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



æ
ç
è 		5
4
 ö
÷
ø 		m

 
 = æ
ç
è 		2
1
 ö
÷
ø 		n

 
 .
This can be written equivalently:
mlog 5
4
 = n log2
or
m
n
 = log2
log5 - log 4
 = 3.10628372     (1).
Now, I shall use Euclid's algorithm to find the first 4 rational approximations of:
log2
log5 - log 4
 .
For the first approximation, I write:
3.10628372 = 3 + 1
1
0.10628372
 » 3 + 1
9.408778692
     (2).
So, the first approximation is
m
n
 » 3 + 1
9
 = 28
9
 = 3.111111111....
Now, for the second approximation I have:
3 + 1
9 + 1
1
0.408778692
 = 3 + 1
9 + 1
2.446311463
The second approximation for m/n is:
m
n
 » 3 + 1
9 + 1
2
 = 59
19
 » 3.105263158.
For the third approximation, I obtain:
3 + 1
9 + 1
2 + 1
1
0.446311463
 = 3 + 1
9 + 1
2 + 1
2.240587757
 .
and consequently
m
n
 » 3 + 1
9 + 1
2 + 1
2
 = 3 + 1
9 + 2
5
 = 146
47
 » 3.106382979.
Then the fourth approximation for m/n is:
m
n
 » 3 + 1
9 + 1
2 + 1
2 + 1
4
 = 643
207
 » 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
n
 » 3 + 1
9.408778692
 < 3 + 1
9
 = 28
9
 .
Now, I shall do the same thing for the second approximation:
m
n
 » 3 + 1
9 + 1
2.446311463
 > 3 + 1
9 + 1
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.