Euclid's algorithm and musical intervals

Here is another beautifully explained solution from Andrei of Tudor Vianu National College, Bucharest, Romania:

$${\left(\frac{5}{4}\right)}^m={\left(\frac{2}{1}\right)}^n.$$ This can be written equivalently: $$m\log \frac{5}{4}=n\log 2$$ or $$\frac{m}{n}=\frac{\log 2}{\log 5-log4}=3.10628372(1).$$ Now, I shall use Euclid's algorithm to find the first 4 rational approximations of: $$\frac{\log 2}{\log 5-log4}.$$ For the first approximation, I write: $$3.10628372=3+\frac{1}{\frac{1}{0.10628372}}\approx 3+\frac{1}{9.408778692}(2).$$ So, the first approximation is $$\frac{m}{n}\approx 3+\frac{1}{9}=\frac{28}{9}=3.111111111....$$ Now, for the second approximation I have: $$3+\frac{1}{{\displaystyle 9+\frac{1}{{\displaystyle \frac{1}{{\displaystyle 0.408778692}}}}}}=3+\frac{1}{{\displaystyle 9+\frac{1}{{\displaystyle 2.446311463}}}}$$ The second approximation for $m/n$ is: $$\frac{m}{n}\approx 3+\frac{1}{9+\frac{1}{{\displaystyle 2}}}=\frac{59}{19}\approx 3.105263158.$$ For the third approximation, I obtain: $$3+\frac{1}{{\displaystyle 9+\frac{1}{{\displaystyle 2+\frac{1}{{\displaystyle \frac{1}{{\displaystyle 0.446311463}}}}}}}}=3+\frac{1}{{\displaystyle 9+\frac{1}{{\displaystyle 2+\frac{1}{{\displaystyle 2.240587757}}}}}}.$$ and consequently $$\frac{m}{n}\approx 3+\frac{1}{{\displaystyle 9+\frac{1}{{\displaystyle 2+\frac{1}{{\displaystyle 2}}}}}}=3+\frac{1}{{\displaystyle 9+\frac{2}{{\displaystyle 5}}}}=\frac{146}{47}\approx 3.106382979.$$ Then the fourth approximation for m/n is: $$\frac{m}{n}\approx 3+\frac{1}{{\displaystyle 9+\frac{1}{{\displaystyle 2+\frac{1}{{\displaystyle 2+\frac{1}{{\displaystyle 4}}}}}}}}=\frac{643}{207}\approx 3.106280193.$$ 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: $$\frac{m}{n}\approx 3+\frac{1}{9.408778692}<3+\frac{1}{9}=\frac{28}{9}.$$ Now, I shall do the same thing for the second approximation: $$\frac{m}{n}\approx 3+\frac{1}{9+\frac{1}{{\displaystyle 2.446311463}}}>3+\frac{1}{9+\frac{1}{{\displaystyle 2}}}.$$ 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.