, \sin (n\theta) is a polynomial about \sin \theta.

We claim that \cos (n+1)\theta = 2\cos \theta \cos n\theta-\cos(n-1)\theta for all n\geq 2. Adding \cos (n-1)\theta and expanding out using the sum-difference identities yields \cos n\theta \cos \theta - \sin n\theta \sin \theta + \cos n\theta \cos \theta + \sin n\theta \sin \theta = 2\cos \theta, which is true.

Now we have a recurrence for \cos n\theta! This general recurrence is called the Chebyshev Polynomial of the First Kind. In fact, generally, the Chebyshev Polynomial of the First Kind states the following:

  1. P_0(x)=1.
  2. P_1(x)=x.
  3. P_{n+1}(x)=2xP_{n}(x)-P_{n-1}(x).

Now we have to prove that \frac{4}{5} is not a root of any of these polynomials. What follows is probably the most ingenious yet simplest step of this problem (and the extension, Niven’s Theorem) – an application of the Rational Root Theorem. We claim that the coefficient of the leading term of P_{n}(x) is 2^{n-1}. The proof is fortunately very obvious; we can discard the P_{n-2}(x) and notice that the leading coefficient of P_{n-1}(x) is multiplied by 2. By the Rational Root Theorem, \frac{4}{5} CANNOT be a root of \cos n\theta for ANY n, because 5 NEVER divides 2^{n-1}. Isn’t it beautiful?


Now let’s talk about Niven’s Theorem. It states that the only non-negative values of k such that \frac{\arcsin k}{\pi} is rational are 0, \frac{1}{2}, 1. To do this, we need to expand the Chebyshev Polynomial of the First Kind a little bit more. The Rational Root Theorem will be more powerful if we have the leading coefficient and the constant value; let’s try to look for a pattern of the constant value.








Actually, let’s make it a bit more obvious.












Again, this is very easy to prove using induction. So what does this mean? By the Rational Root Theorem, the only possible values of k are either 0 or of the form \frac{1}{2^m} for even n of P_n(x), and for odd n we can also have things of the form \frac{n}{2^m}. (Remember k is non-negative; negative values are symmetric anyways.) So now all that remains is to prove that these are not roots.

To utilize modular arithmetic, we’ll divide by \frac{1}{x^n} for P_n(x). If we let y = \frac{1}{x}, we are basically “reversing” the polynomial. Let’s rewrite the polynomials in hope we see a pattern:








Let Q_{n}(x)=\frac{P_{n}(x)}{x^n}. Notice the recursion is Q_{n+1}(x)=2Q_{n}(x)-y^2Q_{n-1}(x). We’ll write the first few Q_{n}x.








The only possible roots are 2^0,2^1\dots 2^{n-1},2^n.

Notice that for odd n, the leading coefficient is \pm n. This then causes an obvious problem \pmod n, so it’s proven for odd n. (It is obvious that this is impossible if x=\frac{n}{2^m}, or y=\frac{2^m}{n}, because we need everything to be an integer if we are to get rid of the constant.)

For even n, notice that the constant term is of the form 2^{k+1} while the terms with a coefficient of y are of the form (2^{k})^{2j}), so it won’t work out either. Thus we have proven this is impossible for all rational numbers except for our three exceptions. (For y=2, Q_6(x) is an exception.)

And that’s Niven’s Theorem!


Stack Exchange: Is arccos 3/5 rational?

Stack Exchange: 1/pi arccos (1/3) is irrational.

Wikipedia: Niven’s Theorem

Wikipedia: Chebyshev Polynomials

Wolfram Mathworld: Chebyshev Polynomials of the First Kind

Brilliant: Chebyshev Polynomials

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