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{\arcsin \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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