Take any positive real number. By rounding down we can see that it consists of an integer plus a bit left over. That extra bit might be zero, in which case we're done, but it might not be. If you subtract off the integer, the bit left over is between 0 and 1. It might be 0, but it can't be 1.

Call the extra x.

Now we can take $\frac{1}{x}.$ That too is an integer plus a bit left over. Again, that extra bit might be zero, in which case we're done, but it might not be. Again, subtract off the integer, and repeat.

• Note: This process will terminate if and only if the starting number is rational.
• Not hard to prove, and an interesting exercise.
Let's do this with $\pi.$

 $\pi$ = 3 + 0.1415926... 1/0.1415926... = 7 + 0.0625133... 1/0.0625133... = 15 + 0.9965944... 1/0.9965944... = 1 + 0.0034172... 1/0.0034172... = 292 + 0.6345908...

$\Large{\pi=3+\frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292+...}}}}}$

We can write this as $\pi=[3;7,15,1,292,...]$

Cutting this off at different stages gives us rational approximations.

• $=3=3.0000...$
• $[3;7]=22/7\approx3.142857$
• $[3;7,15]=333/106\approx3.14151$
• $[3;7,15,1]=355/113\approx3.141593$
• $[3;7,15,1,292]=103993/33102\approx3.1415926530$
Compare these with the "correct" value of 3.1415926535897931...

As you can see, the large number in the expansion causes a sudden jump in the terms used in the rational approximation.

However, a large number in the continued fraction implies a small error in the previous step. That means that cutting off the continued fraction just before a large number will give an unreasonably good approximation.

Hence $\pi\approx\frac{355}{113}$

Using this technique gives an approximation with an error that is "best" given a limit on the size of the denominator.

We can also use continued fractions to give another proof that root two is irrational.
Given a continued fraction, we can compute the successive truncations as follows. Taking $\sqrt{2}=[1;2,2,2,2,2,...]$ as an example, write a table with three rows. Put the CF terms across the top, and start with a sort of "identity matrix", but the wrong way around. Here:

 1 2 2 2 2 2 ... 0 1 1 3 7 17 41 99 ... 1 0 1 2 5 12 29 70 ...
So we have $\frac{41}{29}$ and $\frac{99}{70}$ as approximations to $\sqrt{2}.$

Each term is the product of the quotient above it and the term on its left, plus the term two to the left. Here it is for $\pi$

 3 7 15 1 292 ... 0 1 3 22 333 355 103933 ... 1 0 1 7 106 113 33102 ...
So 333=15*22+3 and 113=1*106+7, and we get $\frac{355}{113}$ as an approximation of $\pi.$

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