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# Euclid's Algorithm and Musical Intervals

##### Age 16 to 18 Challenge Level:

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.