## Wednesday, December 28, 2011

### sin(n) is dense in [-1,1]

Numbers of the form sin(n) get arbitrarily close to every real value from -1 to 1.

Consider the integers mod(2pi).   Equivalently, we can consider the points (cos n, sin n) on the unit circle (a.k.a. using the "wrapping function").   Because we have an infinite set of points in a compact set, there must be an accumulation point.

For any epsilon > 0, we can find distinct integers m and n such that  (cos m, sin m)  and  (cos n, sin n)  are less than epsilon apart on the circle, so  (cos(m-n), sin(m-n))  is within epsilon of (1,0).   Then (by say the Archimedean principle),  every point on the unit circle is within epsilon of some integer multiple of (m-n) wrapped around the unit circle .   Thus the points of the form (cos k, sin k)  form a dense subset of the unit circle.  In particular, numbers of the form sin(k) form a dense subset of [-1,1].

A slightly different argument for density of sin(n) depends on the following lemma.

Lemma:  If x is irrational, then the additive group Z + xZ is dense in R.

If we accept the lemma, then the additive group Z + (2pi)Z is dense in R, so the points (cos t, sin t) for t in Z + (2pi)Z must be dense in the unit circle.  But (cos (n+m2pi), sin (n+m2pi)) = (cos n , sin n), so (cos n , sin n) is dense in the circle.

To prove the lemma, consider the contrapositive and assume that Z + xZ is discrete.  Then Z + xZ must be cyclic:  Z + xZ = dZ for some real number d.  Because Z is contained in dZ, there must be some integer n so that 1 = nd, hence = 1/n is rational.  And then for some m, 1+m/n, so x is rational.

And the converse of the lemma is also true, because of m/n is rational, then Z + xZ =  1/n Z.