Download Musical Mathematics: The Mathematics of Equal Temperament

Musical Mathematics
The Mathematical Structure
of the Pythagorean and Equal Tempered Scale
By Laura Smoyer
Portland State University
Department of Mathematics and Statistics
Fall, 2005
In partial fulfillment of the Masters of Science in Teaching Mathematics
Advisor: John Caughman
Introduction
3
Part One: The Mathematical Structure of the Pythagorean and Equal Tempered Scale
I. The Musical Scale
A. The Diatonic and Twelve-Tone Scale
5
B. Frequency Ratio and Fret Placement
6
C. Measuring Musical Tones with Cents
8
II. The Pythagorean Scale
A. Derivation of the Pythagorean Scale
10
B. Pythagorean Comma
12
C. Fundamental Theorem of Arithmetic
17
III. The Equal Tempered Scale
A. Merging Sharps and Flats
19
B. The Perfect Fifth and the Equal Tempered Scale
21
C. Transposition
28
IV. Fret Placement
A. Constructing Galilei’s Interval
30
B. Strahle’s Construction
30
1. Trigonometric Check
31
2. Algebraic Check
35
3. Fractional Linear Function Check
40
1
Part Two: Lesson Plans for Middle and High School Mathematics
I. The Just Tempered Scale
45
II. Lessons and Reflections
A. How to Build a Monochord
46
B. Instrument Images
1. Piano Keyboard
47
2. Guitar
48
3. Electric Guitar
49
4. Viol
50
C. Lesson One: Cultural Consonance
51
D. Lesson Two: Building the Just Tempered Scale
58
E. Lesson Three: Unequal Intervals of JTS
67
F. Lesson Four: Building the Equal Tempered Scale
76
G. Lesson Five: Transposing Happy Birthday
86
H. Lesson Six: Musical Cents
92
I. Lesson Seven: Fundamental Theorem of Arithmetic
99
J. Lesson Eight: Continued Fractions
103
K. Lesson Nine: Approximating the Perfect Fifth
111
Bibliography
115
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Introduction
Many people seem to know that there is a connection between math and music, but few
seem to be able to explain this connection in any detail. As a middle school math teacher, I have
had students, parents and friends encourage me to use this connection in my teaching to make
math more meaningful, but I have always been at a loss as to how to do this – so I decided to use
my master's project as an opportunity to explore this elusive connection. As I was mentally
gearing up for my project, my oldest son began playing violin – an extracurricular activity we
stumbled upon that quickly became a regular feature of our daily life, with my middle son
joining the fun soon thereafter.
I have little musical training myself and none with a string instrument – but as I became
entrenched in my sons' fledgling violin journeys I became fascinated by the non-fretted violin. I
had very little understanding of how non-fretted instruments worked. All I knew was the piano –
press a key and sound a tone. With the violin, my sons were magically learning where to place
their fingers so that the correct note would sound. And so my project was born – I wanted to
know where this scale came from and hoped it was somehow mathematical. To my delight the
Pythagoreans shared this hope. They were the first (at least in the Western world) to popularize
the mathematical basis of the musical scale, and many other mathematicians and musicians have
continued their work, adjusting the scale to meet the needs of the musical world.
I decided to focus on the beginning (the Pythagorean scale) and end (the equal tempered
scale) of the story of the musical scale, touching only briefly on a few of the scales in between,
and ignoring completely all non-western scales. (Initially that was going to be my "Part Two"-hopefully some other graduate student can explore that part of the story someday.) Following
my explanation of the math behind the musical scale, I have included nine specific lessons that
3
integrate some of the mathematics of the musical scale into standard middle and high school
topics.
In addition to these full-length lessons, I think that with my improved understanding of
the mathematics involved in the musical scale I will be able to connect my teaching of math to
music in little ways here and there. For instance, to me
2
3
will never again be just a common
fraction whose decimal repeats. Instead, I now think of it as the second most consonant musical
interval and the basis of the western musical scale. And in my mind's eye, a geometric
progression is now the placement of my sons' fingers as they move up the scale on the violin. As
my understanding of the rich connection between mathematics and music has developed, I have
come to see that my fascination with the violin is motivated by one of the main things that has
drawn me to mathematics – the challenge of deciphering patterns and structures not immediately
apparent. It is my hope that I am now better prepared to help my students see the mathematics in
music and that any math teachers who read this project will be better able to hear the music in
mathematics.
4
Part One: The Mathematical Structure of the Pythagorean and Equal Tempered Scale
I. The Musical Scale
A. The Diatonic and Twelve-Tone Scale
Music is a universal element of human culture. In the most ancient cultures, music
probably consisted of only rhythm and the human voice, with no “musical tools,” or instruments,
required. Just as humans are driven to invent tools to facilitate their work, musical instruments
seem to have been an almost instinctual goal of the human mind. As instruments advanced, they
became capable of playing a set of tones. Once tones existed, the human mind began its effort to
organize, label and standardize these tones into a scale. Many different scales have developed in
this way, each specific to their culture of origin. The structure of music and sound, like so many
parts of the seemingly random natural world, is not random at all but instead based on complex
mathematical relationships. Western musical scales imitate, and can be analyzed with, specific
mathematical structures.
Western music is based on the seven tone Diatonic scale used by the Ancient Greeks.
Over time, five pairs of tones, known as sharps and flats, were interspersed in this Diatonic scale
to produce the modern twelve-tone scale:
C# D#
F# G# A#
C D E F G A B
Db Eb
Gb Ab Bb
sharps
Diatonic Scale
flats
On an instrument with no fixed notes, such as a violin or trombone, sharps and flats can be
differentiated. On most keyed or fretted instruments, such as a piano, saxophone or guitar,
sharps are not distinguished from flats: C# and Db are the same, D# and Eb are the same, F# and Gb
are the same, etc. On a piano keyboard, there are twelve keys in each octave; the white keys play
5
the Diatonic scale and the black keys play the sharps and flats, one black key for each pair
(Davis and Chinn, 1969, p. 236).
When referring to the musical scale, the terms octave, fourth, fifth etc. are used to
describe the intervals between notes. These are not fractions, but ordinals, that refer to the
original diatonic scale. An octave refers to the eighth note of the diatonic scale, a fourth the
fourth note, and a fifth the fifth note. On the C-scale, an octave is the next higher or lower C, a
fourth is F, and a fifth is G (Osserman, 1993, p. 29).
1st 2nd 3rd 4th 5th 6th 7th 8th
C D E F G A B C
These intervals existed long before they were ever named. The human ear naturally
preferred certain pairing of notes. Over time, a scale was developed grouping a set of musically
harmonious notes (subject to cultural norms), and then the intervals were named. The octave,
fifth, and fourth are perceived as more consonant than any other interval to the western ear. If
two notes, separated by one of these favored intervals, are played simultaneously, the resulting
tone actually sounds louder than a random interval (Fauvel, Flood and Wilson, 2003, p. 62).
B. Frequency Ratio and Fret Placement
The tones of a musical scale played on a stringed instrument are determined by the length
of the string being played. Shorter string lengths produce higher tones than longer string lengths.
Given any string length, half that length will produce a tone one octave higher than the tone
produced by the entire string, and ⅔ of the string length will produce a tone a fifth higher. When
considering the ordinal naming of the notes and string length that produces these notes, it should
be noted that the ordinal names are unrelated to the fractional lengths of the strings. For instance,
"a fifth" is produced when ⅔ of the original string length is played.
6
The frequency ratio that names a note is inversely related to string length. If the entire
string length is thought of as one unit, then the frequency ratio of each note is the reciprocal of its
string length. To determine the frequency ratio of a note relative to a given base note, the
following formula can be used:
the length of the original string
. For instance, if the string
the length of the new string
length of some C is one unit, then the frequency ratio of the C one octave higher, with half the
string length, will be
1
1
2

2
(Schmidt-Jones, 2004, p. 4). At first the fact that the frequency
1
ratio is the reciprocal, instead of the actual fractional string length, seems an unnecessary
complication. In practice, it actually simplifies matters by producing a system where ascending
fractions produce ascending notes and descending fractions produce descending notes. So the
frequency ratios 11 , 12 , 14 , 81 name four notes each ascending an octave and the frequency ratios
1
1
, 32 , 94 , 278 name four notes each ascending a fifth, whereas the frequency ratios 11 , 12 , 14 , 18 name
four notes each descending an octave and 11 , 23 , 94 , 278 name four notes each descending a fifth.
The Ancient Greeks experimented with fractional string lengths on a simple instrument
called a canon, consisting of a single string stretched over two end posts with a movable post in
between that could vary the length of the string (Stewart, 1992, p. 238). Similar experimentation
with string length can be modeled on a monochord (a one-string instrument) by measuring the
entire string length and then finding the fractional lengths that correspond to the different notes.
This is essentially how a beginning violinist finds notes — once the proper placement is
determined, the violinist memorizes approximately where a finger should be placed to produce a
given note. A skilled violinist also learns to hear when a finger is stopping the vibrations at the
correct fractional string length to produce pure notes. Because a violin doesn’t have frets, it is
much more flexible in its ability to find pure notes, and can produce tones matching any of the
7
scales that have been developed over time. However, this requires a well trained musician and a
good ear for music.
The advantage of instruments with keys or frets is that the lengths producing each note
are fixed and so anyone can find the notes of a scale by depressing the proper key or fret of the
instrument. The development of different variations of the western scale was largely driven by
the desire to find the “best” placement for the fixed frets on the viol or lute, and later the
mandolin and guitar, and the proper size for the strings or pipes on the harpsichord, organ and
piano.
C. Measuring Musical Tones with Cents
With so many different scales competing for use, musicians and mathematicians needed a
method to compare the frequency ratios of different scales and judge their success at producing
true notes. Because notes are found by multiplying the base string length, straight forward linear
comparisons are not accurate. To address this problem, Alexander Ellis developed a unit, the
cent, in 1884. The cent is equal to one hundredth of a semitone. There are twelve semitones in
the western musical scale, so there are 1200 cents in an octave. If an octave is divided into
exactly 12 equal parts, the notes are equal to 0, 100, 200, 300, … 1200 cents. This even division
brings to mind the tempting oversimplification of a ruler divided into 12 equal sections.
However, musical notes are produced by a geometric progression of ratios, thus a cent is not like
the standard linear units of measurements that first come to mind – the frequency ratio that
produces a one octave jump, 21 , is not divided into 1200 equal parts. Instead, this frequency
ratio is produced by multiplying 2
1
1200
 1 
by itself 1200 times because  2 1200 


1200
 2 . Each of
1
these factors, 2 1200 , is equal to one cent (Schulter, p. 6).
8
This exponential division results in a logarithmic system for calculating cents. The goal
is to figure out how many cents are needed to produce a given frequency ratio, so that frequency
ratios can be compared using a standardized unit. Rewritten algebraically, where x = cents
needed to produce the given frequency ratio, r:
x
x
 1 
r =  2 1200  or r = 2 1200


Thus, log 2 r 
 log r 
x
 1200 log 2 r  x  1200 10   x  3986.3137  log 10 r  x
1200
 log 10 2 
9
II. The Pythagorean Scale
The Pythagoreans worshipped whole numbers and held a mystical belief that whole
numbers could be used to explain everything in the natural world. Thus, the Pythagoreans were
very pleased to find they could explain the musical scale popular in Ancient Greece using only
whole number ratios. The most harmonious interval is commonly thought to be an octave, but a
scale cannot be based on the interval for an octave because moving up or down an octave would
simply produce the same note over and over, in different octaves. A fifth is generally agreed to
be the second most harmonious interval, so the Pythagoreans based their explanation of the scale
on this interval.
A. Derivation of the Pythagorean Scale
The Pythagoreans showed that if a given base note is multiplied repeatedly by the
frequency ratio used to find a fifth,
3
2
, all the other frequency ratios for the notes in the Diatonic
scale can be created. The goal is to reproduce the eight-note Diatonic scale by starting at a root
note and going up the six intermediate notes before arriving again at the root note transposed up
an octave. Such a scale will have ratios with values between 1 (the root note) and 2 (the root
note transposed up one octave). If the transposed fifth is outside of this range, the product is
multiplied by
1
2
. This transposes the note down an octave, and brings the ratio into the desired
range. This method produces the frequency ratios for five of the six missing notes. The
frequency ratio for the final note, a fourth, is found with a slight variation to this method: the
frequency ratio for the upper octave,
2
1
, must be divided by
3
2
(Fauvel, Flood and Wilson, 2003,
p. 16).
The end result is a scale produced by playing strings whose relative lengths are
determined by the following frequency ratios: 11 ,
9
8
,
81
64
,
4
3
,
3
2
27
, 16
,
243
128
, 12 . For instance, if we
10
start with a base note of C then the string length producing this note will be our unit, 1. To
produce the next note, D,
8
9
of the string is needed. This results in the frequency ratio:
String Length of C 1 9
  . Likewise,
String Length of D 8 9 8
ending with
1
2
64
81
of the original string will produce E and so on,
the string length producing the note C an octave higher than the first C.
Derivation of the Pythagorean scale, ascending from C
1 the first note, C
2
1
Multiplying by
1
1
, or ascending an octave:
   the eighth note, or octave, C
2
1
2
1
Multiplying by 32 , or ascending a fifth:
3
3
1
 the fifth note, G
1 2  2

1

1
1

1
1

1
1
1

3 2
2
3 3
2
3 4
2
3 5
2

9
4

27
8
27
8
 12 
81
 16
81
16
 12  12 

9
4
 12 
243
32
243
32
9
8
ordered frequency ratios
 the second note, D
1
1
9
8
 the sixth note, A
C
D
27
16
81
64
 12  12 
81
64
4
3
3
2
E F G
27
16
A
243
128
B
2
1
C
 the third note, E
243
128
 the seventh note, B
Dividing by 32 , or descending a fifth:
3
2
4
1  2  3  the fourth note, F
These musical intervals can be “added” by multiplying the ratios of their string lengths.
For instance, to add a second to a fifth on the C scale, start at the second note, D, and count up to
the fifth note (relative to D), which is A. So D + G = A, or a second + a fifth = a sixth. However,
to find the frequency ratio of this note, the ratios that create a second and a fifth are multiplied.
The product of the frequency ratios for a second and a fifth is the frequency ratio that creates a
sixth: 89  32 
27
16
.
11
Using this method, the interval between each note on a given scale can be calculated. For
the Pythagorean scale, five of the intervals differ by a factor of
differ by a factor of
256
243
9
8
, a whole tone, (T), and two
, a semitone, (S).
(C
Diatonic Pythagorean Scale with Base Note C
D
E
F)
(G
A
B
9
8
1
1
(  89
(T
81
64
 89
T
3
2
4
3
)
 256
243
S)
 89
T
27
16
(  89
(T
C)
243
128
 89
T
2
1
)
 256
243
S)
The Pythagoreans named their scale diatonic (dia- across, tonic-tone) because it is based on two
tetrachords (four notes that span the interval of a perfect fourth: 43 ) separated by a whole tone.
The tetrachord (C, D, E, F) starts with a ratio of
1
1
and ends with 43 , and 11  43  43 , so (C, D, E, F)
3
2
span the interval of a perfect fourth. Likewise, (G, A, B, C) span a fourth:
tetrachords are separated by a whole tone interval,
9
8
, from F to G: 43  89 
3
2
 43  12 . These two
(Frazer, 2001, p. 3).
B. Pythagorean Comma
The Pythagorean scale can be expanded to include sharps and flats. These intermediate
notes divide the whole tones in the Diatonic scale. In theory, the sharps and flats fall in the
middle of each pair. In practice, the addition of sharps and flats exposes the flaw in the
Pythagorean scale. The scale includes both sharps and flats because they are not exactly in the
middle of the whole tone. If they were in the exact middle, C# and Db, for instance, would be the
same note and would have one name. The problem stems from the fact that two Pythagorean
256
semitones, 256
243  243 , do not quite equal a Pythagorean whole tone,
9
8
(Fauvel, Flood, Wilson,
2003, p.16).
256
2 = 1.109857915
243
9
8
= 1.060660172
9
8
= 1.125
256
243
= 1.053497942
12
The intermediate note cannot be placed exactly in the middle because the semitone is a key
ingredient to the original diatonic scale (T T S, T, T T S) and corrupting this semitone would
change the sound of two of the basic eight intervals. Instead, the frequency ratio for a flat is
calculated by multiplying the whole note that precedes the flat by a semitone,
256
243
. The
frequency ratio for a sharp is calculated by dividing the whole note that follows the sharp by a
semitone,
256
243
.
(See the table of frequency ratios on page 21 for the complete Pythagorean scale
with sharps and flats.)
Through the medieval period the small difference between two semitones and a whole
tone caused little trouble because composers generally wrote music for one key. Instruments
were usually tuned by starting at Eb and moving up by fifths (because this favored interval is the
easiest to “hear”) eleven times to produce a twelve-note scale, Eb, Bb, F, C, G, D, A, E, B, F#, C#,
G#. The calculations producing this scale are shown below:
12-note Pythagorean scale starting with Eb and ascending by fifths:
32
27
32
27
32
27
32
27
32
27
32
27
32
27
32
27
32
27
32
27
32
27
32
27
 Eb
 Bb
 32  169
  32   83
2


















3 3
2
3 4
2
3 5
2
3 6
2
3 7
2
3 8
2
3 9
2
3 10
2
3 11
2

4
1

6
1
 19

27
2
 814

243
8

729
16

2187
32

6561
64
8
3
 12 
F
  11  C
6
   32  G
1
9
9
D
1    8
27
   1627  A
2 
81
   6481  E
4 
243
243
B
   128
8 
729
729
 F#
   512
16 
6
2187
 C#
1   2187
32  2
2048
6
6561
6561
 G#
1   4096
64  2
4
1

4
3
1 2
2
1 2
2
1 3
2
1 3
2
1 4
2
1 4
2
1 5
2
Rearranged in order from least to greatest, the following 12-note scale is produced: C, C#, D, Eb,
E, F, F#, G, G#, A, Bb, B.
13
Imagine a “Pythagorean Piano” is tuned according to this Pythagorean scale. To do this,
the strings of one octave are tuned with the twelve frequency ratios above. The rest of the piano
strings are then tuned to match the twelve original tones by going up or down an octave. For
instance, the first Bb would be used to tune all the other Bb keys on the piano. If every fifth note
on the Pythagorean Piano is played twelve times, seven complete octaves will be played. The
lowest Bb will be in-tune with the highest Bb, but only because the notes of each additional
octave were tuned to match the base octave, and not by calculating successive fifths. The
Pythagoreans’ trouble results from the fact that twelve fifths is not exactly the same as seven
octaves. The lowest and highest Bb are only in tune because every twelfth note was artificially
matched.
If twelve fifths are actually calculated on one string and seven octaves on another
identical string, and then the two resulting notes are played, the sound produced will be harsh or
dissonant because the two tones differ slightly. Specifically, if a Bb seven octaves above a base
Bb is calculated by multiplying the string length by
2
3
twelve times (transposing up twelve fifths)
and the two Bbs are played, their tone will not match. Furthermore, because the two notes are
only slightly off, they will quite obviously clash. This is similar to the dissonance that results
when a young child plays a violin with his fingering slightly off the proper placement. The
musical disagreement between twelve fifths and seven octaves can be shown mathematically by
calculating the exact frequency ratios of each:
32 12  12 7
or
312  219
129.7463  128.
531,441  524,288
14
3 2 12
21 7
The dissonance created by playing two notes with the interval

312 531441

 1.0136 is
219 524288
referred to as the “Pythagorean comma” (Osserman, 1993, p. 55). The comma is equal to 24
cents:
a fifth = 32 = 702 cents
an octave = 2 = 1200 cents
702 x 12 = 8424
1200 x 7 = 8400.
The spiral below shows this lack of alignment between seven octaves and twelve fifths.
The distance between each pair of notes along the spiral represents a fifth. The note reached
from counting twelve successive fifths around the spiral should be the same as the original note,
seven octaves higher. Instead it is off by a “comma.” In the diagram, the notes connected by
straight lines should be the same. Instead they are off by 1, 2, 3 or 4 commas. With each trip
around the spiral, the 12th note becomes off from the base note by an additional comma.
A###
Spiral of fifths
(Fauvel 20)
E###
D###
B#
F#
E#
C
G###
G
F
Dbb
A#
C##
Abb
D
Gbb
Bb
Ebb
Cbb
C###
Bbb
D#
Eb
A
Fbb
Bbbb
G#
E
Db
F###
Fb
Ebbb
Ab
G##
Cb
Gb
D##
B
C#
F#
A##
B##
E##
15
On the Pythagorean Piano described previously, tuned from a base note Eb, the interval
between Eb and G# (produced by the eleventh power of 3/2) would be the most dissonant because
the error resulting from the comma has been compounded eleven times. However, Eb and G# are
rarely played together, so this dissonance is not a practical problem (Schulter, p.1).
Unfortunately for the Pythagoreans, as the medieval period ended, it became increasingly
desirable for musical instruments to be able to change from one key to another. As music
became more complex, the little difference between the two intervals of the Pythagorean scale
became more problematic.
The dissonance created by the comma becomes a practical problem when an instrument
tuned for one key is used to play a piece written for a different key. Suppose a pianist tries to
play a piece in the key of E b on the Pythagorean Piano, (with notes C, C#, D, Eb, E, F, F#, G, G#,
A, Bb, B). The Eb scale is : Eb, F, G, Ab, Bb, C, D. However, the Pythagorean Piano has no Ab,
so G# is used instead. G# is 24 cents higher than Ab (the Pythagorean comma) and will sound
noticeably out of tune. Similarly, there are no keys on the Pythagorean piano tuned from Eb to
sound the tones for A#, D#, Db or Gb. Every Pythagorean tuning is forced to choose one sharp or
flat between each pair of notes, and so every tuning is missing five notes (van Buul, 1995). If a
piece of music is written in a key that includes any of the missing five notes, the pianist is left
with three choices – have the piano completely re-tuned, play the piece out of tune, or invent a
new keyboard where each black key is replaced by two black keys, one for the sharp and one for
the flat. This type of complex keyboard was actually used in the nineteenth century, with one of
each black key “pair” slightly raised so the organist could still play by touch (Dunne, 2000, p.
12).
16
C. Fundamental Theorem of Arithmetic
Given these inadequate options, why not use another scale with different intervals which
align the fifth and the octave? This musical question can be answered definitively with a
mathematical proof. As it turns out, the Pythagoreans stumbled over the Fundamental Theorem
of Arithmetic in their quest to produce a scale based on whole number ratios. There is no way to
reconcile the natural fifth and the natural octave. That is, there is no way to solve the equation:
 32 x  12 y
or 3 x  2 y  x or, more simply, 3 n  2 m
The Fundamental Theorem of Arithmetic states that every natural number can be
uniquely factored as a product of primes. As a result, there is no number, x, whose factors
consist only of the prime number 3, which can equal some other number, y, whose factors consist
only of the prime number 2. In other words, no matter how many times 3 is multiplied by itself,
the product will never equal the product resulting from multiplying any number of 2s.
3  3  3  3  ....  2  2  2  2  ...
The proof for the Fundamental Theorem of Arithmetic consists of two parts. First, every
natural number is shown to be the product of some sequence of primes. In other words, a prime
factorization exists for every number. Second, this prime factorization is shown to be unique.
To show existence, consider any composite number, n, which by definition has factors
other than itself and 1. Break down the factors until they are all prime and what is left is a prime
factorization of n = p1p2p3…pr.
Example:
24
4  6
2 3  2 2
17
To show uniqueness, imagine there are two prime factorizations for n:
n  p1  p2  p3 ...  pr and n  q1  q2  q3 ...  qr . By definition, p1 divides n, thus p1 divides
q1  q2  q3 ...  qr . In addition, all pi and qj are prime, so p1 must equal some qj. These paired
factors can be removed and then this process repeated to show that q1  q2  q3 ...  qr must be
exactly the same as p1  p2  p3 ...  pr (Bogomolny, 2005).
Example:
24  2  2  2  3 and 24  q1  q2  q3 ...  qr
(qj are prime)
2 24  2 q1  q2  q3 ...  qr  2  qj for some j
Repeat this argument to show that q1  q2  q3 ...  qr must equal 2  2  2  3 .
The implications of the Fundamental Theorem of Arithmetic explain the necessity of
abandoning the goal of equating the perfect fifth, on which the divisions of the scale are based,
and the octave, that bounds this scale. Non-fretted instruments, like the violin, are free to
continue to use the Pythagorean scale and play music in varied keys, because the human mind
can change keys at will by minutely adjusting finger placements. Fretted and keyed instruments,
like the guitar and the piano, have opted instead to use a less tonally perfect, but much more
flexible, scale—trading perfect intervals for the ability to play in any key.
18
III. The Equal Tempered Scale
A. Merging Sharps and Flats – Galilei and Marsenne
As the desire to be able to easily switch keys increased, mathematicians and musicians
tried to suggest different ways to divide an octave into equal, or “more equal” parts and avoid, or
reduce, the spiral and resulting dissonance created by the Pythagorean fifth and the Fundamental
Theorem of Arithmetic. In the sixteenth century, Galileo Galilei’s father, Vincenzo Galilei, tried
to promote the use of a different whole number ratio, 1817 , to divide an octave into twelve equal
tones. He chose this ratio because it placed the twelfth fret (an octave) almost at the middle of
the string, and it allowed for eleven equal semitones between octaves –joining sharps and flats so
that instruments could play music in any key with a single tuning.
However, this equally spaced scale does not result in an exact ratio for any of the
important intervals:
an octave (twelve semitones):
a fifth (seven semitones):
a fourth (five semitones):
181712  1.9855...  12
18177  1.4919...  32
18175  1.3308...  43
This tuning system still requires that a piano be tuned by matching successive octaves to one
base octave, and the fact that the octave itself is not perfect is fundamentally problematic. Still,
the ability of Galileo’s ratio to join sharps and flats, increasing the flexibility of the fixed tone
instruments, resulted in his tuning system remaining popular for 200 years (Barbour, 1957, p. 2).
In the seventeenth century, French mathematician Marin Marsenne finally decided to
give up the (impossible) task of creating a scale based on whole number ratios that would divide
the octave into twelve equally spaced notes. Instead, he turned to the flexibility of the irrational
12
2 to create the equal tempered scale. Much to the dismay of the long dead Pythagoreans, this
19
irrational number (deemed “unutterable” in their time) is almost exactly halfway between the
two semitones of the Pythagorean scale:
256
243
12
 1.053498 (or 90 cents)
2 =1.059463 (or 100 cents)
2187
2048
 1.067871 (or 114 cents)
It should be noted that this same scale was explained a century earlier by Chinese scholar, Prince
Chu Tsai-yu (Osserman, 1993, p. 56). Using this ratio, the relative string lengths needed to
produce the twelve tones of the musical scale can be found using a geometric progression with a
= 1 and r = 12 2 . This method divides the scale into twelve exactly equal intervals, each
measuring 100 cents.
The table that follows shows a comparison of the Pythagorean scale based on the perfect
18
fifth, Vincent Galilei’s scale based on the interval 17
, and the equal tempered scale, based on the
interval 12 2 . Because of the Pythagorean comma, the sharp and flat between each note in the
Pythagorean scale are different. In the other two equally divided scales, the sharp and flat are the
same. Also, there are no sharps or flats listed between E and F or B and C because the interval
between these notes is already a semitone, whereas the interval between the other notes of the
original diatonic scale is a whole tone (Taylor, 1965, p.128).
20
PYTHAGOREAN
ordinal
octave
second
third
fourth
fifth
sixth
EQUAL TEMPERAMENT
note Freq. Decimal Cents Freq. Decimal Cents Freq.
ratio
ratio
ratio
1
1
1
1
1
C
0
0
#
2187
C
1.0679
114
2048
18
12
1.0588
99
2
17
b
256
D
1.0535
90
243
9
D
1.1250
204
1718 2 1.1211 198 12 2 2
8
19683
D#
1.2014
318
16384
1718 3 1.1871 297 12 2 3
b
32
E
1.1851
294
27
81
E
1.2656
18 4 1.2569 396 12 2 4
408
64
4
3
1.3333
498
F#
729
512
1.4047
612
Gb
1024
729
1.4047
588
G
3
2
1.5000
702
G#
6561
4096
1.6018
816
Ab
128
81
1.5802
792
A
27
16
1.6875
906
A#
59049
32768
1.8020
1020
Bb
16
9
1.7778
996
243
128
1.8984
1110
2
2
1200
C
0
1.0595
100
1.1225
200
1.1892
300
1.2599
400
5
1.3348
500
6
1.4142
600
7
1.4983
700
8
1.5874
1718 5
1.3308
495
1718 6
1.4091
594
1718 7
1.4920
693
1718 8
1.5798
792
1718 9
1.6727
891
1718 10
1.7711
990
1.8753
1089
1.9856
1188
2
1718 11
1718 12
12
12
12
12
12
12
12
Decimal Cents
1
 
 
 
 2
 2
 2
 2
 2
 2
 2
17
F
seventh B
octave
VINCENT GALILEI
800
9
1.6817
900
10
1.7818
1000
11
1.8877
1100
2
1200
B. The Perfect Fifth and the Equal Tempered Scale
Consider the goal of creating an equal tempered scale that preserves the perfect pitch of
the most important interval, an octave, and comes as close as possible to preserving the perfect
pitch of the second most important interval, a fifth. Choosing to base such a scale on the interval
12
2 can be justified simplistically by noting that the twelfth root of two divides an octave (of
length two) into twelve equal semitones. This is a "good" choice because the previously
established Pythagorean scale also divides the octave into twelve semitones, so using
12
2 allows
21
musicians to continue using a twelve tone scale. However, this reasoning does not consider
whether using the interval
12
2 is the optimal interval for preserving the octave while also best
approximating the fifth. Given that the scale should start with the frequency ratio two so that the
octave sounds perfectly, the question remains: why divide this frequency ratio into twelve
intervals ( 12 2 ) instead of thirteen ( 13 2 ) or eleven ( 11 2 ) or some other reasonable number of
intervals? A mathematical answer supporting the division of the octave into exactly 12 intervals
is found by optimizing a rational solution to the original Pythagorean problem of aligning the
fifth (3/2) with the octave (2).
Recall the unachievable goal of finding a solution to the equation:  32   2 y or 3m  2 n .
x
One approach is to approximate the solution using logarithms:
n
3m  2 n  3  2 m  log 2 3 
n
m

n
m

log 3
= 1.58496…
log 2
This approach results in a complicated non-terminating decimal, and so doesn’t help determine
the best number of intervals for the octave. To determine this, a rational solution for
n
m
is
necessary. In the rational solution, m will be an integer corresponding to the total intervals in the
octave. Further, the (n-m)th interval will be the interval that best approximates a fifth
n
because 3  2 m  3  2  2
nm
m
 32  2
nm
m
. A rational solution for
n
m
can be approximated using
continued fractions (Burr, 1992, p. 5).
A continued fraction is a non-terminating fraction expansion of an irrational number. A
continued fraction can be truncated at any point and the resulting convergent fraction will be a
better rational approximation for the value in question than any other fraction whose
denominator is less than the chosen convergent. In other words, continued fractions can be used
to find the best rational approximation of a desired complexity for an irrational value. In contrast,
22
a truncated non-terminating decimal only offers the best approximation for the power of ten
where it was truncated, not for any denominator less than or equal to that power of ten. Instead
of being able to choose from any denominator, decimal approximations are limited to
x
x
denominators that are powers of ten: 1x , 10x , 100
, 1000
, etc. As a result, they usually yield non-
optimal rational approximations. For instance, 0.625 =
5
8
. Truncated at two decimal places
63
with proper rounding, this decimal becomes 100
, a rational approximation with a much larger
denominator that is not as accurate as the much simpler (and perfectly accurate)
5
8
. This
inefficiency of adding complexity to the fraction without adding accuracy never occurs with
continued fractions.
The method for finding the continued fraction that best approximates an irrational
number is to repeatedly refine the rational approximation by determining the integers that bound
n
successive denominators. Based on the equation in question, 3  2 m or 3  2 x , x must be some
value in between 1 and 2, because 21  3 and 2 2  3 , so the continued fraction solution for x
must start x  1 
1
, where y  1, because this equation guarantees that x will be between 1 and
y
2.
Substituting this value for x in the original equation 3  2 x results in the new equation
1
32
1
y
 2 2
1
1
y
1
y
y
3
3
 2 
 2    . This equation shows that the new denominator y
2
2
1
2
1
3
3
must also be some fraction in between 1 and 2 because    2 and    2 . So y  1  .
z
2
2
Substituting for y in the continued fraction: x  1 
1
1
1
z
.
23
y
3
Substituting this value for y in the new approximating equation 2    results in:
2
3
2 
2
1
1
z
1
1
1
z
3
4  3z
4
 3  3z
     
       . This equation shows that z must be
2
3 2
3
2 2
1
2
1
3
3
4
4
some fraction in between 1 and 2 because    and    . So z  1  and the
w
2
2
3
3
1
continued fraction becomes: x  1 
.
1
1
1
w
1
z
Substituting this value for z in the approximating equation
3 4
in:   
2 3
1
1
w
1
1
1
3 4
   results
2 3
w
4
9  4w
9
 4  4w
     
       . This equation shows that w must be
3
8 3
8
3 3
2
3
1
4
4
9
9
some fraction in between 2 and 3 because    and    . So w  2  and the
t
3
3
8
8
1
continued fraction becomes: x  1 
.
1
1
1
1
2
n
1
t
Continuing in this manner, we find 3  2 m when
n
m
 1;1,1,2,2,3,1,5,1,3,1,1... , where the number
before the semi-colon is an integer value and the numbers after the semi-colon are the successive
denominators for rational approximations of increasing complexity and increasing accuracy.
Imagine a length of string is cut that is exactly
n
m
, or log 2 3 =1.58496... in length. Now
imagine there is a collection of strings each half a unit length that must be used to approximate
the original string. Three of these string pieces will be the best fraction approximation possible.
24
In other words, if only halves are available,
equivalent to
n
m
 1;1,1  1 
65
41
1
41
st of a unit length, then the best fraction approximation would
, the continued fraction truncated as
approximate log 2 3 were each
1
306
n
m
 1;1,1,2,2,3 . If the string pieces used to
th of a unit length, then the best approximation would be
the continued fraction truncated as
n
m
=1.58536…, which is better than
3
2
485
306
,
 1;1,1,2,2,3,1,5 . All of these are continued fraction
485
306
approximations for log 2 3 =1.58496...;
65
41
will be the best approximation for log 2 3 . This is
1
1
1 3
 1
 1   . However, if the string pieces used to
1
1 1
11
2 2
approximate log 2 3 were each
be
3
2
= 1.58496… is a better approximation than
=1.5, but given the size of the denominators for each
fraction, each is the best approximation for any fraction with a denominator its size or smaller.
n
m
Truncating the continued fraction
1
1
1
1
1
1
 1
1
2  12
1
1
1
1
 1
1
1
1
1  52
 1;1,1,2,2,3,1,5,1,3,1,1... at 1;1,1,2,2 results in
 1
1
1
1
 1
1
1
19
. So
 1  12  1  127 
5
1 7
12
7
7
5
2
5
n 19

 1.583 is the best possible rational approximation for
m 12
n
m
with a denominator no greater
than twelve.
If the fraction
19
19
12
is used to approximately equate the fifth to the octave the solution
7
3  2 1 2 can be rewritten as 3  2  2 1 2 or
3
2
7
7
 2 1 2 . That is, 2 12 is the best approximation for any
power whose denominator is twelve or less for the musical fifth,
3
2
. The fact that the
denominator is exactly twelve offers justification for keeping the historical twelve-note division
of the western scale. The difference between the equal tempered and Pythagorean fifth is only
25
two cents, as is the difference between the two different fourths. Thus, the continued fraction
approximation for 3  2 x shows that using
12
2 as the interval for the twelve-toned western
musical scale optimizes the alignment between the three most important intervals, the octave, the
fifth and the fourth, of the equal tempered and the Pythagorean scale, given that the goal is a
scale with a reasonable number of notes.
If having many notes is allowed, the continued fraction can be truncated further along
and produce an even closer alignment. In the seventeenth century, Nicolas Mercator tried to
84
popularize a scale with 53 intervals based on the approximation 3  2 53 , using the continued
fraction 1;1,1,2,2,3,1 . This results in the approximation for a fifth
3
2
2
31
53
 1.4999... The
difference between the Pythagorean and equal tempered fifth in this case is less than one-tenth of
31
a cent ( 2 53  701.89 cents versus
3
2
 701.96 cents) (Coxeter, 1968, p. 319). At least one
keyboard was built to accommodate this 53 note scale -- Bosanquet’s “Generalized Keyboard
Harmonium” built in 1876 had 53 keys per octave (Dunne, 2000, p. 12). Not surprisingly this
was the exception to the rule; most musicians proved unwilling to trade the convenience of
twelve intervals for the more tonally accurate 53 intervals.
Using 312  219 as the approximate solution to 3 m  2 n , the basic interval of the
Pythagorean scale, a fifth, produced by the ratio
3
2
7
, is renamed 2 12 . Because 7 and 12 are
coprime, this new interval for the equal tempered fifth can be repeated 12 times, modulo the
octave, to find all 12 notes of any scale. This is similar to the original method used by the
Pythagoreans to create their scale, with the big difference that the new interval exactly divides 7
octaves. As a result, the equal tempered scale is a closed system in which the original notes are
repeated again and again in different octaves. This fact closes the spiral and results in a true
26
circle of fifths. And so we are left with what Neil Bibby, in his chapter of Music and
Mathematics, describes as the “powerful irony that irrational numbers should come to the rescue
– courtesy of the tolerance of the human ear and cultural conditioning – of the essentially
rationally based system that (the Pythagoreans) originally described for constructing a musical
scale" (Fauvel, Flood and Wilson, 2003, p. 27).
Circle of fifths
(Fauvel, p. 26)
C
F
G
Bb/A#
D
A
Eb/D#
E
Ab/G#
B
Db/C#
Gb/F#
27
7
Multiplying by 2 12 , or ascending a fifth, then multiplying by ½, or descending an octave, to
produce a single scale with intervals r such that 1  r  2 :
7
7
1
1
 2 `12  2 `12  G
1
1
 2 `12
1
1
1
1
1
1
.
.
.
1
1

 2
 2
 2
7
7
`12
7
`1 2
7
`1 2
 2
 2
 2
 2
2
14
`12
3
21
`12
4
28
12
2 `12  12  12  2 `12  E
5
35
12
2 12  12  12  2 12  B
 
7
 2 `1 2
12
84
 2 12
14
2
2 `1 2  12  2 `1 2  D
21
9
2 `1 2  12  2 `1 2  A
28
4
35
11
84
12
2 1 2  12  12  12  12  12  12  12  2 1 2  C (an octave higher than the first C)
C. Significance of the Circle of Fifths vs. Spiral of Fifths: transposition
With the Pythagorean Comma eliminated, this new equal tempered scale makes it
possible for musicians to change from one key to any other without producing discordant
intervals. In the equal tempered scale, every note sacrifices a tiny bit of its perfect pitch to
improve the flexibility of the fixed note instrument. With an equal interval between each key,
the sharps and the flats converge and all notes are now included in any tuning, so music can be
transposed across keys without any problem. (Or, as some violinists or singers will argue, with a
(small) problem everywhere.) The acceptance of the equal tempered scale is credited to Bach,
who wrote the Well-Tempered Clavier in 1720, consisting of 48 preludes and fugues. In a fugue,
a main theme is repeated many times in different, but related keys. The Well-Tempered Clavier
uses fugues to demonstrate how the equal tempered scale allows a piano to be played equally
well in any key (Coxeter, 1968, p. 318).
The equal tempered scale essentially takes the human ear out of the equation. This
mechanical nature is taken to an extreme with modern electronic keyboards, whose tuning is set
28
by computer coding and not corrupted by time, weather and use like real piano strings. Using an
equally tempered keyboard, (or any instrument tuned according to equal temperament) a piece of
music can be transposed to any key according to the following formula:
Given any set, S of tones in an equal tempered Scale, n, one may apply a transposition, Tk ,
where Tk x   x  k mod n for all tones x in S.
(Fauvel, Flood and Wilson 151)
For instance, the first line of “Twinkle, Twinkle Little Star” is A, A, E, E, F#, F#, E
(“twin-kle-twin-kle-lit-tle-star”). This can be transposed three semitones by setting k = 3,
assigning each note a number based on its placement in the equal tempered scale (starting with C
= 0, thus A = 9, E = 4 and F# = 6), and then applying the formula: T3 x   x  3mod 12 .
T3  A  9  3mod 12  0 or C
T3 E   4  3mod 12  7 or G
T3 F #  6  3mod 12  9 or A
So the first line transposed three semitones is C,C, G, G, A, A, G
If this formula for transposition were applied to a non-equal tempered scale, it would fall
apart, because the note indicated by 6 could be F# OR Gb. The human ear, and violinist, can
choose between these notes to produce harmonious intervals. The formula cannot. Even if it
could, any fixed-note instrument with a Pythagorean tuning is missing five sharps or flats and so
sometimes the desired transposed note would not be available.
29
IV. Fret Placement
A. Constructing Galilei’s interval
Using
12
2 to create a twelve-tone scale is only practical if the musical instrument is an
electronic keyboard. Until recently, however, this wasn’t an option and even now the goal is
often to tune an actual guitar or other fretted instrument. Thus, even with our mathematically
perfect interval, the challenge remains to find a practical method to actually place the frets.
One of the advantages of Vincenzo Galilei’s interval, 1817 , is that it offered a practical
method for approximating the equal tempered scale: divide the string length into eighteen equal
parts, and place the first fret at the seventeenth division, then divide the remaining string length
into eighteen parts again, and again place the first fret at the seventeenth division. Repeating this
process twelve times will result in the placement of twelve frets that approximate an equally
tempered scale. This method requires the memorization of a single fraction that produces fairly
manageable calculations.
For example, consider a modern day classical guitar whose string length is generally
1
1
about 66 cm. 66  17
18  62 3 , so the first fret will be placed 62 3 cm from the bridge.
47
47
62 13  17
18  58 54 , so the second fret will be placed 58 54 cm from the bridge. This process could
be continued to approximate the placement of as many equally tempered frets as desired. (Note
47
18
18
 17
 62 13 and 62 13  17
 66 , thus the interval between these
that, working backwards, 58 54
18
string lengths is 17
.)
B. Strahle’s construction
In 1743 Daniel P. Strahle of Sweden explained an ingenious method for placing the frets
to produce an equal tempered scale. Strahle was not a mathematician, but instead a skilled
instrument maker who based his construction on the experience and intuition of a craftsman.
30
The following description of the construction is by J.M. Barbour, who rescued Strahle’s
construction two centuries after it was incorrectly shown to be inaccurate.
Upon the line QR, 12 units in length, erect an isosceles triangle, QOR, its equal
legs being 24 units in length. Join O to each of the 11 points of division in the
base. On QO locate P, 7 units from Q, and draw RP, extending it its own length
M. Then, if RM represents the fundamental pitch and PM its octave, the points of
intersection of RP with the 11 rays from O will be the 11 semitones of the octave.
(Barbour, 1957, p. 2)
O
M
24
13
P
12
1110
9
8
7
6
7
5
4
Q
3
2
1
R
I I III IV V VI VII VIII IX X XI XII XIII
12
1. Trigonometric check
Around 1760, Swedish mathematician, Jacob Faggot attempted to use trigonometry to
check Strahle’s construction. Unfortunately, Faggot calculated one angle wrong, and then used
this wrong angle measure in all the rest of his calculations, discrediting Strahle’s construction for
31
almost two centuries (Barbour, 1957, p. 2). What follows is a correct trigonometric check using
the relationships given by the Law of Cosines and Sines:
Law of Cosines: c 2  a 2  b 2  2ab cos 
a
b
c


sin A sin B sin C
Law of Sines:
C
c
a
b

A
b
a
c
B
First, the measures of the three angles in triangle PQR are calculated.
6
adj
6
cos 1
 OQR  75.52

cos OQR 

24
hyp 24
2
RP  12 2  7 2  2  12  7 
6
= 193  168  0.25  151  RP  12.29
24
sin 75.52 sin QPR

 sin QPR  0.945  QPR  70.98
12.29
12
PRQ  180  70.98  75.52  33.50
O
M
24
P
70.98
12.29
7
Q
33.50
75.52
R
A
12
32
Second, the length of the perpendicular bisector of the isosceles triangle QOR is found.
O
x
tan 75.52 
6  (tan 75.52)  23.23
6
24
23.23
Q
75.52
6
A
Third, this length is used to find the angle measures of the triangles formed by part of PR , part
of QR and each of the 11 lines from O to QR .
O
tan II 
23.23
II  77.85
5
180  (77.85  33.5)  68.65
M
23.23
P
68.65
Q
I II
11.55
33.50
77.85
5
VII
R
6
Then, the Law of Sines is used to find where the fret would be placed along MR.
11
x

sin 68.65 sin 77.85
x II  11.55
Finally, this length is subtracted from the total string length to place the second fret (C#).
Total string length:
Placement of second fret:
12.29  2  24.58
24.58  11.55  13.03 units from the bridge
33
The accuracy of this placement can be compared to the exact equal tempered placement
by dividing the base string length, 24.58, by the 13.03, to find the frequency ratio, and converting
this ratio to cents. As illustrated in the table that follows, Strahle’s method – based not on
mathematics, but on his expertise as a craftsman – gives a practical method for fret placement
extraordinary close to the theoretical equal tempered placement.
STRAHLE (TRIGONOMETRIC SOLUTION)
ordinal
note
base
C
Stahle’s
Placement
x XIII  24.58
C#
Decimal
Cents
24.58
24.58
1
0
Freq.
ratio
1
x XII  23.18
24.58
23.18
1.0604
102
12
D
x XI  21.87
24.58
21.87
1.1239
202
D#
x X  20.65
24.58
20.65
1.1903
302
third
E
xIX  19.49
24.58
19.49
1.2612
402
fourth
F
xVIII  18.41
24.58
18.41
1.3351
500
F#
xVII  17.38
24.58
17.38
1.4141
600
G
xVI  16.42
24.58
16.42
1.4970
698
G#
xV  15.50
24.58
15.50
1.5858
798
A
xVI  14.64
24.58
14.64
1.6790
897
A#
xIII  13.81
24.58
13.81
1.7799
998
seventh
B
x II  13.03
24.58
13.03
1.8864
1099
 2
 2
 2
 2
 2
 2
 2
 2
 2
 2
octave
C
xI  12.29
24.58
12.29
2
1200
2
second
fifth
sixth
Freq.
ratio
EQUAL TEMPERAMENT
Decimal
Cents
1
0
1.0595
100
2
1.1225
200
3
1.1892
300
4
1.2599
400
5
1.3348
500
6
1.4142
600
7
1.4983
700
8
1.5874
800
9
1.6817
900
10
1.7818
1000
11
1.8877
1100
2
1200
2
12
12
12
12
12
12
12
12
12
12
34
2. Algebraic Check
J.M. Barbour discovered Faggot’s error in his trigonometric check, but instead of
correcting it, Barbour presented an Algebraic check that he thought was more powerful because
it could be generalized.
O
M
24
P
B
7
D
R
Q
A
J C
12
Barbour dropped a perpendicular segment, DJ, to create two sets of triangles whose angle
measures are congruent (because they are the same angle, because they are right angles, or
because they are corresponding angles), and take advantage of the resulting similar triangles.
OAC is similar to DJC
and
BAR is similar to DJR
He then used the relationship between the sides of similar triangles to create the following
proportions:
DJ
JC

OA AC
and
DJ
JR DR


BA AR BR
35
Now he solved the first equation for DJ:
DJ into the second equation:
DJ 
JC  OA
and substituted this value for
AC
JC  OA JR

. Next he observed that in the diagram
AC  BA AR
( JR  CR)  JC and ( AR  CR)  AC , and substituted these differences for JC and AC:
JR OA( JR  CR)
. Barbour then cross multiplied and solved for JR:

AR BA( AR  CR)
JR 
CR  OA  AR
and substituted this into the original proportion:
AR(OA  BA)  BA  CR
CR  OA  AR
DR JR
CR  OA
AR(OA  BA)  BA  CR
.



BR AR
AR
AR(OA  BA)  BA  CR
Going back to the original diagram, let xi be the lengths along QR marked off by the 11
lines dropped down from O. Then let
xi
 m , the fraction of the original length at each mark,
QR
and the fractional power needed to approximate the placement of each fret . Using this new
variable, Barbour noted that CR  1  m and AR  12 , and for simplicity sake, he let OA  a and
BA  b . Barbour then substituted these values into the above equation, changing it from a
proportion involving segment lengths to a more general algebraic equation:
DR
(1  m)  a
2a(1  m)
 1
and simplify
a  b  2bm
BR 2 (a  b)  b  (1  m)
Next, he noted that when m = 0, DR = PR and substituted these into the above equation:
PR
2a(1  0)
2a


.
BR a  b  2b  0 a  b
36
Now, Barbour used the fact that in Strahle’s construction dealing with fret placement, the
entire length of MR will sound the base note and so should be two units. Further, MP must equal
PR so that length MP will sound the base note transposed up one octave. So MR = 2, MP = PR
=1, and MD  2 PR  DR . He then combined all these relationships to get:
DR
MD 2 PR  DR
DR
DR BR
2a (1  m) a  b

 2
 2  BR  2 

 2
PR
MP
PR
PR
BR PR
(a  b  2bm) 2a
BR
(1  m)( a  b) 2(a  b  2bm)  (1  m)( a  b) 2a  2b  4bm  a  am  b  bm
 2


a  b  2bm
a  b  2bm
a  b  2bm
a  b  (a  3b)m

a  b  2bm
2m 
O
M
24
P
B
7
D
Q
R
A
J C
12
37
This formula approximates 2 m . Using the values specific to Strahle’s construction, the
values of a and b can be found so that this formula can be used to find Strahle’s specific
1
2
3
12
approximations for 2 m , ( i.e. 2 12 ,2 12 ,2 12 ,...2 12 ) . Strahle states that PQ is 7 units, OR is 24 units
and AR is
12
2
 6 units:
O
M
24
P
B
7
R
Q
E
A
12
Using similar triangles:
7 24
PQ OR
or 
, so x  74  QE . This results in ER  12  74 

x
6
QE AR
41
4
.
PR
2a
41
PR ER 414 41


. So
.

 
BR a  b 24
BR AR 6 24
Further, by similar triangles:
This equation has two unknowns, so the only way to solve it is to find a value for the
ratio of the two unknowns:
ratio
a
b
a
b
. To do this, divide each term in
2a
2 ab
by b:
. Now, let the
a 1
ab
b
equal x and solve for x:
2x
41

 24  2 x  41  ( x  1)  48x  41x  41  7 x  41  x  417  a b .
x  1 24
38
Now the formula for approximating 2 m ,
a
b
41
7
a  b  (a  3b)m
, can be rewritten in terms of x :
a  b  2bm
 1  ( a b  3)m x  1  ( x  3)m
. Finally, x  417 is substituted:

a  1  2m
x  1  2m
b
 1  (417  3)m 41  7  (41  21)m 48  20m 24  10m
(Barbour, 1957, pp. 4-5).



41  1  2m
41  7  14m
48  14m 24  7m
7
This formula, based on Strahle’s construction, results in approximations for 2 m that are as
accurate as the values resulting from the Trigonometric Solution, but much easier to calculate.
There is no evidence that suggests Strahle used anything but trial and error to place the
diagonal line representing the string (RM) so that it crosses the triangle side seven units from the
base. Barbour points out that the placement of the diagonal “string-line” could be improved by
basing its placement on the marking for the sixth fret. Recall that the frequency ratio of the sixth
tone in the equal tempered scale is
 2
12
6
6
1
 2 1 2  2 2  2 -- a constructible number. By
constructing the square root of the entire string length to place the sixth fret, and aligning this
sixth fret with the line extending from the sixth marking on the base, three of the frets would
now be perfect (the first, sixth and thirteenth) and the rest would be a tiny bit more accurate than
Strahle’s method (Barbour, 1957, p. 6). This mathematical analysis could also be used to justify
Strahle’s choice to place the string-line 7 units from the base, as doing this places the sixth fret
almost exactly at 2 .
From a musical perspective, a perfect sixth fret is really not that essential as it is not one
2
4
5
7
9
of the seven major musical intervals: second = 2 12 , third = 2 12 , fourth = 2 12 , fifth = 2 12 , sixth = 2 12 ,
11
12
seventh = 2 12 , and octave = 2 12 . An even more optimal placement of the string-line would be
7
one where the seventh fret – representing the essential fifth interval – was exact. However, 2 12
39
7
is not a constructible number. To overcome this, 2 12 could be approximated with a rational
continued fraction and then constructed to place the seventh fret. This marking would be aligned
with the line extending from the seventh marking on the base, and the remaining frets could then
be approximated. In the end, Strahle’s method based on whole number measurements comes
amazingly close to placing both the sixth and seventh fret perfectly. Perhaps these more
sophisticated mathematical approaches don’t really offer better construction techniques, but
instead justify Strahle’s method by showing that more sophisticated techniques do not
meaningfully improve the accuracy of his method.
3. Fractional Linear Function Check
In examining Barbour’s Algebraic solution, Isaac Schoenberg noticed that Barbour’s
approximation, 2 m 
24  10m
, was a linear fractional function. Using 20/20 hindsight,
24  7 m
Schoenberg explained how Barbour’s formula could be found by approximating an increasing
exponential function with a linear fractional function. Schoenberg started by translating
1
2
Strahle’s musical goal of finding the intervals 2 1 2 , 2 1 2 ,... 2
exponential function: f ( x)  a ba  , 0  x  1 .
x
12
12
into the Algebraic increasing
He then approximated f (x) in the interval
[0,1] by finding a linear fractional function, L( x) 
x  
, that would coincide with the
x  
exponential function, f ( x)  a ba  , at the beginning, 0, middle, ½, and end, 1, of the given
x
interval (Schoenberg 550).
In order to coincide at these three points, f (0)  a  L(0) , f ( 1 2 )  ab  L( 1 2 ) , and
f (1)  b  L(1) . Knowing these three values of f (x) and L (x ) need to be equal implies that the
40
fractional linear function must be in the form: L( x)  R 
x  T 1  x 
because if it is set up this
1x  T 1  x 
S
R
way:
L(0)  R 
 0  T 1  0
R
1  0  T 1  0
S
L(1)  R 
R
 1  T 1  1
S
1  1  T 1  1
S
R
S T
L 1 2 
 12  T 1  12  2
1  T   S R  T  L12   1  T   S  RT

  R R
R
1  12  T 1  12  2
1 T
S
L( 1 2 )  R 
R
 L 12   L 1 2   T  S  RT  L 12   S  R  L 12 T  T 
L 1 2   S
.
R  L 1 2 
Substituting in the known values at 0, ½, and 1, results in the fractional linear function:
L( x )  a 
b a x 
x
1  x
b 1  x 
a
b
a
For Strahle’s construction, a must equal 1 and b must equal 2, so that f ( x)  2 x
0  x  1 .
is: S ( x) 
So the linear fractional function that will approximate f ( x)  2 x in the interval [0,1]
2 x  2 1  x 
x  2 1  x 
. This formula can be used to find proper fret placement by substituting
the fractional power of 2 that corresponds to each fret. So the seventh fret would be placed at
S ( 127 ) 
2  127  2 1  127 
7
12
 2 1  127 
= 1.4975 = 699 cents .
Strahle might be horrified by this complicated interpretation of his practical and elegant
method for placing frets. In contrast, the fact that his craftsman’s approach can be translated into
a fractional linear function delights the mathematician by connecting an abstract formula to an
organically developed method for making instruments. The work of Barbour and Schoenberg
would hopefully make the Ancient Pythagoreans proud, as it offers one more step towards the
goal of explaining the whole world using mathematics.
41
Part Two: Lesson Plans for Middle and High School Mathematics
I am mathematically inclined – organized, comforted by structure and pattern, and eager
to uncover structure and pattern where it is not apparent. I used to consider these qualities to be
the opposite of artists or musicians, who I thought used some magical creativity as inspiration for
their work. The fingers of the violinist magically knew how to find the notes of a scale using a
skill I thought was outside of my realm of talents. Music is more accessible to me now that I see
it is not magic, but mathematics and physics, that at least partially explain the musical scale.
Similarly, I know my more musically inclined friends tend to see mathematics as too dry,
abstract or regimented to be interesting. By reframing their knowledge of the musical scale from
a mathematical perspective, they might see that the patterns and structure in the music they love
are mathematical in nature. Exploring the connections between music and mathematics can help
the musically inclined find meaning in “abstract” mathematics and help the mathematically
inclined see the patterns and structure inherent in music. Furthermore, it is common mathteacher wisdom that any connection we can make for students between what we are teaching and
something concrete, visual or experiential in nature helps student understand and retain concepts.
The nine lessons that follow attempt to make these connections explicit, by teaching topics
commonly covered in middle and high school mathematics through the musical scale. The
sequencing of the lessons parallels the development of some of the key ideas in the mathematical
treatment of the musical scale given in Part I of the paper, thereby aiming to engage students in
drawing or using the math-music connections inherent in the musical scale.
The first lesson gives students an overview of the musical scale by defining consonant
and dissonant intervals, exploring two of the basic intervals of the western scale and their
frequency ratios, and giving students the opportunity to find and play these intervals on a
42
monochord. The second lesson requires students to do more complicated calculations with ratios
and/or fractions and accurately measure string lengths to build the entire diatonic scale on the
monochords. The third lesson examines the intervals between the notes in the scale built in
lesson two by dividing fractions, and introduces sharps and flats as a way to make these intervals
more equal. Students are also introduced to the problem inherent in scales without perfectly
equal intervals and asked to think of ways to solve this problem. In the fourth lesson, students
explore one solution to the problem of unequal intervals, the equal tempered scale. Students are
introduced to the irrational ratio that produces this scale, and then use this ratio to calculate string
lengths and build the twelve tone ETS on the monochords. Students then use this scale to define
and explore geometric and arithmetic progressions.
In lesson five, students use the ETS they built in lesson four to learn some simple songs
and then transpose them, examining the patterns inherent in songs and using modular arithmetic
and basic algebra in the process. In lesson six, students are introduced to the musical cent as a
measurement of tone, and use logarithms to convert the frequency ratios of the JTS and ETS they
found in lesson two and four into cents so that the two scales can be compared. In lesson seven,
students use the Fundamental Theorem of Arithmetic to prove the impossibility of aligning the
two basic western intervals, the fifth and the octave, in their pure form. They are then asked to
articulate the pros and cons of the rational and irrational musical scales explored in the previous
six lessons. Lesson eight is an introduction to continued fractions which is built upon in lesson
nine to show the optimality of a twelve tone scale in best achieving the goal of aligning the fifth
and the octave.
I had the opportunity to try out the first four lessons and a reflection on each is included
after the lesson. The remaining lessons will have to wait for my return to the classroom for their
43
trial run. I have also included background information on the Just Tempered Scale, instructions
on how to build a monochord – which could be done as a class activity in the right setting, and
images of a few instruments that I thought might be helpful in teaching the lessons.
44
I. The Just Tempered Scale
In the lessons that follow, I decided to use a slight variation of the Pythagorean Scale
called the just tempered scale for two reasons: #1) The JTS is based on less complex fractions, so
students' computations are more manageable. #2) The JTS more easily illustrates the difficulty of
changing keys because it is even more unequal than the Pythagorean scale. I did not focus on the
JTS in my paper because I found its mathematical basis less interesting then the Pythagorean
scale. As background for the lessons, a brief overview of the JTS follows.
The Pythagorean scale was adjusted slightly in post-medieval Europe because music
changed from being based primarily on fifths to incorporating more complex harmonies based on
thirds. The new "just tempered scale" was designed to preserve pure fifths and pure thirds, at the
expense of uniform intervals.
Just Tempered Scale
notes: C D E F G A B C
frequency ratios: 11 89 54 43 32 53 158 12
      
interval:
9
8
10
9
16
15
9
8
10
9
9
8
16
15
The just tempered scale has a major whole tone ( 89 ), a slightly smaller minor whole tone ( 109 ),
and a semitone ( 16
) which is not, in fact, exactly half of either whole tone. This lack of
15
uniformity makes changing keys on a fixed note instrument even more difficult that with the
Pythagorean tuning, but because the pure fifth and pure third are particularly pleasing to the well
trained ear, just tuning is still favored by many musicians of more flexible instruments like the
voice or violin. As Catherine Schmidt-Jones explains,
The problem with just intonation is that it matters which steps of the scale are major
whole tones and which are minor whole tones, so an instrument tuned exactly to play
with just intonation in the key of C major will have to retune to play in C sharp major or
D major. For instruments, like voices, that can tune quickly, that is not a problem, but it is
unworkable for piano and other slow-to-tune instruments.
(Schmidt-Jones, 2005, p. 6)
45
II. Lessons and Reflections
A. How to Build a Monochord
A classroom set of monochords is needed for most of the following lessons. Below are
instructions for how I built my set of four 36-inch (with 72 cm string) and six 18-inch (with 40
cm string) monochords. To reduce cost, the tuners can be replaced with string pegs or screws.
Supplies
Approx. Cost
Seven 1-yard long pieces ½ x 2 x 2 pieces of poplar (or other) wood
$13.50
One 1-yard long piece 1 x 2 x 2 pieces of poplar (or other) wood
$2.50
(They come precut in both these sizes at Home Depot.)
10 Guitar Tuners with one-inch stems plus screws
$30.00
10 Guitar strings – G or D (a thicker string that will be easier on fingers)
$24.50
10 little screws
$1.00
100 skinny 1-inch nails
$1.00
10 approx. 1 12 inch screws or nails (whatever you can find)
Total Cost: $72.50 ($7.25 per monochord)
Tools
Saw
Little screwdriver
Hammer
Centimeter tape measure
Drill with a big drill bit a little bigger than tuner stem and a little bit for the screws
Assembly
1. Cut three of the ½ x 2 x 2 pieces in half (for the base of the six 18-inch monochords).
2. Cut the 1 x 2 x 2 piece into at least 38 pieces, each about ¾ inch wide.
3. Drill a hole big enough for the tuner stem to go through about ½-inch from the edge of the ten
monochord bases. The hole should be close enough to the end that the tuner can turn.
4. Insert tuner and mark screw holes. Drill starter holes for screws.
5. Attach tuners. (I used the screws I bought separately for this, reserving 10 of the tuner screws
to attach the strings because I couldn't find small enough screws at the hardware store to attach
the strings.)
6. Nail ¾ inch wide pieces to the bottom of each monochord base – 3 per short base, 5 per long
base. (This raises the base higher than the tuner bottom so that the monochord can stand flat)
7. Drill starter holes for the screws that will hold end of string: approximately 41 cm from tuner
in the short boards, 73 cm from tuners in the long boards.
8. Place screw through end of string and screw into starter holes, then attach string to tuner.
9. Place 1 12 inch screws or nails lengthwise under the string at the opposite end of the tuner at
exactly 40 cm for the short boards and 72 cm for the long boards. (The function of this screw or
nail is to stop the vibration of the string.)
46
B. Instrument Images
I have included four images that may be useful in presenting the following lessons. The first is
of a piano keyboard, the second of a classical guitar, the third of an electric guitar and the fourth
of the one of the original fretted instruments, the viol.
1. Piano Keyboard
Johnson, Charles William. Earth/matrix. 2005 <http://www.earthmatrix.com/piano/a432key.gif>
47
2. Guitar
Luthier, Michael Thames. Guitar sales website.
<http://www.thamesclassicalguitars.com/MFA%202000.635%20Panormo%20guitar.jpg>
48
3. Electric Guitar and its Frets
Build A Recording Studio.com. 2005. <http://www.build-a-recording-studio.com/imagefiles/electric-guitar.jpg>
49
4. Viol
The Early Music Shop, United Kingdom.
<http://www.e-m-s.com/cat/stringinstruments/viols/viol%20-%20bachle.JPG>
50
C. Lesson Plan One: Cultural Consonance and Whole Number Ratios
In this lesson, students are introduced to the basic Western musical intervals, the fifth and
the octave. They are asked to identify the most consonant interval, the octave, on the monochord
first by listening to three intervals and choosing the one that sounds the "best," and then by
finding the octave on a pair of monochords. Students are given examples of culturally consonant
and dissonant intervals by listening to a Western and non-Western musical selection. The names
of the intervals of the musical scale are also highlighted and examined as an example of ordinal
numbers.
Students are then asked to examine these consonant intervals from a mathematical
perspective, exploring the frequency ratios that create these intervals and calculating the string
length that would produce a fifth given different lengths for the base strings. To do this students
either set up an equation with the frequency ratio on one side and the actual string lengths on the
other, or think of the ratio as a relationship between two different lengths and mentally calculate
the string length needed to obtain this relationship. Finally, students are asked to actually
calculate and measure the fifth on a monochord. In addition to using the frequency ratio to find
the string length, students also practice using a ruler to measure a centimeter length longer than
the actual ruler.
51
Lesson One Outline
TOPIC
Cultural Consonance and Whole Number Ratios
GOALS
Skills:
Students will solve equations involving ratios.
Students will understand difference between ordinal and cardinal numbers and connection to
names in Diatonic Scale.
Students will observe how simple ratios of string length produce the common musical intervals.
Thinking:
Students will be able to define consonant and dissonant and understand that these words are
culturally specific.
Students will investigate different methods for creating equivalent ratios.
TIME PERIOD and LEVEL/PREREQUISITES
One 60 minute class periods in middle school math or Algebra.
Students should have experience with ratios and/or multiplying and dividing fractions.
SUPPLIES
Ten monochords – 4 of one length, two of another
Violin, if possible, provided by teacher or a student.
Recording of Chinese music, and western classical piece.
Transparency / Worksheet / Rulers
ASSESSMENT
Students will find length of string for fifth interval given various string lengths.
52
ACTIVITIES
1. The basic musical interval, an octave: (10 minutes)
Ask students if they know of any ways that math and music are related.
Survey class for students who play musical instruments. Explain difference between fretted
and non-fretted instruments.
Ask class a “musical” question: Does anyone know which musical interval is easiest to
recognize?
Have volunteer play whole string on monochord. Play 3 different notes on a different
monochord (tuned to match first). Ask students to identify the most consonant interval.
Agree that this interval is the octave.
Pass out 2 monochords of same length to groups of 6 and have students try to find octave on
the instrument and then measure length of base note and length of octave and compare.
Students should notice the base note is twice (or four times if they go up two octaves) the
octave. Any notes that are an octave apart have frequencies in the ratio of 2:1.
Have students try to play an 8-note scale. (COLLECT INSTRUMENTS)
2. Consonance and Culture: (10 minutes)
Explain that each culture has musical intervals that are agreed to be consonant, that is the
collective ear of that culture has agreed these intervals sound nice.
Play a section of Chinese music and western classical music to illustrate different cultural
understandings of consonance. Have students share reactions to music.
3. Diatonic Scale: (5 minutes)
Play a consonant interval and a dissonant interval. Ask if anyone can name any of the other
intervals popular to western music.
The basic musical intervals in western music are the second, third, fourth, fifth, sixth, seventh
and octave, named for their respective order in the Diatonic Scale.
Review ordinal vs. cardinal numbers (where vs. how many).
Diatonic (dia-across, tonic- tone) so named because it is two tetrachords separated by whole tone.
(A tetrachord is four notes that span the interval of a perfect fourth.)
Ask for a volunteer who can play a scale on the violin. If no violinists, teacher can play.
4. Diatonic Scale Frequency Ratios: (5 minutes)
Pythagoreans first analyzed the mathematics behind the musical scale. They worshiped whole
numbers and attempted to explain everything in the natural world using whole numbers. They
found that the Diatonic Scale could be created using whole ratios. Their scale was further
simplified later on resulting in the Just Tempered Scale (transparency)
frequency ratios: 11 89 54 43 32 53 158 12
notes: C D E F G A B C
Base 2nd 3rd 4th 5th 6th 7th Octave
where each ratio is:
the length of the original string
.
the length of the new string
53
5. Calculating Diatonic Scale: (15 minutes)
Given an instrument with string length 30 cm, calculate the length of string needed to produce
the fifth note (freq ratio = 32 ) to the nearest mm.
Allow students time to explore how to use frequency ratios to calculate actual length. Have
share methods. (Fifth = 20 cm.)
(Divide total string length into 3 parts, take 2 of these parts.)
What if base string were 60cm? (40 cm) 3 cm? (2 cm)
40 cm? 40  3  13.3  13.3  2  26.6
40 3
80
  40  2  3x 
 26 23  26.7mm  x
x
2
3
OR
40  x  32  40  32  x  40  23  803  x
Have students find fifth note given original string length:
48 cm 
90 cm 
21 cm 
72 cm 
150 cm 
32 cm
60 cm
14 cm
48 cm
100 cm
6. Find / Play fifth on Monochord (10 min)
Pass out monochord and rulers and have students find and play fifth on monochord.
Carefully explain how to measure: Use centimeter side of ruler. Start at tuner. For 72 cm
monochord, mark off 30 cm first, then measure 18 more cm (30 +18 = 48 cm). For 40 cm
monochord, mark off 26 cm, plus 7 mm.
Then hold finger on mark to stop vibrations and pluck string by tuner. (The note will be the
same no matter where you pluck between tuner and finger, but it is easier to pluck by tuner
because string is slightly raised.)
54
The Just Tempered Scale
frequency ratios:
1
1
Name __________________
9
8
5
4
4
3
3
2
5
3
15
8
2
1
notes: C D E F G A B C
Base 2nd 3rd 4th 5th 6th 7th Octave
the length of the original string
where each ratio is: the length of the new string
original string length of string needed
length
to produce the fifth interval
30 cm
60 cm
3 cm
40 cm
48 cm
90 cm
21 cm
72 cm
150 cm
55
Reflections on Lesson One
I taught this lesson to a borrowed group of 25 fifth graders at a Catholic school. They
were younger than the targeted audience, but their enthusiasm and cooperation helped bridge the
age gap. The lesson was very well received. All students had some experience playing an
instrument, mostly with the recorder or piano. They loved the monochords and the idea that they
could build one themselves. If I teach this series of lessons in my own classroom, I would pass
out my instructions for building a monochord so that interested students could more easily make
their own at home.
None of the students seemed to have any knowledge of the basic musical intervals,
including the octave, which surprised me. However, they were able to recognize the octave as the
most consonant interval. After listening to three intervals there was murmurs throughout the
classroom that they all sounded bad etc., but when the students voted on which sounded best, the
octave was the clear choice. I was pleased with this as I had been worried that students wouldn’t
be able to hear the monochords well enough, and the even if they could the consonance of the
octave wouldn’t be obvious on such an elementary instrument to relatively untrained ears.
They loved listening to the western and non-western music and expressed a clear
preference for the non-western music. The western music (Bach Cello Suites) was deemed “very
boring” and the non-western music “exciting” “calming” and “interesting.” This surprised me,
although in retrospect it makes sense that to budding adolescents the cultural norm (consonance)
is boring and straying from this norm (dissonance) is exciting. They were interested in the origin
and history of the non-western music, and I was able to work in a nice bit of history of both
musical pieces in this teachable moment. With more knowledge of both, I could have taken
better advantage of the students’ interest. (The Bach Cello Suites, I recently learned, is the first
56
piece of music ever written specifically for the cello in which the cello gets to carry the melody
and not just the play background music.)
The ordinal / cardinal number review was nice because it just slips right in. I think it is
worthwhile for students to understand the difference between the two kinds of numbers, but to
make this a focus of a lesson can be tiresome. Also, the common example for ordinals is
placement in some competition, so I was glad to have a different example in which these
numbers are used in a meaningful way. And because this ordinal naming confused me for so
long (as I didn’t realize they were ordinals), sharing this confusion with the students offers some
evidence for the importance of being able to distinguish between the two types of numbers.
Because the students were 5th graders, not 7th graders, I felt I needed to guide them
through the method for finding equivalent ratios instead of having them figure out and share
different methods on their own. I stuck with the approach of dividing the string into 2 parts and
taking 1 (for the octave) or dividing it into 3 parts and taking 2 (for the fifth). The students were
very agreeable to this method and easily figured out the length of the string needed for the
various fifths, but surely a deeper level of understanding would have been attained if I had
resisted the top down approach.
Overall, I felt this introductory lesson achieved its goals of giving the students a big
picture understanding of consonance and dissonance and allowing them time to explore how
frequency ratios work in relation to an actual instrument string.
57
D. Lesson Plan Two: Building the Diatonic Scale
In this lesson, students use what they learned in lesson one to calculate the string lengths
needed to play the entire diatonic scale, given the frequency ratio for each of its notes. Students
are asked to explore different methods for doing these calculations. The frequency ratios can be
approached as ratios, so that 3/2 indicates that the base string length should be 3 units compared
to 2 units for the new string length. Alternatively, the frequency ratios can be thought of as
fractions, in which case students need to use logic or basic algebra to determine how to use these
fractions to calculate string length. Students then need to use rounding and measuring skills to
mark off the appropriate string lengths accurately. At the end of this lesson, the students will
have built a basic eight tone scale on their monochords.
58
Lesson Two Outline
TOPIC
Building the Diatonic Scale
GOALS
Skills:
Students will multiply or divide whole numbers by fractions.
Students will solve equations involving ratios.
Students will use a centimeter ruler to measure lengths to the nearest millimeter, including
lengths longer than the ruler.
Thinking:.
Students will build Diatonic Scale on a one stringed instrument using ratios.
Students will determine what operation to perform on a whole number to produce a given ratio.
TIME PERIOD and LEVEL/PREREQUISITES
One 60 minute class period in middle school math or Algebra.
Lesson Two should follow Lesson One.
Students should have experience with multiplying and dividing fractions.
SUPPLIES
Ten monochords, 4 of one length, 6 of another.
Violin, if possible, provided by teacher or a student.
Transparency from Lesson One
Group Worksheet / Transparency
Homework Worksheet
Rulers (cm.)
ASSESSMENT
Have students complete worksheet for homework calculating lengths needed for diatonic scale
with a 15 cm base note.
59
ACTIVITIES
1. Review the Just Tempered Scale (5 minutes)
Use transparency from Lesson One to review ratios and ask students to explain how these ratios
are used to find octave and fifth on monochord (demonstrating with monochord).
the length of the original string
frequency ratios: 11 89 54 43 32 53 158 12
freq ratio=
the length of the new string
notes: C D E F G A B C
2. Calculating Diatonic Scale: (20 minutes)
Review methods used to find string length that produces given frequency:
Using Ratios:
Divide whole string into num parts, take den of them
(divide into 3 parts, take 2 of them to produce 32 )
originalst ring
freq

 originalst ring  den  num  x
x
ratio
Using Fractions:
original string length  frequency ratio = new string length
or
original string length  reciprocal of freq ratio = new string length
Check:
the length of the original string
= frequency ratio
the length of the new string
n
 r  n  r  x  nr  x
x
10
 2  10  5  2  10  2  5  10  2  5
Like “fact family”:
5
As a class, determine sting lengths for diatonic JTS with string length 45 (decimals are okay!),
and show how to measure and mark string lengths.
Carefully explain how to measure: Use centimeter side of ruler. Start at tuner. For 72 cm
monochord, mark off 30 cm and 60 cm first. For 40 cm monochord, mark of 30 cm first.
Then hold finger on mark to stop vibrations and pluck string by tuner. (The note will be the
same no matter where you pluck between tuner and finger, but it is easier to pluck by tuner
because string is slightly raised.)
Algebraically:
3. Calculate Actual String Lengths (15 minutes)
Using worksheet, have students calculate string lengths given base length 40cm or 72 cm.
4. Building Diatonic Scale: (20 minutes)
After calculating entire scale, place a mark on the instrument for each note and play scale.
When everyone is finished, have each group play their scale for the class, and play violin scale
again on one string, showing visually that same ratios are being used to produce notes.
60
Building the Diatonic Scale
Name ____________________
HOMEWORK
1. Given a string whose base length is 15 centimeters, calculate the string lengths needed to
produce a diatonic scale. (Round to the nearest millimeter.)
Frequency Ratio =
Note Interval
the length of the original string
the length of the new string
Diatonic Scale
Frequency Ratio
C
base
1
1
D
second
9
8
E
third
5
4
F
fourth
4
3
G
fifth
3
2
A
sixth
5
3
B
seventh
15
8
C'
octave
2
1
Calculation used
to determine string length
Actual Length of String
needed to produce note
2. Mark where the string should be pressed to produce each interval. Label each mark with the
interval name.
_____________________________________________________________________
61
Building the Diatonic Scale
Names ______________________
______________________
______________________
Record your instrument’s letter and base string length.
Calculate the actual length of string needed to produce each note. (Round to the nearest
millimeter.) RECORD how your group performs these calculations in the space below
the table.
Mark this length using ruler on instrument.
Play your scale. If it doesn’t sound right, check your calculations.
If you have time, try to figure out how to play a common song on your instrument, or
make up your own song.
ACTIVITY
1.
2.
3.
4.
5.
Base String Length:
Frequency Ratio =
______
the length of the original string
the length of the new string
Note Interval Diatonic Scale
Actual Length of String
Frequency Ratio needed to produce note
1
C
base
1
D
second
9
8
E
third
5
4
F
fourth
4
3
G
fifth
3
2
A
sixth
5
3
B
seventh
15
8
C'
octave
2
1
CALCULATIONS:
62
Building the Diatonic Scale
Transparency
1. Record your instrument’s letter and base string length.
2. Calculate the actual length of string needed to produce each note. (Round to the nearest
millimeter.) RECORD how your group performs these calculations in the space below
the table.
3. Mark this length using ruler on instrument.
4. Play your scale. If it doesn’t sound right, check your calculations.
5. If you have time, try to figure out how to play a common song on your instrument, or
make up your own song.
Frequency Ratio =
the length of the original string
the length of the new string
Base String Length: 45 cm
Note Interval Diatonic Scale
Actual Length of String
Frequency Ratio needed to produce note
1
C
base
1
D
second
9
8
E
third
5
4
F
fourth
4
3
G
fifth
3
2
A
sixth
5
3
B
seventh
15
8
C'
octave
2
1
Base String Length: 40 cm and 72 cm
Note Interval Diatonic Scale
Actual Length of String
Frequency Ratio needed to produce note
1
C
base
1
D
second
9
8
E
third
5
4
F
fourth
4
3
G
fifth
3
2
A
sixth
5
3
B
seventh
15
8
C'
octave
2
1
63
Building the Diatonic Scale
Key
1. Record your instrument’s letter and base string length.
2. Calculate the actual length of string needed to produce each note. (Round to the nearest
millimeter.) RECORD how your group performs these calculations in the space below
the table.
3. Mark this length using ruler on instrument.
4. Play your scale. If it doesn’t sound right, check your calculations.
5. If you have time, try to figure out how to play a common song on your instrument, or
make up your own song.
Frequency Ratio =
the length of the original string
the length of the new string
Base String Length: 45 cm
Note Interval Diatonic Scale
Actual Length of String
Frequency Ratio needed to produce note
1
C
base
45
1
D
second
9
8
40
E
third
5
4
36
F
fourth
4
3
33.8
G
fifth
3
2
30
A
sixth
5
3
27
B
seventh
15
8
24
C'
octave
2
1
22.5
Base String Length: 40 cm and 72 cm
Note Interval Diatonic Scale
Actual Length of String
Frequency Ratio needed to produce note
1
C
base
40
72
1
D
second
9
8
35.6
64
E
third
5
4
32
57.6
F
fourth
4
3
30
54
G
fifth
3
2
26.7
48
A
sixth
5
3
24
43.2
B
seventh
15
8
21.3
38.4
C'
octave
2
1
20
36
64
Reflections on Lesson Two
I taught this lesson to a borrowed group of 25 fifth graders at a Catholic school. They
were younger than the targeted audience, but very enthusiastic and cooperative. In theory this
lesson would follow Lesson One. In practice, I taught it as a stand alone lesson. This proved
workable, although I am sure the students would have gained a better “big picture”
understanding of the scale had they had Lesson One.
Because the students were younger than the intended audience, I didn’t have them
explore different methods for using the frequency ratios to find string length, but instead stuck
with the least sophisticated method of dividing the string length by the numerator and then
multiplying the quotient by the denominator. I added an example string length of 45 cm to do as
a class after teaching the lesson. The students were able to find the string lengths, but would
have benefited from going through and example together as a class – particularly in my ideal
lesson where the students would be offering several different methods for finding the string
lengths. Also, the students were alarmed at the decimal solutions that came up and doing an
example would have helped them see how to deal with this issue. (I didn’t anticipate this being
an issue! Many students reacted to 13.3 by telling me their calculator was broken, and showing
me this “thing” they clearly did not consider a number at all.)
Doing an example as a class would also be helpful in allowing the teacher to model the
measurement piece. Students had a LOT of trouble with the measuring and in retrospect I should
have given more instruction on how to measure and play the monochords. One group of students
looked at me blankly when I reminded them to use centimeters, not inches. They seemed not to
know what centimeters were, and certainly didn’t know how to measure the additional
millimeters needed for the decimal lengths. Also, there was confusion regarding how to measure
65
lengths longer than the 30 cm on the ruler. Again, several students seemed to see this as in
impossibility; they seemed to feel that if the ruler is 30 cm long, one can only measure lengths
less than or equal to 30 cm. Depending on time constraints, one could provide tape measures
instead, or take advantage of this teachable moment to show how to mark off 30 cm and go from
there. Also, there was understandable confusion regarding which end of the monochord to
measure from because I didn’t explain this clearly. All in all, this turned out to be a lesson in
measurement as much as anything. Based on this group’s measurement skills, giving students a
chance to practice measuring makes this lesson that much more valuable.
Despite these hiccups, the students remained motivated and on task throughout the lesson.
I had imagined that by starting with a box full of actual monochords I couldn’t really go wrong
from the students’ perspective – and this proved to be the case. I wished I had had more time to
go over the final measurements and reinforce that the new string lengths compared to the whole
string length actually produce the desired frequency ratios. Also, I think the students would have
eagerly discussed patterns they saw in how these string lengths changed and how one could use
these patterns to quickly check for calculation errors if using a different string length.
66
E. Lesson Plan Three: Unequal Intervals of the Just Tempered Scale
In this lesson, students practice multiplying and dividing with fractions to figure out the
intervals between each pair of notes in the just tempered scale they built in lessons one and two.
While doing this, students are asked to look for patterns in these intervals. Hopefully, students
will find it motivating to watch the pattern unfold, and be able to use the pattern to self-check
their fraction calculations.
Students are then introduced to sharps and flats and the twelve tone scale. Students use
the observations they made about the interval spacing to understand the placement of sharps and
flats in the JTS. Students are then asked to consider the problems inherent in changing keys in a
scale with irregular intervals. Students observe that with no regular, mathematically sound
pattern, transposition is not possible. Students then suggest ways to build a musical scale that
would solve the problem of the JTS and make transposing to different keys possible.
67
Lesson Plan Three Outline
TOPIC
Unequal Intervals of the Just Tempered Scale
GOALS
Skills:
Students will multiply and divide with fractions to determine intervals between notes.
Students will make observations about the size of these intervals.
Students will be able to define 12-tone scale.
Thinking:
Students will be able to explain why changing keys is limited in Just Temperament
Students will suggest ways to overcome this limitation.
TIME PERIOD and LEVEL/PREREQUISITES
One 50 minute class periods
This lesson should follow Lesson Two in which students build the Just Tempered Scale.
Student should have experience with multiplying and dividing fractions.
SUPPLIES
Violin, if possible, provided by teacher or a student.
Group Worksheet / Transparency
ASSESSMENT
For homework, explain the difference between a non-fretted instrument like a violin, tuba or the
human voice, and a fretted or fixed-note instrument, like a piano or a guitar. Explain why it is
difficult for a fixed-note instrument to change keys if a scale has unequal intervals between notes.
68
ACTIVITIES
1. Intervals of Just Tempered Scale (25 minutes)
Pass back homework from Lesson One and review frequency ratios of JTS and play scale on one
stringed instrument.
frequency ratios: 11 89 54 43 32 53 158 12
notes: C D E F G A B C
where each ratio is:
the length of the original string
.
the length of the new string
Previously defined basic intervals (second, third, fourth etc) in relation to base note only – now
we will look at more general interval – "space" between any two notes.
The interval between any pair of notes is the frequency ratio that separates them.
1
1

9
8
9
8
=
so C and D are separated by the interval
9
8
.
 ? = 54 where ? is the interval between D and E.
To solve, divide 54 by 89 (review fact family 2  ?  10  10  2  ? )
5
9
10
 89  54  89  109 
= 54
4
8  9
9
8
Similarly, have students work in pairs, using worksheet, to find intervals between consecutive
notes of JTS.
notes: C D E F G A B C
frequency ratios: 11 89 54 43 32 53 158 12
      
interval:
9
8
10
9
16
15
9
8
10
9
9
8
16
15
2. Adding Sharps and Flats (10 minutes)
Ask students to look for and share patterns they see in intervals of JTS.
Have students convert to decimals to help examine three JTS intervals.
Note that intervals are not equal: two bigger ones are close, called whole tones.
If you take the smaller interval twice, you get something close to the whole tone so smaller step
is called semitone. Play scale to “hear” different sized steps.
major and minor whole tones:
9
8
 1.125
10
9
 1.11...
semitone x semitone = whole tone:
16
15
 1.066...
16
15
 16
15 
256
225
 1.1378
Because of this, the 12-Tone Scale evolved by adding sharps/flats between whole tones:
C C#/Db D D#/E b E F F#/G b G G#/A b A A#/B b B C
69
3. Problem: Can't change keys on fretted instruments (10 minutes)
In JTS, even with addition of flats and sharps, all intervals are slightly different.
In fact, the reason there are flats AND sharps is because they aren't placed exactly in the middle
– they match one type of semitone or another depending on the key. (Play on violin if possible.)
E b Major has 3 flats (B b E bA b)
A Major has 3 sharps (F# C# G#)
One "tuning" cannot accommodate both keys b.c. G# /A b cannot be tuned simultaneously.
If playing a non-fretted instrument, like violin, can make small adjustments with finger. If
playing a fretted or keyed instrument, like Piano or Guitar, have to choose either G# OR A b.
Also, with two slightly different whole tones, if change keys the placement of the whole tones
changes.
Throughout history, musicians have tried different solutions to this problem. What do you think
they have tried?
Have students discuss possible solutions to this problem in pairs and share.
(Retune between pieces, 2 keys in between whole notes, equal temperament)
"The problem with just intonation is that it matters which steps of the scale are major whole
tones and which are minor whole tones, so an instrument tuned exactly to play with just
intonation in the key of C major will have to retune to play in C sharp major or D major. For
instruments, like voices, that can tune quickly, that is not a problem, but it is unworkable for
piano and other slow-to-tune instruments." Catherine Schmidt Jones
4. Assign homework (5 minutes)
For homework, explain the difference between a non-fretted instrument like a violin, tuba or the
human voice, and a fretted or fixed-note instrument, like a piano or a guitar. Explain why it is
difficult for a fixed-note instrument to change keys if a scale has unequal intervals between notes.
70
The Intervals of the Just Tempered Scale
Names ____________________
____________________
____________________
ACTIVITY
Find the intervals between consecutive notes of the Just Tempered Scale.
notes: C
frequency ratios: 11
interval:
D
9
8
9
8
E
F
G
A
B
5
4
4
3
3
2
5
3
15
8
2
1
___ ___ ___ ___ ___
10
9
STEP ONE
1. Interval between C and D:
C
1
1
STEP TWO
? =
9
8
9
8

1
1
=
9
8
5
4

9
8
=
5
4
2. Interval between D and E:
9
8
? =
5
4
3. Interval between E and F:
5
4
? =
4
3
4. Interval between F and G:
____
?

8
9
=
10
9
= _____
5. Interval between G and A:
6. Interval between A and B:
7. Interval between B and C:
71
The Intervals of the Just Tempered Scale
Names ____________________
____________________
____________________
ACTIVITY: Key
Find the intervals between consecutive notes of the Just Tempered Scale.
notes: C
frequency ratios: 11
interval:
D
9
8
9
8
E
F
G
A
B
5
4
4
3
3
2
5
3
15
8
10
9
16
15
9
8
10
9
1
1
2
1
9
8
16
15
STEP ONE
1. Interval between C and D:
C
STEP TWO
? =
9
8
9
8

1
1
=
9
8
2. Interval between D and E:
9
8
? =
5
4
5
4

9
8
=
5
4

8
9
=
3. Interval between E and F:
5
4
? =
4
3
4
3

5
4
=
4
3

4
5
= 15
4. Interval between F and G:
4
3
? =
3
2
3
2

4
3
=
5. Interval between G and A:
3
2
?
=
5
3
5
3

3
2
=
5
3
6. Interval between A and B:
5
3
?
=
15
8
15
8

5
3
=
15
8
7. Interval between B and C:
15
8
2
1
2
1

15
8
=
? =
2
1
16

3
2


10
9
3
4
10
2
3

8
15
9
=8
=9
3
5
9
=8
16
= 15
72
Reflections on Lesson Three
I taught this lesson to a borrowed group of 25 sixth graders at a Catholic school. In theory
this lesson would follow Lesson One and Two. In practice, I taught it as a stand alone lesson.
All students in the class had experience playing an instrument – mostly piano and recorder. As it
turned out, the 6th grade at this school plays the guitar in music class so they all understood how
frets worked, which was very helpful. And I was able to recruit a string player in this class to
play the A-major scale, much to the delight of her classmates.
I introduced the lesson by showing the monochords and explaining their construction –
and later had a student skeptically ask how much it actually cost to build a monochord. He was
very pleased by my answer that each one cost about $7 (and that’s with the “fancy” tuner –
replacing this with a screw reduces the cost to $4.) I had the students explore the octave and fifth
on the monochord very briefly at the beginning of the lesson, but otherwise this lesson did not
include actual use of the monochords.
Due to constraints of time and student ability level, this lesson had less to do with the
musical scale then the other three lessons in this mini-unit, but proved to be a great review of
multiplying and dividing fractions. Still, the musical scale served as a good motivator in the
beginning, and the big picture concept of the inability to change keys in JTS thrown in at the end.
I ran out of time to give this big picture concept justice and would definitely spend more time on
this in my own classroom where this lesson could in theory be followed by Lesson Four
exploring equal temperament. Despite my rush, I was surprised to find that several students
were able to successfully explain why JTS is only a problem for fretted instruments by the end of
the class.
The students had not seen fractions since the previous school year and were very rusty in
their skills. I had a few students who in the beginning attempted to flat out refuse to do the
73
worksheet because of their distaste for fractions, and others who tried to get me to allow them to
“simply” convert to decimals and then do the calculations. Partially due to the cooperation I was
afforded due to my “guest teacher” status, and partially, I think, because these fraction
computations had some context and produced answers that revealed some type of pattern (so that
as the students worked through a few they became more motivated to see where the other
answers fell in the pattern), students overcame their distaste for the dreaded fraction
computations. The student who seemed most despairing at first very happily explained one
solution when I had the students share their methods and answers at the end of class. As he was
leaving, he commented to me that the fractions made sense after all.
I used a parallel whole number problem ( 2  ?  10 ) to help the students figure out the
necessary operation needed to find the missing fraction, which I then had to refer to numerous
times after the students started working on the problems in pairs. This seemed to help connect
the “overwhelming” fractions to something the students considered “easy.” However, at least
half the class started solving the fraction problems by incorrectly dividing the equivalent of 2 by
10 instead of 10 by 2. I used this as an opportunity to examine the non-commutative quality of
division and subtraction. If time had allowed, I would have liked to diverge on this topic for
much longer as the students didn’t seem to quite understand the significance of their error. I
think many just “did it the other way” because I told them too. It would have also been
interesting to take more time to examine how doing the problems backwards results in the
reciprocals of the correct answers. Several students noticed this, an observation that for many
fed their (false) belief that the order of the division didn’t really matter as the answers were
“basically the same.”
74
Overall I think this lesson offers a motivating context for reviewing multiplication and
division of fractions, with some elementary equation solving and mathematical analysis of the
musical scale as bonus features.
75
F. Lesson Plan Four: Building the Equal Tempered Scale
With the limitations of an unequally spaced scale explored in lesson three as a motivator,
students are introduced to the equal tempered scale in lesson four. The irrational number that
defines this scale ( 12 2 ) is defined as the tangible value that, when multiplied by itself twelve
times, produces the frequency value of the octave (2). Using students' knowledge of frequency
ratios from lessons one to three, connections are made between multiplying an equal frequency
ratio by itself twelve times, exponents and root notation.
Students use their experience in calculating and measuring string lengths for the just
tempered scale to do the same for the more difficult equal tempered scale. At this point, students
have enough experience with the scale to have a good sense of if their calculations are producing
reasonable string lengths and so are hopefully able to come up with a method for calculating
accurate string lengths on their own. In order to measure accurately, students need to round to
the nearest millimeter and measure lengths longer than the given ruler.
Once the scale is built, students are given a chance to experiment playing songs on the
monochords in preparation for the next lesson. Finally, the equal tempered scale is offered as a
visual example of a geometric progression and the distinction between arithmetic and geometric
progressions is explored.
76
Lesson Plan Four Outline
TOPIC
Building the Equal Tempered Scale
GOALS
Skills:
Students will be able to define 12-tone scale.
Students calculate lengths of string for 12 toned ETS.
Students will round to the nearest tenth.
Students will measure lengths to the nearest millimeter, including lengths longer than the 30 cm
ruler provided.
Thinking:
Students will help derive equal tempered scale.
Students will observe visually and "hear" geometric progression.
Students will be introduced to root notation and see how it relates to exponents
TIME PERIOD and LEVEL/PREREQUISITES
One 50 minute class period
This lesson should follow Lesson Three in which students examine the unequal Just Tempered
Scale and Lesson Two in which they build the less complicated JTS.
SUPPLIES
Ten monochords, masking tape below strings
Violin, if possible, provided by teacher or a student
Group Worksheet / Transparency
Homework Worksheet
Rulers with cm
Calculator
ASSESSMENT
Have students complete worksheet for homework calculating lengths needed for equally
tempered diatonic scale with a base length half of original string.
77
ACTIVITIES
1. Equal Temperament (10 minutes)
Share responses to homework re: problem w/ Just Tempered Scale's unequal intervals.
To fix this problem, a new scale was invented that still had 12 notes, but all intervals were equal.
Preserved the most basic interval – octave –with a frequency ratio of 2:1
Goal – Equal intervals between 12 notes that span an octave ( 12 )
C C# D D# E F F# G G# A A# B C
r
r r r r r
r r r
r r r
So frequency ratio times itself twelve times must equal octave:
1
r  r  r  r  r  r  r  r  r  r  r  r=
(Slightly different from
16
15
2
1
 r 12  2  r  12 2  2 12
12
2  1.0595
 1.066... , now every interval is the same.)
2. Build Equal Tempered Diatonic Scale (worksheet) (30 minutes)
Explain how to find string lengths for different notes, then, using worksheet, have students use
calculators to find string lengths and measure notes on monochords.
Carefully explain/review how to measure: Use centimeter side of ruler. Start at tuner. For 72
cm monochord, mark off 30 cm and 60 cm first. For 40 cm monochord, mark off 30 cm first.
Then hold finger on mark to stop vibrations and pluck string by tuner. (The note will be the
same no matter where you pluck between tuner and finger, but it is easier to pluck by tuner
because string is slightly raised.)
base string length
= frequency ratio
new string length
Algebraically:
s
 r  s  rx  s r  x
x
Or numerically, using “fact family”:
10
 2  10  5  2  10  2  5  10  2  5
5
10
 2  10  ?  2  10  2  ?  10  2  ?
?
base string length  frequency ratio = new string length
1
frequency ratios: 11 r1 r 2 r3 r 4 r 5 r6 r 7
notes: C C# D D# E F F# G
r8 r 9 r10 r 11
G# A A# B
( r  12 2  2 12 )
2
1
C
Students who finish early should try to figure out Twinkle or Happy Birthday on their monochord.
78
3. Geometric Series (10 min)
Present both type of series and ask students whether ETS is geometric or arithmetic. Explain
Equal Tempered Scale is a geometric series (also called a geometric progression) because each
term is the previous term multiplied by a given number, r.
The series 2, 4, 6, 8, ... is called an arithmetic series .
Visually (on a ruler), interval the same -- equally spaced.
The series 2, 4, 8, 16 ... is called a geometric series .
Visually (on a ruler), interval gets bigger and bigger: look at scale to see this pattern.
79
Building an Equal Tempered Scale
Names ______________________
______________________
______________________
Record your instrument’s letter and base string length.
Calculate the actual length of string needed to produce each note. (Round to the nearest
millimeter.) RECORD how your group performs these calculations in the space below
the table.
Mark this length using ruler on instrument.
Play your scale. If it doesn’t sound right, check your calculations.
If you have time, try to figure out how to play a common song on your instrument, or
make up your own song.
ACTIVITY
1.
2.
3.
4.
5.
Base String Length:
______
the length of the original string
Frequency Ratio =
the length of the new string
r
12
Label
0
1
2
3
4
5
6
7
8
9
10
11
0'
Note Interval
G
base
#
G
A
A#
B
C
C#
D
D#
E
F
F#
G'
Diatonic Scale
Frequency Ratio
r0 
 2
12
1
second
third
fourth
fifth
sixth
seventh
octave
r =
r2=
r3 =
r4=
r5 =
r6 =
r7 =
r8 =
r9 =
r 10 =
r 11 =
r 12 
0
12
22
1
12
2  1.0595
Calculation used
to determine string length
Actual Length of String
needed to produce note
1
=
=
=
=
=
=
=
=
=
=
=
 2
12
12
2
80
Building an Equal Tempered Scale
Name ____________________
HOMEWORK
1. Given a string whose base length is half your original string = _______, calculate the string
lengths needed to produce a diatonic scale. (Round to the nearest millimeter.)
Frequency Ratio =
the length of the original string
the length of the new string
r  12 2
2  1.0595
Calculation used
Actual Length of String
to determine string length needed to produce note
12
Label
0
1
2
3
4
5
6
7
8
9
10
11
0'
Note Interval
G
base
#
G
A
A#
B
C
C#
D
D#
E
F
F#
G'
Diatonic Scale
Frequency Ratio
r0 
 2
12
1
second
third
fourth
fifth
sixth
seventh
octave
r =
r2=
r3 =
r4=
r5 =
r6 =
r7 =
r8 =
r9 =
r 10 =
r 11 =
r 12 
0
1
=
=
=
=
=
=
=
=
=
=
=
 2
12
12
2
81
Building an Equal Tempered Scale
Names ______________________
______________________
______________________
Record your instrument’s letter and base string length.
Calculate the actual length of string needed to produce each note. (Round to the nearest
millimeter.) RECORD how your group performs these calculations in the space below
the table.
Using a ruler, mark off the lengths for each interval on your instrument.
Play your scale. If it doesn’t sound right, check your calculations.
If you have time, try to figure out how to play a common song on your instrument, or
make up your own song.
ACTIVITY (with ratios calculated)
1.
2.
3.
4.
5.
Base String Length:
base string length
= frequency ratio
new string length
______
base string length  frequency ratio = new string length
1
1
1
2
frequency ratios:
r r
notes: C C# D
4
3
r r
D# E
5
r r
F F#
Label Note Interval Diatonic Scale
Frequency Ratio
0
0
C
base
r 0  12 2  1
1
1
C#
r 1 = 12 2 =1.059
2
2
D second
r 2 = 12 2 =1.122
3
3
D#
r 3 = 12 2 =1.189
4
4
E
third
r 4 = 12 2 =1.260
5
5
F
fourth
r 5 = 12 2 =1.335
6
6
F#
r 6 = 12 2 =1.414
7
7
G fifth
r 7 = 12 2 =1.498
8
8
G#
r 8 = 12 2 =1.587
9
9
A sixth
r 9 = 12 2 =1.682
10
10
A#
r 10 = 12 2 =1.782
11
12
11
B
seventh
r 11 = 2 =1.888
12
0'
C' octave
r 12  12 2  2


























6
7
r
G
8
9
10
11
r r r
r
#
#
G A A B
2
1
Calculation used
to determine string length
(r 
12
1
12
22 )
C
Actual Length of String
needed to produce note
82
Building an Equal Tempered Scale
Names ______________________
______________________
______________________
Record your instrument’s letter and base string length.
Calculate the actual length of string needed to produce each note. (Round to the nearest
millimeter.) RECORD how your group performs these calculations in the space below
the table.
Using a ruler, mark off the lengths for each interval on your instrument.
Play your scale. If it doesn’t sound right, check your calculations.
If you have time, try to figure out how to play a common song on your instrument, or
make up your own song.
ACTIVITY: Key
1.
2.
3.
4.
5.
Base String Length:
base string length
= frequency ratio
new string length
______
base string length  frequency ratio = new string length
1
1
1
2
frequency ratios:
r r
notes: C C# D
4
3
r r
D# E
5
r r
F F#
Label Note Interval Diatonic Scale
Frequency Ratio
0
0
C
base
r 0  12 2  1
1
1
C#
r 1 = 12 2 =1.059
2
2
D second
r 2 = 12 2 =1.122
3
3
D#
r 3 = 12 2 =1.189
4
4
E
third
r 4 = 12 2 =1.260
5
5
F
fourth
r 5 = 12 2 =1.335
6
6
F#
r 6 = 12 2 =1.414
7
7
G fifth
r 7 = 12 2 =1.498
8
8
G#
r 8 = 12 2 =1.587
9
9
A sixth
r 9 = 12 2 =1.682
10
10
A#
r 10 = 12 2 =1.782
11
12
11
B
seventh
r 11 = 2 =1.888
12
0'
C' octave
r 12  12 2  2


























6
7
r
G
8
9
10
11
r r r
r
#
#
G A A B
2
1
Calculation used
to determine string length
(r 
12
1
12
22 )
C
Actual Length of String
needed to produce note
40
72
37.8
68
35.7
64.2
33.6
60.6
31.7
57.1
30
53.9
28.3
50.9
26.7
48.1
25.2
45.4
23.8
42.8
22.4
40.4
21.2
38.1
20
36
83
Reflections on Lesson Four
I taught this lesson to a borrowed group of 25 sixth graders at a Catholic school. They
were younger than the targeted audience, but very enthusiastic and cooperative. In theory this
lesson would follow Lesson One, Two and Three. In practice, I taught it as a stand alone lesson.
To accommodate this and the younger age group, I filled in the decimal equivalents of the twelve
ratios in the student worksheet.
It turned out that without the previous experience of building the just tempered scale, the
students needed a lot more time than I had allowed to build the more complicated equal tempered
scale. Because the students were using calculators and I had figured out the decimal equivalents
to the ratios for them on the worksheet, I thought finding the string lengths would be fairly
straightforward. As it turned out, the decimals were pretty overwhelming to the students and
they needed a LOT of help with the process of rounding to the nearest tenth. The actual
measuring of the lengths was also much more challenging for the students than I anticipated. As
a result, only a few groups actually ended up with correct scales on their monochords by the end
of the class.
In addition, I only had time to very briefly touch on the geometric series aspect of the
lesson – maybe someday when they revisit this subject they will remember their monochords and
understand the connection more clearly. Because many of them didn’t accurately build the scale,
and because all 6th graders at their school conveniently take guitar, I referenced the fret spacing
of the guitar instead of their monochords to visually represent the geometric progression. Even
without the problems with their scale building, I think this guitar reference should be included
because this instrument is so popular and familiar to students.
84
The students seemed sufficiently impressed by the strange number, r, which could be
multiplied by itself 12 times and result in friendly old 2. I think the students hadn’t been
introduced to irrational numbers before and I was pleased to offer this concrete, useful rational as
their first experience. The teacher whose class I was borrowing had told me before my lesson
that she always tells students that irrationals are numbers that don’t follow patterns, that are
“crazy” i.e. irrational. I didn’t tell her that I take issue with this presentation of irrationals
because I think it leads students to the mistaken conclusion that irrationals are “bad” and
confusing numbers. (For similar reasons, I take issue with the name “improper” to fractions
greater than 1.) I prefer the more straightforward definition that they are simply not rational, and
like to reinforce that despite the implications of their name, irrationals can be very useful,
important numbers. As a result, I was glad to provide the musical context that shows the need /
use of
12
2 and also makes some sense of its value.
I think this lesson would be much more effective if taught after the first three lessons.
But even without many of the details falling properly into place, the students were excited about
the monochords and convinced that there was a lot of math involved in music.
85
G. Lesson Plan Five: Transposing Happy Birthday
In this lesson, students are given instruction in actually playing songs on the monochords
whose scale they calculated and examined in lessons one to four. Once students have all
practiced playing Happy Birthday, they are shown how to use a simple transposition formula and
modular arithmetic to change the key of the song. In doing this, students are introduced to a
basic algebraic equation and also shown a practical application of modular arithmetic, as well as
gaining understanding of what it means to change musical keys.
Once students have practiced changing keys several times, they are asked to articulate
exactly what happens when the key of a musical piece is changed by examining what changes
and what stays the same. The essence of Happy Birthday is boiled down to the space between
the notes, with the key being the factor that determines on which note the song begins. This is
similar to the relationship of a line to its slope and y-intercept, respectively.
86
Lesson Plan Five Outline
TOPIC
Transposing Happy Birthday
GOALS
Skills:
Students will use basic modular arithmetic.
Students will use basic transposition formula.
Thinking:
Students will observe that changing keys changes the pitch but retains the melody.
Students will observe that the melody of a song is created by the relationship between notes and
that this relationship between notes can be preserved while moving up and down keyboard.
TIME PERIOD and LEVEL/PREREQUISITES
One 50 minute class period
This lesson should follow Lesson Four and its homework, so that students have done calculations
to build two octave monochord.
SUPPLIES
Ten monochords with C string
Transparency with Happy Birthday Transposed
Practice playing Happy Birthday
ASSESSMENT
Transpose Twinkle, Twinkle Little Star to different keys.
87
ACTIVITIES
1.Construct Second Octave (10 minutes)
Compare homework with other group members and mark measurements on monochord for
second octave, labeling 0', 1', 2'…11', 0'' (one octave higher). Have students check by playing all
25 notes.
2. Happy Birthday (10 minutes) TRANSPARENCY
Play happy birthday on monochord, then teach students to play happy birthday.
G G A G C' B / G G A G D' C' / G G G' E' C' B A / F' F' E' C' D C' / (C major)
(Note: the monochord won't actually be tuned to C-major –unless there is a student or teacher
capable of tuning them properly! Notes and keys are included to show how they change but
melody remains the same.)
3. Transpose: (25 minutes) TRANSPARENCY
Now we will transpose this song , or change its key.
The key that a piece of music is in is the set of notes that are allowed and expected
to be used because they generally sound "good" together.
Given any set, S of tones in an equal tempered scale, n, one may apply a transposition, Tk ,
where Tk x   x  k mod n for all tones x in S.
We have n = 12 notes, mod 12 means that the only the numbers 0 to 11 are allowed, as in our
scale – modular arithmetic.
To transpose up k = 2 notes, formula is T2 x   x  2mod 12 (D major)
Have students play on monochords.
To transpose up k = 7 notes, formula is T7 x   x  7 mod 12 (G major)
Have students play on monochords.
Ask students to transpose k = 10 notes. Formula is T10 x   x  10mod 12 (B major)
Have students play on monochords.
*
C Major
D Major (F# C#)
G Major (F#)
B Major (Bb Eb)
C D
D E
G A
Bb C
*
E F GA
F# G A B
B C DE
D Eb F G
(* = SEMITONE)
B C
C# D k = 2
F# G k = 7
A Bb k = 10
4. Patterns / Melody: (5 minutes)
Ask students what makes the melody of Happy Birthday? How could we recreate it ?
The interval between consecutive notes (# semitones) is constant in any key.
0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5 /
0 +2 -2 -5 –1 -4 0 +2 –2 +7 –2 –5 0 +12 –3 –4 –1 –2 +8 0 -1 –4 +2 –2
To transpose must have equal tempered scale
All transpositions won't work with Just Tempered Scale because not equally spaced.
88
Lesson Five: Transposing Happy Birthday
Transparency
HAPPY BIRTHDAY
_____________________________________________________________
0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5 /
G G A G C B / G G A G D C / G G G' E C B A / F F E C D C /
(C major)
_____________________________________________________________
0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5 /
T2 
2 2 4 2 7 6 / 2 2 4 2 9 7 / 2 2 2' 11 7 6 4 / 0' 0' 11 7 9 7 /
A A B A D C#/ A A B A E D / A A A' F# D C# B/G G F# D E D /
(D Major: C# F#)
_____________________________________________________________
0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5 /
T7 
7 7 9 7 12 11/ 7 7 9 7 14 12/ 7 7 7' 16 12 11 9/ 17 17 16 12 14 12/
mod127 7 9 7 0' 11/ 7 7 9 7 2' 0' / 7 7 7' 4' 0' 11 9/ 5' 5' 4' 0' 2' 0'/
DD E D G F#/DD E D A G / D D D' B G F# E / C' C' B G A G /
(G major: F#)
_____________________________________________________________
0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5 /
T10 10 10 12 10 15 14/10 10 12 10 17 15/10 10 10' 19 15 14 12 / 20 20 19 15 17 15/
mod1210 10 0' 10 3' 2'/ 10 10 0' 10 5' 3' /10 10 10' 7' 3' 2' 0' / 8' 8' 7' 3' 5' 3'/
F F G' F A#'A'/F F G' F C' A# /F F F' D' A#' A' G'/D#' D#'D'A#' C'A#'/
(B Major: Bb Eb)
_____________________________________________________________
89
Transposing Happy Birthday
Activity
Name __________________
HAPPY BIRTHDAY
C major:
0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5 /
G G A G C B / G G A G D C / G G G' E C B A / F F E C D C /
Transpose to D Major (C# F#):
C Major: 0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5/
T2 
Transpose to G major (F#):
C Major: 0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5/
T7 
mod12
Transpose to B Major (Bb Eb):
C Major: 0 0 2 0 5 4 / 0 0 2 0 7 5 / 0 0 0' 9 5 4 2 / 10 10 9 5 7 5/
T10
mod12
90
Transposing Twinkle Twinkle Little Star
Homework
Name __________________
Twinkle Twinkle Little Star
C major:
C C G G A A G/ F F E E D D C / G G F F E E D/
Transpose to D Major (C# F#):
C Major:
T2 
D D A A B B A/ G G F# F# E E D/ A A G G F# F# E
Transpose to G major (F#):
C Major:
T7 
mod12
G G D D E E D/ C C B B A A G / D D C C B B A
Transpose to B Major (Bb Eb):
C Major:
T10
mod12
Bb Bb F F G G F/Eb Eb D D C C Bb/F F Eb Eb D D C
List the spacing between consecutive notes needed to produce the melody for Twinkle, Twinkle
Little Star in ANY key.
+7 0 +2 0 –2 / -2 0 –1 0 –2 0 –2 / +7 0 –2 0 –1 0 –2
91
H. Lesson Plan Six: Musical Cents
Building on the students experience with different musical scales gained in lessons one to
four, students are given the tools needed to compare these musical scales in this lesson. The
measurement unit, Cents, is introduced along with the concept that with units of measurement
"necessity is the mother of invention." Building on students' knowledge of the equal tempered
scale, the cent is defined as one hundredth of an equal tempered semitone. Students are then
shown how logarithms are used to simplify the process of calculating the exponent based cent.
Instead of just being an abstract inverse of exponents, the need for and meaning of logarithms is
motivated by this connection to musical cents.
Students then perform the necessary calculations to change the linear frequency ratios
into the geometric cents so that accurate comparison of tone is possible. After converting the
scales to their cent equivalents, students are asked to compare the different scales, and based on
the concepts explored in lessons one to five, explain their advantages and disadvantages.
Students should think of the scale in terms of their mathematical and musical value. Hopefully
some will argue for the mathematical simplicity of the equal tempered scale and its musical
flexibility, while others will remember that without the constraints of keys or frets, the just
tempered scale and the human mind can play pure tones and change keys at the same time.
92
Lesson Plan Six Outline
TOPIC
Musical Cents
GOALS
Skills:
Students will use formula to change decimal value into musical cents.
Students will get basic introduction to relationship between exponents and logarithms.
Thinking:
Students will compare different tuning systems: similarities, advantages, disadvantages.
Students will see application of logarithms to simplify calculations.
Students will consider how measurement units develop.
TIME PERIOD and LEVEL/PREREQUISITES
One 45 minute class period
Students should have some familiarity with more than one tuning (Just Tempered: Lesson Two,
and Equal Tempered: Lesson Four)
SUPPLIES
String Instrument if available or Monochord
Transparencies of PS vs. JTS vs. ETS and VGS vs. ETS: Freq Ratio and Decimal , Cents
Worksheets for activity and homework.
Calculators with logarithm function identified.
ASSESSMENT
For homework, students will translate another tuning system (Vincent Galilei's) to cents.
93
ACTIVITIES
1. Examine A in different between tuning systems (5 minutes)
Play Just Tempered vs. Equal Tempered A
Assuming a base note of C: Just Tempered A has a frequency ratio of
Equal Tempered A has freq. ratio of
5
3
9
 2
12
 1.6  1.67
9
 2 12  1.6817
2. Illustrate problem with using decimals to compare tuning systems (5 minutes)
Show transparency of table comparing five tuning systems and all messy decimals – hard to
make sense of all the numbers!
AND b/c scale is a geometric progression, can’t compare arithmetic differences.
Alexander Ellis invented a uniform measure, the cent, to compare scales in 1884.
12 notes in standard western scale, equally or quasi-equally spaced. Ellis gave each interval a
value of 100. In ETS, truly equally spaced,
C = 0 C# = 100 D = 200, D# = 300, E = 400, F = 500, etc. with 1200 cents in an octave.
3. Calculating a Cent (5 minutes)
To translate other tuning systems to cents, must first understand how scales are built.
In ETS, where r = 12 2 , to get from C =1 to C# MULTIPLY by r (geometric progression.)
Ellis's unit must also function as geometric progression:
Given a complete octave with freq. ratio 2, 2
If we multiply 2
1
1200
1
1200
is 1 out of 1200 parts of the octave.
1
 1200


by itself 1200 times we get a complete octave because  2 


1200
 2 . So
1
2 1200 is a cent.
4. Translating Frequency Ratios to Cents (30 minutes)
x
x
 1 
Any Freq. Ratio, r, can be thought of as some number of cents: r =  2 1200  or r = 2 1200


We can use logarithms to find the value of x because logarithms "undo" exponents:
8 = 2 3  log 2 8  3 . Taking the log produces the exponent needed to solve 2 x  8 .
Similarly, with the help of the change of base law:
 log r 
x
log 2 r 
 1200 log 2 r  x  1200 10   x  LOG r  3986  x
1200
 log 10 2 
In small groups, change PS, VGS or JTS into cents using formula, recording results on
transparency.
NOW we can compare the different tunings. Throughout history, people have invented units of
measurement specific to a given situation to help make sense of that situation.
94
Making Musical Cents
Name ____________________
ACTIVITY
1. Translate the frequency ratios in the table below into cents using the formula:
LOG r  3986  x
PYTHAGOREAN
ordinal
octave
second
third
fourth
fifth
sixth
JUST TEMPERAMENT
EQUAL TEMPERAMENT
note Freq. Decimal Cents Freq. Decimal Cents Freq.
ratio
ratio
ratio
1
1
1
1
1
C
2187
25
C#
1.0679
1.0417
2048
24
12
2
b
256
16
D
1.0535
1.0667
243
15
2
9
9
D
12
1.1250
1.1250
2
8
8
19683
75
D#
1.2014
1.1719
3
16384
64
12
2
b
32
6
E
1.1851
1.2000
27
5
4
81
5
E
12
1.2656
1.2500
2
F
4
3
1.3333
4
3
1.3333
F#
729
512
1.4047
45
32
1.4063
Gb
1024
729
1.4047
36
25
1.4400
G
3
2
1.5000
3
2
1.5000
G#
6561
4096
1.6018
25
16
1.5625
Ab
128
81
1.5802
8
5
1.6000
A
27
16
1.6875
5
3
1.6667
A#
59049
32768
1.8020
225
128
1.7578
Bb
16
9
1.7778
9
5
1.8000
243
128
1.8984
15
8
1.8750
 
 
 
 2
 2
 2
 2
 2
 2
 2
2
2
2
2
2
64
seventh B
octave
(Round to the nearest cent.)
C
4
12
12
12
12
12
12
12
Decimal Cents
1
1.0595
1.1225
1.1892
1.2599
5
1.3348
6
1.4142
7
1.4983
8
1.5874
9
1.6817
10
1.7818
11
1.8877
2
2. Do you think the Pythagorean or Just Tempered scale is closer to the Equal Tempered scale?
Justify your answer.
95
Making Musical Cents
Name ____________________
ACTIVITY: KEY
1. Translate the frequency ratios in the table below into cents using the formula:
LOG r  3986  x
PYTHAGOREAN
ordinal
octave
second
third
fourth
fifth
sixth
JUST TEMPERAMENT
EQUAL TEMPERAMENT
note Freq. Decimal Cents Freq. Decimal Cents Freq.
ratio
ratio
ratio
1
1
1
1
1
C
0
0
2187
25
C#
1.0679
1.0417
114
71
2048
24
12
2
b
256
16
D
1.0535
1.0667
90
112
243
15
2
9
9
D
12
1.1250
1.1250
204
204
2
8
8
19683
75
D#
1.2014
1.1719
318
275
3
16384
64
12
2
b
32
6
E
1.1851
1.2000
294
316
27
5
4
81
5
E
12
1.2656
1.2500
408
386
2
0
1.0595
100
1.1225
200
1.1892
300
1.2599
400
5
1.3348
500
6
1.4142
600
7
1.4983
700
8
1.5874
F
4
3
1.3333
498
4
3
1.3333
498
F#
729
512
1.4047
612
45
32
1.4063
590
Gb
1024
729
1.4047
588
36
25
1.4400
631
G
3
2
1.5000
702
3
2
1.5000
702
G#
6561
4096
1.6018
816
25
16
1.5625
773
Ab
128
81
1.5802
792
8
5
1.6000
814
A
27
16
1.6875
906
5
3
1.6667
884
A#
59049
32768
1.8020
1020
225
128
1.7578
977
Bb
16
9
1.7778
996
9
5
1.8000
1018
243
128
1.8984
1110
15
8
1.8750
1088
2
2
1200
2
2
1200
2
64
C
4
12
12
12
12
12
12
12
Decimal Cents
1
 
 
 
 2
 2
 2
 2
 2
 2
 2
seventh B
octave
(Round to the nearest cent.)
800
9
1.6817
900
10
1.7818
1000
11
1.8877
1100
2
1200
2. Do you think the Pythagorean or Just Tempered scale is closer to the Equal Tempered scale?
Justify your answer.
96
Making Musical Cents
Name ____________________
Homework
1. Translate the frequency ratios in the table below into cents using the formula:
LOG r  3986  x
(Round to the nearest cent.)
VINCENT GALILEI
ordinal note
octave
C
C# / Db
second
D
D#
ETS
Freq. Decimal Cents
Cents ratio
1
1
0
18
100
1.0588
17
2
18
200
  1.1211
17
/
Eb
300
third
E
400
fourth
F
500
F#
fifth
/
Gb
600
G
700
G# / Ab
800
A
900
A# / Bb
1000
seventh
B
1100
octave
C
1200
sixth
1718 3
1718 4
1718 5
1718 6
1718 7
1718 8
1718 9
1718 10
1718 11
1718 12
1.1871
1.2569
1.3308
1.4091
1.4920
1.5798
1.6727
1.7711
1.8753
1.9856
2. In your opinion:
What are the musical and/or mathematical advantages to Vincent Galilei's scale?
What are the musical and/or mathematical disadvantages to Vincent Galilei's scale?
97
Making Musical Cents
Name ____________________
Homework: KEY
1. Translate the frequency ratios in the table below into cents using the formula:
LOG r  3986  x
(Round to the nearest cent.)
VINCENT GALILEI
ordinal note
octave
C
C# / Db
second
D
D# / Eb
ETS
Freq. Decimal Cents
Cents ratio
1
1
0
0
18
100
1.0588
99
17
2
18
200
17  1.1211 198
300
18 3 1.1871 297
third
E
400
fourth
F
500
F#
fifth
/
Gb
600
G
700
G# / Ab
800
A
900
A# / Bb
1000
seventh
B
1100
octave
C
1200
sixth
17
18 4
17
18 5
17
18 6
17
18 7
17
18 8
17
18 9
17
18 10
17
18 11
17
18 12
17


















1.2569
396
1.3308
495
1.4091
594
1.4920
693
1.5798
792
1.6727
891
1.7711
990
1.8753
1089
1.9856
1188
2. In your opinion:
What are the musical and/or mathematical advantages to Vincent Galilei's scale?
What are the musical and/or mathematical disadvantages to Vincent Galilei's scale?
98
I. Lesson Plan Seven: Fundamental Theorem of Arithmetic
In this lesson students reexamine the use of an irrational ratio as the only way to produce
an equal tempered scale. Based on their knowledge of the musical scale gained in previous
lessons, students define in mathematical terms the musical problem of aligning the fifth interval
with the octave. Students use their knowledge of fractions and exponents to simplify this
problem to equating some power of two with a power of three.
Students then explore and try to explain why this task is impossible, in so doing laying
the groundwork for a proof of the fundamental theorem of arithmetic. Students are challenged to
explain how they are sure that no power of two will ever equal any power of three, and through
this effort students should see the need for mathematical proof versus argument by example.
After this exploration, students are walked through the proof of the fundamental theorem
of arithmetic, which is based on their knowledge of prime factorization. For homework, students
are asked to explain why the fifth and the octave cannot be perfectly aligned and in this
explanation to restate the proof of the FTA in their own words.
99
Lesson Plan Seven Outline
TOPIC
Fundamental Theorem of Arithmetic
GOALS
Skills:
Students will translate a musical situation into mathematical language.
Students will solve equation involving fractions and exponents.
Students will define consonance and dissonance and hear how these concepts are culturally
specific.
Students will name the seven basic musical intervals and see their derivation as ordinal (vs.
cardinal) numbers.
Thinking:
Students will explore application of Fundamental Theorem of Arithmetic.
Students will practice articulating proof of the FTA.
Students will consider more formal proof and summarize in their own words.
TIME PERIOD and LEVEL/PREREQUISITES
One 70 minute block.
Pre-calculus students with familiarity with fractions, exponents
SUPPLIES
Ten monochords tuned to match each other.
Violin or other string instrument, or keyboard if possible.
Identify any student who is a musician and willing to share skill during demonstrations.
ASSESSMENT
Ask students to explain in paragraph form why the octave and the fifth can't be matched up
exactly, and how the Fundamental Theorem of Arithmetic explains this impossibility. Students
should be sure to explain what the FTA is in their answer.
Answers should be one to two pages.
100
ACTIVITIES
1. The basic musical interval, an octave: (10 minutes)
Survey class for students who play musical instruments. Ask class a “musical” question:
Does anyone know which musical interval is easiest to recognize?
Have volunteer play A on monochord. Play 3 different notes on another monochord – one A
an octave higher. Ask students to identify the most consonant interval. Agree that this
interval is the octave.
(If possible place successive octaves on violin or keyboard)
Pass out 2 monochords of same length to groups of 6 and have students try to find octave on
the instrument and then measure length of base note and length of octave and compare.
Students should notice the base note is twice the octave. Any notes that are an octave apart
have frequencies in the ratio of 2:1.
Have students try to play an 8-note scale. (COLLECT INSTRUMENTS)
2. Consonance and Culture: (5 minutes)
Explain that each culture has musical intervals that are agreed to be consonant, that is the
collective ear of that culture has agreed these intervals sound nice.
Play a section of Chinese music and western classical music to illustrate different cultural
understandings of consonance. Have students share reactions to music.
3. Diatonic Scale: (10 minutes)
Play a consonant interval and a dissonant interval. Ask if anyone can name any of the other
intervals popular to western music.
The basic musical intervals in western music are the second, third, fourth, fifth, sixth, seventh
and octave, named for their respective order in the Diatonic Scale.
Review ordinal vs. cardinal numbers (where vs. how many).
Diatonic (dia-across, tonic- tone) so named because it is two tetrachords separated by whole tone.
(A tetrachord is four notes that span the interval of a perfect fourth.)
Ask for a volunteer who can play a scale on the violin. If no violinists, teacher can play.
4. Aligning the Octave and the Fifth (15 minutes)
Can't base a scale on an octave, b/c would get a scale with only one tone in successive octaves
(A, A', A'', A'''). After the octave, the fifth is agreed to be the most consonant. If you divide a
base note into 2/3 successively, get all different notes.
Demonstrate on monochord / violin and or keyboard.
Ask students what operation are we doing when we divide by 2/3 successively?
3
3
3
3
2
2
2
2
3 x
3  3  3  3 ... or 2  2  2  2 ... or  2 
Similarly, to get successive octaves, operation is
 12 x
The scale-makers goal is to figure out how many fifths will equal some number of octaves. For
instance, if 5 fifths equaled 3 octaves, then the musical scale would have 5 notes once the fifths
had been transposed down into a single octave.
DIAGRAM**
101
How many fifths do we need to get a perfect number of octaves? i.e. solve:
 32 x  12 y
5. IMPOSSIBLE! (10 minutes)
Hopefully, students can explain why and discuss – no power of 3 can equal a power of 2
32 x  2 y or 3m  2n
6. Proof (10 minutes)
The proof for the Fundamental Theorem of Arithmetic consists of two parts. First, every
natural number is shown to be the product of some sequence of primes. In other words, a prime
factorization exists for every number. Second, this prime factorization is shown to be unique.
To show existence, consider any composite number, n, which by definition has factors
other than itself and 1. Break down the factors until they are all prime and what is left is a prime
factorization of n = p1p2p3…pr.
24
4  6
2 3  2 2
To show uniqueness, imagine there are two prime factorizations for n:
n  p1  p2  p3 ...  pr and n  q1  q2  q3 ...  qr . By definition, p1 divides n, thus p1 divides
q1  q2  q3 ...  qr . In addition, all pi and qj are prime, so p1 must equal some qj. These paired
factors can be removed and then this process repeated to show that q1  q2  q3 ...  qr must be
exactly the same as p1  p2  p3 ...  pr .
24  2  2  2  3 and 24  q1  q2  q3 ...  qr (qj are prime)
2 24  2 q1  q2  q3 ...  qr  2  qj for some j
Repeat this argument to show that q1  q2  q3 ...  qr must equal 2  2  2  3 .
7. Summary and Homework (10 minutes)
And so…many scales, of two types, have been used over the years. One that preserves the fifth,
but has unequal intervals and is hard to transpose, another that fudges the fifth a bit to have equal
intervals.
See ASSESSMENT for homework.
102
J. Lesson Plan Eight and Nine: Continued Fractions
Lesson eight and nine build on the concept explored in lesson seven -- that the fifth
interval and the octave cannot be aligned perfectly -- by having the students explore the best
approximate rational solution to this problem. In doing this, students refresh their understanding
of irrational and rational numbers and then learn how to find the best rational approximation for
an irrational number using continued fractions. Students are given an overview of the meaning
and use of continued fractions in which the complexity of a fraction is defined and the idea of
"best approximation" examined. Students then explore how continued fractions are calculated
and how this process can be simplified into a fairly straight forward algorithm. Finally, students
interpret continued fraction notation and simplify to calculate different convergents.
In lesson nine, students build on this knowledge of continued fractions to find the actual
solution to the musical problem that motivated the exploration of continued fractions. Students
calculate the continued fraction approximation and examine different convergents to see what
they mean in terms of the musical scale. In doing this, the idea of "best approximation" is further
explored in the practical, concrete context of building musical instruments. Through the
continued fraction convergents, students see the mathematical justification for the use of a twelve
tone musical scale to align the fifth interval and the octave.
103
Lesson Plan Eight Outline
TOPIC
Continued Fractions
GOALS
Skills:
Students will be able to define irrational and rational number.
Students will be able to change an irrational decimal into a continued fraction.
Students will be able to interpret continued fraction notation.
Students will be able to find continued fraction convergents.
Thinking:
Students will understand use of continued fraction as approximation technique.
Students will consider concept of "best approximation."
Students will look for patterns to simplify cont fraction algorithm.
Students will consider a fraction's complexity.
TIME PERIOD and LEVEL/PREREQUISITES
One 50 minute class period
Algebra II / Pre-calculus students
SUPPLIES
Calculators with square root and reciprocal button.
ASSESSMENT
For homework, have students find first 5 numbers in the continued fraction for 3 . Write out
complete continued fraction, short hand form, and find first 5 rational approximations for 3 .
Explain which of these convergents is the most accurate and why someone might choose to use
one of the less accurate fractions.
104
ACTIVITIES
1. Review Irrational (10 minutes)
Ask students to volunteer some irrational numbers, and review definition:
An irrational number is a number that is "not rational" – it cannot be expressed as a
fraction. In its decimal form, an irrational number is non-terminating and does not repeat.
Ask students for an approximate value as decimal and fraction for 2 and  :
17
7
2  1.412135... 
 1.416   1.4
12
5
22
 3.142857142857143
  3.141592653589793 
7
2. Continued Fraction Algorithm (15 minutes)
Ask students if they have ever wondered how someone figured out these fractions.
Explain that using continued fractions, it is possible to find the best rational approximation (of a
given complexity) for any irrational number.
Go through algorithm:
Start with what we know: 1< 2 <2 so in rational form must be 1 + ? where ? is a fraction
1
1
1
1
 2=1+
 2 - 1 =  .4142135…=  x =
 x = 2.414213562
x
x
x
.414...
1
1
SOO 2 < x < 3 or x = 2 +
 2=1+
2  1y
y
1
1
1
 .414213562…=
y=
.414...
y
y
1
1
 y = 2.414213562 so 2 < y < 3 or y = 2 +
 2=1+
z
2  11
To continue, 2.414213562…= 2 +
2
z
As it turns out, with 2 , there is a pattern and the continued fraction repeats. Instead of writing
out whole fraction, write [1; 2, 2, 2, 2…]
Then, can truncate where ever we want to get rational equivalent of a certain complexity:
1
1
2 7
1 3
[1;2] = 1  = [1:2,2] = 1 
=1  5  1  
(Have students try some.)
1
2 2
5 5
2 2
2
3. Simplify Algorithm (15 minutes)
Have students try same method to get first 4 numbers in continued fraction for  .
Point out, or have student point out, that we can skip steps: whole number portion is first number
in cont. fraction. Subtract whole number from decimal, take reciprocal – whole number portion
is second number of cont. frac. Repeat.
105
4. Fraction Complexity / Accuracy in Approximating (10 minutes)
Using transparency of continued fraction for
2 and
 , explain "of a given complexity" -- 7
5
17
is the best with a
12
denominator less than 29. If you are willing to deal with unwieldy denominators, you can get
8119
super accurate -= 1.4142135516460547 is the best rational approx for denominators less
5741
than or equal to 13860 – and is accurate to the 7th decimal place (ten-millionth)!
is the best rational approximation for
Similarly with
2 with a denominator less than 12.
 , 22 is the best with a denominator up to 105 , 333 is the best up to 112 (VS
7
106
314 157

using the decimal to get a fraction 3.14 
which has a higher, and therefore less
100 50
22
manageable, denominator then
, but is less accurate.)
7
Continued fraction calculator on the WWW:
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfCALC.html
106
Square Root of Two:
2 = 1.4142135623730951 = [1; 2,2,2…]
Convergents:
1: 1/1 = 1
2: 3/2 = 1.5
2: 7/5 = 1.4
2: 17/12 = 1.4166666666666667
2: 41/29 = 1.4137931034482758
2: 99/70 = 1.4142857142857143
2: 239/169 = 1.4142011834319525
2: 577/408 = 1.4142156862745098
2: 1393/985 = 1.4142131979695431
2: 3363/2378 = 1.4142136248948695
2: 8119/5741 = 1.4142135516460547
2: 19601/13860 = 1.4142135642135642
2: 47321/33461 = 1.4142135620573204
2: 114243/80782 = 1.4142135624272733
2: 275807/195025 = 1.4142135623637994
2: 665857/470832 = 1.4142135623746898
2: 1607521/1136689 = 1.4142135623728213
2: 3880899/2744210 = 1.414213562373142
2: 9369319/6625109 = 1.414213562373087
2: 22619537/15994428 = 1.4142135623730964
2: 54608393/38613965 = 1.4142135623730947
2: 131836323/93222358 = 1.4142135623730951
Pi
pi = 3.141592653589793 = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 3 ...]
Convergents:
3: 3/1 = 3
7: 22/7 = 3.142857142857143
15: 333/106 = 3.141509433962264
1: 355/113 = 3.1415929203539825
292: 103993/33102 = 3.1415926530119025
1: 104348/33215 = 3.141592653921421
1: 208341/66317 = 3.1415926534674367
1: 312689/99532 = 3.1415926536189364
2: 833719/265381 = 3.141592653581078
1: 1146408/364913 = 3.141592653591404
3: 4272943/1360120 = 3.141592653589389
1: 5419351/1725033 = 3.1415926535898153
14: 80143857/25510582 = 3.1415926535897926
3: 245850922/78256779 = 3.141592653589793
107
What follows are the first 2 million digits of the square root of 2.
Actually, slightly more than 2 million digits are given here. These digits
were computed by Robert Nemiroff (George Mason University and NASA Goddard
Space Flight Center) and checked by Jerry Bonnell (University Space Research
Association and NASA Goddard Space Flight Center). There were computed
during spare time on a VAX alpha class machine over the course of a week. We
do NOT guarantee the accuracy of these digits. Although these digits have
been checked once we encourage others to check them as well. We believe
these are the most digits ever computed for the square root of two on or
before 1 April 1994,. If anyone is aware of more digits we ask them to
please alert us of their existence. We have computed at least 10 million
digits of the square root of two as well as several digits of the number e
and the square roots of other numbers. These are available on this mosaic
server (URL: http://antwrp.gsfc.nasa.gov/htmltest/rjn.html). We welcome
comments.
- Robert Nemiroff and Jerry Bonnell
The square root of two =
1.414213562373095048801688724209698078569671875376948073176679737990732478462
10703885038753432764157273501384623091229702492483605585073721264412149709993
58314132226659275055927557999505011527820605714701095599716059702745345968620
14728517418640889198609552329230484308714321450839762603627995251407989687253
39654633180882964062061525835239505474575028775996172983557522033753185701135
43746034084988471603868999706990048150305440277903164542478230684929369186215
80578463111596668713013015618568987237235288509264861249497715421833420428568
60601468247207714358548741556570696776537202264854470158588016207584749226572
2600208558446652145839
88939443709265918003113882464681570826301005948587040031864803421948972782906
41045072636881313739855256117322040245091227700226941127573627280495738108967
50401836986836845072579936472906076299694138047565482372899718032680247442062
92691248590521810044598421505911202494413417285314781058036033710773091828693
14710171111683916581726889419758716582152128229518488472089694633862891562882
76595263514054226765323969461751129160240871551013515045538128756005263146801
71274026539694702403005174953188629256313851881634780015693691768818523786840
52287837629389214300655869568685964595155501644724509836896036887323114389415
57665104088391429233811320605243362948531704991577175622854974143899918802176
24309652065642118273167262575395947172559346372386322614827426222086711558395
99926521176252698917540988159348640083457085181472231814204070426509056532333
39843645786579679651926729239987536661721598257886026336361782749599421940377
77536814262177387991945513972312740668983299898953867288228563786977496625199
66583525776198939322845344735694794962952168891485492538904755828834526096524
09654288939453864662574492755638196441031697983306185201937938494005715633372
05480685405758679996701213722394758214263065851322174088323829472876173936474
67837431960001592188807347857617252211867490424977366929207311096369721608933
70866115673458533483329525467585164471075784860246360083444911481858765555428
64551233142199263113325179706084365597043528564100879185007603610091594656706
76883605571740076756905096136719401324935605240185999105062108163597726431380
60546701029356997104242510578174953105725593498445112692278034491350663756874
77602831628296055324224269575345290288387684464291732827708883180870253398523
38122749990812371892540726475367850304821591801886167108972869229201197599880
70381854333253646021108229927929307287178079988809917674177410898306080032631
18164279882311715436386966170299993416161487868601804550555398691311518601038
63753250045581860448040750241195184305674533683613674597374423988553285179308
960373898915173195874134428817842125021916951875593444387
108
A page of Pi
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164
0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172
5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975
6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482
1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436
7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381
8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277
0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342
7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235
4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837
2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035
2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904
2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787
6611195909 2164201989 3809525720 1065485863 2788659361 5338182796 8230301952
0353018529 6899577362 2599413891 2497217752 8347913151 5574857242 4541506959
5082953311 6861727855 8890750983 8175463746 4939319255 0604009277 0167113900
9848824012 8583616035 6370766010 4710181942 9555961989 4676783744 9448255379
7747268471 0404753464 6208046684 2590694912 9331367702 8989152104 7521620569
6602405803 8150193511 2533824300 3558764024 7496473263 9141992726 0426992279
6782354781 6360093417 2164121992 4586315030 2861829745 5570674983 8505494588
5869269956 9092721079 7509302955 3211653449 8720275596 0236480665 4991198818
3479775356 6369807426 5425278625 5181841757 4672890977 7727938000 8164706001
6145249192 1732172147 7235014144 1973568548 1613611573 5255213347 5741849468
4385233239 0739414333 4547762416 8625189835 6948556209 9219222184 2725502542
5688767179 0494601653 4668049886 2723279178 6085784383 8279679766 8145410095
3883786360 9506800642 2512520511 7392984896 0841284886 2694560424 1965285022
2106611863 0674427862 2039194945 0471237137 8696095636 4371917287 4677646575
7396241389 0865832645 9958133904 7802759009 9465764078 9512694683 9835259570
9825822620 5224894077 2671947826 8482601476 9909026401 3639443745 5305068203
4962524517 4939965143 1429809190 6592509372 2169646151 5709858387 4105978859
5977297549 8930161753 9284681382 6868386894 2774155991 8559252459 5395943104
9972524680 8459872736 4469584865 3836736222 6260991246 0805124388 4390451244
1365497627 8079771569 1435997700 1296160894 4169486855 5848406353 4220722258
2848864815 8456028506 0168427394 5226746767 8895252138 5225499546 6672782398
6456596116 3548862305 7745649803 5593634568 1743241125 1507606947 9451096596
109
Homework
3 = 1.7320508075688772 = [1; 1, 2, 1, 2…]
This CF ends with a repeating pattern: 1, then 1, 2 repeating for ever
1
1
1
1
1
2
1
1
2
Convergents:
1: 1/1 = 1
1: 2/1 = 2
2: 5/3 = 1.6666666666666667
1: 7/4 = 1.75
2: 19/11 = 1.7272727272727273
1: 26/15 = 1.7333333333333334
2: 71/41 = 1.7317073170731707
1: 97/56 = 1.7321428571428572
2: 265/153 = 1.7320261437908497
1: 362/209 = 1.7320574162679425
2: 989/571 = 1.7320490367775831
1: 1351/780 = 1.7320512820512821
2: 3691/2131 = 1.7320506804317221
1: 5042/2911 = 1.732050841635177
2: 13775/7953 = 1.73205079844084
1: 18817/10864 = 1.7320508100147276
110
K. Lesson Plan Nine Outline
TOPIC
Approximating the Perfect Fifth
GOALS
Skills:
Students will translate musical problem into algebraic equation.
Students will use continued fractions to approximate the perfect fifth.
Students will find convergents of a given continued fraction.
Students will use this fraction approximation to determine the number of notes in the scale and
which notes is the best approximate for the fifth.
Thinking:
Students will consider what it means to be the "best" approximation.
Students will consider how the scale is based on the musical fifth.
TIME PERIOD and LEVEL/PREREQUISITES
One 55 minute class period
Algebra II / Pre-calculus students: experience with manipulating exponents
This lesson should follow Lesson Eight about continued fractions, and Lesson Seven about the
Fundamental Theorem of Arithmetic and irreconcilable octave and fifth
(See pages 17 to 24 for background.)
SUPPLIES
monochords
Worksheet
ASSESSMENT
For homework, students will determine the number of notes that would result from different
convergent fractions and which of these notes is the musical fifth.
111
ACTIVITIES
1. Review problem with Octave and Fifth (5 minutes)
Ask students to explain the problem with basing the musical scale on the perfect octave and fifth.
(Play octave and fifth on monochord)
(Successive fifths can never equal successive octaves b/c 2 x  3 y b/c FTA)
To solve this problem, the fifth was tampered with a little bit and the octave split into 12 equal
notes. (Play scale)
But why 12? Why not 11 or 13?
Before Equal Temperament, 12 had already become the accepted number of notes – so 12 was
the logical choice of musicians – but is it also the logical choice of mathematicians seeking the
best approximate solution to the problem?
2. Set up problem (review) (10 minutes)
The frequency ratio for an octave is 12 (half the original string sounds the octave—show on
monochord).
The frequency ratio for a musical fifth is 32 ( 23 of the original string sounds a fifth – show on
monochord).
The unachievable goal is to find some number of successive fifths that will equal some other
number of successive octave: 23  23  23  23 ...  12  12  12  12 ...
Then, each of the successive fifths will be a different note until you get back to the first one, and
the number of fifths needed will be the number of different notes in the scale.
Show on imaginary musical string where 13 "fifths" equal 8 "octaves"
____F____F____F____F____F____F____F____F____F____F____F____F____F
_______O_______O_______O_______O_______O_______O_______O_______O
3. What equation can we write to represent this problem? (5 minutes)
x
3
2
   
2
1
y
simplified: 3 x  2 y 1 or 3  2
y 1
x
m
or 3  2 n
112
4. Use continued fractions to find the BEST approximation: (15 minutes)
1
1  mn  2  mn  1 
y
1
Substituting this value for x in 3  2  3  2
x
So 1< y <2  y  1 
1
y
 2 2
1
1
1
y
y
3
3

 2y  2    .
2
2
1
. Substituting for y in the continued fraction: x  1 
z
1
y
1
1
1
1
1
1
1
z
.
z
3
4  3z
3
4
 3 z  3  3z
Substituting for y in 2     2          
       . So
2
3 2
2
3
2
2 2
1
1
1< z < 2  z  1  and the continued fraction becomes: x  1 
etc.
1
w
1
1
1
w
5. Have students find first 6 convergents of [1; 1, 1, 2, 2, 2, 3, 1, 5, 1, 3, 1, 1…] (15 min)
1: 1/1 = 1
1: 2/1 = 2
1: 3/2 = 1.5
2: 8/5 = 1.6
2: 19/12 = 1.5833333333333332
3: 65/41 = 1.5853658536585366
1: 84/53 = 1.5849056603773585
5: 485/306 = 1.5849673202614378
1: 569/359 = 1.584958217270195
3: 2192/1383 = 1.5849602313810555
1: 2761/1742 = 1.5849598163030997
1: 4953/3125 = 1.58496
6. What can we conclude? (5 minutes)
19
12
12
12
7
12
7
12
7
12
3  2  3  2  2  3  2 2   2
The seventh (of twelve) note is almost a perfect fifth.
3
2
Turns out 12 IS the best number of tones for a scale in which the goal is to equate the octave
with the fifth as well as possible. To approximate any better, we would need 41 tones which is
unrealistic (although musicians have tried)…
113
Lesson Nine: Aligning the Octave and the Fifth
HOMEWORK
Name __________________
A picture of how 12 fifths equal 7 octaves:
_____F_____F_____F_____F_____F_____F_____F_____F_____F_____F_____F_____F
_________O_________O_________O_________O_________O_________O_________O
m
If the continued fraction solution for 3  2 n is truncated at mn  [1; 1, 1, 2, 2], the
convergent fraction is 19
. This results in a 12 tone scale with the seventh note
12
approximating a perfect fifth..
If the continued fraction is truncated at a different point, we could make a different
equal tempered scale.
For each convergent listed below, determine how many notes would be in the
resulting scale and which of these notes would best approximate a perfect fifth.
32
m
n
m
n
 [1; 1, 1, 2, 2, 2, 3, 1, 5, 1, 3, 1, 1…]
Convergent Fraction
Total Notes
in Scale
Best approximate for
a Fifth
12
7/12  seventh note
[1]  1/1
[1; 1]  2/1
[1; 1, 1]  3/2
[1; 1, 1, 2]  8/5
[1; 1, 1, 2, 2]  19/12
[1; 1, 1, 2, 2, 2, 3]  65/41
[1; 1, 1, 2, 2, 2, 3, 1]  84/53
Extra Fun: (OPTIONAL)
What is the next convergent for this continued fraction? [1; 1, 1, 2, 2, 2, 3, 1, 5]
How many notes would this scale have and which note would approximate the fifth?
114
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117