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



( 5 4 )m = ( 2 1 )n .

This can be written equivalently:
mlog 5 4 =nlog2

or
m n = log2 log5-log4 =3.10628372   (1).

Now, I shall use Euclid's algorithm to find the first 4 rational approximations of:
log2 log5-log4 .

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.