Euclid's Algorithm and Musical Intervals
Use Euclid's algorithm to get a rational approximation to the
number of major thirds in an octave.
Age
16 to 18
Challenge level
Problem
How many major thirds are there in an octave on a musical scale?
Going back in history, before the discovery of logarithms, some theorists used Euclid's algorithm to find the answer to this question.
A rational approximation ${m\over n}$ for the relationship between 5/4, the musical interval called the major third, and the octave 2/1, is given by $$ \left({5\over 4}\right)^m \approx \left({2\over 1}\right)^n, $$ where $m$ and $n$ are integers. Using Euclid's algorithm show that ${m\over n}={28\over 9}$ gives a first approximation and find three closer rational approximations.
In the articles
Euclid's Algorithm and
Approximations, Euclid's Algorithm and Continued Fractions you
can find out about this method and also that Euclid's algorithm can
be used not only for integers but for any numbers.
[See also the problems
Tuning and Ratio and
Rarity. The set of three problems on mathematics and music was
devised by Benjamin Wardaugh who used to be a member of the NRICH
team. Benjamin is now doing research into the history of
mathematics and music at Oxford University and his article Music
and Euclid's Algorithm should help you with this
problem.]
Getting Started
Euclid's algorithm can be used not only for integers but for any two numbers.
You could use the same method for this problem as given in the article Approximations, Euclid's Algorithm and Continued Fractions for finding rational approximations to $\pi$. In this method you write the process down in terms of continued fractions.
Alternatively, see the article by Benjamin Wardhaugh in the Plus magazine entitled Euclid's Algorithm and Music which shows a method for solving this problem. The examples given there use different numbers. If you use this method it might be easier if you write down your working using the factorizations of the numbers involved, rather than writing them out in full.
Student Solutions
Here is another beautifully explained solution from Andrei of Tudor Vianu National College, Bucharest, Romania:
$$ \left({5\over 4}\right)^m = \left({2\over 1}\right)^n .$$ This can be written equivalently: $$m\log {5\over 4} = n \log 2$$ or $${m\over n} = {\log 2 \over \log 5 - log 4} = 3.10628372 \quad (1).$$ Now, I shall use Euclid's algorithm to find the first 4 rational approximations of: $${\log 2 \over \log 5 - log 4}.$$ For the first approximation, I write: $$3.10628372 = 3 + {1\over {1\over 0.10628372}} \approx 3 + {1\over 9.408778692}\quad (2).$$ So, the first approximation is $${m\over n} \approx 3 + {1\over 9} = {28\over 9} = 3.111111111....$$ Now, for the second approximation I have: $$3 + {1 \over \displaystyle 9 + {1\over \displaystyle {1\over \displaystyle 0.408778692}}} = 3 + {1 \over \displaystyle 9 + {1\over \displaystyle 2.446311463}}$$ The second approximation for $m/n$ is: $${m\over n} \approx 3 + {1\over {9 + {1 \over \displaystyle 2}}} = {59\over 19} \approx 3.105263158.$$ For the third approximation, I obtain: $$3 + {1 \over \displaystyle 9 + {1\over \displaystyle 2 + {1\over \displaystyle {1\over \displaystyle 0.446311463}}}} = 3 + {1 \over \displaystyle 9 + {1\over \displaystyle 2 + {1\over \displaystyle 2.240587757}}}.$$ and consequently $${m\over n} \approx 3 + {1\over \displaystyle 9 + {1\over \displaystyle 2 + {1\over \displaystyle 2}}} = 3 + {1\over \displaystyle 9 + {2\over \displaystyle 5}}= {146\over 47} \approx 3.106382979.$$ Then the fourth approximation for m/n is: $${m\over n} \approx 3 + {1 \over \displaystyle 9 + {1\over \displaystyle 2 + {1\over \displaystyle 2 + {1\over \displaystyle 4}}}} = {643\over 207} \approx 3.106280193.$$ I see that using continued fractions I come nearer to the given real number by rational numbers greater and smaller than the number: the first and third approximations are greater than $m/n$ and the second and the fourth are smaller than the initial number.
This is a natural thing. I arrived to the first approximation considering, in relation (2) a smaller denominator: $${m\over n}\approx 3 + {1\over 9.408778692} < 3 + {1\over 9} = {28\over 9}.$$ Now, I shall do the same thing for the second approximation: $${m\over n} \approx 3 + {1\over 9 + {1\over \displaystyle 2.446311463}} > 3 + {1\over 9 + {1\over \displaystyle 2}}.$$ So, the second approximation is smaller than the initial number.
In a similar manner, the odd-order approximations are greater than $m/n$, but they form a decreasing series. The even-order approximations are smaller than $m/n$, and they form an increasing series.
Teachers' Resources
See the problem Tuning and Ratio in which you have to find a decimal approximation for this ratio using logarithms. Here you must find an approximation in the form of a ratio of two integers without using logarithms.
You could use the same method as given for finding rational approximations to $\pi$ in the article Approximations, Euclid's Algorithm and Continued Fractions.