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Difference between revisions of "1983 USAMO Problems/Problem 5"

(I'll leave my progress up for anyone who wants to finish the problem. I have a solution sketch in mind, but I need sleep now.)
 
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== Solution ==
 
== Solution ==
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This problem references the [[Farey Sequence]] of order <math>n</math>. We wish to show that no open interval of length <math>1/n</math> contains more than <math>(n+1)/2</math> consecutive terms of the <math>n</math>th Farey sequence.
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This problem references the [[Farey Sequence]] of order <math>n</math>. We wish to show that no open interval of length <math>1/n</math> contains more than <math>(n+1)/2</math> consecutive terms of the <math>n</math>th Farey sequence. To do this, we provide a construction of the Farey Sequences of order at most <math>n</math>, prove that this construction yields the desired sequences, and then use properties exhibited from the construction to prove the result.
  
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'''Lemma:''' If <math>a/b</math> and <math>c/d</math> are consecutive terms of the <math>n</math>th Farey sequence and <math>c/d>a/b</math>, then <math>(c/d)-(a/b)=1/bd</math>.
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'''Lemma 1:''' Let the sequence <math>\{F_1(i)\}</math> be the sequence of integers: <math>F_1(i)=i</math>. Then for defined <math>F_k</math>, define <math>F_{k+1}</math> as follows: we first include every number in <math>F_k</math> in <math>F_{k+1}</math> in order, and then for every pair of adjacent, reduced elements <math>a/b</math> and <math>a_1/b_1</math> in <math>F_k</math>, we include <math>(a+a_1)/(b+b_1)</math> in <math>F_{k+1}</math> in between the two fractions if <math>b+b_1=k+1</math>. Then <math>F_k</math> is the Farey sequence of order <math>k</math>. In addition, if <math>a/b</math> and <math>c/d</math> are adjacent terms in any Farey sequence, then <math>bc-ad=1</math>.
  
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'''Proof:''' It suffices to show that, for any reduced fraction <math>a/b</math> with <math>1\leq b\leq n</math>, we can find an ordered pair of integers <math>(c,d)</math> with <math>1\leq d\leq n</math> such that <math>bc-ad=1</math>. Then <math>(c/d)-(a/b)=1/bd</math>.
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'''Proof:''' We go about this using strong induction on <math>k</math>. If <math>k=1</math>, then it is clear that <math>F_1</math> is the Farey sequence of order 1. In addition, the <math>bc-ad=1</math> invariant is clear here. Now say that for <math>i=1</math> through <math>k-1</math>, <math>F_i</math> is the Farey sequence of order <math>i</math>, and each Farey sequence has the <math>bc-ad=1</math> invariant. Then consider <math>F_k</math>. First, we know that <math>F_k</math> is strictly increasing, because the elements that are in <math>F_{k-1}</math> are increasing, while any new fractions <math>(a+a_1)/(b+b_1)</math> are strictly between <math>a/b</math> and <math>a_1/b_1</math>. In addition, <math>F_k</math> contains every fraction that can be expressed as <math>a/b</math> with <math>b\leq k-1</math>, and it only contains fractions that can be expressed as <math>a/b</math> with <math>b\leq k</math>. It only remains to be shown that <math>F_k</math> contains ''every'' such fraction, and the <math>bc-ad=1</math> invariant still holds. Now consider any fraction that can be expressed as <math>c/k</math>. Note that if this fraction can be reduced, then we have already shown that it is in <math>F_k</math>.
  
 
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Revision as of 21:12, 18 July 2016

Problem

Consider an open interval of length $1/n$ on the real number line, where $n$ is a positive integer. Prove that the number of irreducible fractions $p/q$, with $1\le q\le n$, contained in the given interval is at most $(n+1)/2$.

Solution

This problem references the Farey Sequence of order $n$. We wish to show that no open interval of length $1/n$ contains more than $(n+1)/2$ consecutive terms of the $n$th Farey sequence. To do this, we provide a construction of the Farey Sequences of order at most $n$, 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 $\{F_1(i)\}$ be the sequence of integers: $F_1(i)=i$. Then for defined $F_k$, define $F_{k+1}$ as follows: we first include every number in $F_k$ in $F_{k+1}$ in order, and then for every pair of adjacent, reduced elements $a/b$ and $a_1/b_1$ in $F_k$, we include $(a+a_1)/(b+b_1)$ in $F_{k+1}$ in between the two fractions if $b+b_1=k+1$. Then $F_k$ is the Farey sequence of order $k$. In addition, if $a/b$ and $c/d$ are adjacent terms in any Farey sequence, then $bc-ad=1$.

Proof: We go about this using strong induction on $k$. If $k=1$, then it is clear that $F_1$ is the Farey sequence of order 1. In addition, the $bc-ad=1$ invariant is clear here. Now say that for $i=1$ through $k-1$, $F_i$ is the Farey sequence of order $i$, and each Farey sequence has the $bc-ad=1$ invariant. Then consider $F_k$. First, we know that $F_k$ is strictly increasing, because the elements that are in $F_{k-1}$ are increasing, while any new fractions $(a+a_1)/(b+b_1)$ are strictly between $a/b$ and $a_1/b_1$. In addition, $F_k$ contains every fraction that can be expressed as $a/b$ with $b\leq k-1$, and it only contains fractions that can be expressed as $a/b$ with $b\leq k$. It only remains to be shown that $F_k$ contains every such fraction, and the $bc-ad=1$ invariant still holds. Now consider any fraction that can be expressed as $c/k$. Note that if this fraction can be reduced, then we have already shown that it is in $F_k$.

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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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