How can we prove this sequence is dense in [0, 1].?

Question

I've been thinking about this but couldn't finish up a proof yet:

For each x >0, define frac(x) as the fractionary part of x, that is, f(x) = x - [x], where {x} is the largest integer <= x.

Let p>0 be an irrational. Show that the sequence (frac(n*p), n=1,2,3....) is dense in [0, 1].

Thank you
Steiner

I mean, "where [x] is the largest integer <= x"

Answer 1

THis is fairly standard. Think of everything as happening (mod 1). Since the sequence has infinitely many values (since p is irrational), there are m and n with mp and np closer than e (mod 1). Suppose m>n. Then by looking at multiples of (m-n)p, we get an e-covering of [0,1].

Answer 2

Consider the sequence:
z(n) = exp(i*2*pi*p*n)

Note that
z(n) = exp(i*2*pi*frac(n*p)) , because the integer part of n*p just multiplies the number by factors of exp(i*2*pi) = 1.

Now, suppose that the sequence z(n) is NOT dense on the unit circle. Then there is some value alpha and some positive value epsilon, such that for x satisfying:
alpha - epsilon < x < alpha + epsilon, x is NOT the fractional part of any n*p: in other words, there is a sector on the unit circle from which the sequence of fractional parts is completely excluded.

But this also implies that the related sector is also excluded:
(alpha - epsilon)/2 < x < (alpha + epsilon)/2
Why? Because if there were an integer m such that x = frac(m*p) were in that range, so that z(m) is in the second sector, then z(2m) would be in the first range, because z(2m) = z(m)^2.

In other words, if z(2m) is not to be in the first forbidden sector, z(m) cannot be in the second forbidden sector.

In the same way, if z(3m) is not to be in the first forbidden sector, z(m) cannot be in the sector:
(alpha - epsilon)/3 < x < (alpha + epsilon)/3
and in general, z(m) cannot inhabit
(alpha - epsilon)/N < x < (alpha + epsilon)/N

But notice what is happening: we are taking chunks out of the unit circle, for which the sequence frac(n*p) is not allowed to inhabit. But when N gets large enough, the upper limit of one forbidden range will overlap the lower limit of another forbidden range: that is,
(alpha + epsilon)/(N+1) > (alpha - epsilon)/N
This happens for
N > (alpha/epsilon - 1)/2
and this will continue to happen going onward to the next larger iinteger

At that point, you are forbidding ALL the values of x with fractional part less than:
2*(alpha + epsilon)/(alpha/epsilon + 1)

OK, this is clearly getting ridiculous: We have excluded a sector of the circle from angle 0 to a small but positive angle. That means there must be a minimum positive angle among the z(n), say delta. Take the smallest multiple, m, of delta such that:
m*delta > 1
But then:
(m-1) *delta < 1 < m * delta , so:
(m-1)*delta - 1 < 0 < m*delta - 1

Now, either:
m*delta - 1 < delta or else:
m*delta - 1 > delta; but then (m-1)*delta would have been chosen instead of m*delta; or else
m*delta - 1 = 0 ; but in that case delta = 1/m = rational.

So it must then be that:
m*delta - 1 < delta. But in that case, there is a z(n) with an angle smaller than delta, which contradicts the definition of delta.

So the result is that if there were an open sector of the unit circle that were free of the sequence z(n) = exp(i*2*pi*p*n) = exp(i*2*pi*frac(p*n)), this would imply that there is an open sector from angle 0 to delta (from 1 to exp(i*2*pi*delta) which would also be free of the sequence. And this can only be true if delta is a fraction of the form 1/m, which would in turn imply that p is rational.

OK, this proof is a little messy, but the main ideas are here. It would take more time to tidy everything up, and I'm out of time.

Answer 3

It sure is a dense micro-array.