Download Mediants, Contiguous Numbers, and Farey`s Sequence

MODULE 4
Mediants, Contiguous Numbers, and Farey’s Sequence
1. Victoria’s Theorem
Victoria was asked to find a fraction between
2
3
and
5
7
Fifth Grade Textbook Solution: Find the midpoint of these two fractions:
12 3
+
2 5 7
1 14 15 =
+
2 35 35
1 29
=
2 35
29
=
70
P =
Victoria’s Dad’s Solution: Add the numerator and denominator to get
M=
2+3
5
=
5+7
12
Is this answer correct?
A little help from a calculator shows that
5
3
2
= .40000 · · · <
= .41666 · · · < = .428571 · · ·
5
12
7
When her teacher marked her answer incorrent (because it was not the answer she
expected) her father contacted the teacher and explained that since there is more than one
2
3
5
fraction between and , there is more than just one answer; with 12
being correct.
5
7
It turns out that this method always works and that, in some sense, it is the very best
possible answer.
37
38
4. MEDIANTS, CONTIGUOUS NUMBERS, AND FAREY’S SEQUENCE
Exercise #79. Show that if
Definition 4.1. If
a
b
a
c
< are two positive fractions, then
b
d
a
a+c
c
<
< .
b
b+d
d
and
c
d
∈ Q, then
a+c
b+d
is called their mediant. We will write
a
c
a+c
⊕ =
.
b
d
b+d
This seems a bit naughty—it looks like we are adding fractions by the totally wrong
way of adding numerator and denominator. The point is that ab ⊕ dc is not the sum of the
fractions; it is the mediant. We must be careful never to confuse ⊕ with +.
2. Contiguous Fractions
In working with fractions we shall always abide by the following two conventions:
(i) we will assume that a given fraction
a
b
is reduced, that is, gcd(a, b) = 1, and
a
is negative, then we will assume that the numerator a is negative,
b
3
−3
3
that is, we will think of − as
and not as
.
7
7
−7
(ii) if a fraction
Definition 4.2. Let S be a set of reduced rational numbers with distinct denominators
greater than zero. Then the element of S with least denominator is said to be the simplest
element of S.
2
108 23107
Example: S = { 109
83 , 91 , 1237 , 30 } has simplest element
Definition 4.3. Two fractions
a
b
<
tiguous.
Note that
a
b
+
1
bd
=
c
d
∈ Q which satisfy ad + 1 = bc are called con-
c
d
Exercise #80. Show that the fractions
a
2
c
3
= and = are contiguous.
b
5
d
7
Discussion Problem #81. How are the fractions
of Water Problem for 5 and 7?
Theorem #82. Suppose
a
b
<
p
q
<
c
d
23107
30
a
b
and
c
d
a
2
c
3
= and = related to the Pails
b
5
d
7
are contiguous. Then any rational number
has denominator q greater than max(b, d).
p
q
satisfying
3. THE FAREY SEQUENCE
39
2
c
3
a
c
a
Exercise #83. If = and = , show that the mediant ⊕ is contiguous to both
b
5
d
7
b
d
a
c
and .
b
d
It turns out that this behavior is not an accident. It is always the case that the mediant
of two contiguous fractions is contiguous to both of them.
a
c
Theorem #84. Suppose < are contiguous. Then the simplest rational number between
b
d
a+c
a
c
a+c
. Moreover,
is contiguous to both and .
them is
b+d
b+d
b
d
3. The Farey Sequence
How do you go about finding contiguous numbers, say in [0, 1]?
Constructive Method: Filtered by the size of the denominators.
Note that
0
1
and
1
1
are contiguous:
Start with the set F1 =
0
1
1
+
=
1 1·1
1
0 1
.
,
1 1
What is the simplest number between them?
Answer: By the previous Theorem, it is
Put F2 =
1
0+1
= .
1+1
2
0 1 1
.
, ,
1 2 1
1
2
is contguous, with numbers around it.
0 1 1 2 1
.
, , , ,
So F3 =
1 3 2 3 1
Now
Exercise #85. Find F4 , F5 , F6 , F7 , F8 , F9 , F10 .
Definition 4.4. Fn is called the Farey sequence of order n.
It is formed by taking all mediants between successive elements of Fn−1 so long as the
resulting denominator is ≤ n.
Theorem Fn consists of the sequence of fractions
increasing order.
a
b
∈ [0, 1], b ≤ n, (a, b) = 1, arranged in
40
4. MEDIANTS, CONTIGUOUS NUMBERS, AND FAREY’S SEQUENCE
a
c
Exercise #86. Find the fraction
immediately before and the fraction
immediately
b
d
61
after
in the Farey sequence F100 .
79
[Hint: Pails of Water]
a
b
Exercise #87. Let
1
2
and
c
d
be the fractions immediately to the left and right of the fraction
in Fn . Prove that
jn − 1k
b=d=1+2
,
2
where ⌊x⌋ (the floor function) is the greatest integer less than or equal to x. (⌊π⌋ = 3,
√
⌊ 20⌋ = 4) Also, prove that a + c = b.
Exercise #88. Let
a c
b, d
that
min
c
d
run through all pairs of adjacent fractions in Fn , n > 1. Prove
−
1
a
=
b
n(n − 1)
and
max
c
d
−
1
a
= .
b
n
Definition 4.5. The infinite Farey sequence of order n is the sequence of all reduced
fractions with denominators ≤ n listed in order of size. Notation: Fn
Example. For n = 3, F3 is
−2 −4 −1 −1 0 1 1 2 1 4 3 5 2
,
,
,
, , , , , , , , , ,...
...,
1 3 2 3 1 3 2 3 1 3 2 3 1
Question #89. Explain how this sequence can be obtained from the Farey Sequence F3 .
Question #90. How do we know the consecutive terms are contiguous?
Theorem #91. If
a c
b, d
∈ Fn and are consecutive, then
a a + c
1
1
=
−
≤
.
b
b+d
b(b + d)
b(n + 1)
Theorem #92. If n ∈ N, x ∈ R, then there exists ab ∈ Q such that 0 < b ≤ n and
1
a .
x − ≤
b
b(n + 1)
Exercise #93. Suppose in the previous theorem that x = π = 3.1415926 · · · and n = 10,
1
?
what fraction ab satisfies the inequality x − ab ≤ b(n+1)