1983 USAMO Problems/Problem 5
Problem
Consider an open interval of length on the real number line, where
is a positive integer. Prove that the number of irreducible fractions
, with
, contained in the given interval is at most
.
Solution
This problem references the Farey Sequence of order . We wish to show that no open interval of length
contains more than
consecutive terms of the
th Farey sequence. To do this, we provide a construction of the Farey Sequences of order at most
, prove that this construction yields the desired sequences, and then use properties exhibited from the construction to prove the result.
Lemma 1: Let the sequence be the sequence of integers:
. Then for defined
, define
as follows: we first include every number in
in
in order, and then for every pair of adjacent, reduced elements
and
in
, we include
in
in between the two fractions if
. Then
is the Farey sequence of order
. In addition, if
and
are adjacent terms in any Farey sequence, then
.
Proof: We go about this using strong induction on . If
, then it is clear that
is the Farey sequence of order 1. In addition, the
invariant is clear here. Now say that for
through
,
is the Farey sequence of order
, and each Farey sequence has the
invariant. Then consider
. First, we know that
is strictly increasing, because the elements that are in
are increasing, while any new fractions
are strictly between
and
. In addition, the
invariant is preserved: if we insert a new fraction
in between two fractions
and
, we calculate the invariants:
, and
. And also,
contains every fraction that can be expressed as
with
, and it only contains fractions that can be expressed as
with
. It only remains to be shown that
contains every such fraction. Now consider any fraction that can be expressed as
. Note that if this fraction can be reduced, then we have already shown that it is in
.
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See Also
1983 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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