Given an irrational number α we can for each q find a p such that
p
1
0 < |α − | <
q
2q
, because of all the fractions with denominator q we can find integers p, p + 1
such that pq < α < p+1
q . (As α is irrational, equality can never occur.)
Can we do better if we choose q appropriately? In fact we can. We have the
following basic theorem.
Theorem For any irrational α there is an infinite number of q such that for
some p (depending on q) we have that
p
1
|α − | < 2
q
q
Proof;. The proof is somewhat subtle. The first idea is based on the
Dirichlet box principle.
Choose a (large) number N and consider the N + 1 numbers 0, α, 2α, . . . N α
modulo 1. Dividing the unit interval [0, 1) (of fractional parts) into N equal
parts, there will be two distinct integers 0 ≤ a, b ≤ N such that their difference
of fractional parts will be < N1 . Thus we can find pa , pn such that |(aα − pa ) −
(bα − pb )| < N1 setting q = (a − b) we hence get that
|α −
bpa − apb
1
1
|<
< 2
q
Nq
q
Now comes the second step. The number q could be quite small, in fact it
could be 1 in which case the inequality |α − pq | < q12 says very little.
But now for any (large) number N0 we get that m = minq<N0 |α − pq | > 0
because there is only a finite number of cases p, q sucht that |α − pq | < 1. Now
choose N such that N1 < m we will then find a q as above such that |α − pq | < q12
but note that q by construction (of N ) must satisfy q > N0 .
Application: The infinite set {N α}N ≥0 modulo 1 is dense.
Choose a q such that α − pq | < q12 and p, q relatively prime. Then consider
{N α}0≤N ≤q and the corresponding {N pq }0≤N ≤q the latter consists of the raNp
N
1
tional numbers 0, 1q 2q . . . q−1
q . Furthermore |N α − q | < q 2 ≤ q thus a gap
between two consequetive numbers N α in the interval is bounded by 3q . As q
can be chosen arbitrarily large, we are done.
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